Excel in Mathematics Book 10 Final_CTP (2080)_compressed (1)

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 201 Vedanta Excel in Mathematics - Book 10 Exponential Equations Solution: Here, 5x + 1 5x = 25 1 25 or, 5x + 1 5x = 626 25 … (i) Let 5x = a, then equation (i) becomes a + 1 a = 626 25 Now, a + 1 a = 626 25 or, a2 + 1 a = 626 25 or, 25a2 + 25 = 626a or, 25a2 – 626a + 25 = 0 or, 25a2 – (625 + 1) a + 25 = 0 or, 25a2 – 625a – a + 25 = 0 or, 25a (a – 25) – 1 (a – 25) = 0 or, (a – 25) (25a – 1) = 0 Either a – 25 = 0 i.e. a = 25 or, 5x = 52 ∴x = 2 or, 25a – 1 = 0 i.e. a = 1 25 or, 5x = 1 52 = 5–2 ∴x = – 2 Hence, x = ±2. Example 5: Given exponential equation is 4 × 3x+1 – 9x = 27. (i) Solve the equation. (ii) Prove that the value of x obtained by solving the given equation also satisfy the equation 7x + 343 × 7 -x = 56. Solution: Here, 4 × 3x+1 – 9x = 27 or, 4 × 3x × 31 – (3x ) 2 = 27 or, 12 × 3x – (3x ) 2 = 27 … (A) (i) Let 3x = a, then equation (A) becomes 12a – a2 = 27 Now, 12a – a2 = 27 or, a2 – 12 a + 27 = 0 or, a2 – (9 + 3) a + 27 = 0 or, a2 – 9a – 3a + 27 = 0 or, a (a – 9) – 3 (a – 9) = 0 or, (a – 9) (a – 3) = 0 Either a – 9= 0 i.e. a = 9 or, 3x = 32 ∴x = 2 or, a – 3 = 0 i.e. a = 3 or, 3x = 31 ∴x = 1 Hence, x = 1, 2 (ii) Also, putting x = 1 in the equation 7x + 343 × 7 -x = 56,we get 71 + 343 × 7 – 1 = 7+ 343 7 = 56 which is true. Again, putting x = 2 in the equation 7x + 343 × 7 -x = 56, we get 72 + 343 × 7 – 2 = 72 + 343 72 = 56 which is true. Hence, the value of x obtained by solving the equation 4 × 3x+1 – 9x = 27 also satisfy the equation 7x + 343 × 7 -x = 56. Checking: When x = 2, then 52 + 1 52 = 25 1 25 or, 25 + 1 25 = 626 25 or, 626 25 = 626 25 which is true. When x = – 2, then 5-2 + 1 5–2 = 25 1 25 or, 1 25 + 52 = 626 25 or, 1 25 + 25=626 25 or, 626 25 = 626 25 which is true.

Vedanta Excel in Mathematics - Book 10 202 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Exponential Equations Example 6: The area of a window is 4x square meter. The curtain of size 3 m by 2x m is used in the window. Given that the area of the curtain is 2 square meters more than the area of window. (i) Make an exponential equation (ii) Solve the equation for x. (iii) Find the possible area of the window. (iv) Find the possible area of the curtain. Solution: Here, the area of a window = 4x m2 And, the area of a curtain =3 × 2x m2 (i) According to question, area of curtain = area of window + 2 m2 or, 3 × 2x = 4x + 2 or, 3 × 2x = (2x ) 2 + 2 (ii) Let 2x = a, then the above equation reduces to 3a = a2 + 2 or, a2 – 3a + 2 = 0 or, a2 – (2 + 1) a + 2 = 0 or, a2 – 2a – a + 2 = 0 or, a (a – 2) – 1 (a – 2) = 0 or, (a – 2) (a – 1) = 0 Either a – 2 = 0 i.e. a = 2 or, 2x = 21 ∴x = 1 or, a – 1 = 0 i.e. a = 1 or, 2x = 20 ∴x = 0 (iii) When x = 1, the area of the window = 4x m2 = 41 m2 = 4 m2 When x = 0, the area of the window = 4x m2 = 40 m2 = 1 m2 (iv) When x = 1, the area of the curtain = 3 × 21 m2 = 6 m2 When x = 0, the area of the curtain = 3 × 20 m2 = 3 m2 . Hence, the possible area of the window is 1 m2 or 4 m2 and the possible area of the curtain is 6 m2 or 3 m2 . EXERCISE 10.1 General section 1. a) If ax =am × an then express x in terms of m and n. b) If ax =(ap × aq ) ÷ ar write x in terms of p, q and r. c) If (bm ÷ bn) × bp = by , find the value of y in terms of m, n and p. d If (xm) n = xm ×x n, express m in terms of n. 2. a) If 3x = 9, what is the value of x? b) If 23x = 64, what is the value of x? c) If am = 1, is the value of m? d) For what value of x is 53 – x = 1 satisfied? 3. a) If 2x × 23 = 1, find the value of x. b) If 3m × 3 - 4 = 27, find the value of m. c) If If ap ÷ a4 = 1, find the value of p. d) If 5y ÷ 52 = 125, find the value of y. 4. Solve the following exponential equations. a) 2x + 2 = 4x – 3 b) 32x + 1 = 92x – 1 c) 42x – 1 = 83x – 4 d) 252y – 1 = 125y + 1 e) 25x + 3 = 1 0.04 f) 103y – 3 = 1 0.001 g) 3 2 x =27 8 h) 13 31 x = 961 169 i) 3 × 81x = 9x + 4 j) 7 × 49x =343 x + 1 k) 9x + 1 = 1 27x l) 25x + 1 = 1 625x

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 203 Vedanta Excel in Mathematics - Book 10 Exponential Equations 5. Solve. a) 2x+5 = 16 b) ( 3 5) x–1 = 25 c) ( 2) 3x – 1 = ( 4) x – 2 d) ( 9) x – 3 = ( 3) x +2 e) (0.5) x 2 = 0.25 f) (0.3) 3x 5 = 0.027 g) 1 2 × 2x = 2x h) 1 9 × 32x = 27–x 6. Solve. a) 2x + 1 – 2x = 8 b) 3x + 1 – 3x = 54 c) 2x + 2x + 2 = 5 d) 7x + 7x+1 = 56 e) 11x + 1 + 11x = 12 f) 3x + 2 + 3x + 1 = 11 3 g) 2x + 2x – 1 = 3 h) 3x – 3x – 2 = 8 i) 3x + 5 = 3x + 3 + 8 3 j) 2x + 2 + 2x + 3 2 = 1 k) 3x + 3 + 3x + 4 3 = 162 l) 5x + 5x + 1 + 5x + 2 = 155 m) 32x + 3 – 2.9x + 1 = 1 9 n) 9x – 2 + 2 × 32x – 3 = 63 o) 23x – 1 + 3 × 8x – 1 = 56 7. Solve. a) 2x + 3 × 3x + 4 = 18 b) 2x + 3 × 3x + 2 = 432 c) 2x – 5 × 5x – 4 = 5 d) 72x + 1 × 52x – 1 = 7 5 e) 23x – 5 × ax – 2 = 2x – 2 ×a1 – x f) 75x – 4 × a4x – 3 = 72x – 3 × ax – 2 Creative section-A 8. Solve each of the given exponential equation. a) 2x + 1 2x = 41 4 b) 3x + 1 3x = 91 9 c) 5x + 5 –x = 25 1 25 d) 7x + 7 –x = 71 7 e) 2x + 16 2x = 10 f) 7x + 343 7x = 56 g) 3x + 3 + 1 3x - 28 = 0 h) 3x + 2+ 1 3x–2 = 30 9. Solve the following exponential equations. a) 4x – 6 × 2x + 1 + 32 = 0 b) 9x – 4 × 3x + 1 + 27 = 0 c) 3.2x + 1 – 4x = 8 d) 5.4x + 1 – 16x = 64 e) 100x + 1000 = 11.10x + 1 f) 626.5x – 2 – 25x = 1 g) 2x + 21– x = 3 h) 3x – 1 + 32–x = 4 i) 5a + 1 + 52 – a = 126 10. a) In how many years does Rs 2,000 amount to Rs 2,420 at 10% p.a. compound interest? b) In how many years does Rs 8,000 amount to Rs 9,261 at 5% p.a. compound interest? Creative section- B 11. a) Given exponential equation is 4 × 3x+1 – 9x = 27. (i) Solve the given equation. (ii) Prove that the values of x obtained by solving the given equation also satisfy the equation 2x +23 – x = 6. b) Given exponential equation is 49x + 7 = 8.7x (i) Solve the given equation. (ii) Prove that the values of x obtained by solving the given equation also satisfy the equation 9x +3 = 4.3x 12. a) The area of a window is 4x square meter. The curtain of size 5 m by 2x m is used in the window. Given that the area of the curtain is 4 square meters more than the area of window.

Vedanta Excel in Mathematics - Book 10 204 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Exponential Equations (i) Make an exponential equation. (ii) Solve equation for x. (iii) Find the possible area of the window. (iv) Find the possible area of the curtain. b) Mr. Uprety bought a plot of 5 Aana land in a municipality. Now, he has built a house covering the space of 42 – x Aana and the kitchen garden covers remaining 4x – 1 Aana of his plot. (i) Make the exponential equation. (ii) Solve equation for x. (iii) Find the area of the kitchen garden. (iv) Find the area of plot occupied by the house. c) A manufacturing company produces the cartons each of having volume (5m + 2 – 5m) cm3 to keep the boxes with irons, each of having volume (5m – 5m – 1) cm3 . (i) How many boxes with iron can be kept inside a carton? (ii) If the volume of each carton is 15,000 cm3 , find the value of m. 10.6 Solving exponential equations with the given condition Let’s study the following examples. Example 1: If a = bx , b = cy and c = az , prove that xyz = 1 Solution: Here, a = bx , b = cy and c = az Now, a = bx or, a = (cy ) x [∵b = cy ] or, a = cxy or, a = (az ) xy [∵c = az ] or, a1 = axyz ∴ xyz = 1 proved. Example 2: If ap.aq = (ap) q, prove that p(q – 2) + q (p – 2) = 0. Solution: Here, ap .aq = (ap ) q or, ap+q = apq ∴p + q = pq … (i) Now, LHS = p(q – 2) + q (p – 2) = pq – 2p + pq – 2q = 2pq – 2 (p + q) = 2pq – 2pq [From (i)] = 0 = R.H.S. proved Alternative process b = cy or, b = (az ) y [ c = az ] or, b = ayz or, b = (bx ) yz [ a = bx ] or, b = bxyz ∴ xyz = 1

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 205 Vedanta Excel in Mathematics - Book 10 Exponential Equations Example 3: If 2a = 3b = 6 –c, show that 1 a + 1 b + 1 c = 0 Solution: Let, 2a = 3b = 6 –c = k Then, 2a = k ∴2 = k 1 a 3b = k ∴3 = k 1 b and 6-c = k ∴6 = k– 1 c Now, 2 × 3 = 6 or, k 1 a × k 1 b = k– 1 c or, k 1 a + 1 b = k– 1 c or, 1 a + 1 b = – 1 c ∴ 1 a + 1 b + 1 c =0 proved Example 4: For 2x + y = 8, solve the equation 2x + 2y = 6. Solution: Here, 2x + y = 8 or, 2x + y = 23 or, x + y = 3 … (i) And 2x + 2y = 6 … (ii) Multiplying equation (ii) by 2x , we get (2x ) 2 + 2x + y = 6 × 2x or, (2x ) 2 + 23 = 6 × 2x From equation (i) or, (2x ) 2 + 8 = 6 × 2x Let 2x = a. Then, a2 + 8 = 6a or, a2 – 6a + 8 = 0 or, a2 – (4 + 2) a + 8 = 0 or, a2 – 4a – 2a + 8 = 0 or, a (a – 4) – 2 (a – 4) = 0 or, (a – 4) (a – 2) = 0 Either a – 4 = 0 i.e. a = 4 or, 2x = 22 ∴ x = 2 or, a – 2 = 0 i.e. a = 2 or, 2x = 21 ∴ x = 1 Also, putting x = 1 in equation (i), we get 1 + y = 3 ∴y = 2 Again, putting x = 2 in equation (i), we get 2+ y = 3 ∴y = 1 Hence, the required solutions are x = 1 and y = 2 or x = 2 and y = 1. Example 5: If x2 + 2 = 3 2 3 + 3 2 3 , show that 3x(x2 + 3) = 8. Alternative process 2a = 3b ∴2 = 3 b a and 6 – c = 3b ∴6 = 3– b c Now, 2 × 3 = 6 or, 3 b a × 3 = 3– b c or, 3 b a + 1 = 3– b c or, b a + 1 = – b c or, b a + 1 + b c =0 or, b(1 a + 1 b + 1 c ) =0 ∴ 1 a + 1 b + 1 c = 0

Vedanta Excel in Mathematics - Book 10 206 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Exponential Equations Solution: Here, x2 + 2 = 3 2 3 + 3 2 3 or, x2 = 3 1 3 2 + 3 1 3 2 – 2 or, x2 = 3 1 3 2 + 3 1 3 2 – 2.3 1 3 .3 1 3 or, x2 = 3 1 3 – 3 1 3 2 or, x = 3 1 3 – 3 1 3 or, x3 = 3 1 3 – 3 1 3 3 or, x3 = 3 1 3 3 – 3 1 3 3 – 3.3 1 3 .3 1 3 3 1 3 – 3 1 3 or, x3 = 3 – 1 3 – 3x or, x3 + 3x = 8 3 or, 3x3 + 9x = 8 or, 3x (x2 + 3) = 8 Proved. EXERCISE 10.2 General section 1. a) If a = bc , b = ca and c = ab , prove that abc = 1. b) If xa = y, yb = z and zc = x, prove that abc = 1. c) If x = 10a , y = 10b and xb .ya = 100, prove that: ab = 1. d) If a = 5x , b = 5y and ay .bx = 25, prove that: xy = 1. 2. a) If ax = by and b = a2 , show that x – 2y = 0. b) If a 1 x = b 1 3 and ab = 1, prove that x + 3 = 0 c) If xa .xb = (xa ) b , prove that: a (b – 2) + b (a – 2) = 0. d) If am.an = (am) n, prove that: m n + n m= mn - 2 e) If ax = by and bx = ay , prove that: x = y Creative section-A 3. a) If 3x = 5y = 15 – z , show that 1 x + 1 y + 1 z = 0 . b) If 2a = 3b = 12c , show that 1 c = 1 b + 2 a . c) If ap = bq = cr and b2 = ac, prove that q = 2pr p + r . d) If xa = yb = zc and y3 = xz, prove that 3 b = 1 a + 1 c e) If a x = b y = c z and abc = 1, prove that x + y + z = 0. f) If ap = bc, bq = ca and cr = ab, prove that pqr = p + q + r + 2.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 207 Vedanta Excel in Mathematics - Book 10 Exponential Equations 4. a) If 2x + y = 8, solve the equation 4x + 4y = 20. b) If 2m + n = 16 and 3m + 3n = 30, find the value of m and n. 5. a) If x2 + 2 = 2 2 3 + 2– 2 3, show that 2x (x2 + 3) = 3 b) If x = 3 1 3 + 3 2 3 , prove that x3 – 9x – 12 = 0 Project Work and Activity Section 6. Write any five exponential equations of your own in the form of ax = b, where a is positive integer greater than 1 and b is the integer of power of a. Then, solve your equations. OBJECTIVE QUESTIONS Tick the correct alternative. 1. Which of the following equations is an exponential equation? (A) 2x = 4 (B) 2x = 4 (C) x2 = 4 (D) x3 = 8 2. If ax = ab then (A) x = b (B) x = a (C) x = ab (D) x = ab 3. What value of x satisfies the equation 3x = 9? (A) 2 (B) – 2 (C) ± 2 (D) 3 4. If ax = 1 then (A) x = 0 (B) x = 1 (C) x = – 1 (D) x = ± 1 5. For what value of y the equation 6y – 2 equals to 1? (A) 0 (B) 1 (C) 2 (D) 3 6. What is the solution of the equation 11x + 1 + 11x = 12? (A) x = 11 (B) x = 12 (C) x = 1 (D) x = 0 7. If 2x + 1 × 3x – 2 = 48, what is the value of x? (A) – 2 (B) 2 (C) – 3 (D) 3 8. What is the value of x if 2x =3 x ? (A) 0 (B) 1 (C) 2 (D) 3 9. Which of the following values of x satisfy the equation 2x +1 2x = 8 18 ? (A) ± 1 (B) ± 2 (C) ± 3 (D) ± 4 10. The solution of the equation 9x + 3x = 90 is (A) x = 3 and – 10 (B) x = 2 and – 10 (C) x = 3 only (D) x = – 10 only

Vedanta Excel in Mathematics - Book 10 208 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Exponential Equations Assessment -IV 1. Jitendra and Yankee are working in a company. The following table shows their monthly savings of 5 months. Month Baishakh Jestha Asar Shrawan Bhadra Jitendra Rs 2000 Rs 2500 Rs 3000 Rs 3500 Rs 4000 Yankee Rs 400 Rs 800 Rs 1600 Rs 3200 Rs 6400 Study the table above and write the answers to the questions below. (a) Which employee made the savings in arithmetic sequence? Write with reason. (b) Find the mean saving of both the employees between the months of Baisakh to Asar. (c) By using formula, calculate their total saving for five months. Also, find who made more savings in 5 months and by how much? 2. Simplify: ax2 + b 2x – 1 + ax2 – b 2x + 1 + 4ax2 1 – 4x2 3. In a rectangular field, the longer side is 10 m more than the shorter side and the diagonal is 10 m more than the longer side. (a) Assume the shorter side of the field by x and make the equation according to the given context. (b) Find the length and breadth of the rectangle. (c) By what percentage should the length of the filed be decreased so that it becomes a square? 4. A bus travels a journey of 360 km at a uniform speed. If the speed of the bus had been 5 km/hr more, it would have been taken 1 h less for the same journey. Answer the following questions. (a) Assume the speed of the bus by x and make the equation for the context given above. (b) Find the speed of the bus. 5. The students studying in Class 10 organized a picnic with a total budget of Rs 66,000. They decided to collect an equal amount for the budget. But on the picnic day, 5 students could not go to the picnic and each of the participated student paid Rs 100 more. (a) Suppose the total number of students studying in class 10 by x and make the equation for the given context. (b) How many students were participated in the picnic? (c) Calculate the amount paid by each of the participated students. 6. The area of a window is 4x square meter. The curtain of size 3 m by 2x m is used in the window. If the area of the curtain is 2 square meters more than the area of window, answer the following questions. (a) Make the equation for the question. (b) Find the possible areas of the window. (c) Find the possible areas of the curtain.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 209 Vedanta Excel in Mathematics - Book 10 Unit 11 Area of Triangles and Quadrilaterals Euclid of Alexandria, around 300 BCE) was a Greek mathematician and is often called the father of geometry. His book The Elements first introduced Euclidean geometry, defines its five axioms, and contains many important proofs in geometry, is one of the most influential books ever published, and was used as textbook in mathematics until the 19th century. 11.1 Area of triangles and quadrilaterals – Looking back The table given below helps to recall the area of triangles and quadrilaterals. Name of figure Diagram Formula to find area Triangle A = 1 2 × b × h or A = s (s – a) (s – b) (s – c) Rectangle A = l × b Square l d l A = l 2 or A = 1 2 × d2 Parallelogram A = b × h Rhombus d1 d2 A = b × h or A = 1 2 × d1 × d2 Trapezium a A = 1 2 h (a + b)

Vedanta Excel in Mathematics - Book 10 210 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Area of Triangles and Quadrilaterals 11.2 Relation between the area of triangles and quadrilaterals In this chapter, we will discuss about the relation of area of triangles and quadrilaterals which are formed between parallel lines. Theorem 1 (Concept only) Diagonal of a parallelogram bisect the parallelogram. ‘OR’ The area of each of the triangles formed by drawing a diagonal of a parallelogram is half of the area of the parallelogram. Activity Steps: (i) Take a rectangular sheet of paper or a chart paper and draw a parallelogram on it. (ii) Cut out the parallelogram and name it as ABCD. (iii) Join a diagonal AC (or BD). B B B A A A A A D D D C C C C C (iv) Cut out the parallelogram along diagonal AC (or BD). (v) Place the triangle ADC on triangle ABC such that DA coincides with BC ad CD coincides with AB. (vi) Observe whether the triangles ABC and ADC exactly fit each other or not. Thus, a diagonal divides the parallelogram into two congruent triangles. Theoretical Proof Given: ABCD is a parallelogram in which AC is its diagonal. To prove: D ABC = 1 2 ABCD Proof Statements Reasons 1. In ∆ ABC and ∆ ACD (i) AB = DC (S) (ii) BC = AD (S) (iii) AC = AC (S) (iv) ∆ ABC ≅ ∆ ACD 1. (i) Sides of the parallelogram (ii) Same as (i) (iii) Common side (iv) S.S.S. axiom 2. Area of ∆ ABC = Area of ∆ ACD 2. Area of congruent triangles 3. ∆ ABC + ∆ ACD = ABCD 3. Whole part axiom 4. 2 ∆ ABC = ABCD 4. From statements (2) and (3) 5. Area of ∆ ABC = 1 2 area of ABCD 5. From statement (4) Proved www.geogebra.org/classroom/mpdcsyfs Classroom code: MPDC SYFS Vedanta ICT Corner Please! Scan this QR code or browse the link given below: A B C D

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 211 Vedanta Excel in Mathematics - Book 10 Area of Triangles and Quadrilaterals Facts to remember 1. The diagonal of a square (or a rectangle or a rhombus) bisects it. 2. The congruent figures (triangles) are equal in area. Theorem (Theoretical proof only) Parallelograms on the same base and between the same parallels are equal in area. Activity 1 Test I: Finding the area of parallelograms on the same base with different altitudes. Steps: (i) Draw a parallelogram ABCD on a graph paper. (ii) Count the number of complete squares (C), the number of squares having more than half (M) of their parts and the number of squares having half (H) of their parts enclosed by the figure. Ignore the squares which are less than half parts. Find the area of ABCD by adding C + M + H. Here, the area of ABCD is 12 sq. units. (iii) Draw another parallelogram ABEF on the same base but the altitude is different than the altitude of parallelogram ABCD. Similarly, find its area. Here, the area of ABEF is 15 sq. units. Test II: Finding the area of parallelograms on the different bases but lying between same parallel lines. Steps: (i) Draw a parallelogram ABCD on a graph paper and find its area. Here, the area of ABCD is 12 sq. units. (ii) Draw another parallelogram PQRS on the different base but lying with parallelogram ABCD between the same parallel lines. Find area of parallelogram PQRS. Here, the area of PQRS is 16 sq. units. Test III: Finding the area of parallelograms on the same base and between same parallel lines. Steps: (i) Draw a parallelogram ABCD on a graph paper and find its area. Here, the area of ABCD is 20 sq. units. (iii) Draw another parallelogram ABEF on the same base and between the same parallel lines. Find its area. Here, the area of ABEF is also 20 sq. units. From the above tests, we came to know that the parallelograms on the same base and between the same parallel lines are equal in area. A B C F E D A P B Q C R S D A B C F D E

Vedanta Excel in Mathematics - Book 10 212 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Area of Triangles and Quadrilaterals Experimental verification Step 1: Three parallelograms ABCD with different sizes of bases and heights are drawn. Another parallelogram BCEF on the same base BC and between the same parallels BC and AE is drawn in each figure. Step 2: DM ⊥ BC is drawn in each figure, where DM is the height (altitude) of the parallelograms. Step 3: The base BC and the height DM are measured and the areas of ABCD and BCEF are calculated. Fig. Base BC Height DM Area of ABCD (BC × DM) Area of BCEF (BC × DM) Result (i) ABCD = BCEF (ii) ABCD = BCEF (iii) ABCD = BCEF Conclusion: Parallelograms on the same base and between the same parallels are equal in area. Theoretical Proof Given: Parallelograms ABCD and EBCF are standing on the same base BC and between the same parallel lines AF and BC. To prove: Area of ABCD = Area of EBCF Proof: S.N. Statements S.N. Reasons 1. In ∆ABE and ∆DCF (i) ∠AEB = ∠DFC (A) (ii) ∠BAE = ∠CDF (A) (iii) AB = DC (S) 1. (i) BE//CF and corresponding angles. (ii) AB//DC and corresponding angles. (iii) Being the opposite sides of ABCD. 2. ∆ABE ≅ ∆DCF 2. By A.A.S. axiom 3. ∆ABE = ∆DCF 3. Being the area of congruent triangles equal. 4. ∆ABE + Trapezium EBCD = ∆DCF + Trapezium EBCD 4. Adding the same Trapezium EBCD to both sides of statement (3). 5. ABCD = EBCF 5. By whole part axiom Proved A A A B C B C B C D D D E E E F F F M M M (i) (ii) (iii) A B C E D F

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 213 Vedanta Excel in Mathematics - Book 10 Area of Triangles and Quadrilaterals Alternative process-I Given: Parallelograms ABCD and EBCF are standing on the same base BC and between the same parallel lines AF and BC. To prove: Area of ABCD = Area of EBCF Proof: S.N. Statements S.N. Reasons 1. In ∆ABE and ∆DCF (i) ∠AEB = ∠DFC (A) (ii) ∠BAE = ∠CDF (A) (iii) AB = DC (S) 1. (i) BE//CF and corresponding angles. (ii) AB//DC and corresponding angles. (iii) Being the opposite sides of ABCD. 2. ∆ABE ≅ ∆DCF 2. By A.A.S. axiom 3. ∆ABE = ∆DCF 3. Being the area of congruent triangles equal. 4. Trapezium ABCF – ∆ABE = Trapezium ABCF – ∆DCF 4. Subtracting statement (3) from the same Trapezium ABCF. 5. EBCF = ABCD i.e., ABCD = EBCF 5. The remaining parts from statement (4). Proved Alternative process-II Given: Parallelograms ABCD and EBCF are standing on the same base BC and between the same parallel lines AF and BC. To prove: Area of ABCD = Area of EBCF Construction: EM ⊥ BC is drawn. EM is the height of the parallelograms ABCD and EBCF. Proof: S.N. Statements S.N. Reasons 1. Area of ABCD = BC × EM 1. Area of parallelogram is base × height. 2. Area of EBCF = BC × EM 2. Same as (1). 3. ABCD = EBCF 3. From statements (1) and (2). Facts to remember Corollary 1: The area of a parallelogram is equal to the area of a rectangle (or square or rhombus) standing on the same base and between the same parallels. Corollary 2: The area of a rectangle is equal to the area of a rhombus standing on the same base and between the same parallels. Corollary 3: Parallelograms on the equal bases and between the same parallels are equal in area. A B C E D F A B M C E D F www.geogebra.org/classroom/dcf6btfz Classroom code: DCF6 BTFZ Vedanta ICT Corner Please! Scan this QR code or browse the link given below:

Vedanta Excel in Mathematics - Book 10 214 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Area of Triangles and Quadrilaterals Theorem 3 (Theoretical proof only) The area of a triangle is equal to half of the area of a parallelogram standing on the same base and between the same parallels. Activity 1 Steps: (i) Draw two congruent parallelograms ABCD and EFGH on a chart paper and cut out them. (ii) Cut out the parallelogram EFGH along the diagonal FH. Then, the triangles EFH and FGH are congruent to each other and each of the triangles is equal to half of the area of the EFGH. B A D E H C F G F E H G G B A H D F C (iii) Place the triangle HFG on ABCD so that GF coincides with BC and H lies on the line DA produced. Thus, the area of triangle HGF is equal to half of the area of the parallelogram ABCD. Experimental verification Step 1: Three different figures with a triangle BCD and a parallelogram ABCE on the same base BC and between the same parallels AD and BC are drawn. Step 2: AM ⊥ BC is drawn in each figure. AM is the height of ∆ BCD and ABCE. Step 3: The base BC and the height AM in each figure are measured. The areas of ∆ BCD and ABCE are calculated. Fig. Base BC Height AM Area of ∆ BCD Area of ABCE Result (i) ∆ BCD = 1 2 ABCE (ii) ∆ BCD = 1 2 ABCE (iii) ∆ BCD = 1 2 ABCE Conclusion: The area of triangle is equal to half of the area of a parallelogram on the same base and between the same parallels. A A A B (i) (ii) (iii) B B M C M C M C D D E D E E

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 215 Vedanta Excel in Mathematics - Book 10 Area of Triangles and Quadrilaterals Theoretical Proof Given: ∆ BCD and parallelogram ABCE are on the base BC and between the same parallels AD and BC. To prove: ∆ BCD = 1 2 ABCE Construction: BF // CD is drawn. BF meets AD at F. Proof Statements Reasons 1. BCDF is a parallelogram 1. BC // FD and BF // CD 2. ∆ BCD = 1 2 BCDF 2. Diagonal bisects a parallelogram. 3. BCDF = ABCE 3. They are on the same base and between the same parallels. 4. ∆ BCD = 1 2 ABCE 4. From statements (2) and (3) Proved Alternative process Construction: EF ⊥ BC is drawn. EF is the height of the parallelogram ABCE and ∆ BCD. Proof Statements Reasons 1. ABCE = BC × EF 1. Area of a parallelogram is base × height 2. ∆ BCD = 1 2 BC × EF 2. Area of a triangle is 1 2 base × height 3. ∆ BCD = 1 2 BCEA 3. From statements (1) and (2) Proved Facts to remember Corollary 1: The area of a triangle is equal to half of the area of a parallelogram standing on the same base and between the same parallels. Corollary 2: The area of a triangle is equal to half of the area of a rectangle or square or rhombus standing on the same base and between the same parallels. Corollary 3: The area of a triangle is equal to half of the area of a parallelogram or rectangle or square or rhombus standing on the equal base and between the same parallels. A B C D F E A E B C D F www.geogebra.org/classroom/dwmvefjj Classroom code: DWMV EFJJ Vedanta ICT Corner Please! Scan this QR code or browse the link given below:

Vedanta Excel in Mathematics - Book 10 216 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Area of Triangles and Quadrilaterals Theorem 4 (Theoretical proof only) Triangles on the same base and between the same parallels are equal in area. Activity 1 Steps: (i) Make a pair of parallel lines using rubber bands on a Geo-board. (ii) Construct two triangles ABC and BCD on the same base and between the same parallels. (iii) Count the number of dots to find the base and height of the triangles. Here, in ∆ABC; base (b) = 5 units and height (h) = 4 units. ∴Area of ∆ABC = 1 2 × b × h = 1 2 × 5 × 4 = 10 sq. units. Here, in ∆BCD; base (b) = 5 units and height (h) = 4 units. ∴Area of ∆BCD= 1 2 × b × h = 1 2 × 5 × 4 = 10 sq. units. Experimental verification Step 1: Three pairs of triangles ABC and BCD are drawn on the same base BC and between the same parallels AD and BC. Step 2: AM ⊥ BC and DN ⊥ BC are drawn. Step 3: The heights AM and DN and the base BC of each figure are measured. The areas of ∆ ABC and ∆ BCD are calculated. Fig. Base BC Height AM Area of ∆ ABC Base BC Height DN Area of ∆ BCD Result (i) ∆ ABC = ∆ BCD (ii) ∆ ABC = ∆ BCD (iii) ∆ ABC = ∆ BCD Conclusion: Triangles on the same base and between the same parallels are equal in area. Theoretical Proof Given: ∆ ABC and ∆ BCD are on the same base BC and between the same parallels BC and AD. A B C D A A A B B C B M N M C N M C N (i) (ii) (iii) D D D D B C E A

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 217 Vedanta Excel in Mathematics - Book 10 Area of Triangles and Quadrilaterals To prove: ∆ ABC = ∆ BCD Construction: BE // CD is drawn. BE meets AD at E. Proof Statements Reasons 1. BCDE is a parallelogram 1. Opposite sides are parallel 2. ∆ ABC = 1 2 BCDE 2. They are on the same base and between the same parallels. 3. ∆ BCD = 1 2 BCDE 3. Diagonal bisects a parallelogram. 4. ∆ ABC = ∆ BCD 4. From statements (2) and (3) Proved Alternative process Construction: AM ⊥ BC is drawn. AM is the height of ∆ ABC and ∆ BCD. Statements Reasons 1. Area of ∆ ABC = 1 2 BC × AM 1. Area of a triangle is 1 2 (base × height) 2. Area of ∆ BCD = 1 2 BC × AM 2. Area of a triangle is 1 2 (base × height) 3. ∆ ABC = ∆ BCD 3. From statements (1) and (2) Proved Facts to remember Corollary 1: The area of a triangles standing on the same base and between the same parallels are equal. Corollary 2: Triangles on the equal base and between the same parallels are equal in area. Theorem 5 (Concept only) Median of a triangle bisects the triangle. ‘OR’ The area of each of the triangles formed by drawing a median of a triangle is half of the area of the triangle. Activity 1 Steps: (i) Draw a triangle ABC on a chart paper and cut it out. (ii) Fold the side BC in such a way that vertex B meets the vertex C. www.geogebra.org/classroom/a53yatar Classroom code: A53Y ATAR Vedanta ICT Corner Please! Scan this QR code or browse the link given below: A B C D M A A B C B D C

Vedanta Excel in Mathematics - Book 10 218 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Area of Triangles and Quadrilaterals (iii) Find the mid-point D of BC and join A and D. Here, AD is called a median of triangle. (iv) Measure the base and height of each of triangles ABD and ADC by using ruler or using a graph paper. Calculate the area of ∆ABD and ∆ADC. Here, in ∆ABD; base (b) = 3 units and height (h) = 4 units. ∴Area of ∆ABD = 1 2 × b × h = 1 2 × 3 × 4 = 6 sq. units. Also, in ∆ADC; base (b) = 3 units and height (h) = 4 units. ∴Area of ∆ADC = 1 2 × b × h = 1 2 × 3 × 4 = 6 sq. units. Thus, the median of a triangle divides it into two smaller triangles with equal area. Worked-out Examples Example 1: In the figure, the area of D PQT is 28 cm2 . Find the area of D PRS. Solution: (i) Area of PQRS = 2 area of D PQT = 2 × 28 cm2 = 56 cm2 (ii) Area of D PRS = 1 2 Area of PQRS = 1 2 × 56 cm2 = 28 cm2 So, the required area of D PRS is 28 cm2 . Example 2: In the given figure, E is the mid point of DC. If the area of ABCD is 24 cm2 , find the area of the quadrilateral ABCE. Solution: (i) Area of D ABE = 1 2 Area of ABCD = 1 2 × 24 cm2 = 12 cm2 (ii) Area of D BCE = Area of D AED [Both the triangles are standing on the equal base and between the same parallels.] ∴ Area of D AED = 1 2 × 12 cm2 = 6 cm2 [Area of (D BCE + D AED) = 12 cm2 ] (iii) Area of quadrilateral ABCE = Area of (D ABE + D BCE) = 12 cm2 + 6 cm2 = 18 cm2 So, the required area of the quadrilateral ABCE is 18 cm2 . S P Q R T [D PQT and PQRS are standing on the same base and between the same parallels.] [Diagonal PR bisects PQRS] D A B C E A B C D

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 219 Vedanta Excel in Mathematics - Book 10 Area of Triangles and Quadrilaterals Example 3: In the figure, ABCD is a rhombus in which the diagonals AC = 20 cm and BD = 28 cm. If AD is produced to E, find the area of D BCE. Solution: (i) Area of rhombus ABCD = 1 2 × AC × BD = 1 2 × 20 × 28 cm2 = 280 cm2 (ii) Area of D BCE = 1 2 area of rhombus ABCD = 1 2 × 280 cm2 = 140 cm2 So, the required area of D BCE is 140 cm2 . Example 4: In the given figure, PQRS is a square with its diagonal PR = 4 3 cm. If PS is produced to T, find the area of D TQR. Solution: (i) Area of the square PQRS = 1 2 × PR2 = 1 2 × (4 3 cm)2 = 24 cm2 (ii) Area of D TQR = 1 2 area of PQRS = 1 2 × 24 cm2 = 12 cm2 So, the required area of D TQR is 12cm2 . Example 5: In the given parallelogram ABCD; AD is produced to a point E, BF // CE and EF // CD produced to meet BF at H. BF cuts AE at G. Prove that: area of parallelogram ABCD = area of parallelogram CEFH. Solution: Given: In parallelogram ABCD, AD is produced to a point E, BF // CE and EF // CH. BF cuts AE at G. To prove: Area of parallelogram ABCD = Area of parallelogram CEFH. Proof: S.N. Statements S.N. Reasons 1. ABCD = GBCE 1. Both of them are standing on the same base BC and between AE // BC. 2. GBCE = CEFH 2. Both of them are standing on the same base CE and between BF // CE. 3. ABCD = CEFH 3. From statements (1) and (2). Proved D E B C A S T Q R P D A B C E G H F

Vedanta Excel in Mathematics - Book 10 220 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Area of Triangles and Quadrilaterals Example 6: In the given figure, PQRS is a trapezium in which PS // QR. M and N are the points on PS such that QM // RS and NR // PQ. QM produced and RN produced meet at T. Explore the relationship between the triangles PQT and RST in area. Solution: Given: In trapezium PQRS, PS // QR, M and N are the points on PS such that QM // RS and NR // PQ. QM produced and RN produced meet at T. To explore: Relationship between ∆PQT and ∆RST. Proof: S.N. Statements S.N. Reasons 1. PQRN = MQRS 1. Both of them are standing on the same base QR and between PS // QR. 2. ∆PQT = 1 2 PQRN 2. Both of them are standing on the same base PQ and between TR // PQ. 3. ∆RST = 1 2 MQRS 3. Both of them are standing on the same base RS and between QT // RS. 4. ∆PQT = ∆RST 4. From statements (1), (2) and (3). Example 7: In the quadrilateral ABCD given alongside; AB//DC and BC//AD. P is any point on the side AD and BP is produced to meet CD produced at Q. Prove that the area of triangles APQ and PDC are equal. Solution: Given: In quadrilateral ABCD; AB//DC and BC//AD. P is any point on the side AD and BP is produced to meet CD produced at Q. To prove: Area of DAPQ = Area of ∆PDC Proof: S.N. Statements S.N. Reasons 1. ABCD is a parallelogram 1. AB // CD and AD // BC 2. ∆PBC = 1 2 ABCD 2. Both of them are standing on the same base BC and between AD // BC. 3. ∆ABP + ∆PCD =1 2 ABCD 3. Remaining parts of ABCD. 4. ∆ABQ = 1 2 ABCD 4. Both of them are standing on the same base AB and between AB // QC. 5. ∆ABP + ∆PCD = ∆ABQ 5. From statements (2) and (3). 6. ∆ABP + ∆PCD = ∆APQ + ∆ABP 6. ∆ABQ = ∆APQ + ∆ABP 7. Area of ∆APQ = Area of ∆PDC 7. From statement (5) Proved A B C D P Q P Q R T S M N

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 221 Vedanta Excel in Mathematics - Book 10 Area of Triangles and Quadrilaterals Example 8: In the given figure, PARM is a parallelogram. B is any point on diagonal AM. Prove that DPMB and D RMB are equal in area. Solution : Given: PARM is a parallelogram. AM is its diagonal and B is any point on the diagonal. To prove: Area of DPMB = Area of DRMB Construction: P and R are joined. Diagonals AM and PR bisect each other at X. Proof Statements Reasons 1 X is the mid-point of diagonals AM and PR 1. From construction and given 2. DBPX = DBRX 2. BX is the median of D BPR 3. DPMX = DRMX 3. MX is the median of D PMR 4. DBPX + DPMX = DBRX + DRMX 4. Adding the statements (2) and (3) 5. DPMB = DRMB 5. Whole parts axiom 6. ∴ DPMB and DRMB are equal in area 6. From statement (5) Proved Example 9: In the given figure, D is the mid-point of side BC of D ABC, E is the mid-point of AD, F is the mid-point of AB and G is any point of BD. Prove that DABC = 8DEFG. Solution : Given: D, E and Fare the mid-points of BC, AD and AB respectively. G is any point of BD. To prove: DABC = 8 DEFG Construction: D and F are joined. Proof Statements Reasons 1 EF//BD 1. E and F are the mid-points of the sides AD and AB of D ABD. 2. DEFD = DEFG 2. Both of them are standing on the same base and between the same parallels. 3. DEFD = 1 2 DAFD 3. Median EF bisects D AFD. 4. DAFD = 1 2 DABD 4. Median FD bisects DABD 5. DABD = 1 2 DABC 5. Median AD bisects DABC 6. D EFG = 1 2 ×1 2 ×1 2 D ABC 1 8 D ABC ∴ D ABC = 8 D EFG 6. From statements (2), (3), (4) and (5) Proved M R A B X P A C B G D F E A C B G D E F

Vedanta Excel in Mathematics - Book 10 222 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Area of Triangles and Quadrilaterals Example 10: In parallelogram TWIN; side WI is produced to E such that WI = IE. TE and NI intersect at O. Prove that the area of triangle TWO is equal to twice the area of triangle ONE. Solution: Given: In parallelogram TWIN, WI = IE. TE and NI intersect at O. To prove: ∆TWO = 2∆ONE. Proof: S.N. Statements S.N. Reasons 1. ∆TWO = 1 2 TWIN 1. Both are standing on same base TW and between NI // TW. 2. ∆TEN = 1 2 TWIN 2. Both are standing on same base TN and between TN//WI. 3. ∆TWO = ∆TEN 3. From statements (1) and (2) 4. O is the mid-point of TE. 4. In ∆TWE; OI//TW and WI = IE. 5. ∆ONE = 1 2 ∆TEN 5. The median NO bisects ∆TEN. 6. ∆ONE = 1 2 ∆TWO i.e., ∆TWO = 2∆ONE 6. From statements (3) and (5) Proved Example 11: In ∆ABC, D is the mid-point of side AB. If P is any point on BC and Q is any point on AD such that CQ//PD, prove that the area of ∆BPQ is half of the area of ∆ABC. Solution: Given: In ∆ABC, D is the mid-point of side AB. P is any point on BC and Q is any point on AD such that CQ//PD. To prove: Area of ∆BPQ = 1 2 Area of ∆ABC Construction: D and C are joined. Proof: S.N. Statements S.N. Reasons 1. ∆QDP = ∆CDP 1. Both are standing on the same base DP and between QC//DP. 2. ∆QDP +∆BPD = ∆CDP+∆BPD 2. Adding ∆BPD to statement (1) 3. ∆BPQ = ∆BDC 3. By whole part axiom 4. ∆BDC = 1 2 ∆ABC 4. The median DC bisects the ∆ABC. 5. ∆BPQ = 1 2 ∆ABC 5. From statements (3) and (4). Proved T N W I E O A B C D Q P

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 223 Vedanta Excel in Mathematics - Book 10 Area of Triangles and Quadrilaterals Example 12: In ∆ABC, P and Q are the mid-points of sides AB and AC respectively. If S and R are the points on BC such that PS//QR, prove that the area of ∆ABC is double of the area of quadrilateral PQRS. Solution: Given: In ∆ABC, P and Q are the mid-points of sides AB and AC respectively, S and R are the points on BC such that PS//QR. To prove: Area of ∆ABC = 2×Area of quadrilateral PQRS. Construction: B and Q are joined. Proof: S.N. Statements S.N. Reasons 1. PQ//BC 1. In ∆ABC, PQ joins the mid-points P and Q of AB and AC respectively. 2. ∆PBQ = 1 2 PQRS 2. Both are standing on the same base PQ and between PQ//BC. 3. ∆PBQ = 1 2 ∆ABQ 3. The median QP bisects the ∆ABQ. 4. ∆ABQ = PQRS 4. From statements (2) and (3). 5. ∆ABQ = 1 2 ∆ABC 5. The median BQ bisects the ∆ABC. 6. PQRS = 1 2 ∆ABC ∴∆ABC = 2 PQRS 6. From statements (4) and (5). Proved Example 13: In trapezium PQRS, M and N are the middle points of diagonals QS and PR respectively. Prove that the area of triangle PMR is equal to the area of triangle SNQ. Solution: Given: In trapezium PQRS, M and N are the middle points of diagonals QS and PR respectively. To prove: Area of ∆PMR= Area of ∆SNQ. Construction: M and N are joined. Proof: S.N. Statements S.N. Reasons 1. ∆PQR = ∆SQR 1. Both are standing on the same base QR and between PS//QR. 2. ∆PQR = 2∆NQR and ∆SQR = 2∆MQR 2. The median QN bisects ∆PQR and the median MR bisects ∆SQR. 3. ∆NQR = ∆MQR 3. From statements (1) and (2). 4. MN//QR 4. From statement (3). 5. ∆PMN = ∆SMN 5. Both are standing on the same base MN and between PS//MN. 6. ∆MNR = ∆MNQ 6. Both are standing on the same base MN and between QR//MN. 7. ∆PMN + ∆MNR = ∆SMN + ∆MNQ 7. Adding statements (5) and (6) 8. ∆PMR= ∆SNQ 8. By whole part axiom. Proved A B C P Q S R P S R M N Q

Vedanta Excel in Mathematics - Book 10 224 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Area of Triangles and Quadrilaterals Example 14: In the adjoining figure, PS // QR. Answer the following questions. a) Name the triangles which are standing on the same base QR and between PS // QR. b) Write the relation between the area of triangles PQS and PSR. c) If T is the mid-point of QR and the area of ∆PQT is 60 cm2 , what is the area of ∆SQR? d) If the triangles PQS and RST are equal in area, show that PT // SR. Solution: Here, PS // QR. a) ∆PQR and ∆SQR are standing on the same base QR and between PS // QR. b) Area of ∆PQS = Area of ∆PSR [Both are on the same base PS and between PS // QR] c) T is the mid-point of QR and the area of ∆PQT = 60 cm2 Now, area of ∆PQR = 2 × area of ∆PQT [Median PT bisects ∆PQR] = 2 × 60 cm2 = 120 cm2 Also, area of ∆SQR = area of ∆PQR [Both are on the same base PS and between PS // QR] ∴ The area of ∆SQR =120 cm2 d) Given: area of ∆PQS = area of ∆RST To prove: PT // SR Proof: S.N. Statements S.N. Reasons 1. Area of ∆PQS = Area of ∆PRS 1. Both are standing on the same base PS and between PS//QR. 2. Area of ∆PQS = Area of ∆RST 2. Given 3. Area of ∆PRS = Area of ∆RST 3. From statements (1) and (2). 4. PT // SR 4. From statement (3). EXERCISE 11.1 General section 1. a) Write down the relation between the area of a parallelogram and a rectangle standing on the same base and between the same parallels. b) In the given figure; AE // BC, AB // DC and FB // EC. What is the relation between the area of quadrilaterals ABCD and FBCE? Give reason. c) What is the relation between the area of a rectangle and a rhombus standing on the same base and between the same parallels? d) In the given figure, write down the relation between area of PQRS and ∆TQR. e) What is the relation between area of square MNOP and triangle MAN standing on the same base MN and between the same parallels PA and MN? P S R Q T A D F E B C P S T Q R

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 225 Vedanta Excel in Mathematics - Book 10 Area of Triangles and Quadrilaterals f) From the given figure, name the triangle that is equal to the triangle XYZ in area. 2. a) In the given figure, ABCD is a rectangle and EBCF is a parallelogram. If EH ⊥ CF, CF = 16 cm and EH = 7 cm, find the area of rectangle ABCD. b) In the given diagram, AF // DC, DE // CF and ABCD is a square. If AC = 8 cm, find the area of parallelogram CDEF. c) A rectangle AQRB and a rhombus PQRS are on the same base QR and between AS // QR. If the AQ = 20 cm and QR = 12 cm, find the area of rhombus. d) In the given figure, MNOP is a rhombus and MQRN is a rectangle. If the diagonals MO and PN of the rhombus are 18 cm and 15 cm respectively, find the area of rectangle MQRN. 3. a) In the figure, WXYZ is a rhombus in which XW is produced to the point P. If WY = 10 cm and XZ = 9 cm, find the area of DPYZ. b) In the given figure, PQRS is a square whose each side is 4 cm. Find the area of D QRT. c) In the figure, AE // BC, square ABCD and D EBC are standing on the same base BC and between the same parallels. If AC = 6 cm, find the area of D EBC. d) In the given figure, if ABCE is a rectangle, find the area of D ADE. W X Y Z A B C E D F H 16 cm 7 cm D C A E B F M N O Q P R W Z P Y X S T Q R P D E B C A E D 10 cm C 8 cm B A

Vedanta Excel in Mathematics - Book 10 226 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Area of Triangles and Quadrilaterals e) In the adjoining figure, PS = 5 cm and SM = 8 cm. Calculate the area of D PQN. f) In the figure, the area of trapezium ABCD is 100 sq.cm and the area of D ADC is 40 sq.cm. Find the area of D DEC. g) In the given figure, ABCD is a square whose perimeter is 40 cm and AB is produced to the point F. If E is the mid point of DC, find the area of D EFC. h) In the given figure, WXYZ is a parallelogram. If ∠WAZ = 90o , WA = 9 cm and AZ = 6 cm, find the area of WXYZ. 4. a) In the figure, parallelogram PQSR and D PQT are standing on the same base and between the same parallel lines. The area of D PQT is 50 cm2 . Find the area of D PRS. b) In the djoinin figure, parallelogram ABCD and ∆ADE are on the same base AD and between AD//BE. If the area of ∆ADE is 50cm2 , find the area of ∆BCD. c) In the given parallelogram PQRS, M is the mid-point of side QR and PT ⊥ RS. If PQ = 20 cm and PT = 7 cm; find the area of ∆PQM. Creative section-A 5. a) In the adjoining figure, ABCD and BCEF are parallelograms, AE // BC. Prove that ABCD = BCEF. b) In the given figure, rectangle PQRS and parallelogram AQRB are on the same base QR and between the same parallels PB and QR. Prove that (i) D PQA ≅ D SRB (ii) Area of rectangle PQRS = Area of AQRB. Q N S R P M A B E C D B F D C A E W Z X A Y 9 cm 6 cm A B C E D P Q M R T 7 cm 20 cm S B C A D F E P Q R A S B R S T P Q

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 227 Vedanta Excel in Mathematics - Book 10 Area of Triangles and Quadrilaterals c) In the given figure, if AB // DC // EF, AD // BE and AF // DE, prove that parm. DEFH = parm . ABCD. d) In the given figure, AH // BG, AB // DC, EF // HG and BC = FG. Prove that the area of parallelograms ABCD and EFGH are equal. [Hints: Join B, E and C, H] 6. a) In the given figure, DX // AB. Prove that area of D ABX = 1 2 area of ABCD. b) In the given parallelogram ABCD, X and Y are any points on CD and AD respectively. Prove that the area of D AXB = the area of D BYC. c) In the given figure, ABCD is a parallelogram. If M and N are any points on CD and DA respectively, prove that D AMB = D CDN + DANB. d) In the given diagram, ABCD and PQRD are two parallelograms. Prove that ABCD = PQRD. e) In the adjoining parallelogram ABCD, A is joined to any point E on BC. AE and DC produced meet at F. Prove that the area of DBEF = the area of D CDE D C B E F G A H A B C F G H D E D A C B X B C X Y D A B C M N D A D C Q B P R A B C D E F A

Vedanta Excel in Mathematics - Book 10 228 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Area of Triangles and Quadrilaterals 7. a) In the figure alongside, QR // PS. Prove that D PQR = D QRS. b) In the adjoining figure, triangles AMB and ANB are standing on the same base AB and between the same parallel lines AB and MN. Prove that area of D AOM = area of DBON. c) In the given figure, ABCD is a parallelogram. X is any point within it. Prove that the sum of the areas of D XCD and D XAB is equal to half of the area of ABCD. d) In the given figure, DE // BC. Prove that (i) D BOD = D COE (ii) D BAE = D CAD. e) In the adjoining figure, it is given that AD // BC and BD // CE. Prove that, area of D ABC = area of D BDE. f) The line drawn through the vertex C of the quadrilateral ABCD parallel to the diagonal DB meets AB produced at E. Prove that, the area of quad. ABCD = the area of D DAE. 8. a) In the given DABC, P is any point on the median AD. Prove that D APB = D APC. b) In the adjoining D ABC, P and Q are the mid-points of the sides AB and AC respectively and R be any point on BC. Prove that, area of D PQR = 1 4 area of D ABC. Q R P S B O A M N B X C D A A B C E O D A B C E D A B C D E A B C P D A B C R P Q

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 229 Vedanta Excel in Mathematics - Book 10 Area of Triangles and Quadrilaterals c) In the figure, PQRS is a parallelogram. Q and S are joined to any point M on the diagonal PR. Prove that, the area of D PQM = the area of D PSM. d) In the given figure, M is the mid-point of AE, prove that area of DABE is equal to the area of parallelogram ABCD. Creative section-B 9. a) In the given figure, AD // BC. If the area of DABE and DACF are equal, prove that EF // AC. b) In the given figure, AD // BE // GF and AB // DG // EF. If the area of parallelograms ABCD and CEFG are equal, then prove that DE // BG. c) In the adjoining figure, PQRS and LQMN are two parallelograms of equal in area. Prove that LR // SN. 10. a) In the adjoining figure, ABCD is a rectangle. Answer the following questions. (i) Name the triangles which are standing on the same base BC and between AD // BE. (ii) Write the relation between the area of triangles ABD and ACD. (iii) If the area of ∆BCD is 20 cm2 , what is the area of ∆AED? (iv) If the triangles ACE and ABD are equal in area, show that AC // DE. b) In the adjoining figure, PQRS is a parallelogram and U is the mid-point of QT. Answer the following questions. (i) Write the relation between the area of triangles PQU and PUT. (ii) If the area of ∆PUT is 35 cm2 , what is the area of parallelogram PQRS? (iii) Prove that: area of PQRS = area of ∆PQT. (iv) Show that: area of PQRS = 4 × area of ∆VUT. R M S P Q M C E D A B A D F E B C D C B E G F A S L N Q R M P A D B C E P Q R U T V S

Vedanta Excel in Mathematics - Book 10 230 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Area of Triangles and Quadrilaterals Objective Questions Tick the correct alternatives. 1. Which of the following statement is true? (A) The median of a triangle bisects it. (B) The diagonal of parallelogram bisects it. (C) The diameter of circle bisects it. (D) All of the above 2. Which of the following statement is NOT true? (A) Parallelograms standing on the same base and between the same parallels are equal in area. (B) Triangles standing on the same base and between the same parallels are equal in area. (C) The area of triangle and the parallelogram standing on the same base and between the same parallels are equal. (D) The rectangle and rhombus standing on the same base and between the same parallels are equal in area. 3. Parallelogram ABCD and DBCE are on the same base BC and between AE // BC, the relation between their areas is (A) Ar. ( ABCD) = Ar. (DBCE) (B) Ar. ( ABCD) = 2 × Ar. (DBCE) (C) Ar. (DBCE) = 2 × Ar. ( ABCD) (D) Ar. (DBCE) > Ar. ( ABCD) 4. The following pairs of quadrilaterals are on the same base and between the same paralles, which pair have unequal area? (A) Square and parallelogram (B) Rectangle and rhombus (C) Rhombus and parallelogram (D) Trapezium and rectangle 5. A farmer has a field ABCD in the form of parallelogram. If he joins the corners A and B to any point P on the side CD by thread, what is the ratio of areas of ∆ABP to that of his field? (A) 1:2 (B) 2:1 (C) 1:4 (D) 4:1 6. In the given figure, ABCD is a parallelogram. Which one of the triangle is equal in area to ∆BCD? (A) ∆ABD (B) ∆BED (C) ∆ABE (D) both (A) and (C) 7. In the given diagram, PQRS is a square. If PQ//ST and SQ = 3 2 cm then the area of ∆PQT is (A) 9 cm2 (B) 12 cm2 (C) 18 cm2 (D) 36 cm2 8. In the given diagram, GOAT is a parallelogram in which AE = ET. What percent of area of parallelogram GOAT is occupied by the trapezium GOAE? (A) 20% (B) 25% (C) 75% (D) 80% 9. DABC and DBCD lying between the lines AD and BC have equal area, the relation between AD and BC is ... (A) AD = BC (B) AD // BC (C) AD ⊥ BC (D) AD ≤ BC 10. In hexagon ABCDEF; AB // FC // ED, BC // AE, AF // BE // CD and BE cuts CF at O. Which of the following relations is true? (A) ABOF = OCDE (B) ABOF = ∆ FBC (C) OCDE = ∆ BCE (D) ∆ BOF = ∆ BOC E D C A B S P Q R T G O A T E

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 231 Vedanta Excel in Mathematics - Book 10 Unit 12 Geometry - Construction 12.1 Construction of a triangle equal in area to the given triangle Pre-knowledge A. To construct the triangle from the given measurements. B. The triangles standing on the same (or equal) base and between the same parallels are equal in area. Checking theoretically: Area of ∆PQR = Area of ∆SPQ [Both are standing on the same base PQ and between YZ // PQ] Example: Construct a triangle PQR in which PQ = 5.2 cm, QR = 6 cm and ∠Q = 60°. Construct another triangle SPQ which is equal in area to the ∆PQR and having a side PS = 8.5 cm. Solution: R S P 5.2 cm Q 6 cm 8.5 cm Rough Sketch Y Z 5.2 cm 6 cm P Q X R 60° 5.2 cm 6 cm P Q X R 60° Y Z S 8.5 cm 2nd phase: Construction of ∆SPQ Steps i) Through the vertex R, draw YZ // PQ. ii) From the point P, draw an arc with radius PS = 8.5 cm to cut YZ at S. iii) Join P, S and Q, S. Thus SPQ is the required triangle. 1st phase: Construction of ∆PQR Steps i) Draw PQ = 5.2 cm ii) At Q, construct ∠PQX = 60° iii) Take an arc of 6 cm and from Q cut QX at R iv) Join P and R. Thus, PQR is the given triangle.

Vedanta Excel in Mathematics - Book 10 232 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Geometry - Construction 12.2 Construction of a parallelogram equal in area to the given triangle Pre-knowledge A. To construct the triangle from the given measurements. B. The area of a parallelogram is equal to the area of given triangle if they lie between the same parallels and the base of the parallelogram is equal to half of the base of the triangle. Example 2: Construct a triangle ABC in which a = 5.8 cm, b = 3.5 cm and c = 4.7 cm. Construct a parallelogram with a side 6.5 cm and equal in area to the area of triangle. Solution: 5.8 cm 4.7 cm 6.5 cm 3.5 cm B D C A E F X Y Rough Sketch B 5.8 cm C A 4.7 cm 3.5 cm B 5.8 cm C A 4.7 cm 3.5 cm E F 6.5 cm 6.5 cm D 1st phase: Construction of ∆ABC. Steps i) Draw BC = 5.8 cm ii) From C, draw an arc with radius CA = 3.5 cm and from B draw another arc with radius AB = 4.7 cm. These two arc intersect at A. iii) Join A, B and A, C. Thus, ABC is the given triangle 2nd Phase: Construction of parallelogram CDEF. Steps i) Through the vertex A, draw XY // BC. ii) With the help of perpendicular bisector of BC, mark the mid-point D of BC. iii) Take an arc with radius 6.5 cm and from D cut XY at E. iv) From the point E, draw an arc with radius EF = DC to cut XY at F. v) Join D, E and C, F. Checking theoretically: Area of CDEF = CD × height (h) = 1 2 BC × height (h) [∆ and have equal height] = Area of ∆ABC ∴ Area of CDEF = Area of ∆ABC Thus, CDEF is the required parallelogram. X Y

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 233 Vedanta Excel in Mathematics - Book 10 Geometry - Construction Example 3: Construct a triangle ABC in which AB = 5cm, ∠ABC = 75° and ∠ACB = 60°. Then construct a rectangle equal in area to the triangle ABC. Solution: A D B E C F P Q 45° 5 cm 75° Rough Sketch Y X C A 45° 5 cm B 75° Y X C A 45° 5 cm B 75° D E F 1st phase: Construction of ∆ABC. Steps i) Find the measure of ∠BAC = 45°. [∠A + ∠B + ∠C = 180°] = [∠A + 75° + 60° = 180°] ii) Draw AB = 5cm iii) At A, draw ∠BAC = 45° and at B, draw ∠ABC = 75°. These two lines intersect at C. Thus, ABC is the given triangle. 2nd phase: Construction of rectangle Steps i) Through the vertex C, draw PQ // AB. ii) With the help of perpendicular bisector of AB, mark the mid-point D of AB. Fix that the perpendicular bisector DE within the parallel lines as a side of rectangle BDEF. iii) From the point E, draw an arc with radius EF = DB to cut PQ at F. iv) Join F and B. Thus, BDEF is the required rectangle. Checking theoretically: Area of rectangle BDEF = BD × DE = 1 2 AB × DE [ ∴ BD = 1 2 AB] = ∆ABC [ ∴ ∆ABC = 1 2 AB × DE] ∴ Area of rectangle BDEF = Area of ∆ABC 12.3 Construction of a parallelogram equal in area to the given parallelogram Pre-knowledge A. To construct the parallelogram from the given measurements. B. The parallelograms standing on the same (or equal) base and between the same parallels are equal in area. P Q

Vedanta Excel in Mathematics - Book 10 234 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Geometry - Construction Example 4: Construct a parallelogram ABCD in which AB = 3.5 cm, AC = 5.9 cm and BC = 4.2 cm. Construct another parallelogram ABEF equal in area to the parallelogram ABCD with BE = 6.5 cm. Solution: 2nd phase: Construction of ABEF. Steps i) Produce DC to X. ii) From B, draw an arc with radius BE = 6.5 cm to cut DX at E. iii) From A, draw another arc with radius AF = BE = 6.5 cm to cut DX at F. iv) Join B, E and A, F Thus, ABEF is the required parallelogram. Checking theoretically: Area of ABCD=Area of ABEF [Both are standing on the same AB between DC // AB] 3.5 cm 3.5 cm 4.2 cm 5.9 cm 4.2 cm D C A B 3.5 cm 3.5 cm 5.9 cm 4.2 cm 4.2 cm D C A B F E X 6.5 cm D C F E R 3.5 cm B 5.9 cm 4.2 cm 6.5 cm A Rough Sketch 1st phase: Construction of ABCD Steps i) Draw AB = 3.5 cm ii) From A, draw an arc with radius AC = 5.9 cm and from B draw another arc with radius BC = 4.2 cm. These two arcs intersect each other at C. iii) Join A, C and B, C. iv) From C, draw an arc with radius CD = AB = 3.5 cm and from A, draw another arc with radius AD = BC = 4.2 cm. These two arcs intersect at D. v) Join A, D and C, D. Thus, ABCD is the given parallelogram. 12.4 Construction of a triangle equal in area to the given parallelogram Pre-knowledge A. To construct the parallelogram from the given measurements. B. The area of a triangle is equal to the area of given parallelogram if they lie between the same parallels and the base of the triangle is double of the base of the parallelogram.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 235 Vedanta Excel in Mathematics - Book 10 Geometry - Construction Example 5: Construct a parallelogram PQRS in which PQ = 5 cm, diagonal PR = 6 cm and diagonal QS = 8 cm. Construct a ∆PSA whose area is equal to the area of parallelogram. Solution: 5 cm 3cm 3 cm 4 cm 4 cm P O Q R Y X S 5 cm 3cm 3 cm 4 cm 4 cm P O Q R Y X S A S R P Q O 5 cm 4 cm 4 cm 3 cm 3 cm Rough Sketch 1st Phase: Construction of parallelogram PQRS. Steps i) Draw PQ = 5 cm ii) From P, draw an arc with radius PO = 1 2 PR = 1 2 × 6 cm = 3 cm and from Q, draw another arc with radius QO = 1 2 QS = 1 2 × 8 cm = 4 cm. These two arcs intersects at O. iii) Join P and O and produce to X, take an radius OR = 3 cm and cut OX at R. iv) Join Q and O and produce to Y, take an arc with radius OS = 4 cm and cut OY at S. v) Join P and S, Q and R and R and S. Thus PQRS is the given parallelogram. 2nd phase Steps i) Produce PQ to A such that PQ = QA ii) Join S and A. Thus, PSA is the required triangle. Checking theoretically: Area of ∆PSA = 1 2 PA × height [ ∴ ∆ = 1 2 b × h] = 1 2 × 2 PQ × height [ ∴ PA = 2 PQ] = PQ × height [ ∴ ∆ and have same height] = Area of PQRS [ ∴ = b × h] ∴ Area of ∆PSA = Area of PQRS

Vedanta Excel in Mathematics - Book 10 236 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Geometry - Construction 12.5 Construction of a triangle equal in area to the given quadrilateral Pre-knowledge A. To construct the quadrilateral from the given measurements. B. The triangles standing on the same base and between the same parallels are equal in area. C. Whole part axiom E Y 60° 5.2 cm 4 cm 5 cm 4.2 cm A D C X B 60° 5.2 cm 4 cm 5 cm 4.2 cm A D C X B Example 6: Construct a quadrilateral ABCD having AB = 5.2 cm, BC = 5 cm, CD = 4.2 cm, AD = 4 cm and ∠ABC = 60°. Construct ∆ADE whose area is equal to the quadrilateral ABCD. Solution: 5.2 cm D x E 60° C B A 4.2 cm 4 cm Rough Sketch Steps 5 cm i) Draw AB = 5.2 cm ii) At B, draw ∠ABX = 60° iii) From B, draw an arc with radius BC = 5 cm to cut BX at C. iv) From C, draw an arc with radius CD=4.2 cm and from A, draw another arc with radius AD = 4 cm. These two arcs intersect each other at D. Join C, D and D,A. Thus ABCD is the given quadrilateral. 1st phase: Construction of quadrilateral ABCD 2nd Phase: Construction of ∆ADE. Steps i) Join D and B. ii) Measure ∠DBC and copy the angle at C so that DB // CE and CE meets AB produced at E. iii) Join D and E and get a triangle ADE.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 237 Vedanta Excel in Mathematics - Book 10 Geometry - Construction Thus, DADE is the required triangle. Checking theoretically: Quad. ABCD =∆ABD + ∆BCD = ∆ABD + ∆BED [ ∴ ∆ BCD = ∆ BED] = ∆ ADE EXERCISE 12.1 General section 1. Construct triangle ABC by using the following measurements: a) AB = 4.2 cm, BC = 3.6 cm, and CA = 5.5 cm b) AB = 5.4 cm., BC = 6.3 cm and CA = 4.5 cm c) AB = 5.7 cm, ∠ A = 45°, and AC = 4.2 cm d) BC = 4.8 cm, ∠ B = 60° and CA = 6.8 cm e) BC = 3.6 cm, ∠ B = 30°, and ∠ C = 45°. 2. Construct quadrilaterals ABCD by using the following measurements: a) AB = 3.5 cm, BC = 4.2 cm, AC = 6.3 cm, CD = 2.5 cm, and AD = 5.2 cm b) AB = 4.6 cm, BC = 3.9 cm, AC = 6.5 cm, CD = 3.7 cm, and AD = 4.8 cm c) AB = 4.5 cm, BC = 4.2 cm, AC = 5.8 cm, BD = 6.4 cm, and AD = 5.3 cm d) AB = 3.8 cm, ∠ B = 120°, BC = 4.6 cm, CD = 5.4 cm, and AD = 5 cm e) AB = BC = 5.5 cm, CD = AD = 6.2 cm, and ∠ BAD = 75°. 3. Construct parallelograms ABCD by using the following measurements: a) AB = 5.6 cm, BC = 4.5 cm, and ∠ ABC = 60° b) AB = 4.3 cm, BC = 3.5 cm, and ∠ BAD = 120° c) AB = 4.5 cm, BC = 5.5 cm, and AC = 6.7 cm d) AB = 3.8 cm, ∠ ABD = 60°, and AD = 5.2 cm e) AB = 5.5 cm, diagonal AC = 7.6 cm, and diagonal BD = 5.6 cm. Creative section 4. a) Construct a triangle ABC having sides a = 6.4 cm, b = 6cm, and c = 5.6 cm. Also construct another triangle equal in area to the DABC and having a side 7 cm. b) Construct a triangle PQR in which PQ = 5.2 cm, QR = 6 cm and ∠Q = 60°. Construct another triangle SPQ which is equal in area to the DPQR and side PS = 8.5 cm.

Vedanta Excel in Mathematics - Book 10 238 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Geometry - Construction c) Construct a triangle XYZ in which XY = 6.3 cm, ∠X = 30°, and ∠Y = 45°. Construct another triangle WXY equal in area to D XYZ and a side WY = 7.5 cm. 5. a) Construct a parallelogram ABCD in which AB = 3.5 cm, AC = 5.9 cm, and BC = 4.2 cm. Construct another parallelogram ABEF where BE = 6.5 cm. b) Construct a parallelogram PQRS in which PQ = 4 cm, ∠P = 45°, and PS = 4.5 cm. Construct another parallelogram PQXY where ∠PQX = 30°. 6. a) Construct a triangle in which a = 7 cm, b = 6 cm and c = 5 cm and also construct a parallelogram whose area is equal to the area of triangle ABC and one of the angles being 60°. b) Construct a triangle ABC in which a = 7.8 cm, b = 7.2 cm and C = 6.3 cm. Then construct a parallelogram DBEF equal in area to D ABC and ∠ DBC = 75°. c) Construct a triangle PQR in which PQ = 7.5 cm, QR = 6.8 cm, and RP = 6 cm. Then construct a parallelogram TQSU equal in area to D PQR having a side TQ = 6.4 cm. d) Construct a triangle ABC from the following data. Also construct a parallelogram having one side 7.5 cm and equal in area to the triangle. AB = 7.1 cm, ∠ BAC = 60° and AC = 5.7 cm e) Construct a triangle ABC in which AB = 5 cm, ∠ABC = 30°and ∠ ACB = 60°. Then, construct a rectangle equal in area to the triangle ABC. f) Construct a triangle ABC having sides AB = 4 cm, BC = 6.8 cm, and CA = 6.5 cm. Then, construct a rectangle equal in area of D ABC. 7. a) Construct a rectangle with length 7.1 cm and breadth 6.1 cm. Also, construct a triangle having one angle 60° and equal to the area of the rectangle. b) Construct a rectangle ABCD with length 5.5 cm and breadth 4.5 cm. Also, construct a triangle EBF having one angle 45° and equal to the area of the rectangle. 8. a) Construct a parallelogram ABCD in which AB = 6 cm, BC = 4 cm and ∠BAD = 45°. Construct a triangle APQ having one angle 60° and equal in area of the parallelogram. b) Construct a parallelogram ABCD having AB = 5 cm, BC = 4 cm, and ∠B = 120°. Also construct a triangle with the area equal to the parallelogram.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 239 Vedanta Excel in Mathematics - Book 10 Geometry - Construction c) Construct a parallelogram ABCD in which AB = 6 cm, BC = 4 cm and ∠ BAD = 60°. Then construct a triangle AEF equal in area to the parallelogram and the side AE = 7.5 cm. d) Construct a parallelogram PQRS in which PQ = 5 cm, diagonal PR = 6 cm, and diagonal QS = 8cm. Construct a triangle PSA whose area is equal to the area of the parallelogram. 9. a) Construct a quadrilateral PQRS in which PQ = QR = 5 cm, RS = SP = 4 cm, and ∠P = 60°. Then, construct a triangle whose area is equal to the area of the quadrilateral. b) Construct a quadrilateral PQRS having PQ = QR = 5.9 cm, RS = PS = 6.1 cm and ∠QPS = 75°. Then, construct D PST which is equal in area to the given quadrilateral. c) Construct a quadrilateral MNOP in which MN = 4 cm, NO = 5 cm, OP = 5.5 cm, PM = 3.5 cm, and ∠PMN = 30°. Construct a triangle MPQ whose area is equal to the quadrilateral MNOP. d) Construct a quadrilateral ABCD having AB = 3.5 cm, BC = 4.8 cm, CD = 2.6 cm, AD = 4 cm, and AC = 5.5 cm. Construct D ABE whose area is equal to the quadrilateral ABCD. e) Construct a quadrilateral ABCD in which AB = 4 cm. BC = 5 cm, CD = 5.5 cm, DA = 4.5 cm, and diagonal AC = 6 cm. Also construct a triangle ADE equal in area to the quadrilateral. f) Construct a quadrilateral ABCD in which AB = 5.4 cm, BC = 5.1 cm, CD = 4.9 cm, AD = 6.1 cm, and the diagonal BD = 5.7 cm. Also construct a triangle equal in area to the quadrilateral ABCD. g) Construct a quadrilateral PQRS in which QR = 5.3 cm, RS = 5 cm PS = 5.7 cm, PQ = 6.2 cm, and PR = 5.6 cm. Then, construct a triangle RQT equal in area to the quadrilateral PQRS. h) Construct a DWXB equal in area to the quadrilateral WXYZ having WX = XY = 5.5 cm, YZ = ZW = 4.5 cm, and ∠WXY = 75°.

Vedanta Excel in Mathematics - Book 10 240 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Unit 13 Geometry - Circle Classwork - Exercise 13.1 Circle-Looking back 1. Study the given circle, tick (√) the correct alternative for the questions from the brackets. a) O is the …………… of the circle. (circle, centre, circumference) b) OA, OB and OC are………………… (radii, diameters, chords) c) AB and BC are ………………… (chords, radii, diameters) d) The chord AB is a ………………… (radius, diameter) e) The shaded region AOC is a ………………… (sector, segment) f) The shaded region BCD is a ………………… (segment, sector) 2. In the circle with centre O; the perpendicular line segments OM and ON drawn from centre to the chords PQ and RS respectively are equal. Answer the following questions. a) What is the relation between PM and MQ? b) What is the relation between the chords PQ and RS? 13.2 Definition of terms related to circle A circle is the locus traced out by a moving point which is equidistant from a fixed point called the centre of the circle. In the figure, O is the centre of a circle. (i) Circumference The perimeter of a circle is called its circumference. It is the total length of the curved line of the circle. (ii) Radius It is the line segment that joins the centre of a circle and any point on its circumference. In the figure, OA is the radius. The plural form of radius is radii. All the radii of a circle are equal, i.e., OA = OB = OC = OD = ... A D B C O Q R P O M N S P1 P2 P4 P3 O B A O C D

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 241 Vedanta Excel in Mathematics - Book 10 Geometry - Circle (iii) Diameter A line segment that passes through the centre of a circle and joins any two points on its circumference is called the diameter of the circle. In the figure PQ, RS, etc., are the diameters. The length of a diameter is two times radius. So, PQ = 2OQ or 2OP and RS = 2OR or 2OS. A diameter divides a circle into two halves. (iv) Semi-circle A diameter divides a circle into two halves and each half is called the semi-circle. In the figure, PXQ and PYQ are the semi-circles. (v) Arc A part of the circumference of a circle is called an arc. It is denoted by a symbol ‘ ‘. In the figure, PQR is the minor and PSR is major arcs of the circle. A minor arc is less than half of the circumference and a major arc is greater than half of the circumference. (vi) Chord The line segment that joins any two points on the circumference of a circle is called the chord of the circle. In the figure, PQ and RS are any two chords of the circle. Diameter is the longest chord of a circle. (vii) Segment The region enclosed by a chord and corresponding arc is called the segment of a circle. In the figure, the shaded region is the minor segment and non-shaded region is the major segment of the circle. A minor segment is less than half of a circle whereas the major segment is greater than half of the circle. The minor and the major segments of a circle are known as the alternate segments to each other. (viii) Sector The region enclosed between any two radii and the corresponding arc of a circle is called a sector of the circle. In the figure, OPQ is a sector of a circle. (ix) Concentric circles Two or more circles are said to be concentric circles if they have the same centre but different radii. In the figure, circles PQR, XYZ, and ABC are three concentric circles. P Q R S O P Q X Y O P R Q S O P Q R S O P Q O P Q O Q P X A O C Z R B Y

Vedanta Excel in Mathematics - Book 10 242 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Geometry - Circle (x) Intersecting circles If two circles intersect each other at two points, they are said to be intersecting circles. In the figure, two circles are intersecting each other at P and Q. PQ is the common chord of these two intersecting circles. xi) Concylic points and cyclic quadrilateral The points lying on the circumference of a circle are called concylic points. In the given circle, A, B, C, D, E and F are concyclic points. When all the vertices of a quadrilateral lie on the circumference of a circle, the quadrilateral is called cyclic quadrilateral. In other words, the quadrilateral formed by joining any four concylic points in order is called cyclic quadrilateral. In the given circle; ABCD is a cyclic quadrilateral. xii) Central angle An angle at the centre of a circle formed by any two radii is called central angle. In the given figure, ∠AOB is a central angle. xiii) Inscribed angle An angle at the circumference of a circle formed by any two chords is called inscribed angle. In the given figure, ∠ACB is an inscribed angle. 13.3 Theorems related to arcs and the angles subtended by them In the figure alongside, O is the centre of a circle, OP and OQ are the radii, and PQ is the arc of the circle. Here, PQ subtends an ∠POQ at O. So, ∠POQ is the angle at the centre of the circle. Of course, ∠POQ is standing on PQ. So, we may call the opposite arc of ∠POQ is PQ. Furthermore, if we consider the radius OP starts rotating from the point P to the point Q, it describes an arc PQ and ∠POQ. Thus, the degree measurements of PQ and ∠POQ are exactly the same. ∴ ∠POQ PQ Thus, the degree measurement of the angle at the centre of a circle is equal to the degree measurement of its opposite arc. O P Q O' B C A O D E F A B O B C A O P Q O

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 243 Vedanta Excel in Mathematics - Book 10 Geometry - Circle Again, ∠PAQ is subtended by the arc PQ at the circumference of the circle. ∠PAQ is said to be an angle at the circumference of the circle. It is also called inscribed angle. Here, the opposite arc of ∠PAQ is PQ. Now, let's establish the relation between an inscribed angle and its opposite arc. In ∆OAP, ∠OAP = ∠OPA [∴ OA = OP] ∴ 2∠OAP = 180° – ∠AOP [From sum of the angles of ∆OAP] In ∆OAQ, ∠OAQ = ∠OQA [OA = OQ] ∴ 2∠OAQ = 180° – ∠AOQ [From sum of the angles of ∆OAQ] Now, 2∠OAP + 2∠OAQ = 180° – ∠AOP + 180° – ∠AOQ or, 2∠PAQ = 360° – (∠AOP + ∠AOQ) or, 2∠PAQ = 360° – Ref. ∠POQ or, 2∠PAQ circumference – PAQ [Circumference 360° and PAQ Ref. ∠POQ] or, 2∠PAQ PQ ∴ ∠PAQ 1 2 PQ. Thus, the degree measurement of the angle at the circumference of a circle is half of the degree measurement of its opposite arc. Theorem 6 (Concept only) Equal arcs of a circle subtend equal angles at the centre of the circle. Activity Steps: (i) Draw a circle with a suitable radius on a chart paper and cut it out. (ii) Mark the equal arcs AB and CD by using compass. (iii) Cut out the sectors having equal arc. A B C D O A B O A B O D C O D C O A D B C O (iv) Place one sector COD above another sector AOB. (v) Observe the central angles ∠AOB and ∠COD coinciding each other. P Q A P Q O A Activity

Vedanta Excel in Mathematics - Book 10 244 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Geometry - Circle Experimental verification Step 1: Three circles with centre O and different radii are drawn. Step 2: Two equal arcs AB and CD are drawn with the help of pencil and compasses. Step 3: ∠AOB and ∠COD are measured and the results are tabulated. Figure ∠AOB ∠COD Result (i) ∠AOB = ∠ COD (ii) ∠AOB = ∠ COD (iii) ∠AOB = ∠ COD Conclusion: Equal arcs of a circle subtend equal angles at the centre of the circle. Converse of Theorem 6 (Concept only) If two arcs of a circle subtend equal angles at the centre of the circle, the arcs are equal. Activity Steps: (i) Draw a circle with a suitable radius on a chart paper and cut it out. (ii) Draw two angles ∠AOB and ∠COD with equal measurements at the centre O of circle. C D A B O A B O D C O D C O A D B C O A B O (iii) Cut out the sectors having equal central angles. (iv) Overlap the sector COD on the sector AOB then observe the arcs AB and CD. Experimental verification Step 1: Three circles with centre O and different radii are drawn. Step 2: Two equal angles AOB and COD are drawn at the centre of each circle. O O O A D B D B (i) (ii) (iii) C B D A C A C Activity O O O A D D D B (i) (ii) (iii) B B A A C C C

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 245 Vedanta Excel in Mathematics - Book 10 Geometry - Circle Step 3: The lengths of AB and CD are measured with the help of thread and ruler and the results are tabulated. Figure arc AB arc CD Result (i) AB = CD (ii) AB = CD (iii) AB = CD Conclusion: If two arcs of a circle subtend equal angles at the centre of the circle, the arcs are equal. Theorem 7 (Concept only) Equal chords of a circle form equal arcs in the circle. Activity Steps: (i) Draw a circle with a suitable radius on a chart paper and cut it out. (ii) Draw equal chords in the circle. (iii) Cut out the segments having equal chords. (iv) Overlap the sectors one over another and observe the corresponding arcs. A C B D A C B D B = D A C A = C B D Experimental verification Step 1: Three circles with centre O and different radii are drawn. Step 2: Two equal chords AB and CD are drawn in each circle. Step 3: The length of arcs AB and CD are measured with the help of thread and ruler and the results are tabulated. Figure arc AB arc CD Result (i) AB = CD (ii) AB = CD (iii) AB = CD Conclusion: Equal chords of a circle form equal arcs in the circle. A D D D B (i) (ii) (iii) B B A A C C C

Vedanta Excel in Mathematics - Book 10 246 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Geometry - Circle Converse of Theorem 7 (Concept only) If the arcs formed by the chords of a circle are equal, then the chords are equal. Experimental verification Step 1: Three circles with centre O and different radii are drawn. Step 2: Two equal arcs AB and CD are drawn with the help of pencil and compasses in each circle. Step 3: A, B and C, D are joined to form chords AB and CD. Step 4: The lengths of chords AB and CD are measured and the results are tabulated. Figure AB CD Result (i) AB = CD (ii) AB = CD (iii) AB = CD Conclusion:If the arcs formed by the chords of a circle are equal, then the chords are equal. Theorem 8 (Concept only) If two chords are parallel, then the arcs between them are equal. Activity Steps: (i) Draw a circle with a suitable radius on a chart paper and cut it out. (ii) Draw two parallel chords AB and CD in the circle, fold it and cut out the segments ABP and CDQ. A C B D O A C B P Q D O A C B M N D O B, A M N D, C O (iii) Fold the remaining portion ABCD along MN through the centre O so that A coincides to B and C coincides to D. We observe that the arcs between the parallel chords are equal. Converse of theorem 8 (Concept only) If two arcs are equal, then the chords between them are parallel. Steps: (i) Draw a circle with a suitable radius on a chart paper and cut it out. (ii) By using thread or compass, take two equal arcs AB and CD on it. As shown in the figure. Then, Join AD and BC. A D D D O O O B (i) (ii) (iii) B B A A C C C A B C D O

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 247 Vedanta Excel in Mathematics - Book 10 Geometry - Circle (iii) Join BD and measure the alternate angles ADB and CBD. (iv) The alternate angles ADB and CBD are equal. Thus, we can conclude that AD // BC. Facts to remember 1. Equal arcs of a circle subtend equal angles at the centre. 2. Equal arcs of a circle subtend equal angles at the circumference. 3. In a circle, arcs subtending equal angles at the centre are equal. 4. In a circle, arcs subtending equal angles at the circumference are equal. 5. Equal chords of a circle form equal arcs. 6. Chords formed by equal arcs of a circle are equal. 7. If two chords are parallel, then the arcs between them are equal and vice versa. Worked-out Examples Example 1: In the given figure, O is the centre of circle. If ∠AOB = 95o , find the degree measures of arc AB and arc ACB. Solution: Here, O is the centre of circle and ∠AOB = 95o . Now, degree measure of arc AB ∠AOB [Relation between the central angle and its corresponding arc] = 95o Again, degree measure of arc ACB degree measure of circumference – degree measure of arc AB 360o – 95o = 265o Thus, the degree measures of arc AB is 95o and arc ACB is 265o . Example 2: In the adjoining figure, AB // CD and ∠BED = 25o , find the size of angle x. Solution: Here, AB // CD and ∠BED = 25o Now, arc AC = arc BD [∵AB// CD and they subtend equal arcs] Again, ∠AEC = ∠BED [The inscribed angles subtended by equal arcs are equal] or, x = 25o Thus, the value of x is 25o . A O B C 95° A D x 25° E B C

Vedanta Excel in Mathematics - Book 10 248 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Geometry - Circle Example 3: In the adjoining circle, AD // BC. Prove that ∠AYC = ∠BXD. Solution: Given: In the given circle, AD // BC. To prove: ∠AYC = ∠BXD Proof Statements Reasons 1. AB = CD 1. AD // BC and they subtend equal arcs. 2. AB + BC = BC + CD 2. Adding BCto both sides of statement (1). 3. ABC = BCD 3. Whole part axiom 4. ∠AYC= ∠BXD 4. Circumference angles of the circle standing on the equal arcs. Proved Example 4: In a circle centered at O; AB is a diameter. C and D are two points on the circumference lying on the same side of the diameter AB such BC = CD. Prove that: (i) AD // OC. (ii) Area of DAOC = Area of DDOC Solution: Given: O is the centre of circle, AB is the diameter and BC = CD To prove: AD // OC Proof: S.N. Statements S.N. Reasons 1. ∠BOC BC 1. Relation between the centre angle and its opposite arc. 2. ∠BAD 1 2 BCD 2. Relation between the inscribed angle and its opposite arc. 3. ∠BAD 1 2 × 2BC = BC 3. From statement(2) and BCD = 2BC 4. ∠BAD = ∠BOC 4. From statements (1) and (3) 5. AD//OC 5. From statement (4), the corresponding angles are equal. 6. DAOC = DDOC 6. Both are standing on the same base OC and between AD // OC. Proved Example 5: In the figure alongside; the chords MN and RS intersect at an external point X. Prove that ∠MXR 1 2 (MR – NS). A D X B C Y A O B C D M N X S R

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 249 Vedanta Excel in Mathematics - Book 10 Geometry - Circle Solution: Given: Two chords MN and RS intersect at an external point X. To prove: ∠MXR 1 2 (MR – NS) Proof: S.N. Statements S.N. Reasons 1. ∠MNR 1 2 MR and ∠NRX 1 2 NS 1. Relation between the inscribed angles and their opposite arcs. 2. ∠MNR = ∠NRX + ∠MXR 2. The exterior angle of ∆NRX is equal to the sum of its two opposite interior angles. 3. 1 2 MR 1 2 NS + ∠MXR 3. From statements (1) and (2). 4. ∠MXR 1 2 MR – 1 2 NS 1 2 (MR – NS) 4. From statement (3). Example 6: In the given figure, two chords PQ and RS intersect at right angle at the point X. Prove that: arc (PS – QS) = arc (PR – QR) Solution: Given: Chords PQ and RS intersect at right angle at the point X. To prove: arc (PS – QS) = arc (PR – QR) Construction: P, S are joined. Proof Statements Reasons 1. ∠SPQ 1 2 QS 1. Relation between inscribed angle and its opposite arc. 2. ∠PSR 1 2 PR 2. Same reason as above 3. ∠SPQ + ∠PSR 1 2 (QS + PR ) 3. Adding statements (1) and (2) 4. 90° 1 2 (QS + PR ) i.e. QS + PR 180° 4. ∆PXS is a right angled triangle and ∠X is a right angle and the sum of two acute angles of a right angled ∆ 5. PS + QR 180° 5. PS + SQ + QR + RP 360° and from statement (4) 6. PS + QR = QS + PR i.e. arc (PS – QS) = arc (PR – QR) 6. From statements (4) and (5) Proved P Q R S X P Q R S X

Vedanta Excel in Mathematics - Book 10 250 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Geometry - Circle Example 7: Points P, Q, R and S are concyclic such that arc PQ = arc SR. Prove that: (i) Area of ∆PQR = Area of ∆SQR (ii) chord PR = chord QS Solution: Given: Arc PQ = Arc SR To prove: (i) Area of DPQR = Area of ∆SQR (ii) chord PR = chord QS Proof: S.N. Statements S.N. Reasons 1. PS // QR 1. PQ = SR 2. Area of ∆PQR = Area of ∆SQR 2. They are standing on the same base QR and between PS // QR. 3. PQ = SR 3. Given 4. PQ + QR = QR + SR 4. Adding QR in statement (3) 5. PQR= QRS 5. By whole part axiom. 6. chord PR = chord QS 6. From statement (5), equal arcs subtend equal chords. Proved EXERCISE 13.1 General section 1. In the given figure, O is the centre of circle. Answer the following questions. a) Write the relation between the central angle AOB and its corresponding arc AB. b) Write the relation between the inscribed angle ACB and its corresponding arc AB. c) If ∠AOB = 92o , what is the degree measure of arc AB? d) If ∠ACB = 46o , what is the degree measure of arc AB? 2. Study the given circle centered at O. Answer the following questions. a) Which are the equal arcs? b) Which are the equal chords? c) If the length of chord PQ is 2.4 cm, what is the length of the chord RS? d) If the length of arc PQ were 3 cm, what would be the length of arc RS? 3. In the given circle, O is the centre and arcs MP and NQ are equal. a) Identify and write the name of angle which is equal to ∠MOP. b) Name the arc which is equal to arc NQ. c) Write the relation between the chords MN and PQ. d) If the degree measure of arc MP is 60o , what is the measure of ∠NOQ? P S R Q O C B A O R S 70° 70° Q P O N P Q M