Oasis School Mathematics-10 197 (b) 4x – 6×2x + 1 + 32 = 0 (d) 4x + 1 4x = 16 1 16 (e) 52x + 2 – 126×5x + 5 = 0 (f) 4×3x+1 – 9x = 27 (g) 2x+3 + 1 2x – 9 = 0 (h) 5×4x+1 – 16x = 64 (i) 7x + 343 7x = 56 (j) 3x+2 + 1 3x-2 = 30 (k) 5x–1 + 52–x = 6 (l) 4x–1 + 42–x = 5 (m) 16x – 3×22x + 2 = 0 (n) 5x+1 + 52–x = 126 6. If a = bx , b = cy and c = az , prove that xyz = 1. 7. If ax = by and bx = ay , show that x = y. 8. If a b c x y z 1 1 1 = = and abc = 1, prove that x + y + z = 0 . 9. If ax = by = cz and b2 = ac, prove that y = 2xz x+z . 10. If 2x = 3y = (12)z , prove that, 1 z – 1 y = 2 x . 11. If 8712 = 2x × 3y × 11z , find x, y and z. 12. If 2400 = 2x .3y .5z , find x, y and z. 13. (a) Why the equation 7x + 1 7x = 49 1 49 is an exponential equation? (b) Assuming 7x = a, convert the above equation, quadratic in ‘a’. (c) Solve the quadratic equation to get values of x. (d) Prove that these value of x also satisfy the equation 2x – 2² = 2–2 – 1 2x . 14. (a) What type of equation is 3x + 3–x = 10 3 ? (b) Which two values of x, satisfy the above equation? (c) Prove that these values also satisfy the equation 5x + 1 5x = 5 1 5 . Answers 1. Consult your teacher 2. (a) 6 (b) –3 (c) 7 (d) 4 (e) 1 (f) 1 3. (a) 3 2 (b) 5 (c) 3 2 (d) -1 11 (e) 1, -3 2 (f) 2 (g) 4 (h) -2 (i) 6 (j) 6, -2 (k) ± 2 4. (a) 3 2 (b) 1 3 (c) 2 (d) -2 (e) 1 (f) 0 (g) 0 (h) -3 2 (i) 5 (j) -2 5. (a) 0, 1 (b) 2, 3 (c) ± 2 (d) ± 2 (e) 1, -2 (f) 1, 2 (g) 0, -3 (h) 1, 2 (i) 1, 2 (j) ± 1 (k) 1, 2 (l) 1, 2 (m) 1 2 , 0 (n) -1, 2 11. 3, 2, 2 12. 5, 1, 2 13. (c) 2, –2 14. (b) 1, –1 Project Work Collect some problems related to the indices in our daily life and present them in your class room.
198 Oasis School Mathematics-10 Miscellaneous exercise Sequence and Series 1. There is an arithmetic means between 4 and 24. (a) What is the A.M. between 4 and 24? (b) If there are n AM’s between them such that third mean to the last mean is 4:5, find the value of d. (c) Find the number of means (n). (d) Find the other means also. [Ans: (a) 14 (b) 4 (c) 4 (d) 8, 12, 16, 20] 2. A G.P has 8 terms where common ratio is positive. (a) If the sum of first 4 terms is 45 and the sum of last 4 terms is 720, what is the sum of all 8 terms? (b) Using the conditions given in (a), find the value of r. (c) Find the first term (a). [ Ans: (a) 765 (b) 3 (c) 2] 3. The sum of 3 numbers in A.P is 15. (a) If 2 is subtracted from the second term, the number are G.P; find the numbers. (b) If 1 is added to the first and second term and third term is multiplied by 2. What type of series is formed? Explain with reason. 4. A sum of Rs. 2750 is distributed among 10 students according to their performance in the examination; If the second prize is Rs. 50 less than the first price, third prize is Rs. 50 less than the third prize and so on. (a) Which type of sequence is formed in the above information? Write all the informations given in the questions. (b) How much prize does the first student get? (c) How much prize does the student of 10th rank get? (d) If the same format of prize is distributed among 10 students with first prize of 800, how much money is required? [ Ans: (b) Rs.500 (c) Rs. 50 (d) Rs. 5750] Quadratic equation 1. The difference of the present ages of a brother and his sister in 2 years. (a) Convert this statement into equation. (b) If the product of their ages is equal to six times the sum of their ages, find their present ages. [ Ans: 10 yrs, 15 yrs] 2. The product of the present ages of two brothers in 160. (a) Convert this statement into equation. (b) 4 years ago the elder brother was twice an old as his younger brother, find their ag es. [Ans: 16 yrs, 10 yrs]
Oasis School Mathematics-10 199 3. A said to B, “I was twice as old as you were when I was as old as you are now”. (a) How many years ago A was as old as B? Convert this statement into equation. (b) If B replied to A “10years ago product of ages was 200 years”. Convert this statement into equation. (c) Solve above two equations to find the age of A and B. [Ans: (c) 30 years, 20 years] 4. Present ages of father and his son are x years and y years respectively. When the father was as old as his son is now, (a) How many years ago father was as old as his on is now? (b) What was the ages of father and his son, when the father was as old as his son now? (c) Convert the above statement into the equation. (d) 20 years ago product of their ages was 800, find their present ages. [Ans: (d) 60 years, 40 years] 5. In a two – digit number, if the digit in the place of tens is x and digit in the place of ones is y, (a) What is the number? Write in terms of x and y. (b) If the number is 3 times the product of the digits and if 18 is added to the number, the digits are reversed, convert above two statement into equations.’ (c) Solve two equations to find the required number. [Ans: (c) 76] 6. A number of two – digits is 8 less than 2 times the product of the digits, (a) Convert above statement into equation. (b) If the number is added to the number formed by reversing its digit, the sum is 143. Convert this statement into equation. (c) Solve above two equation to get the required number. [Ans: (c) 76] 7. A number of two – digits is obtained by two ways. (i) Multiplying the sum of the digits by 8 and adding 3. (ii) Multiplying the product of the digits by 10 adding 1. a) Convert above two statements into equation. (b) Solve the equations of get the required number. [Ans: (b) 51] 8. Length and breadth of a rectangular field is 20m and 10m. It is surrounded by a path of uniform width ‘d’ cm where area is136cm². (a) Write the formula to get the area of a path around the rectangular field. (b) Solve the equation to get value of ‘d’. (c) What would be the area of path if 2m wide path is running inside the field? [(b) 2m (c) 104 m²] 9. A rectangular field has dimension 22m × 12m. If a path of uniform width ‘d’ is running inside it, (a) Write the formula to calculate the area of path inside it. (b) If the area of the path is 64m², find the value of ‘d’.
200 Oasis School Mathematics-10 (c) By what percent area of path increased if the path of same width is running outside the field? [Ans: (b) 1m (c) 12 1 2 %) 10. The perimeter of a rectangular field is 32m. Another rectangular field has same perimeter but the breadth is 2m less than the first field. (a) How many meters more or less, the length of second field is? (b) Calculate the area of second field in terms of length and breadth. (c) If the area of second field is 12m² less than the area of the first field, convert this statement into equation. (d) Find the length and breadth of first field. [ Ans: 12m. 4m] 11. Some students of class X has reserved the bus for their picnic with total budget of Rs. 800. If 10 students didn’t attend the picnic, each would have to contribute Rs. 30 more. (a) Find the number of students and the amount that each has to contribute. (b) What amount should each of them have to contribute if 6 more students join the picnic? [Ans: (a) 30 students, Rs. 60 (b) Rs. 50] 12. The capacity of a rectangular hall is 600. (a) If the number of rows of a hall is 10 less than the number of columns, convert the above statement into equation. (b) Solve the equation to get the number of rows and columns in the hall. [Ans: Rows: 30, Columns: 20] 13. First bus starts journey from ‘A’ and second bus starts its journey from point B. (a) If they move in opposite direction, they will meet after 4 hours. Length of AB = 400 km, speed of the first bus is x km/hr and that of second in y km/hr. convert this statement into equation. (b) If first bus move towards second bus and second bus move to the same direction, first bus will overtake the second bus after 16 hours. Convert this statement into equation. (c) Solve the equations obtained from (a) and (b) to get the speed of both bus. [Ans: (c) 62.5km/hr and 37.5 km/hr ] 14. Distance between two places ‘A’ and ‘B’ is 600 km. If the speed of a car is x km/hr and that of bus is y km/hr and speed of the car is 20km/hr more than that bus (a) Convert this statement into equaton.
Oasis School Mathematics-10 201 (b) Can takes 5 hours less time to cover that distance, Convert this statement into equation. (c) Solve the equations obtained in (a) and (b) to find the speed of car and bus. [Ans: (c) 40 km/hr, 40 km/hr] rational Fractions 1. (a) Prove that 1 (a – b) (a – c) = 1 (a – b) (b – c) – 1 (b – c) (a – c). (b) Using above relation, simplify the expressions a² (a – b) (a – c) + b² (b – c) (b – a) + c² (c – a) (c – b) [Ans: 1] (c) Using relation of (a), prove that bc (a – b) (a – c) + ca (b – a) (b – c) + ab (c – a) (c – b) [Ans: 1] 2. (a) Simplify the rational fraction a² – (2b – 3c)² (a + 3c)² – 4b² (b) Following the same method, simplify 4b² – (3c – a)² (a + 2b)² – 9c² and 9c² – (a – 2b)² (2b + 3c)² – a². (c) Simplify the expression a² – (2b – 3c)² (a + 3c)² – 4b² + 4b² – (3c – a)² (a + 2b)² – 9c² + 9c² – (a – 2b)² (2b + 3c)² – a² [Ans: 1] 3. (a) Simplify the rational fraction x – 3 x² – x – 6 . (b) Using similar method, simplify 2x – 1 2x² + 5x – 3 and 2x – 1 x² – x – 6 (c) Simplify the expression x – 3 x² – x – 6 + 2x –1 2x² + 5x – 3 – 2x –1 x² – x – 6 [ Ans: 0]
202 Oasis School Mathematics-10 Full marks : 23 1. A Sequence 3, 5, 7, 9, .............is given. (a) What are the values of the first term and the common difference? (1) (b) Calculate its 10th term. (1) (c) Calculate the sum of first 10 terms. (3) (d) If the terms are in the order 3, 6, 12, 24, .............. . What type of sequence is this? (1) 2. The product of the ages of father and his son is 96 years. (a) Convert this statement into mathematical sentence. (1) (b) After how many years, son will be as old as his father now? (1) (c) At that time, if the sum of the ages of father and son is 68 years, find their present ages. (3) (d) What will be the product of their ages after 5 years? (1) 3. (a) Factorise the algebraic expression 16a4 + 4a2 b2 + b4 . (1) (b) What is the L.C.M. of 4a2 + 2ab + b2 and 4a2 - 2ab + b2 ? (1) (c) Using above two results. Simplify the rational expressions. (3) 2a + 6 4a2 + 2ab + b2 + 2a – b 4a2 – 2ab + b2 – 2b3 16a4 + 4a2 b2 + b4 4. (a) What is the meaning of exponential equation? (1) (b) Convert the equation 3x+2 + 1 3x - 2 = 30, into quadratic in 3x . (1) (c) Assuming 3x = a; solve the above equation. (3) (d) Check whether both values satisfy the equation or not. (1)
Oasis School Mathematics-10 203 Geometry Contents • Area of triangle and quadrilateral • Construction • Circle expected Learning outcomes At the end of this unit, students will be able to develop the following competencies: • To find the area of triangles and quadrilaterals • To prove theorems related to area of triangle and parallelograms • To solve the problems using theorems related to area of triangle and parallelogram • To construct a triangle and different types of quadrilateral equal in area with given triangle or quadrilateral • To verify the theorems on a circle experimentally • To prove the theorems on a circle • To solve problems using theorems of circle • To solve the problems of tangents and circles using the properties of a tangent materials required • set square, compass, scale, protractor, pencils, etc. Estimated Teaching Hours 28
204 Oasis School Mathematics-10 12.1 Warm-up Activities • What is the formula to calculate the area and perimeter of a rectangle? • What is the formula to calculate the area and perimeter of a square? • If 'd' is the diagonal of a square, what is its area in terms of 'd'? • Is every rhombus a parallelogram? • Is every rectangle a parallelogram? • If d1 and d2 are the diagonals of a rhombus, what is its area? Discuss the following in your class and draw a conclusion. 12.2 Base and Altitude of triangle and Parallelogram Any side of a triangle can be considered the base. The perpendicular drawn from the vertex to the base is the altitude. In ∆ABC, AD⊥BC, BE⊥AC and CF ⊥AB. If AC is the base, BE is the altitude. If AB is the base, CF is the altitude and if BC is the base, AD is the altitude. Similarly, any side of a parallelogram can be considered the base. The perpendicular drawn from the vertex to the base is the altitude. The perpendicular distance between two parallel sides is the altitude of the parallelogram. In parallelogram ABCD, AE⊥DC and AF ⊥ BC. If DC is the base, AE is the altitude. If BC is the base, AF is the altitude. 12.3 Area of triangle (i) When the base and altitude are given, In this case the area of a triangle = 1 2 base × altitude. A B D C P Q R X W Y Z A E O C D B F A B F C E D Unit 12 Area of triangles and Quadrilaterals
Oasis School Mathematics-10 205 Area of ∆ABC = 1 2 BC × AD, Area of ∆PQR = 1 2 QR × PQ Area of ∆XYZ = 1 2 YZ × XW In a right angled triangle, two perpendicular sides are taken as the base and altitude. (ii) When three sides of a triangle are given: In this case, area of a triangle = s s( ) − − a s( ) b s( ) − c Where, s = a + b + c 2 = Semi perimeter of a triangle. 12.4 Area of Parallelogram In the given parallelogram ABCD, if DC is the base, AE is the altitude. Area of a parallelogram = base × altitude = DC × AE remember ! • Areas of two congruent figures are equal. • Any side of a parallelogram can be taken as its base. • A line passing through any vertex and perpendicular to the base of a parallelogram is called the altitude of a parallelogram. In the given parallelogram ABCD, • If BC is the base, AE is the altitude. • If CD is the base, AF is the altitude. A F D B C E theorem 12.1 D A C B A diagonal of a parallelogram divides it into two triangles of equal area. Given : BD is the diagonal of the parallelogram ABCD. to prove : Area of ∆ABD = Area of ∆DBC. Proof: S.N. Statements S.N. reasons 1. In ∆ABD and ∆BDC, (i) AB = DC (S) (ii) ∠ABD = ∠BDC (A) (iii) BD = BD (S) 1. (i) Being opposite sides of a parallelogram (ii) Being alternate angles on AB//DC (iii) Being a common side 2. ∆ABD ≅ ∆BDC 2. By Side-Angle-Side axiom 3. ∆ABD = ∆ BDC 3. Area of congruent triangles Hence, the diagonal of a paralleogram divides it into two triangles of equal area. A c B a b C B A D E C Base altitude
206 Oasis School Mathematics-10 Alternative method Given : ABCD is a parallelogram in which AC is the diagonal. to prove : ∆ABC = ∆ACD Construction : Draw AH⊥BC, CK⊥AD. Proof: S.N. Statements S.N. reasons 1. ∆ABC = 1 2 AH×BC 1. Area of triangle = 1 2 base × height. 2. ∆ACD = 1 2 AD×CK 2. Area of triangle = 1 2 base × height. 3. AH = CK 3. Distance between the same parallel lines. 4. ∆ABC = ∆ACD 4. From (1), (2) & (3). Hence, a diagonal of a paralleogram divides it into two triangles of equal area. theorem 12.2 Parallelograms on the same base and between the same parallel lines are equal in area. Given : Parallelogram ABCD and parallelogram EBCF are standing on the same base BC and between the same parallel lines AF and BC. to prove : Area of ||gm ABCD = Area of ||gm EBCF. Proof : Hence, parallelogams on the same base and between the same parallel lines are equal in area. A K D C B H A B C E D F S.N. Statements S.N. reasons 1. In ∆ABE and ∆DCF, (i) ∠AEB = ∠DFC (A) (ii) ∠EAB = ∠CDF (A) (iii) AB = DC (S) 1. (i) Being corresponding angles on parallel lines (EBІІFC) (ii) Being corresponding angles on parallel lines (ABІІDC) (iii) Opposite sides of a parallelogram are equal. 2. ∆ABE ≅ ∆DCF 2. By A.A.S. axiom 3. ∆ABE = ∆ DCF 3. Area of congruent triangles. 4. ∆ABE + Trap. EBCD = ∆ DCF+Trap EBCD 4. Adding common trapezium EBCD in (3). 5. //gm ABCD = //gm EBCF 5. From (4), using whole part axiom.
Oasis School Mathematics-10 207 Alternative method Given : ABCD and BCEF are two parallelograms standing on the same base BC and between the parallel lines AE and BC. to prove : Area of ||gm ABCD = Area of ||gm BCEF Construction : Draw DH⊥BC and FG⊥BC Proof: S.N. Statements S.N. reasons 1. ||gm ABCD = BC×DH 1. Area of parallelogram = base × height 2. ||gm BCEF = BC×FG 2. Area of parallelogram = base × height 3. DH = FG 3. Distance between the same parallel lines 4. ||gm ABCD = ||gm BCEF 4. From (1), (2) & (3) Hence, a parallelograms on the same base and between the same parallel lines are equal in area. Corollary: (i) A parallelogram and a rectangle on the same base and between the same parallel lines are equal in area. (ii) Parallelograms on equal base and between the same parallel lines are equal in area. (iii) A rhombus and a square on the same base and between the same parallel lines are equal in area. theorem 12.3 the area of a triangle is one half the area of a parallelogram standing on the same base and between the same parallel lines. Given : Triangle ABC and parallelogram DBCF both stand on the same base BC and between the parallel lines BC and AF. to prove : Area of ∆ ABC = 1 2 Area of ||gm DBCF. Construction : Draw a line CH parallel to BA where H is a point on AF. Proof: S.N. Statements S.N. reasons 1. ABCH is a parallelogram 1. By construction 2. ∆ ABC = 1 2||gmABCH 2. Diagonal bisects the parallelogram. 3. ||gm ABCH = ||gm DBCF 3. Parallelograms on the same base BC and between the parallel lines BC and AF. 4. ∆ ABC = 1 2||gm DBCF 4. From (2) and (3). Proved. A F D H E B G C A D H F B C
208 Oasis School Mathematics-10 Hence, area of a triangle is half the area of a parallelogram standing on the same base and between the same parallel lines. Alternative method Given : Parallelogram ABCD and triangle EBC both stand on the same base BC and lie between the parallel lines BC and AD. to prove : ∆BEC = 1 2 ||gm ABCD. Construction : Draw AK⊥BC and EH⊥BC. Proof: S.N. Statements S.N. reasons 1. Area of ||gmABCD = BC×AK 1. Area of parallelogram = base × height 2. Area of ∆BEC= 1 2 BC×EH 2. Area of triangle = 1 2 base × height 3. AK = EH 3. Distance between the same parallel lines 4. ∆EBC = 1 2 ||gm ABCD. 4. From (2) and (3). Hence, the area of a triangle is one half the area of parallelogram standing an same base and between same parallelograms. theorem 12.4 Triangles on the same base and between the same parallel lines are equal in area. Given : Triangles ABC and DBC are standing on the same base BC and lie between the parallel lines BC and AD. to prove : ∆ABC = ∆DBC. Construction : Draw CE||BA, where E is a point on AD. Proof : S.N. Statements S.N. reasons 1. ABCE is a parallelogram 1. By construction (ABllCE) and given (AEllBC). 2. ∆ABC = 1 2 parallelogram ABCE 2. Diagonal bisects the parallelogram 3. ∆DBC = 1 2 parallelogram ABCE 3. Triangle and parallelogram on same base BC and between the parallel lines BC and AD 4. ∆ABC = ∆DBC 4. From (2) and (3). Conclusion: Hence, triangles on the same base and between the same parallel lines are equal in area. A E D B K H C D A E B C
Oasis School Mathematics-10 209 Alternative method Given : Triangles ABC and DBC both lie on the same base BC and between the parallel lines AD and BC. to prove : Area of ∆ABC = Area of ∆DBC Construction: From A and D, draw AM and DN perpendiculars to BC. Proof : S.N. Statements S.N. reasons 1. Area of ∆ABC = 1 2 BC × AM 1. Area of a triangle = 1 2 base × height. 2. Area of ∆DBC = 1 2 BC × DN 2. Area of a triangle = 1 2 base × height. 3. AM = DN 3. Distance between two parallel lines. 4. Area of ∆ABC = Area of ∆DBC 4. From statements (1), (2) and (3). Hence proved. Some important results: 1. Area of trapezium = 1 2 height × sum of the parallel sides. Area of trap ABCD = 1 2 AH×(AD + BC). 2. Area of rhombus = 1 2 product of the diagonals. Area of rhombus ABCD = 1 2 AC × BD = 1 2 d1 × d2 3. Area of quadrilateral = 1 2 diagonal (sum of altitude from opposite vertices to the diagonal) Area of quadrilateral ABCD = 1 2 d ×(h1 + h2 ) 4. Area of kite = 1 2 product of the diagonals = 1 2 AC × BD A H B C D A B C O D C A D B h2 h1 d A B D C A M B C D N
210 Oasis School Mathematics-10 remember ! C D A B • Median bisects a triangle. In the given figure, AD is the median. ∴ ∆ABD = ∆ADC • Diagonal bisects a parallelogram. • Parallelograms on the same base and between the same parallel lines are equal in area. • Area of a triangle is half the area of a parallelogram standing on the same base and between the same parallel lines. • Triangles on the same base and between the same parallel lines are equal in area. Worked out examples example 1 In the given figure, ABCD is a square whose perimeter is 40 cm. Find the area of ∆eBC. Solution: Given, Perimeter of square = 40 cm 4a = 40 cm [a = a side of a square] a = 10 cm We have, Area of a square = a2 = (10cm)2 = 100 cm2 . Area of ∆EBC = 1 2 area of square ABCD [Area of a triangle is half of the area of a square standing on the same base and between the same parallel lines]. or, Area of ∆EBC = 1 2 × 100 cm2 = 50 cm2 example 2 In the given figure, ABCD is a parallelogram. DF = FE and area of parallelogram ABCD = 60 cm2 . Find the area of ∆EDC. Solution: Given, area of parallelogram ABCD = 60 cm2 Area of ∆FDC = 1 2 area of ||gm ABCD or, Area of ∆FDC = 1 2 × 60 cm2 = 30 cm2 Again, ∆EDC = 2∆FDC [Median bisects the triangle] = 2×30 cm2 = 60 cm2 Hence, area of ∆EDC = 60 cm2 . A D E B C A D C B F E
Oasis School Mathematics-10 211 example 3 In a quadrilateral ABCD, it is given that M is the mid-point of AC. Prove that quadrilateralral ABMD = quadrilateral DMBC. Solution: Given : ABCD is a quadrilateral. M is the mid-point of AC. to prove : Quadrilateral ABMD = Quadrilateral DMBC Proof : S.N. Statements S.N. reasons 1. ∆ADM = ∆DMC 1. Median bisects the triangle 2. ∆AMB = ∆BMC 2. Same as (1) 3. ∆ADM + ∆AMB = ∆DMC + ∆BMC 3. Adding (1) and (2) 4. Quadrilateral ABMD = Quadrilateral DMBC 4. From (3), whole part axiom. Hence, proved. example: 4 If O is a point inside the parallelogram ABCD, prove that ∆AOB + ∆COD = ∆AOD + ∆BOC. Given : ABCD is a parallelogram. O is a point interior of it and OC, OA, OB, OD are joined. to prove : ∆AOB + ∆COD = ∆AOD + ∆BOC Construction : Draw XOY ||AB ||DC. Proof: S.N. Statements S.N. reasons 1. XABY is a parallelogram 1. Being AX//BY and XOY // AB. 2. ∆OAB = 1 2||gm XABY 2. They stand on the same base AB and between the parallel lines AB and XY. 3. DXYC is a parallelogram 3. Being DX //CY at XOY // AB. 4. ∆DOC = 1 2||gm DXYC 4. Same as (2) 5. ∆OAB+∆DOC =1 2||gm XABY+1 2||gmDXYC 5. Adding (2) and (4) 6. ||gm ABCD = ||gm XABY + ||gm DXYC 6. Whole part axiom 7. ∆OAB + ∆DOC = 1 2 ||gm ABCD 7. From (3) and (4) 8. Similarly ∆DOA + ∆COB = 1 2||gm ABCD 8. Same as above by drawing line parallel to DA and BC through O. 9. ∆OAB + ∆DOC=∆DOA + ∆COB 9. From (7) and (8). Hence, proved. D C B M A D C Y O X A B
212 Oasis School Mathematics-10 example 5 In the given figure, P and Q are the mid-points of AB and AC respectively. Prove that. (i) ∆POB = ∆QOC (ii) Quad. APOQ = ∆BOC Given : In ABC, P and Q are the mid-points of AB and AC respectively. to prove : ∆POB = Quad. APOQ, (ii) Quad. APOQ = ∆BOC Construction : Join PQ Proof: S.N. Statements S.N. reasons 1. PQ || BC 1. The line joining the mid points of two sides of a triangle is parallel to the third side. 2. ∆PBC = ∆QBC 2. Triangles on the same base and between the same parallel lines 3. ∆POB + ∆OBC = ∆QOC + ∆OBC 3. Whole part axioms 4. ∆POB = ∆QOC 4. From (3) 5. ∆BAQ = ∆BQC 5. Median bisects the triangle 6. ∆BAQ–∆POB = ∆BQC–∆QOC 6. Subtracting (4) from (5) 7. Quad. APOQ = ∆BOC. 7. From (6) Hence, proved. Example 6 In the adjoining figure, ABCD is a parallelogram whose diagonals AC and BD intersect at O, a line segment through O meets AB at P and DC at Q. Prove that the area of the quadrilateral APQD = 1 2 parallelogram ABCD. Given : ABCD is a parallelogram, diagonals AC and BD intersect at OP and Q, and P and Q are points on sides AB and DC respectively. to prove : Area of quad. APQD = 1 2 (Area of ||gm ABCD) Proof : S.N. Statements S.N. reasons 1. In ∆APO and ∆QOC (i) ∠OAP = ∠QCO (A) (ii) AO = OC (S) (iii) ∠AOP = ∠QOC (A) 1. (i) Being alternate angles on AB ||DC (ii) Diagonals of a parallelogram bisect each other (iii) Vertically opposite angles A P Q B C O D Q C B A P O
Oasis School Mathematics-10 213 2. ∆APO ≅ ∆OQC 2. By A.S.A. axioms 3. ∆APO = ∆OQC 3. Area of congruent triangles 4. ∆DAC = Quad. DAOQ + ∆QOC 4. Whole part axiom 5. Quad. APQD = Quad. DAOQ + ∆AOP 5. Whole part axiom 6. ∆DAC = Quad. APQD 6. From (3), (4) and (5) 7. ∆DAC = 1 2||gm ABCD 7. They stand on the same base DA and between DA || CB 8. Quad. APQD = 1 2 ||gm ABCD 8. From (6) and (7) Hence, proved. example 7 In the adjoining figure, PABC and PQRS are two parallelograms of equal area. Prove that QC // BR. Given : In the given figure, ||gm PABC = ||gm PQRS to prove : QC // BR Construction : Join QB and CR. Proof : S.N. Statements S.N. reasons 1. ∆QBC = 1 2 ||gmPABC 1. Area of a triangle is half of the area of parallelogram standing on the same base and between the same parallel lines 2. ∆CQR = 1 2||gm PQRS 2. As statement (1) 3. ∆QBC = ∆CQR 3. From (1) and (2) 4. CQ ||BR 4. From statements 1, being area of two triangles on the same base is equal Hence,, proved. exercise 12.1 1. (a) What is the relation between the area of two parallelograms standing on the same base and between the same parallel lines? (b) What is the relation between the area of a triangle and a parallelogram standing on the same base and between the same parallel lines? P C S R B O A Q
214 Oasis School Mathematics-10 (d) In the given figure, which two triangles are equal in area? Why? (e) In the given figure, write two pairs of triangles which are equal in area. 3. (a) In the figure alongside, ABCD is a parallelogram. P is a point on CD. If ∆DPA = 15 cm2 , ∆BPC = 20 cm2 , find the area of ∆APB. (b) In the given figure, ABCD is a parallelogram, ∆APD = 7 cm2 and ∆BCP = 5 cm2 . Find the area of parallelogram ABCD. (c) In the given figure, PQRS is a parallelogram in which T is the mid-point of PS. If the area of ∆PTQ is 6 cm2 , find the area of quad. TQRS. A B D C A B D C A B C P D Q R P S T A P B D C (c) If a parallelogram and a rectangle are on the same base and between the same parallel lines, what is their relation? (d) What is the relation of two triangles standing on the same base and between the same parallel lines? (e) If a triangle and a parallelogram are standing between the same parallel lines and the base of a triangle is twice the base of a parallelogram, what is their relation? 2. (a) In the given figure, name two parallelograms which are equal in area. Why? (b) In the given figure, what is the relation between ∆PQR and parallelogram TQRS? (c) In the given figure, what is PS called? What is the relation between ∆PQS and ∆PRS? Why? A F D E B C T S Q R P P Q S R
Oasis School Mathematics-10 215 (d) In the adjoining figure, if the area of the parallelogram ABCD = 88 cm2 , find the area of ∆BQC. (e) In the given figure, ABCD is a parallelogram and EBC is a triangle. If the area of ∆EBC is 7 sq. cm, find the area of the parallelogram ABCD. (f) ABCD is a parallelogram. If ∆ ABF = 21 cm2 , find the area of ∆EBC. (g) In the given figure, ABCD is a parallelogram and AH⊥BC. If BC = 8 cm, AH = 8 cm, find the area of ∆AED. (h) In the given figure, ABCD is a rhombus whose diagonals AC and BD are 6 cm and 8 cm respectively. Find the area of ∆EBC. (i) In the given figure, ABCD is a square with its diagonal AC = 8 2 cm. Find the area of ∆EBC. (j) In the given figure, ABCD is a square and DEC is a triangle. If the area of triangle DEC = 50 cm2 , find the length of diagonal AC. (k) 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 middle point of CD, find the area of ∆EFC. (l) In the adjoining figure, AF//DC, ED//FC and ABCD is a square. If AC = 5 2 cm, find the area of the parallelogram DEFC. A P D C Q B A E D F B C A D B C H E 8cm 8cm A D E B C A E D C B A B F D E C A D B C E D C F A E B A E B C D F
216 Oasis School Mathematics-10 (m) In the adjoining figure, PS = 5 cm and SM = 8 cm. Calculate the area of ∆PQN. (n) Find the area of the quadrilateral PQRS given in the adjoining figure in which 3RB = 2PA = QS = 6 cm. (o) In the given figure, ABCD is a trapezium. If AB = 10 cm, BC = 22 cm, AD = 16 cm, AD||BC and DC⊥BC, calculate the area of ∆ADC. (p) In ∆ABC, D is the mid-point of BC. If AC = 10 cm and DE = 6 cm, find the area of ∆ABC. (q) In the given figure, PT = TQ and TS||QR. If the area of ∆TSP = 15cm2 , find the area of (i) ∆TQS (ii) ∆ TQR (iii) ∆PSQ (r) In the given figure, AE||DB and EB = BC. If area of ∆ABD = 20cm2 , find the area of (i) ∆DBC (ii) Quad. ABCD. 4. In the adjoining figure, ABCD is a trapezium in which AB||DC. Prove that: ∆AOD = ∆BOC. 5. In the given figure, D and E are the points on AB and AC of ∆ABC respectively. If DE||BC, prove that: (i) ∆ACD = ∆ABE (ii) ∆OCE = ∆DOB. 6. In the adjoining figure, O is any point inside a parallelogram ABCD. Prove that: (i) ∆OAB + ∆OCD = 1 2 ||gm ABCD (ii) ∆OAC + ∆OBD = 1 2 ||gm ABCD P Q R N S M S P R B A Q A D B C 10cm 16cm 22cm A C D E B A B E C D C A B D O C E A D B O A C B D O R Q T S P
Oasis School Mathematics-10 217 7. In the given figure, R is the mid point of AB and PQCB is a parallelogram. Prove that area of ∆ABC = area of parallelogram PQCB. 8. In a quadrilateral ABCD, a line through D, parallel to AC, meets BC produced at P. Prove that: ∆ABP = quad. ABCD. 9. In the given figure, PQRS is a trapezium and PQ||MN||SR. Prove that area of ∆PSN = area of ∆QRM. 10. In the adjoining figure, ACDE is a parallelogram. F is the mid-point of BC. Prove that ∆ABC = Parallelogram ACDE 11. ABC is a triangle in which D is the mid–point of BC and E is the mid-point of AD. Prove that ∆ABC = 4 ∆ABE. 12. The vertex A of ∆ABC is joined to a point D on the side BC. The mid-point of AD is E. Prove that: ∆BEC = 1 2 ∆ABC. 13. D is the mid-point of side BC of ∆ABC and E is the mid-point of BD. If O is the mid–point of AE, Prove that: 8 ∆BOE = ∆ABC. 14. In the given figure, PQRS is a parallelogram, T is any point on the line RP produced. Prove that ∆TPS = ∆TQP. A B E D C O T Q S R P A P B C R Q P Q N S R M E A C D B F A D B P C A C D E B A C D B E
218 Oasis School Mathematics-10 A F B G D C E A D B M C N A P B D R C Q 15. In a parallelogram ABCD, O is any point on the diagonal AC, show that (i) ∆AOB = ∆AOD, (ii) ∆COB = ∆DOC. 16. P, Q, R and S are respectively the mid–points of the sides AB, BC, CD and DA of the parallelogram ABCD. Show that the area of parallelogram PQRS = 1 2 area of ||gmABCD. 17. In the given figure AD//BE, AB//DC, prove that area of ∆AFC = Area of ∆DFE. 18. In the figure, PQRS is a parallelogram. PM and SR are produced to meet at N. N and Q are joined. Prove that area of ∆SMR = area of ∆QMN. 19. In the adjoining figure, it is given that AD//BC and BD//CE. Prove that area of ∆ABC = area of ∆BDE. 20. In the given figure, ABCD is a parallelogram. If MN||BD, then, prove that area of ∆BAM = area of ∆AND. 21. In the given ∆ABC, D is the mid-point of BC. E is the mid–point of AD. F is the mid–point of AB and G is any point on BD. Prove that: ∆ABC = 8 ∆EFG. 22. In the given diagram, ABCD and PQRD are two parallelograms. Prove that: //gm ABCD = //gm PQRD. A D E B C A B F C E D P S R N M Q C B O A D C B P Q R S A D
Oasis School Mathematics-10 219 Answers 1. Consult your teacher 2. Consult your teacher 3. (a) 35 cm2 (b) 24 cm2 (c) 18 cm2 (d) 22 cm2 (e) 14 cm2 (f) 21 cm2 (g) 32 cm2 (h) 12 cm2 (i) 32 cm2 (j) 10 2cm (k) 25 cm2 (l) 25 cm2 (m) 20 cm2 (n) 15 cm2 (o) 64 cm2 (p) 60 cm2 (q) (i) 15 cm2 (ii) 15 cm2 (iii) 30 cm2 (r) (i) 20 cm2 (ii) 40 cm2 Project Work On a graph paper, draw two given figures on the same base and between the same parallel lines. From the graph, find the length of their base and height. Find their area and draw and draw out conclusion. (i) two parallelograms (ii) a parallelogram and a rectangle (iii) a parallelogram and a triangle (iv) two triangles
220 Oasis School Mathematics-10 13.1 Warm-up Activities • What are the features of a parallelogram? • Is it possible to construct a parallelogram with two of its adjacent sides and angle between them given? • Is it possible to construct a parallelogram if the length of its two diagonals and angle between them are given? • From the above figure name four triangles which are equal in area with ABC. Find out the reason why they are equal in area. • Is parallelogram ABCD = parallelogram BCED? why? Discuss the following in your class and draw a conclusion. A D E B C 13.2 Construction of a parallelogram equal in area with given parallelogram Concept: Parallelograms on the same base and between the same parallel lines are equal in area. example 1 Construct a parallelogram whose adjacent sides are 4 cm and 6 cm and contained angle is 45°. Construct another parallelogram equal in area whose one angle is 60°. Q rough Sketch C P D B 6cm 4cm A B C P A Q 450 600 4cm 6cm D Unit 13 Construction
Oasis School Mathematics-10 221 Steps: (i) Draw a line segment BC = 6 cm. (ii) Construct an angle of 45° at B, such that ∠ABC = 45°. (iii) From B cut by an arc of 4cm to get A. (iv) From A, cut by an arc of 6 cm and from C, cut by an arc of 4cm to get point D. (v) Draw an angle of 60° at B such that ∠PBC = 60°. (vi) Take an arc equal to PB and from C cut at Q. (vii) Join PB and QC. Hence, PBCQ is the required parallelogram. example 2 Construct a parallelogram ABCD in which AB = 3 cm, AD = 4 cm, BD = 3.1 cm. Construct another parallelogram equal in area whose one side is 3.5 cm. Steps: (i) Draw a line segment AD = 4 cm. (ii) From A and D cut by an arc of 3 cm and 3.4 cm respectively to get point B. (iii) From D and B cut by an arc 3 cm and 4 cm to get C. (iv) Now, ABCD is a parallelogram. (v) Take an arc of 3.5 cm and cut from the points A and D to get points P and Q. PADQ is the a required parallelogram exercise 13.1 1. a) Construct a parallelogram ABCD where AB = 6.3 cm, BC = 5.6 cm and ∠B = 600 . Also construct another parallelogram equal in area to the parallelogram ABCD, whose one side is 6.5 cm. How is the area of the parallelogram ABCD equal to the area of the parallelogram PBCQ? • The base of both a parallelograms is BC. • Both lie between the same parallel lines PD and BC. • Parallelograms on the same base and between the same parallel lines are equal in area. Analysis B P C Q A 4cm D 3cm 3.5cm 3.1cm 4cm B P 4cm C Q A D 3.1cm 3cm rough Sketch
222 Oasis School Mathematics-10 b) Construct a parallelogram PQRS where PQ = 6.8 cm, QR= 5.9 cm, PR = 6.2 cm. Also construct another parallelogram equal in area to the parallelogram PQRS, whose one angle is 450 . c) Construct a rhombus ABCD whose one side is 6.5 cm and one angle is 450 . Also construct another parallelogram equal in area to the rhombus ABCD, whose one side is 7 cm. d) Construct a rectangle equal in area with parallelogram ABCD where, BC = 5.2 cm, ∠ACB = 600 and AC = 5.6cm. e) Construct a parallelogram PQRS where QR = 5.5 cm, diagonal PR = 6.4 cm, diagonal QS = 6cm. Also construct another parallelogram equal in area with to the parallelogram PQRS, whose one side is 5.9cm. f) Construct a parallelogram ABCD where diagonal AC = 7cm, diagonal BD = 6cm and angle between them is 600 . Also construct a parallelogram equal in area with to the parallelogram ABCD, whose one side is 6.8cm. 2. a) Construct a parallelogram PQRS where SR = 6.2 cm, QR = 5.6 cm, ∠PSR = 600 . Also construct a parallelogram SRTU where RT = 7cm. b) Construct a parallelogram ABCD where AB = 5.4 cm, BC = 6cm, ∠ABC = 600 . Also construct a rectangle FBCE equal in area to the parallelogram ABCD. c) Construct a parallelogram QXYP where ∠QXY = 300 equal in area to the parallelogram WXYZ where WX = XY = WY = 6cm. 13.3 Construction of triangle equal in area to given parallelogram Concept: The area of a triangle is half of the area of parallelogram standing on the same base and between the same parallel lines. In other words, the area of a triangle is equal to the area of a parallelogram between the same parallel lines if the base of the triangle is twice the base of the parallelogram. example 1 Construct a rectangle whose length and breadth are respectively 5 cm and 3.5 cm. then construct a triangle which is equal to the area of the given rectangle, in which one of the sides is 8.5 cm. A B C P D Q 8.5cm 5cm 5cm rough Sketch A B C 8.5cm 900 D Q P 3.5cm 3.5cm 5cm 3.5cm
Oasis School Mathematics-10 223 Steps: (i) Draw a line BC = 5 cm. (ii) Construct ∠B = 900 and cut AB = 3.5 cm. (iii) Similarly, construct ∠C = 900 and cut CD = 3.5 cm. (iv) Now, ABCD is a rectangle. (v) Select the point P such that BC = CP i.e. BP = 2BC. (vi) Cut an arc of 8.5 cm from B, to get Q. (vii) Join BQ and QP. ∆QBP is the required triangle. exercise 13.2 1. a) Construct a parallelogram ABCD where AB = 6.5 cm, BC = 6 cm and ∠ABC = 600 . Also construct a triangle equal in area to the parallelogram ABCD, whose one side is 7.6 cm. b) Construct a parallelogram PQRS in which PQ = 5cm, diagonal PR = 6 cm and QR = 5.6 cm. Also construct a triangle equal in area to the parallelogram PQRS, whose one angle is 450 . c) Construct a right-angled triangle equal in area to a parallelogram whose two diagonals are 6.8 cm and 7.4 cm, and the angle between the diagonals is 600 . 2. a) Construct a parallelogram ABCD where BC = 6cm, AC = 6.4 cm, ∠ACB = 600 . Also construct a ∆FEC equal in area to the parallelogram ABCD, where FC = 6cm. b) Construct a parallelogram PQRS, where PQ = 6.4 cm, QR = 5.6 cm and PR = 6cm. Also construct a ∆SQT equal in area to the parallelogram PQRS. c) Construct a parallelogram WXYZ, where XY = 5.8 cm, YZ = 6.5 cm, ∠XYZ = 1200 . Also construct a ∆NXM equal in area to the parallelogram WXYZ where ∠NXY = 450 . 12.4 Construction of a triangle equal in area to given triangle Concept: Triangles on the same base and between the same parallel lines are equal in area. example 1 Construct a ∆ABC in which a = 6 cm, b = 5.5 cm and c = 7.5 cm. Construct a triangle equal to ∆ABC and having an angle of 60°. • Area of rectangle ABCD = Area of ∆QBP. • Base of ∆QBP = 2 (base of rectangle ABCD) Analysis
224 Oasis School Mathematics-10 A B C P A 6cm 7.5cm 5.5cm P A B C 7.5cm 600 6cm 5.5cm rough Sketch 600 Steps: (i) Draw a line segment BC = 6 cm. (ii) From B and C cut an arc length of 7.5 cm and 5.5 cm respectively to get point A. (iii) Join AB and AC to get ABC. (iv) From A draw a line parallel to BC. (v) Draw an angle ∠PBC = 600 at B. (vi) Join PC. ∆PBC is the required triangle. exercise 13.3 1. (a) Construct a ∆ABC having AB = 6.2 cm, BC = 5.4 cm and AC = 6.5 cm. Also construct a triangle equal in area to ABC, whose one angle is 600 . (b) Construct a ∆ABC in which a = 6.3 cm, ∠B = 300 , ∠C = 600 . Also construct a triangle equal in area to a triangle, whose one side is 5.6 cm. (c) Construct a ∆XYZ in which XY = 5.6cm, YZ = 6.2 cm, ∠XYZ = 600 . Also construct another triangle equal in area to XYZ, whose one side is 6.8 cm. 2. (a) Construct an equilateral triangle PQR having a side 6 cm. Also construct an equivalent ∆SQR where SQ = 6.5 cm. (b) Construct a ∆ABC in which AB = 6 cm, BC = 6.8 cm, ∠ABC = 600 . Also construct a ∆DBC where ∠DBC = 300 . (c) Construct a ∆GEF where ∠GEF = 900 equal in area to ∆DEF having DE = EF = 6.8 cm and DF = 7.2 cm. ∆ABC and ∆PBC both lie on the same base BC and between the parallel lines PA and BC. Analysis
Oasis School Mathematics-10 225 13.5 Construction of a parallelogram equal in area to given triangle Concept: The area of a triangle is half the area of a parallelogram standing on the same base and between the same parallel lines. Or, The area of a triangle is equal to the area of a parallelogram if they lie between the same parallel lines and the base of the triangle is twice the base of the parallelogram. example 1 Construct a triangle ABC in which a = 7.8 cm, b = 7.2 cm, c = 6.3 cm. Construct a parallelogram equal in area to ∆ABC and having a side = 8.5 cm. Steps: (i) Draw a line segment BC = 7.8 cm. (ii) Cut an arc of length AB = 6.3 cm and AC = 7.2 cm. respectively to get point A. (iii) From A, draw a line parallel to BC. (iv) Draw the perpendicular bisector of BC which meets BC at M. (v) From M and C cut an arc 8.5 cm to get P and Q. (vi) Join MP and CQ. Hence, PMCQ is the required parallelogram. example 2 Construct a triangle PQR in which PQ = 6.5 cm QR = 7.6 cm, PR = 6.9 cm. Construct a rectangle equal in area to ∆PQr. Q 8.5cm 8.5cm 7.2cm 6.3cm A P B M C 7.8cm rough Sketch 8.5cm 7.2cm 6.3cm 7.8cm A P Q B M C Area ∆ABC= area of parallelogramPMCQ. Base of ∆ABC = 2 (base of parallelogram PMCQ) Analysis
226 Oasis School Mathematics-10 P N O Q 7.6cm M R 6.5cm 6.9cm rough Sketch P N O Q M R 6.5cm 6.9cm 7.6cm Steps: (i) Draw a line segment QR = 7.6 cm. (ii) From Q and R cut an arc of length 6.5cm and 6.9 cm, to get point P. (iii) Join PQ and PR. (iv) Draw a perpendicular bisector of QR which meets QR at M. (v) From P, draw a line parallel to QR. (vi) Take an arc equal to MR and cut from N to get O. (vii) Join OR. Hence, MNOR is the required rectangle. exercise 13.4 1. (a) Construct a triangle ABC where AB = 5.7 cm, BC = 5.5 cm and ∠ABC = 600 . Also construct a parallelogram equal in area to ∆ABC whose one side is 6 cm. (b) Construct a triangle XYZ where XY = YZ = 5.2cm and XZ = 5.9 cm. Also construct a parallelogram equal in area to ∆XYZ whose one angle is 450 . (c) Construct a triangle PQR having PQ = 6.8 cm, ∠P = 600 , ∠Q = 450 . Also construct a rectangle equal in area to given triangle. 2. (a) Construct a ∆ABC having AB = 6 cm, BC = 7cm and AC = 7.2 cm. Also construct a parallelogram CFGH equal in area to the ∆ABC, where FG = 6.9 cm. (b) Construct a right-angled isosceles triangle PQR where PQ = PR, QR = 6 cm and ∠QPR = 900 . Also construct a parallelogram TSRU equal in area with ∆PQR where RU = 5 cm. (c) Construct a ∆ABC in which a = 6.5 cm, b = 7.5 cm, c = 6 cm. Reduce the triangle ABC into an equivalent rectangle DEFC.
Oasis School Mathematics-10 227 13.6 Construction of a triangle equal in area to given quadrilateral Concept : Triangles on the same base and between the same parallel lines are equal in area. example 8 Construct a quadrilateral ABCD in which AB = 4.2 cm, BC = 4.8 cm, CD = 5.4 cm, DA = 5.8 cm and diagonal BD = 5.8 cm. Construct a triangle equal in area to the quadrilateral ABCD. Steps: (i) Draw a line segment BC = 4.8 cm. (ii) Cut an arc of 5.8 cm and an arc of 5.4 cm from B and C respectively to get the point of intersection at D. (iii) Cut an arc of length 4.2 cm and 5.8 cm respectively from B and D to get point A. (iv) Join AD and AB, then ABCD is a quadrilateral. (v) From A, draw AE//DB which meets CB produced at E. (vii) Join DE. ∆DEC is the required triangle. D C 5.4cm 5.8cm 4.2cm 4.8cm 5.8cm E B rough Sketch D A C 5.4cm 5.8cm 4.2cm 4.8cm 5.8cm E B A ∆EBD = ∆ABD ('Triangles on the same base BC and between the parallel lines AE and DB) ∆EBD + ∆DBC = ∆ABD + ∆DBC ∴ ∆DEC = quadrilateral ABCD Analysis
228 Oasis School Mathematics-10 exercise 13.5 1. (a) Construct a quadrilateral ABCD where AB = 5.5 cm, BC = 5.1 cm, CD = 6cm, AD = 4.9 cm and diagonal BD = 5.8 cm. Also construct a triangle equal in area with quadrilateral ABCD. (b) Construct a quadrilateral PQRS having PQ = 6 cm, PS = 6.4 cm, QR = 5.2 cm, RS = 5.6 cm and ∠QPS = 600 . Also construct a triangle equal in area with quadrilateral PQRS. (c) Construct a quadrilateral ABCD in which AB = 6.4 cm, BC = 5.5 cm, CD = 4.5 cm, ∠B = 1200 and ∠C = 600 . Also construct a triangle equal in area to the given quadrilateral. 2. (a) Construct a quadrilateral PQRS where PQ = QR = 5.2 cm PR = 5.5 cm, PS = RS = 6.7 cm. Also construct a ∆PQT equal in area to the quadrilateral PQRS. (b) Construct a quadrilateral ABCD having AB = 5.8 cm, AD = 5.2 cm, CD = 5.6 cm, BC = 6 cm and ∠ADC = 600 . Also construct a ∆BEC equal in area with quadrilateral ABCD. (c) Construct a quadrilateral MNOP having MN = 5.6 cm, ON = 6.2 cm, OP = 5.8 cm, ∠MNO = 600 , ∠PON = 1200 . Also construct a ∆MNQ equal in area to the quadrilateral MNOP. 13.7 Construction of a quadrilateral equal in area to given triangle Concept Triangles on the same base and between the same parallel lines are equal in area. example 1 Construct a quadrilateral equal to the area of ∆ABC such that the sides of the triangle are AB = 4 cm, BC = 4.5 cm and AC = 5 cm. Q D B P C A 4.5cm 5cm 4cm rough Sketch Q D B P C A 4.5Cm 5Cm 4Cm
Oasis School Mathematics-10 229 Steps: (i) Draw a line segment BC = 4.5 cm. (ii) From B and C cut an arc of 4 cm and 5 cm respectively to get A. (iii) Join AB and AC. (iv) Choose any point P on BC and join AP. (v) From C, CQ //PA. (vi) Take any point D on CQ and join PD. ABPD is the required quadrilateral. ∆APD = ∆APC [Triangles on the same base and between the same parallel lines] ∆APD + ∆ABP = ∆ APC + ∆ABP Quad. ABPD = ∆ABC. Analysis exercise 13.6 1. (a) Construct a ∆ABC in which ∠ABC = 600 , AB = 6 cm, BC = 5.5 cm. Also construct a quadrilateral equal in area to ∆ABC. (b) Construct a ∆PQR in which PQ = 7 cm, QR = 6 cm and PR = 6.5 cm. Also construct a quadrilateral equal in area to ∆PQR. 2. (a) Construct a ∆ABC in which AB = 7cm, ∠A = 600 and ∠B = 450 . Also construct a quadrilateral BDEC equal in area to ∆ABC where DE = 7.2 cm. (b) Construct a ∆PQR in which PQ = 5.7 cm, QR = 5.6 cm. ∠Q = 600 . Also construct a quadrilateral PTSQ equal in area to ∆PQR where TS = 6cm. Project Work On a chart paper, draw a model of each type of construction mentioned above.
230 Oasis School Mathematics-10 1. ABCD is a parallelogram and ∆BCE is a triangle standing on the same base BC and between the parallel lines BC and AE. (a) What is the relation of ∆BCE and parallelogram ABCD? (b) If AH⊥BC, E is any point on BC, BC = 12cm and AH = 8cm, find the area of ∆AED. [Ans: (b) 48cm²] (c) Construct a parallelogram ABCD where AB = 6.5cm, BC = 7.2cm, ∠B = 60°. Also construct a triangle BCE whose area is half of the area of parallelogram ABCD. 2. Study the given questions and answer the following. (a) What is relation of a square and a parallelogram standing on the same base and between the same parallel lines? (b) ABCD is a square and CDEF is a parallelogram. If AC = 10 2cm, find the area of parallelogram CDEF. [Ans: 100cm2 ] (c) Construct a parallelogram ABCD where AB = 6.8cm, BC = 6.2cm and diagonal AB = 5.6cm. Also construct a rectangle PBCQ equal in area with parallelogram ABCD. 3. In the given figure, AC DE and AB ⊥ BC. (a) Write name of two pairs of triangles which are equal in area. (b) If the area of ∆ADC = 32cm², find the area of ∆ABF. [Ans: 56cm2 ] (c) Construct a quadrilateral ABCD where AB = 5cm, BC = 6cm, CD = 7.2cm, AD = 6.2cm and ∠B = 900 . Also construct a ∆ABED equal in area with quadrilateral ABCD. 4. Study the given questions and answer the following. (a) Write the name of a pair of triangle which are equal in area. (b) If BF = 12cm, CE = 5cm, find the area of rectangle ABCD and length of AB. [Ans: 60cm2 , 6.67cm] (c) Construct ∆ABC where AB = 6.8cm, BC = 6.2cm and AC = 5.6cm. Also construct a right angled triangle ∆PAC equal to area with ∆ABC. 5. Study the given questions and answer the following. (a) Write the name of a pair of triangle which are equal to area. (b) If the area of ∆BEF is equal to 18cm², find the area of ∆BEC and parallelogram ABCD. [Ans: 36cm2 , 36cm] A B C 8cm 6cm E D A D F C E B 9cm A B C B F E miscellaneous exercise
Oasis School Mathematics-10 231 (c) Construct a parallelogram ABCD where DC = 8cm, BC = 6cm and ∠BCD = 60°. Also construct a triangle BCE equal to area with parallelogram ABCD. 6. In the given figure, D is the mid point of BC. (a) What is the relation of ∆ABC with ∆ABD? (b) If E is the mid point of AD, and area of ∆ABE = 12cm², then what is the area of ∆ABC? [Ans: 48 cm2 ] (c) Construct a ∆ABC where AB = 6cm, BC = 7cm, AC = 6.5cm. Take any paint D on BC, draw a line from C which is parallel to DA. Construct a quadrilateral equal in area with ∆ABC. 7. ABCD is a square. ∆FDC is the triangle on the same base DC and between the parallel line AF and DC. (a) What is the relation of ∆FDC with square ABCD? (b) E is the mid point of DC, and the perimeter of square ABCD is 60cm, find the area of ∆EFC. [Ans: 56.25 cm2 ] (c) Construct a square ABCD whose one side is 6cm. Also construct a triangle FDE equal in area with square ABCD where ∠FDE = 60°. 8. Study the given questions and answer the following. (a) What is the relation of ∆ABC with parallelogram FBCE? (b) If FX is drawn perpendicular to CE, FB = 12cm and FX = 8cm, find the area of ∆ABC. [Ans: 48 cm2 ] (c) Construct a ∆ABC where AB = 6.5cm, AC = 7.2cm, ∠A = 75°. Also construct a parallelogram PAQR equal in area with ∆ABC where PR = 7.5cm. 9. Study the given questions and answer the following. (a) Write the name of two triangles whose area is half of the area of parallelogram PQRS. (b) If the area of ∆PST = 30cm², what is the area of ∆PSM + ∆QRM? Is ∆SMT = ∆OMR? Justify your answer. [Ans: 30 cm2 ] (c) Construct a parallelogram PQRS where PR = 7cm, QS = 6cm angle between PR and QS = 60°. Also construct a triangle equal in area with given parallelogram. 10. Study the given questions and answer the following. (a) Write the name of parallelogram which are equal to parallelogram ABQP and parallelogram PQCD? (b) If the area of parallelogram ABQP = 32cm² and the area of parallelogram PQCD = 28cm², find the area of parallelogram ABCD. [Ans: 60 cm2 ] (c) Construct a parallelogram PQRS where PQ = 5.9cm, QR = 6.2cm and PR = 6.6cm. Also construct another parallelogram RSTU where ST = 7cm. A F E X B C M P S Q R T P A D B S C Q R A C D B E
232 Oasis School Mathematics-10 14.1 Warm-up Activities • Draw a circle and show the following parts, • Circumference • Radius • Diameter • Chord • Arc • Semi circle • Sector • Segment • If chord AB = 6cm, then find the length of AM and MB. What is the length of OB? • If two chords are equidistant from the center, are they equal in length? • Draw a circle and show its different parts. Discuss the following in your class and draw a conclusion. A B M O 14.2 Circle Basic definitions If a point moves on a plane in such a way that its distance from a fixed point is always the same, the locus of the point is called a circle. It is denoted by . . The fixed point which is equidistant from any point of the circle is called the center of the circle. In the given figure, points P, Q, R and S are equidistant from the point O, the center of the circle. Circumference The total length of the rim of a circle is its circumference. In other words the perimeter of the circle is the circumference of the circle. radius A line segment which joins the center of a circle to any point of the circumference is the radius. It is denoted by 'r'. Chord A line segment which joins any two points of the circumference is a chord. AB and CD are chords. S P R Q O O Q Circumference O Q Radius A B D O C Chord Unit 14 Circle
Oasis School Mathematics-10 233 Diameter A chord which passes through the center of a circle is called the diameter, i.e., a diameter is the longest chord. AC is the diameter. Diameter = 2 × Radius = 2r. Arc Any part of the circumference of a circle is called an arc. It is denoted by ' '. If the arc is less than half of the circumference, then it is called minor arc and if it is more than half of the circumference, then it is a major arc. In the given figure, AB is the minor arc and ACB the major arc. If a minor and a major arc together make up the whole circumference, then they are called conjugate arcs. Semi–circle A diameter of a circle divides a circle into two equal parts, and each part is called a semi–circle. In the figure, DAB and DCB are two semi–circles of the circle. Sector A region enclosed by any two radii and an arc of a circle is called the sector of the circle. If the sector is less than a semi-circle, then it is a minor sector, and if it is more than a semi-circle, then it is a major sector. Segment A region enclosed by an arc and a chord of a circle is called the segment of the circle. If the segment is less than half the circle then it is called a minor segment and if it is more than half the circle then it is called a major segment. Concentric circles Two or more than two circles having the same centre are called concentric circles, and each circle is called a family of circles. The portion between two circles is called the annulus. A O D B C major sector minor sector O P Q minor segment major segment O A B annular region (annulus) O II III O I Diameter Chord A C D C O A B Minor arc Major arc
234 Oasis School Mathematics-10 Line of centers The straight line joining the centers of two circles is called the line of centers. In the adjoining figure, OO' is the line of centers. Intersecting circles When two circles intersect each other at two different points, they are said to be intersecting circles. The line which joins their points of intersection is their common chord. In the given figure, two circles intersect at A and B, and AB is the common chord. Concyclic points The points which lie on the circumference of the circle are called concyclic points. In the figure, A, B, C and D are concyclic points. Central angle An angle at the center of a circle is central angle. The centre is the vertex for the angle. Here, AOB is the central angle. Some points to remember about a circle: 1. A line cannot intersect a circle at more than two points. 2 The radii of the same circle or equal circles are equal. 3. Two circles are equal if their radii are equal. 4. The greatest chord of a circle is its diameter. 5. In general, the sector of a circle is referred to the area of this region. Some important theorems related to the circle 1. A perpendicular drawn from the center of a circle to a chord bisects the chord. 2. A line, passing through the center of a circle and bisecting a chord, is perpendicular to the chord. 3. Equal chords of a circle are equidistant from the center. 4. Chords which are equidistant from the center of a circle are equal. The theorems mentioned above have already been explained in Grade IX. Central angle and inscribed angles Any angle whose vertex is at the center of a circle is called a central angle (i.e., angle at the center of a circle). In the figure alongside, ∠AOB standing on the APB and reflex ∠AOB standing on the arc AQB are the central angle with the center at O. O O' O O' B A A D C B O O P A B Central angle O P A B Q
Oasis School Mathematics-10 235 Any angle subtended by an arc or chord of a circle at its circumference (i.e. angles at the circumference of a circle) is called an inscribed angle. In the figure alongside, ∠ABC is an inscribed angle subtended by the arc AC. theorem related to arcs and angle subtended by them theorem 15.1 In a circle, arcs that subtend equal angles at the centre are equal. Experimental Verification: • Draw two circles of different radii with centre O. • Draw two radii OA and OC in each circle. • Draw two equal angles AOB and COD at point O. to verify : AB = CD Measure the lengths of AB and CD with the help of a thread and a ruler and tabulate them as follows. Figure AB CD remarks (i) ..... cm ...... cm AB = CD (ii) ..... cm ..... cm AB = CD Conclusion : Hence, if two arcs of a circle subtend equal angles at the center of the circle, they are equal. Arcs that subtend equal angles at the circumference are equal. theorem 15.2 equal arcs of a circle subtend equal angles at the center Experimental Verification: Draw two circles with centre O with different radii and also draw two equal arcs AB and CD with the help of a pencil and a compass. B A D O C D C B A O Fig. I Fig. II A C O B
236 Oasis School Mathematics-10 B A D O C D C B A O Fig. I Fig. II to verify : ∠AOB = ∠COD Measure ∠AOB and ∠COD with the help of a protractor and tabulate them as follows: Figure ∠AOB ∠COD remarks (i) ...... 0 ...... 0 ∠AOB = ∠COD (ii) ...... 0 ...... 0 ∠AOB = ∠COD Conclusion: Thus, equal arcs of a circle subtend equal angles at the center of the circle. Equal arcs subtend equal angles at the circumference. theorem 15.3 equal chords of a circle form equal arcs in the circle. Experimental Verification: Draw two circles with centre O and having different lengths of radii. Also draw two equal chords AB and CD in each circle. A B D O D C C B A O Fig. I Fig. II to verify : AB = CD Measure the lengths of AB and CD with the help of a thread and a ruler and tabulate them as follows: Figure AB CD remarks (i) ..... cm ...... cm AB = CD (ii) ..... cm ..... cm AB = CD Conclusion: Equal chords of a circle form equal arcs on the circle.
Oasis School Mathematics-10 237 theorem 15.4 Chords formed by the equal arcs of a circle are equal. Experimental Verification: Draw two circles with centre O and different length of radii. With the help of pencil and compass, draw two equal arcs AB and CD in each figure. to verify : AB = CD Measure the lengths of AB and CD with the help of scale and tabulate them as follows: Figure AB CD remarks (i) ..... cm ...... cm AB = CD (ii) ..... cm ..... cm AB = CD Conclusion: Thus, chords formed by the equal arcs of a circle are equal. theorem 15.5 If two chords are parallel, then the arc between them are equal. Experimental Verification: Draw two circles of different radii. Draw two chords AB and CD, which are parallel to each other. to verify : AC = BD Measure the length of arc AC and arc BD with the help of a ruler and a thread and tabulate as given below. D C B A O Fig. I A B D O C Fig. II A B D C B D A C Fig. I Fig. II
238 Oasis School Mathematics-10 observation: Figure Arc AC Arc BD remarks (i) ..... cm ...... cm Arc AC = Arc BD (ii) ..... cm ..... cm Arc AC = Arc BD Conclusion: Hence, if two chords are parallel, then the arcs between them are equal. theorems related to arc and central angles/inscribed angles a. Central angle and corresponding arc: If two radii of the same circle meet at the center and they make an angle at the center, it is called a central angle and the arc opposite to the central angle is called corresponding arc. In the adjoining figure, ∠AOB = central angle AB = corresponding arc. Write : AB ≗ ∠AOB Read : Degree measure of AB is ∠AOB relations between central angle and corresponding arc The degree measure of an arc is always equal to the central angle in degree. i.e., ∠AOB ≗ AB b. Inscribed angle and opposite arc: If two chords of the same circle meet at a point on the circumference of a circle and they make an angle, such an angle is called an inscribed angle, and the arc opposite to it is called an opposite arc. In the adjoining figure, ∠ACB is an inscribed angle and AB is an opposite arc. relation between inscribed angle and opposite arc O is the center of a circle, ∠ACB is an angle at the circumference, i.e., an inscribed angle and APB is its opposite arc. Half the degree measure of the arc APB is equal to the inscribed angle ACB, i.e. 1 2 AB ≗ ∠ACB Write : 1 2 AB ≗ ∠ACB Read : Degree measure of 1 2 AB is ∠ACB A B O C O P A B
Oasis School Mathematics-10 239 [Degree measure of an arc] • minor arc APB ≗ ∠AOB • major arc ACB ≗ 3600 – ∠AOB • semi circle ≗ 1800 • circumference ≗ 3600 • 1 2 arc APB ≗ ∠ACB B P A C O Worked out examples example 1 Find the value of x in the given figure. Solution: Given AB//CD then AC = BD [Arcs between parallel chords are equal] ∠AFC = ∠BED [Inscribed angles standing on equal arcs] or x = 150 . example 2 In the given figure, if ∠ABC = 800 , find the degree measure of arc ABC. Solution: Given ∠ABC = 800 AC ≗ 2∠ABC = 2 × 800 = 1600 Then degree measure of arc ABC = 3600 – 1600 = 2000 example 3 In the given figure, DE is the diameter, if Be = Ce prove that ∠AED = 1 2 (∠ABC – ∠ACB). Given : In the given figure, DE is the diameter and BE = CE to prove : ∠AED = 1 2 (∠ABC – ∠ACB) C A E F D B 150 x C B 800 O A C A E O D B
240 Oasis School Mathematics-10 Proof S.N. Statements S.N. reasons 1. DABE = DCE 1. DE being the diameter 2. AD + AB + BE = DC +CE 2. From (1) 3. BE = CE 3. Given 4. AD + AB = DC 4. From (1), (2) and (3) 5. 2AD + AB = DC + AD 5. Adding AD on both sides 6. 2 AD + AB = ADC 6. From (5) 7. 2 AD = ADC – AB 7. From (6) 8. 2∠AED = ∠ABC–∠ACB 8. From (6), using the relation of inscribed angle with its opposite arc length 9. ∠AED = 1 2 (∠ABC – ∠ACB) 9. From (8) Hence, proved. example 4 In the given figure, PQ//RS and QT = RS, prove that ∠PQR = ∠rQt. Solution: Given : In the given figure, PQ //RS and QT = RS to prove : ∠PQR = ∠RQT Proof S.N. Statements S.N. reasons 1. PQ//RS 1. Given 2. PR = QS 2. Arc between two parallel chords are equal 3. QT = RS 3. Given 4. QT = RS 4. Corresponding arcs of two equal chords 5. QS + ST = RT + ST 5. Using whole part axiom on statement 4 6. QS = RT 6. From statement 5 7. PR = RT 7. From statements 2 and 6 8. ∠PQR = ∠RQT 8. Inscribed angles on the equal arcs PR and RT. Hence, proved. T P S R Q
Oasis School Mathematics-10 241 example 5 In the given figure, ABCD is a square and ∆AEF is an equilateral triangle, prove that DB//EF. Solution: Given : In the given figure, ABCD is a square and AEF is an equilateral triangle. to prove : DB // EF Proof S.N. Statements S.N. reasons 1. AB = AD 1. Being the sides of a square 2. AF = AE 2. Being the sides of an equilateral triangle 3. AB = AD 3. Corresponding arcs of two equal chords 4. AF = AE 4. Same as statement 3 5. AF – AB = AE – AD 5. Subtracting statement 3 from 4 6. BF = DE 6. From statement 5 7. DB//EF 7. From statement 6, being arcs between two chords equal Hence, proved. Example 6 In the given figure, AD = BC then, prove that AC = BD. Solution: Given : In the given figure, AD = BC to prove : AC = BD Proof S.N. Statements S.N. reasons 1. AD = BC 1. Given 2. AD = BC 2. Corresponding arcs of two equal chords 3. AD + DC = BC + DC 3. Adding common arc DC on statement 2 4. ADC = BCD 4. From statement 4 5. AC = BD 5. From 4, corresponding chords of two equal arcs Hence, proved. C A E B F D A B D C
242 Oasis School Mathematics-10 example: 7 In the given figure, two chords AB and CD are perpendicular to each other, prove that AC + BD = AD + BC Solution: Given : Two chords AB and CD are perpendicular each other at E. to prove : AC + BD = AD + BC Construction : Join AC and AD Proof S.N. Statements S.N. reasons 1. ∠ACE = ∠CAE = ∠AED 1. Sum of two angles of triangle is equal to opposite exterior angle. 2. ∠ADE + ∠DAE = ∠AEC 2. As statement 1. 3. ∠AED = ∠AEC 3. Both being 900 . 4. ∠ACE + ∠CAE = ∠ADE + ∠DAE 4. From statement 1, 2 and 3. 5. AC + BC = AC + BD 5. From statement 4, using the relation of inscribed angle with corresponding arc. Hence, proved. exercise 14.1 1. (a) In the given figure, write the relation between ∠BAC and arc BC. (b) In the given figure, write the relation between ∠BOC and arc BC. (c) In the given figure, AB//CD. Which two arcs are equal? (d) In the given figure, PR = QS , write the relation between chord PQ and RS. A D B C E P R Q S C A D B B C A B C O
Oasis School Mathematics-10 243 (e) In the given figure, BC = QR , which two angles are equal? (f) In the given figure, O is the center of the circle and ∠AOB = ∠BOC. Which two arcs are equal. 2. Find the value of x in the given figures. 3. In the given figure ∠BAC = 60°. Find the degree measure of arc BC and arc BAC. 4. (a) In the adjoining figure, AB is the diameter with center O. If arc BC = arc CD, prove that AD//OC. (b) In the adjoining figure, AB is the diameter. If AD//OC, prove that BC = DC . 5. (a) In the given circle, ∠WAN = ∠MBZ. Prove that WZ||MN. (b) In the given figure, if AB||CD prove that ∠APC = ∠BQD. A P B C Q R A B C O (a) (c) (b) (d) (e) x O D B x 600 A Q P C B R 200 E A C B G 450 H D P Q 350 R S B A C 200 x D E x x Q B C 600 A C D A B O C D B O A A N W Z M B P Q B C D A
244 Oasis School Mathematics-10 6. (a) In the given figure, if AC= BD prove that ∠AEC = ∠DFB. (b) In the given figure, ∠AEC = ∠BFD. Prove that AC= BD. 7. (a) In the adjoining figure, POQ and ROT are two diameters of the circle with center at O. If Q is the mid-point of arc TQS and ∠QOR is an obtuse, prove that PQ//RS. (b) In the given figure, O is the center of the circle. Prove that: TQ = QS . 8. (a) In the given figure, chords MN and RS of the circle intersect externally at X. Prove that ∠MXR ≡ 1 2 ( MR – NS ). (b) In the given figure, chords MN and RS meet at X. Prove that ∠MXR ≡ 1 2 ( MR + NS ). 9. In the given figure, O is the center of the circle. If ∠AOC = 2 ∠BED, prove that AB//CD. 10. In the given figure, chord PQ is perpendicular to chord RS at M. Prove that: PQ + QS = PS + RQ . 11. In the given figure, PT is the bisector of ∠QPS and UR is the bisector of ∠SRQ. Prove that UT is a diameter of the circle. E O B C D A F E O B C D A F T O Q R S P T O Q R S P X S N R M R P S M Q M S X N R B C D O A E U P Q S T R
Oasis School Mathematics-10 245 12. In the given figure, chord PS and QR intersect at T, prove that: (i) PQ = RS, (ii) PS = QR. 13. In the given figure, PQ//RS and ∠PQR = ∠RQT, prove that QT = RS. 14. In the given figure, AB//CD then, prove that AD = BC. 15. In the given figure, AB = AC and AD = AE, then prove that: BC//DE. 16. In the given figure, If AQ = PB, prove that (i) AP = BQ, (ii) AB //PQ. 17. In the given figure, AOB is a diameter of the circle. AOB bisects ∠CAD prove that AC = AD. Answers 1. Consult your teacher 2. (a) 600 (b) 200 (c) 450 (d) 350 (e) 200 3. 1200 , 2400 theorem 14.5 the angle at the center of a circle is twice the angle at its circumference standing on the same arc. Experimental Verification: Draw two circles of different radii, with centre O and also draw an angle BOC at the centre and ∠BAC at the circumference standing on the same arc BC. A C O B C B A O Fig. I Fig. II Q T P R S Q T P R S C D A B A C D E B B P Q A C B O D A
246 Oasis School Mathematics-10 to verify : ∠BOC = 2∠BAC observation: Figure ∠BAC ∠BoC remarks I ..... 0 ...... 0 ∠BOC = 2∠BAC II ..... 0 ..... 0 ∠BOC = 2∠BAC Conclusion : Hence, the central angle is twice the angle at its circumference standing on the same arc. theoretical Proof: Given : O is the center of the circle. Central angle BOC and inscribed angle BAC are standing on the same arc BC. to prove : ∠BOC= 2∠BAC Construction : Join AO and produce it to K. Proof S.N. Statements S.N. reasons 1. In AOB, AO = OB 1. Radii of same circle 2. ∠OAB = ∠OBA 2. Base angles of an isosceles triangle 3. ∠BOK = ∠OAB + ∠OBA 3. An exterior angle is equal to sum of its opposite interior angles. 4. ∴ ∠BOK = 2∠OAB 4. From (2) and (1) 5. Similarly, in ∆AOC, ∠COK = 2∠CAO 5. Same as above 6. ∠BOK + ∠COK = 2(∠OAB +∠CAO) 6. Adding (4) and (5) 7. ∠BOC = 2∠BAC 7. From (6) whole part axiom Hence, proved. Hence, the angle at the center of a circle is twice the angle at the circumference standing on the same arc. C B K A O Given : O is the center of the circle. ∠BOC and ∠BAC are the central angle and inscribed angle standing on the same arc BC. to prove : ∠BOC = 2∠BAC Proof S.N. Statements S.N. reasons 1. ∠BOC ≗ BC 1. Relation of central angle with arc length 2. ∠BAC ≗ 1 2 BC 2. Relation of inscribed angle with arc length 3. ∠BOC = 2∠BAC 3. From (1) and (2) Hence, proved. Alternative method