Excel in Mathematics - BOOK 8 Final (2080)_compressed

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 201 Vedanta Excel in Mathematics - Book 8 EXERCISE 14.1 General Section – Classwork 1. Let's tell and write the correct answers as quickly as possible. a) If a°, b°, and c° are the angles, of a triangle, a° + b° + c° = ............................ b) If x°, 40°, and 60° are the angles of a triangle, x° = ............................ c) If x°, y°, and z° are the angles of an equilateral triangle, x° = y° = z° = ............. d) If m° and n° are the base angles of an isosceles right-angled triangle m° = n° = ........................ e) If one of the acute angles of a right-angled triangle is 50°, another acute angle is .............. f) x° is the exterior angle of a triangle, and a° and b° are its opposite interior angles x° = ............................ Creative Section - A 2. a) If x°, 50° and 30° are the angles of a triangle, find x°. b) If x°, 2x°, and 30° are the angles of a triangle, find x° and 2x°. c) If 3x°, 4x°, and 5x° are the angles of a triangle, find these angles. d) If the angles of a triangle are in the ratio 2:3:5, find them. e) If two acute angles of a right angled triangle are in the ratio 1:2, find them. f) If x° is the exterior angle and 70° and 50° are its opposite interior angle of a triangle find x°. 3. Find the unknown sizes of angles in the following figures. 4. Find the unknown sizes of angles in the following figures. X Y Z E F G H A B C D P Q T S R M e) b a z y y x y x x 36° 86° 130° 124° f) g) h) A x B C P Q R M K L F D E a) x 60° 30° 3x x x+5° 52° x y x+7° 40° b) c) d) A B C P Q a b c 45° 52° x y z 58° 40° 87° 2x y x y x z 85° 65° Q Y X A B C D O P R Q S O P R a) b) c) d) Geometry: Triangle

Vedanta Excel in Mathematics - Book 8 202 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Creative Section - B 5. a) In the adjoining figure, show that a = 90° – z 2 b) In the given figure, show that AB//CD c) In the figure alongside, GP bisects ∠AGH and HP bisects ∠GHC. Show that ∠GPH = 1 right angle. d) In the adjoining figure, AP and BP are the angular bisectors of ∠BAD and ∠ABC respectively. Show that ∠APB = 1 right angle. e) In the given figure, ABC is an equilateral triangle. If AB//CE, prove that ∠ACE = ∠ECD. 6. a) Let's draw two equilateral triangles ABC of different size. Verify experimentally that each angle of an equilateral triangle is 60°. b) Let's draw two isosceles triangle PQR of different measurements. From the vertex P, draw a straight line to meet the mid-point of QR at S. Verify experimentally that PS is perpendicular to QR. c) Let's draw two isosceles triangles XYZ of different measurements. From the vertex X draw XP perpendicular to YZ at P. Verify experimentally that YP = PZ. A x x x x y y y y z z z E P K A G B D E F C L O A B Q R S B C D 60° 65° 65° 42° 72° 75° 30° e) f) g) h) i) A A B B P P X B P Z Q Y d a x z y c b e A P S T R Q Q Q R R 54° 72° 48° 52° 100° 140° 40° 57° 80° 50° x q r p p 25° C j) k) l) y B C 88° 46° A D A B P C D H G P a b z x y Q A B O A E B C D D A B P C Geometry: Triangle

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 203 Vedanta Excel in Mathematics - Book 8 14.3 Congruent triangles Let's take a rectangular sheet of paper. Fold it along its diagonal. Cut the folded edge. Place two triangles, one above another as shown in the figure alongside. Here, the two triangular pieces can be exactly fitted to one above another. In such a case, the triangular pieces are said to congruent. The parts (sides and angles) fitting one above another are called the corresponding parts. Thus, DABD and DB'CD' are congruent triangles. It is written as DABD @ DB'CD'. Corresponding sides of AB, AD and BD are CD', B'C and B'D' respectively. Similarly, the corresponding angles of ∠A, ∠B, and ∠D are ∠C, ∠D', and ∠B' respectively. In congruent triangles corresponding sides are equal and corresponding angles are also equal. 14.4 Experiment on the conditions of congruency of triangles by construction. Construction 1: D PQR is a given triangle. (i) Measure the length of PQ with the help of a compass and draw P'Q' = PQ. (ii) With the centres at P' and Q' and the radii equal to PR and QR respectively draw two arcs to intersect each other at R'. (iii) Join P', R', and Q'R'. Here, three sides of DPQR are equal to three sides of DP'Q'R'. So, DPQR and DP'Q'R are congruent triangles. Construction 2: D ABC is a given triangle. (i) Measure AB with the help of a compass and draw A'B' = AB. (ii) Measure ∠A with the help of a compass and draw the same angle at A' by drawing A'X. (iii) Measure the length of AC by a compassed and cut the same length A'C' on A'X. (iv) Join B' and C'. Here two side of DABC are equal to two sides of A'B'C and angle made by them ∠A and ∠A' are also equal. So, DABC and DA'B'C' are congruent triangles. DD' D D' BB' A A B B' C C D' B' D B C A R P Q R' P' Q' A B C X A' B' C' Geometry: Triangle

Vedanta Excel in Mathematics - Book 8 204 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur A P B C Q R Construction 3: D XYZ is a given triangle. (i) Measure XY with the help of a compass and draw X'Y' = XY. (ii) Measure ∠X and ∠Y by a compass and draw the same angles at X' and Y' respectively such that two straight line segments intersect at Z'. Here, two angles of DXYZ are equal to two angles of DX'Y'Z'. Also, the sides between these angles are also equal. So, DXYZ and DX'Y'Z' are congruent triangles. Construction 4: DPQR is a given right angled triangle, right angled at Q. (i) Measure QR by a compass and draw Q'R' = QR. (ii) Construct an angle of 90° at Q' such that ∠ R'Q'X = 90°. (iii) Take an arc equal to PR and cut Q'X from R' at P'. (iv) Join P' and R'. Here, both the triangles are right angled triangles. Hypotenuse of DPQR is equal to the hypotenuse of DP'Q'R'. Also a side QR of DPQR is equal to the side Q'R' of DP'Q'R'. So, DPQR and DP'Q'R' are congruent triangles. 14.5 Conditions of congruency of triangles There are 3 sides and 3 angles in a triangle. Two triangles can be congruent if 3 parts (out of 6 parts) of one triangle are equal to 3 corresponding parts of another triangle under the following conditions. These conditions are used as axioms. (i) S.S.S. axiom When three sides of a triangle are respectively equal to three corresponding sides of another triangle, they are said to be congruent. In ∆ABC and ∆PQR, a) AB = PQ (S) b) BC = QR (S) c) CA = RP (S) d) ∴ ∆ABC ≅ ∆PQR (S.S.S. axiom) Corresponding parts of congruent triangles are also equal. ∴ ∠A = ∠P, ∠B = ∠Q and ∠C = ∠R Note:The parts opposite to the equal parts of congruent triangles are called the corresponding parts. www.geogebra.org/classroom/hjwdhkkx Classroom code: HJWD HKKX Vedanta ICT Corner Please! Scan this QR code or browse the link given below: X Y Z X' Y' Z' P Q R P' X Q' R' Geometry: Triangle

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 205 Vedanta Excel in Mathematics - Book 8 A P B C Q R A P B C Q R (ii) S.A.S. axiom When two sides of one triangle and angle made by them are respectively equal to the corresponding sides and angle of another triangle, they are said to be congruent triangles. In ∆ABC and ∆PQR, a) AB = PQ (S) b) ∠B = ∠Q (A) c) BC = QR (S) d) ∴ ∆ABC ≅ ∆PQR (S.A.S. axiom) Now, ∠C = ∠R and ∠A = ∠P [Corresponding angles of congruent triangles] Also, CA = RP [Corresponding sides of congruent triangles] (iii) A.S.A. axiom When two angles and their adjacent side of one triangle are respectively equal to the corresponding angles and sides of another triangle, they are said to be congruent triangles. In ∆ABC and ∆PQR, a) ∠B = ∠Q (A) b) BC = QR (S) c) ∠C = ∠R (A) d) ∆ABC ≅ ∆PQR (A.S.A. axiom). Now, AC = PR and AB = PQ [Corresponding sides of congruent triangles] Also, ∠A = ∠P [Corresponding angles of congruent triangles] (iv) R.H.S. axiom In two right angled triangles, when the hypotenuse and one of the two remaining sides are respectively equal, they are said to be congruent triangles. In right angled ∆s ABC and PQR, a) ∠B = ∠Q (R) b) AC = PR (H) c) BC = QR (S) d) ∴ ∆ABC ≅ ∆PQR (R.H.S. axiom) Now, ∠C = ∠R and ∠A = ∠P [Corresponding angles of congruent triangles.] Also, AB = PQ [Corresponding sides of congruent triangles.] A P B C Q R Geometry: Triangle

Vedanta Excel in Mathematics - Book 8 206 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Worked-out Examples Example 1: Find the value of x in the following pairs of congruent triangles. Also write the sizes of corresponding angles and corresponding sides. Solution: (i) Here, ∆ABC ≅ ∆PQR ∠A = ∠P = 55° ∴ BC = QR x + 1 = 3.5 or, x = 3.5 – 1= 2.5 cm Now, ∠C = ∠R = 45° ∴ AB = PQ = (x + 0.5) cm = (2.5 + 0.5 cm) = 3 cm Also, ∠B = ∠Q = 80° ∴ AC = PR = 4.2 cm (ii) Here, ∆DEF ≅ ∆XYZ ∠E = ∠Y = 90° ∴ DF = XZ or, 3x – 1.5 = 4.5 or, 3x = 4.5 + 1.5 = 6 or, x = 6 3 = 2 cm Now, ∠D = ∠X = 40° ∴ EF = YZ = (x + 0.5) cm = (2 + 0.5) cm = 2.5 cm Also, ∠F = ∠Z = 50° ∴ DE = XY = (2x – 1) cm = (2 × 2 – 1) cm = 3 cm Example 2: In the figure alongside, AB = CD and AB // DC, show that a) AD = BC b) AD // BC Solution: In ∆ABC and ∆ACD, (i) AB = CD [Given] (ii) ∠BAC = ∠ACD [Being alternate angles] (iii) AC = AC [Common side] (iv) ∴ ∆ABC ≅ ∆ACD [S.A.S. axiom] (v) ∴AD = BC [Corresponding sides of congruent triangles] (vi) ∠ACB = ∠CAD [Corresponding angles of congruent triangles] (vii) AD // BC [Being equal alternate angles] Proved In ∆ABC, ∠A = 180° – (80° + 45°) = 55° In ∆PQR, ∠Q = 180° – (45° + 55°) = 80° A P D E F Y X Z R Q B C (i) (ii) 80° 45° 45° 55° 40° 50° (2x-1)cm (x+0.5)cm 3.5cm (x+0.5)cm 4.2cm (x+1)cm 4.5cm (3x-1.5)cm A B D C Geometry: Triangle

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 207 Vedanta Excel in Mathematics - Book 8 EXERCISE 14.2 General Section 1. From these pairs of congruent triangles. Let's tell and write the necessary conditions of congruency. 2. From these pairs of congruent triangles. Let's tell and write the corresponding sides and corresponding angles. Corresponding angle of ∠A is .................... Corresponding angle of ∠R is .................... Corresponding angle of ∠B is .................... Corresponding side of FG is ....................... Corresponding angle of ∠Z is .................... Corresponding angle of ∠F is .................... Corresponding side of DF is ....................... Corresponding side of DE is ....................... Corresponding angle of ∠B is .................... Corresponding side of AC is ...................... Corresponding angle of ∠C is .................... Corresponding angle of ∠R is .................... ......................................... axiom ......................................... axiom ......................................... axiom ......................................... axiom a) b) c) d) a) A B C Q R P b) G E F Z X Y d) A B C R Q S c) E F D C B A Geometry: Triangle

Vedanta Excel in Mathematics - Book 8 208 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 3. What are the necessary conditions to be included in the following conditions so that DABC @ D DEF? a) AB =DE, ∠B = ∠E ....................... b) AB = DE, BC = EF, ....................... c) ∠A = ∠D, ∠B = ∠E ....................... Creative Section - A 4. Use the necessary axioms and show that the following pairs of triangles are congruent. Also, write their corresponding sides and angles 5. Use the necessary axioms and show that the following pairs of triangles are congruent. Also, write their corresponding sides and angles. 6. Mention the necessary axioms to show the following pairs of triangles congruent. Also, find the value of x in each case. a) b) c) d) A D B C E F a) A P Q R 3.8cm 3.8cm 4.2cm 2.5cm 2.5cm 4.2cm B C b) X F E D Y Z 3cm 3.5cm 3.5cm 3cm 120° 120° c) 3.6cm 3.6cm 6.4cm 6.4cm K L M T S R d) 4.7cm 4.7cm 40° 40° 60° 80° W X Y G E F a) A B C R Q P b) 55° 60° 55° 65° W D E F Y X c) d) F E Z A B C D E F Y X G 5 cm x A B C D A B D C x 50° A B C D A B C D E 45° x x 6.5 cm Geometry: Triangle

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 209 Vedanta Excel in Mathematics - Book 8 Creative Section - B 8. a) In the figure alongside, AB// CD and AB = CD. Prove that i) DAOB @ DCOD ii) AO = OD and BO = OC. b) In the adjoining figure, PQ = SR and PQ // SR, show that i) PS = QR ii) PS // QR c) In the given figure, ABC is an isosceles triangle in which AB = AC. If AP ^ BC, prove that (i) DAPB @ DAPC (ii) ∠B = ∠C (iii) BP = PC d). In the adjoining figure, AP ^ CD, BQ ^ CD and OP = OQ. Prove that AP = BQ. It's your time - Project work and Activity Section 9. Let's take a rectangular chart paper. Fold it diagonally into two halves and cut out the halves using scissors. Now, put one half on the another half and identify the corresponding angles and corresponding sides of two congruent triangles. 14.6 Similar triangles Although the adjoining triangles are not equal in size, they have exactly the same shape. Because of their similar shape, they are similar triangles. Thus, if two triangles are equiangular, they are the similar triangles. 7. Find the value of x in the following pairs of congruent triangles. Also, write the sizes of corresponding angles and sides. A (x+1.3)cm (x+3.5)cm P B 5.5cm C 35° 85° 35° 60° x cm 2.5cm (2x–1.5)cm 20° 45° 45° 115° x cm 3.2cm Q R X Y D F E W a) b) L K D A (x+0.8)cm (x+3.2)cm (x+2.3)cm (2x–2)cm 3.1cm 6.5cm 3.1cm (x+2)cm P M B C E F 5.2cm 38° 52° 86° 64° 30° 30° Q R c) d) B C A P S R P Q A C D O B A B C D Q P O 75° 45° 60° 75° 45° 60° A B C Q R P Geometry: Triangle

Vedanta Excel in Mathematics - Book 8 210 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Example 1: In the given figure, ∆PQR ~ ∆XYZ, find the sizes of ∠R and ∠Y. Also find the length of side PQ. P X Q R Y Z 65° 95° 5 cm 7 cm 10 cm The similar triangles ABC and PQR is written as ∆ABC ~ ∆PQR. The symbol ~ is used to denote the similarity of triangles. Furthermore, if two triangles are similar, the ratios of their corresponding sides are equal. It means the corresponding sides of similar triangles are proportional. ∴ AB PQ BC QR AC PR = = Solution Here, ∠R = ∠Z = 95° [Corresponding angles of similar triangles are equal.] ∠Y = ∠Q =65° Again, ∆PQR ~ ∆XYZ ∴ PQ XY PR XZ = [Corresponding sides of similar triangles are proportional] or, PQ 7 10 5 = =2 or, PQ = 7 × 2 = 14 cm Example 2: In the given figure, ∆ABC ~ ∆PQC. Find the value of x and y. Solution Here, ∆ABC ~ ∆PQC ∴ AB and PQ are corresponding sides and their ratio is AB PQ . AC and PC are corresponding sides and their ratio is AC PC . BC and QC are corresponding sides and their ratio is BC QC . Now, AB PQ BC QC AC PC = = [Corresponding sides of similar triangles are proportional] or, y 5 24 x 20 10 = = Taking the first and the third equal ratios, y 5 20 10 = = 2 or, y = 5 × 2 = 10 cm Also, taking the second and the third equal ratios, www.geogebra.org/classroom/y3gktyer Classroom code: Y3GK TYER Vedanta ICT Corner Please! Scan this QR code or browse the link given below: Worked-out Examples y cm A B P Q 20 cm 10 cm C 24 cm x cm 5 cm Geometry: Triangle

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 211 Vedanta Excel in Mathematics - Book 8 EXERCISE 14.3 General Section – Classwork 1. Let's say and write the corresponding sides of these similar triangles. Also write the equal ratios of the corresponding sides. a) C A B F D E b) P Q A B R Corresponding side of BC is .............. Corresponding side of AC is .............. Corresponding side of AB is .............. Equal ratios of corresponding sides are ........................ Corresponding side of QR is ................. Corresponding side of PQ is ................. Corresponding side of PR is ................. Equal ratios of corresponding sides are ...................... Creative Section 2. a) In the given figure ∆ABC ~ ∆PQR. If ∠C = ∠R = 50° and AC = 12 cm, find the length of side PQ and also find the sizes of ∠A and ∠Q. b) In the adjoining figure, ∆DEF ~ ∆XYZ. If DE = 10 cm, EF = 6 cm, DF = 8 cm, YZ = 12 cm , ∠E = ∠Y and ∠D = ∠X, find the value of x and y. 6 cm 10 cm y cm 8 cm x cm X Z Y D E F 24 x 20 10 = = 2 or, 2x = 24 or, x 24 2 = = 12 cm ∴ x = 12 and y = 10 cm. A B C P R 12 cm Q 9 cm 8 cm 55° 50° 50° 75° Geometry: Triangle

Vedanta Excel in Mathematics - Book 8 212 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur c) In the adjoining figure, ∆ABC ~ ∆ADE. AD = 15 cm, DE = 10 cm, BC = 25 cm and AC = 20 cm. Find length of AB and AE. d) In the figure alongside, ∠BAD = ∠ACD and ∆ABD ~ ∆ABC. Find the length of BD. e) In the given figure, ∆PQR ~ ∆XQY. If ∠Q = 40°, ∠R = 60°, PR:XY = 5 :3 and PQ = 15 cm, find ∠QXY and the length of QX. f) In the given figure, ∆ABC ~ ∆AXC, AX ⊥ BC, AB = 3 cm and AC = 4 cm. Find the length of CX. 3. a) In the given figure, ∆ABC ~ ∆XCY, BC = 9 cm, BY = 3 cm and XY = 4 cm. Find the length of AB. b) In the given figure, ∆CDE ~ ∆ABC. Find the value of x with suitable reasons. A B C x 10 cm 8 cm 7 cm 6 cm D E A X B Y C 15 cm 10 cm 20 cm 25 cm D E A B C 40° 60° P X Q Y R 5 cm 4 cm 3 cm A X B C 10 cm 4 cm 8 cm A B D C 14.7 Conditions of similarity of triangles Two triangles can be similar under the following conditions: (i) If two angles of one triangle are respectively equal to two angles of another triangle, they are said to be similar. Geometry: Triangle

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 213 Vedanta Excel in Mathematics - Book 8 A P B C Q R (ii) When the ratios of the corresponding sides of two triangles are equal, i.e. the corresponding sides of triangles are proportional, the triangles are said to be similar. In ∆KLM and ∆XYZ, KL XY = LM YZ = MK ZX ∴ ∆KLM ~ ∆XYZ Here, a) KL is the opposite side of ∠M and ∠M = ∠Z; the opposite side of ∠Z is XY. So, KL and XY are corresponding sides. b) LM is the opposite side of ∠K and ∠K = ∠X; the opposite side of ∠X is YZ. So, LM and YZ are corresponding sides. c) MK is the opposite side of ∠L and ∠L = ∠Y; the opposite side of ∠Y is ZX. So, MK and ZX are the corresponding sides. (iii) If any two corresponding sides of triangles are proportional and the angles included by them are equal, the triangles are said to be similar. In ∆DEF and ∆RST, a) DE RS = EF ST b) ∠DEF = ∠RST c) ∴ ∆DEF ~ ∆RST. Worked-out Examples Example 1: In the adjoining figure, ∠PXY = ∠PQR = 40°. Show that a) ∆PQR ~ ∆PXY b) XY QR = PY PR Solution a) In ∆PQR and ∆PXY, (i) ∠PQR = ∠PXY [Both of them are 40°] (ii) ∠QPR = ∠XPY [Being common angles] (iii) ∆PQR ~ ∆PXY [Two pairs of angles of two triangles are equal] b) XY QR PY PR = [Corresponding sides of similar triangles are proportional] K L M X Y Z D E F S T R In ∆ABC and ∆PQR a) ∠ABC = ∠PQR b) ∠ACB = ∠PRQ c) ∴ ∆ABC ~ ∆PQR. Note:When two angles of one triangle are equal to two angles of another triangle, their remaining angles are equal. P Y Q R X 40° 40° Geometry: Triangle

Vedanta Excel in Mathematics - Book 8 214 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Example 2: In the given figure, ABC is a right angled triangle where ∠A = 90°. If AD ^ BC, prove that a) DABC ~ DADC b) AC2 = BC.CD Solution a) In ∆ABC and ∆ADC, (i) ∠BAC = ∠ADC [Both of them are right angle (90°)] (ii) ∠ACB = ∠ACD [Being common angle] (iii) ∠ABC = ∠DAC [Remaining angles of the triangles] (iv) ∴ ∆ABC ~ ∆ADC b) AC CD BC AC = or, AC2 = BC.CD Proved A D B C EXERCISE 14.4 General Section – Classwork 1. Find out the pairs of similar triangles from the following figures. (i) ....................... and ............ are similar (ii) ...................... and ............are similar (iii) ....................... and ............ are similar (iv) ...................... and ............are similar A 4 cm 2.1 cm 1.8 cm 20° 20° 120° 120° 70° 7 cm 5 cm P Q R X Y Z F G E B C M K L F D S R T W X Y E 8 cm 10 cm 14 cm 8 cm 15 cm 10 cm 12 cm 2.8 cm 2.4 cm 70° 150° 150° Creative Section - A 2. a) In the adjoining figure, ∠ABC = ∠APQ = 50°. Show that (i) ∆ABC ~ ∆APQ (ii) BC PQ = AC AQ . A P B C 50° 50° Q Geometry: Triangle

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 215 Vedanta Excel in Mathematics - Book 8 b) In the given figure, ∠DEF = ∠DHG = 50°. Prove that (i) ∆DHG ~ ∆DEF (ii) DG DF = GH EF c) In the figure alongside, PQ//WY. Show that (i) ∆WXY ~ ∆PXQ (ii) WY.PX = WX.PQ d) In the given figure, AB//CD. Prove that (i) ∆AOB ~ ∆COD (ii) AO.OC = BO.OD. e) In the adjoining figure, ABC is a right angled triangle where ∠A = 90°. If AD ⊥ BC, prove that (i) ∆ABC ~ ∆ABD (ii) ∆ABC ~ ∆ACD (iii) ∆ABD ~ ∆ACD (iv) AB2 = BC.BD (v) AC2 = BC.CD (vi) AC.BD = AB.AD Creative Section - B 3. a) In the adjoining figure, ∠BED = ∠BAC prove that (i) ∆ABC ~ ∆BDE (ii) Find the value of x. b) In the figure given alongside, prove that (i) ∆ABC ~ ∆BDC (ii) Find the length of AC c) In the given figure, PQ ⊥ QR, ST ⊥ QR. Show that (i) ∆PQR ~ ∆SRT (ii) Find the length of the flag. A D B C A C D B 8 cm 6 cm 4 cm D G E F 50° 50° H W P X Q Y A O C D B P Q S T R 6 m 6 m 8 m A 8 cm 6 cm 6 cm x cm 9 cm C B E D 4 cm 3 cm Geometry: Triangle

Vedanta Excel in Mathematics - Book 8 216 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 15.1 Quadrilaterals - Looking back Classroom - Exercise Let’s say and write the answers as quickly as possible. 1. a) Name of the given quadrilateral is ........................... b) Vertices of the given quadrilateral are ........................... c) Sides of the given quadrilateral are ........................... d) Angles of the given quadrilateral are ........................... 2. a) In the quadrilateral PQRS,∠P + ∠Q + ∠R + ∠S = ........................... b) If ∠P + ∠Q + ∠S = 300°, then ∠R = ........................... The plane figure bounded by four line segments is called a quadrilateral. In the given figure, ABCD is a quadrilateral. AB, BC, CD, and DA are the sides of the quadrilateral ABCD. Similarly, ∠A, ∠B, ∠C, and ∠D are the angles of the quadrilateral. In any quadrilateral, the total sum of its angle is 360°. ∴ ∠A + ∠B + ∠C + ∠D = 360°. 15.2 Some special types of quadrilaterals On the basis of certain special properties, there are some special types of quadrilaterals. Parallelogram, rectangle, square, rhombus, etc. are some special types of quadrilaterals. 1. Parallelogram In the adjoining figure, quadrilateral ABCD is a parallelogram. It’s opposite sides are equal and parallel. A quadrilateral can be a parallelogram, if it satisfies any one of the following conditions. (i) Its two pairs of opposite sides are parallel (ii) Its two pairs of opposite angles are equal (iii) Any two of its opposite sides are equal and parallel (iv) Its diagonals bisect each other. D C A B www.geogebra.org/classroom/d5chbt4m Classroom code: D5CH BT4M Vedanta ICT Corner Please! Scan this QR code or browse the link given below: A B C D C D A B R S P Q Unit 15 Geometry: Quadrilateral and Regular Polygon

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 217 Vedanta Excel in Mathematics - Book 8 15.3 Construction of parallelogram Let's use the properties of parallelograms to construct them under the given conditions. (A) When two adjacent sides and angle made by them are given: Example 1: Construct a parallelogram ABCD in which the sides AB = 5.2 cm, AD = 4.5 cm and ∠BAD = 45°. Solution: Steps of construction (i) Draw a line segment AB = 5.2 cm (ii) At A, draw ∠BAX = 45°. (iii) With the centre at A and radius AD = 4.5 cm, draw an arc to intersect AC at D. (iv) With centre at D and radius DC = AB = 5.2 cm, draw and arc. (v) With centre at B and radius B BC = AD = 4.5 cm, draw another arc to cut the previous arc at C. (vi) Join D, C and B, C. Thus ABCD is the required parallelogram. (B) When two adjacent sides and a diagonal are given: Example 2: Construct a parallelogram PQRS in which PQ = 4.8 cm, QR = 37 cm and PR = 5.6 cm. Solution: Steps of construction (i) Draw a line segment PQ = 4.8 cm. (ii) With centre at Q and radius QR = 3.7 cm, draw an arc. (iii) With centre at P and radius PR = 5.6 cm draw another arc to cut previous arc at R. Then, join Q and R. A 5.2 cm B 4.5 cm 4.5 cm D 5.2 cm C X 45° P Q R 4.8 cm 5.6 cm 3.7 cm Geometry: Quadrilateral and Regular Polygon

Vedanta Excel in Mathematics - Book 8 218 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur (iv) With centre at R and radius RS = PQ = 4.8, draw an arc. (v) With centre at P and radius PS = QR = 3.7, draw another arc to cut previous arc at S. (vi) Join R, S and S, P. Thus, PQRS is the required parallelogram. (c) When one side, a diagonal and angle made by them ae given: Example 3: Construct a parallelogram PREM in which side PR = 5.4 cm, diagonal PE = 6.3 cm and ∠RPE = 30°. Solution: Steps of construction (i) Draw a line segment PR = 5.4 cm (ii) At P, draw = ∠RPX = 30° (iii) With centre at P and radius PE = 6.3 cm draw an arc to cut PX at E. Join E and R. (iv) With centre at E and radius EM = PR = 5.4 cm, draw an arc. (v) With centre at P and radius PM = Re, draw another arc to cut previous arc at M. (vi) Join E, M and P. Hence, PREM is the required parallelogram. Properties of parallelogram Property 1: The opposite sides of a parallelogram are equal. Experimental verification Step 1: Draw three parallelograms ABCD of different sizes with the help of set-squares. P Q S R 4.8 cm 3.7 cm 5.6 cm 3.7 cm 4.8 cm P 5.4 cm R 6.3 cm E 30° P 5.4 cm Q 6.3 cm M 30° D C A B Fig (i) C D Fig (iii) A B D C Fig (ii) A B Geometry: Quadrilateral and Regular Polygon

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 219 Vedanta Excel in Mathematics - Book 8 Step 2: Measure the sides of each parallelogram. Write the measurements in the table. Fig. Length of opposite sides Length of opposite sides Result AB CD AD BC (i) AB = CD and AD = BC (ii) AB = CD and AD = BC (iii) AD = CD and AD = BC Conclusion: The opposite sides of a parallelogram are equal. Property 2: The opposite angles of a parallelogram are equal. Experimental verification Step 1: Draw three parallelograms ABCD of different sizes with the help of set squares. Step 2: Measure the angles of each parallelogram with the help of protractor. Write the measurements in the table. Fig. Size of opposite angles Size of opposite angles Result ∠A ∠C ∠B ∠D (i) ∠A = ∠C and ∠B = ∠D (ii) ∠A = ∠C and ∠B = ∠D (iii) ∠A = ∠C and ∠B = ∠D Conclusion: The opposite angles of a parallelogram are equal. Property 3: Diagonals of parallelogram bisect each other. Experimental verification Step 1: Draw three parallelograms ABCD of different sizes with the help of set squares. D A C B Fig (i) D C B Fig (ii) A D C B Fig (iii) A D A C B Fig (iii) D A C B Fig (i) D A C B Fig (ii) O O O Geometry: Quadrilateral and Regular Polygon

Vedanta Excel in Mathematics - Book 8 220 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Step 2: Draw diagonals in each parallelogram. Step 3: Measure the lengths of AO, OC, BO, and OD in each parallelogram. Write the measurements in the table. Fig. Parts of AC Parts of BD Result AO OC BO OD (i) AO = OC and BO = OD (ii) AO = OC and BO = OD (iii) AO = OC and BO = OD Conclusion: Diagonals of parallelogram bisect each other. 2. Rectangle The adjoining quadrilateral is a rectangle. A rectangle is a parallelogram in which each of its angle is 90° and opposite sides are equal. Following are some of the important properties of rectangles. (i) Each interior angle is 90°. (ii) Opposite sides are equal and parallel. (iii)The diagonals of rectangle are equal. (iv)The diagonals bisects each other. 15.4 Construction of rectangle (A) When adjacent sides are given: Example 4: Construct a rectangle ABCD in which AB = 5 cm, AD = 4.3 cm. Solution: Steps of construction (i) Draw a line segment AB =5 cm. (ii) At the points A and B construct ∠BAX = 90° and ∠ABY = 90° respectively. (iii) With the centre at B and radius BC = 4.3 cm, draw an arc to cut BY at C. (iv) Wit the centre at A andradius AD = BC = 4.3 cm, draw another arc to cut AXat D. (v) Join C and D. Hence, ABCD is the required rectangle. A B C D A B D C X Y 5 cm 4.3 cm D A 90° C B Geometry: Quadrilateral and Regular Polygon

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 221 Vedanta Excel in Mathematics - Book 8 Properties of rectangle Property 1: All the angles of a rectangle are equal and are right angle. Experimental verification Step 1: Draw three rectangles ABCD of different measurements. fig (i) fig (ii) fig(iii) Step 2: Measure ∠A, ∠B, ∠C, and ∠D with the help of a protractor and write the measurements in the table. Fig. No. ∠A ∠B ∠C ∠D Result (i) (ii) ∠A = ∠B = ∠C = ∠D = 90° (iii) Conclusion: All the angles of a rectangle are equal and are right angle. Property 2: The diagonals of a rectangle are equal. Experimental verification Step 1: Draw three rectangles ABCD of different sizes by using set squares. Fig (iii) Fig (i) Fig (ii) D D A C B D A C B A C B O O O Step 2: Draw diagonals in each rectangle. Step 3: Measure the lengths of AC and BD in each rectangle. Write the measurements in the table. Fig. AC BD Result (i) AC = BD (ii) AC = BD (iii) AC = BD Conclusion: The diagonals of a rectangle are equal. C C C D D D A B A B A B Geometry: Quadrilateral and Regular Polygon

Vedanta Excel in Mathematics - Book 8 222 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 3. Square The quadrilateral given alongside is a square. It is a rectangle in which its adjacent sides are equal. In other words, all sides of a square are equal. Following are some of the important properties of square. (i) Each interior angle of square is 90°. (ii) All sides of square are equal. (iii)Diagonals of square are equal. (iv)Diagonals of square bisect each other perpendicularly. 15.5 Construction of square Example 5: Construct a square of a side 4.5 cm. Solution: Steps of construction (i) Draw a line segment EF =4.5 cm. (ii) At the points E and F, construct ∠FEX = 90° and ∠EFY = 90° respectively. (iii) With the centre at E and radius EH = EF = 4.5 cm draw an arc ot cut EX at H. (iv) Wit the centre at F and radius FG = EF = 4.5 cm draw another arc to cut FY at G. (v) Join H and G. Hence, EFGH is the required square. Properties of square Property 1: Diagonals of a square bisect each other perpendicularly. Experimental verification Step 1: Draw three squares ABCD of different measurements by using set squares. Step 2: Draw diagonals in each square. Step 3: Measure AO, OC, BO, OD and ∠AOB, ∠BOC, ∠COD, ∠AOD. Write the measurements in the table. Fig. AO OC BO OD ∠AOB ∠BOC ∠COD ∠AOD Result (i) AO = OC, BO = OD, ∠AOB = ∠BOC = ∠COD = ∠AOD (ii) AO = OC, BO = OD, ∠AOB = ∠BOC = ∠COD = ∠AOD (iii) AO = OC, BO = OD, ∠AOB = ∠BOC = ∠COD = ∠AOD Conclusion: Diagonals of a square bisect each other perpendicularly. A B D C E F G H X Y 4.5 cm 4.5 cm Fig (i) D A C B O Fig (ii) D A C B O Fig (iii) D A C B O D A C B Geometry: Quadrilateral and Regular Polygon

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 223 Vedanta Excel in Mathematics - Book 8 Property 2: The diagonals of a square bisect the vertical angles. Experimental verification Step 1: Draw a square ABCD. Draw diagonals AC and BD. Step 2: Measure the vertical angles and the angles formed due to the division of each vertical angle by the diagonals. Write the measurements in the table. Vertical angles Other angles formed by diagonals at each vertex Result ∠BAD = ∠BAC = ∠CAD = ∠BAC = ∠CAD ∠ABC = ∠ABD = ∠CBD = ∠ABD = ∠CBD ∠BCD = ∠BCA = ∠ACD = ∠BCA = ∠ACD ∠ADC = ∠ADB = ∠BDC = ∠ADB = ∠BDC Conclusion: The diagonals of a square bisect the vertical angles. 4. Rhombus A rhombus is a parallelogram in which all sides are equal. (i) The opposite angles of a rhombus are equal. (ii) Each diagonal bisects the vertical angles. (iii)The diagonals are not equal and bisect each other perpendicularly. (iv)The triangles formed by the diagonals are congruent. Properties of rhombus Property 1: The diagonals of a rhombus bisect each other perpendicularly. Experimental verification Step 1: Draw a rhombus ABCD. Draw diagonals AC and BD. Step 2: Measure OA, OC, OB and OD. Also measure ∠AOB and write all the measurements in the table. OA OC OB ∠AOB Result OA = OC, OB = OD, ∠AOB = 90° Conclusion: The diagonals of a rhombus bisect each other perpendicularly. D C A B A B C D O A B C D Geometry: Quadrilateral and Regular Polygon

Vedanta Excel in Mathematics - Book 8 224 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Property 2: The diagonals of a rhombus bisect the vertical angles. Experimental verification Step 1: Draw a rhombus ABCD. Draw diagonals AC and BD. Step 2: Measure the vertical angles and the angles formed due to the division of each vertical angle by the diagonals. Write the measurements in the table. Vertical angles Other angles formed by diagonals at each vertex Result ∠BAD = ∠BAC = ∠CAD = ∠BAC = ∠CAD ∠ABC = ∠ABD = ∠CBD = ∠ABD = ∠CBD ∠BCD = ∠BCA = ∠ACD = ∠BCA = ∠ACD ∠ADC = ∠ADB = ∠BDC = ∠ADB = ∠BDC Conclusion: The diagonals of a rhombus bisect the vertical angles. 5. Kite The adjoining quadrilateral is a kite ABCD. The properties of a kite are as following. (i) Two pairs of adjacent sides are equal. AB = AD and BC = CD (ii) One pair of opposite angle are equal ∠ABC = ∠ADC (iii)The diagonals of a kite intersect each other perpendicularly. (iv)The longer diagonal of a kite bisect the shorter diagonal. OD = OB Properties of kite Property 1: The adjacent sides of a kite are equal and angle made by the longer and shorter sides are also equal. Experimental verification Step 1: Draw a kite ABCD. Step 2: Measure the sides AB, BC, CD and DA. Also measure ∠ABC ∠ADC. Write the measurements in the table. AB BC CD DA ∠ABC ∠ADC Result AB = AD, BC = CD, ∠ABC = ∠ADC Conclusion: The adjacent sides of a kite are equal and angle made by the longer and shorter sides are also equal. A B C D O A B C D B D C A Geometry: Quadrilateral and Regular Polygon

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 225 Vedanta Excel in Mathematics - Book 8 Property 2: The longer diagonal of a kite bisect the shorter diagonal perpendicularly. Experimental verification Step 1: Draw three kites ABCD of different measurements. B D A O C B D O A C B D O C A Fig (i) Fig (ii) Fig (iii) Step 2: Measure OB and OD. Also measure ∠AOB and write all the measurements in the table. Fig. OB OD ∠AOB Result (i) OB = OD, ∠AOB = 90° (ii) OB = OD, ∠AOB = 90° (iii) OB = OD, ∠AOB = 90° Conclusion: The longer diagonal of a kite bisect the shorter diagonal perpendicularly. 6. Trapezium The adjoining quadrilateral is a trapezium ABCD. The properties of a trapezium are as following. (i) A pair of opposite sides are parallel. DC // AB. (ii) The sum of co-interior angles formed on nonparallel sides is 180°. Properties of trapezium Property 1: The sum of co-interior angles formed between parallel and non-parallel sides of a trapezium is 180°. Experimental verification Step 1: Draw three trapezium ABCD. B D C A B D C A B D C A Fig (i) Fig (ii) Fig (iii) Step 2: Measure ∠ABC, ∠BCD and ∠BAD, ∠ADC in each figure. Write the measurements in the table. Fig ∠ABC ∠BCD ∠BAD ∠ADC Result (i) ∠ABC + ∠BCD = 180°, ∠BAD + ∠ADC = 180° (ii) ∠ABC + ∠BCD = 180°, ∠BAD + ∠ADC = 180° (iii) ∠ABC + ∠BCD = 180°, ∠BAD + ∠ADC = 180° Conclusion: The sum of co-interior angles formed on non-parallel sides is 180°. A B D C Geometry: Quadrilateral and Regular Polygon

Vedanta Excel in Mathematics - Book 8 226 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Worked-out Examples Example 1: In the following figures, find the unknown sizes of angles. Solution: (i) x = 115° [Being opposite angles of a parallelogram] y + 115° = 180° [Being the sum of co-interior angles] or, y = 180° – 115° = 65° z = y = 65° [Being opposite angles of a parallelogram] ∴ x = 115° = z = 65° (ii) y = 110° [Being alternate angles] z = 110° [Being corresponding angles] w + 110° = 180° [Being the sum a straight angle] or, w = 180° – 110° or, w = 70° x = w = 70° [Being opposite angles of a parallelogram] ∴ w = x = 70° and y = z = 110° (iii) a + 73° + 37° = 180° [Being the sum of the angles of a triangle] or, a = 180° – 110° or, a = 70° a = b = 70° [Being opposite angles of a parallelogram] c = 73° [Being alternate angles] d = 37° [Being alternate angles] ∴a = b = 70°, c = 73° and d = 37°. Example 2: In the adjoining parallelogram, find the length of AB. Solution: In the parallelogram ABCD, AB = CD [Being opposite sides of the parallelogram] or, 2x + 3 = 4x – 1 or, 2x – 4x = – 1 – 3 or, – 2x = – 4 or, x = 4 2 = 2 ∴ Length of AB = 2x + 3 = 2 × 2 + 3 = 4 + 3 = 7 cm. Example 3: In the adjoining quadrilateral ABCD, AB // CD and BC // AD. Prove that: a) ∆ABC ≅ ∆ACD b) AB = CD c) AD = BC (i) 115° x y z c b d a 37° 73° (iii) x y z w 110° (ii) D C A B (4x–1) cm (2x+3) cm D C A B Geometry: Quadrilateral and Regular Polygon

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 227 Vedanta Excel in Mathematics - Book 8 Solution: a) In ∆ABC and ∆ACD, (i) ∠BAC = ∠ACD (A) [Being alternate angles] (ii) AC = AC (S) [Being common side] (iii) ∠ACB = ∠CAD (A) [Being alternate angles] (iv) ∴∆ABC ≅ ∆ACD [A.S.A. axiom] b) AB = CD [Corresponding sides of congruent triangles] c) AD = BC [Corresponding sides of congruent triangles] EXERCISE 15.1 General section - Classwork 1. Let's say and write the answer as quickly as possible. Ramesh constructed a quadrilateral WXYZ in which XY = 4.7 cm, WX = 3.8 cm and ∠W = ∠X = ∠Y = ∠Z = 90° a) What type of quadrilateral did Ramesh construct? .......................................... b) What are the measurements of sides YZ and WZ of the quadrilateral? .................. YZ= ............................... WZ = ............................... c) What is the relation between the diagonals WY and XZ? ...................................... 2. Pramila constructed a quadrilateral KLMN in which KL = 4.8 cm, LM = 3.6 cm, ∠K = ∠M = 60° and ∠L = ∠N = 120°. a) What type of quadrilateral did Pramila construct? ............................... b) What are the measurements of the sides MN and KN? MN = ............................... KN = ............................... c) Are KL // MN and KN // LM? ............................... 3. Sunayana constructed a quadrilateral ABCD in which AB = 5.4 cm, diagonal BD = 7.2 cm, ∠ABD = 30°, AB // CD and AD //BC. a) What type of quadrilateral did she construct? ............................... b) What is the measurement of side CD? CD =.............................. c) Are the sides AD and BC equal? ............................... Geometry: Quadrilateral and Regular Polygon

Vedanta Excel in Mathematics - Book 8 228 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 4. From the following given sketches, construct parallelograms, rectangles and squares. a) b) c) d) e) Creative section - A 5. Let's construct parallelograms of the following measurements. a) Parm. ABCD in which AB = 5.8 cm, ∠BAD = 60°, AD = 4.5 cm. b) Parm. PQRS in which PQ = 6.3 cm, ∠PQR = 105°, QR = 5.4 cm. c) Parm. EFGH in which EF = 4.9 cm, FG = 3.8 cm and EG = 5.7 cm. d) Parm. MNOP in which MN = 5.2 cm, MP = 4.3 cm and NP = 6.6 cm e) Parm. QRST in which QR = 4.5 cm, ∠RQS = 30° and QS = 6.3 cm. f) Parm. WXYZ in which WX = 5.1 cm, ∠WXZ = 45° and XZ= 7 cm. 6. Let's construct rectangles of the following measurements. a) Rectangle ABCD in which AB= 4.7 cm and BC = 3.6 cm. b) Rectangle KLMN in which KL= 5.5 cm and KN = 4.5 cm. 7. Let's construct squares of the following measurements. a) Square PQRS in which PQ = 3.6 cm b) Square EFGH in which EF = 6.3 cm. Creative section - B 8. Prakash constructed a quadrilateral QRST in which QR = ST = 5 cm, ∠QRS = ∠QTS = 120° and RS = QT = 4.2 cm a) What type of quadrilateral is QRST? b) Draw an appropriate sketch of QRST. c) Construct the quadrilateral QRST according to your sketch. 5.4 cm D C A B 75° 4.2 cm 4.6 cm 3.5 cm S R Q 6 cm P E F H G 45° 5 cm 3.3 cm 4.8 cm 5.7 cm 4.3 cm W X Z Y Geometry: Quadrilateral and Regular Polygon

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 229 Vedanta Excel in Mathematics - Book 8 9. Teacher asked Nirmala to construct a quadrilateral ABCD in which AB = CD = 4.8 cm, AC = 6 cm, ∠BAC = 45° and BC = AD. a) What type of quadrilateral is ABCD? b) Draw an appropriate sketch of ABCD. c) Construct the quadrilateral ABCD according to your sketch. 10.A swimming pool is exactly in the shape of a rectangle of length 45 m and breadth 30 m. a) Taking the scale of 1 cm to represent 10 m, express the length and breadth of the rectangular pool in centimetres. b) Draw an appropriate sketch of the pool. c) Construct a rectangle according to your sketch. 11. Sunita Magar has a vegetable garden which is exactly in the shape of a square of length 55 m. a) Taking the scale of 1 cm to represent 10 m, express the length of the garden in centimetres. b) Draw an appropriate sketch of the garden. c) Construct a square according to your sketch. It's your time - Project work and Activities section 12. Let's draw parallelograms of your own measurements according to the following conditions. a) Take the lengths of two adjacent sides and angle made by them. From your construction show that (i) opposite sides of a parallelograms are equal (ii) opposite angles of a parallelograms are equal. b) Take the lengths of two adjacent sides and a diagonal. From your construction show that: (i) Opposite sides of a parallelogram are equal. (ii) Opposite angles of a parallelogram are equal. (iii) Diagonals of a parallelogram bisect each other. (iv) Sum of the interior angles of a parallelogram is 360°. 13. a) Let's construct a rectangle or your own measurements. From your construction show that: (i) Opposite sides of a rectangle are equal. (ii) Diagonals of a rectangle bisect each other. (iii) Sum of the interior angles of a rectangle is 360°. Geometry: Quadrilateral and Regular Polygon

Vedanta Excel in Mathematics - Book 8 230 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur b) Let's construct a square of your own measurements. From your construction, show that: (i) All sides of a square are equal. (ii) Diagonals of a square bisect each other perpendicularly. (iii) Sum of the interior angles of a square is 360°. 14. a) Let's measure the length and breadth of the floor of your classroom. Taking an appropriate scale, express the length and breadth in centimetres and construct the shape. b) Let's measure the length and breadth of a wall of your classroom. Taking an appropriate scale, express them in centimeters and construct the shape. EXERCISE 15.2 General Section - Classwork 1. Let’s say and write the unknown angles as quickly as possible. x = .............., y = .................. z = ............. x = ............., y = .............., z= ....... a) y x 45° z x z y 120° b) a = ................., b = ..................... c = ................., d = ..................... p = ................, q = ..................... r = ................., s = ..................... c a b d 110° c) d) r q s p 120° 2. a) In the adjoining rectangle, if AO = 2 cm, BD = .................... b) In the given square, if SQ = 6 cm , PO = ...................... 2cm D A B O C S P Q O R Geometry: Quadrilateral and Regular Polygon

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 231 Vedanta Excel in Mathematics - Book 8 c) In the given parallelogram, if GP = 2.5 cm and HF = 3 cm, EG = ............... and PF = ....................... Creative Section - A 3. Let’s calculate the unknown sizes of angles in the following figures. H E P F G a) y x x y z z y x w a b c d 55° z 120° 105° 110° b) c) d) e) z x y 55° f) g) h) z w y y y x a b 30° 4x-5 z 35° 5x+5 5x 4x 30° 45° 150° 110° 40° i) j) k) x x a c b x y x z z y y l) 80° 40° 70° 56° 40° 150° m) n) o) x x b c d z x y 3x a y y z p) 30° 4. Find the values of x. Then, find the lengths of the sides of the following parallelograms. a) D (2x–3) cm (x+7) cm C A B b) S P Q R (x+4) cm (3x–2) cm c) Z W X Y (4x–1) cm (2x+3) cm d) H E F G (x+5) cm (3x–5) cm Geometry: Quadrilateral and Regular Polygon

Vedanta Excel in Mathematics - Book 8 232 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Creative Section -B 5. a) In the adjoining quadrilateral ABCD, ∠A = ∠C and ∠B = ∠D. Prove that, quadrilateral ABCD is a parallelogram. Hint: Sum of the angles of quad. ABCD is 360° b) In the adjoining quadrilateral PQRS, PQ // SR and PS // QR. Prove that (i) ∆PQR ≅ ∆PRS (ii) PQ = SR (iii) PS = QR c) In the given figure, ABCD is a parallelogram. Diagonals AC and BD are intersecting at O. Prove that (i) ∆AOB ≅ ∆COD (ii) AO = OC and BO = OD. d) In the adjoining figure, ABCD is a rectangle. Prove that (i) ∆BAD ≅ ∆ABC (ii) AC = BD e) In the given figure, PQRS is a square. Prove that (i) ∆POQ ≅ ∆QOR (ii) QO ⊥ PR. 6. a) Draw two parallelograms ABCD of different measurements. Then explore experimentally the relationship between their opposite sides. D A C B S P R Q O D A C B S P Q R D A B O C b) Draw two rectangles PQRS of different measurements and explore experimentally the relationship between their diagonals. c) Explore experimentally the relationship between the diagonals of square. (Two figures of different measurement are necessary) It’s your time - Project Work and Activity Section 7. a) Let’s take a rectangular sheet of paper and fold it diagonally. Then, explore the relationship between the diagonals of a rectangle. b) Let’s fold a rectangular sheet of paper to make a square. Cut out the square. Now, fold the square sheet of paper diagonally. Then, explore the relationship between the diagonals of a square. Geometry: Quadrilateral and Regular Polygon

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 233 Vedanta Excel in Mathematics - Book 8 15.6 Regular polygons A polygon is a closed plane figure bounded by three or more than three line segments. Triangles, quadrilateral, pentagon, hexagon, etc. are a few examples of polygons. A polygon having all sides and angles equal is called a regular polygon. Equilateral triangle, square, regular pentagon, etc. are a few examples of regular polygons. Equilateral triangle Square Regular pentagon Regular hexagon Interior angles of a polygon The angles formed by the sides of a polygon inside the polygon are known as interior angles of the polygon. In the given figure, ∠ABC, ∠BAC, and ∠ACB are the interior angles of the ∆ABC. Exterior angles of a polygon When a side of a polygon is produced, it substends an angle with another side of the polygon outside the polygon. The angle so formed is called an exterior angle of the polygon. In the given figure, side AB of the quadrilateral ABCD is produced to E. So, ∠CBE is the exterior angle of the quadrilateral. A B C A B E C D 8. a) Let’s take a rectangular sheet of paper and fold it as shown in the diagram to make a parallelogram. Cut out the parallelogram. Now, fold the parallelogram diagonally and explore the relationship between the diagonals of a parallelogram. 15.7 Sum of the interior angles of a polygon (i) The figure alongside is a triangle. The sum of the interior angles of the triangle is 180°. ∴ ∠A + ∠B + ∠C = 180°. B A C Geometry: Quadrilateral and Regular Polygon

Vedanta Excel in Mathematics - Book 8 234 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur On the basis of this knowledge, we can generalise a formula to find out the sum of the angles of any polygon having ‘n’ number of sides. Polygons No. of sides (n) No. of triangles formed Sum of the interior angles of polygon Triangle 3 1 1 × 180° = (3 – 2) × 180° Quadrilateral 4 2 2 × 180° = (4 – 2) × 180° Pentagon 5 3 3 × 180° = (5 – 2) × 180° Hexagon 6 4 4 × 180° = (6 – 4) × 180° …… …… …… …… n – gon n n – 2 (n – 2) × 180° Thus, the sum of the interior angles of a polygon = (n – 2) × 180° In the case of a regular polygon, its n number of interior angles are equal. (ii) The figure alongside is a quadrilateral. The number of sides of the quadrilateral (n) = 4 The sum of the angles of a quadrilateral = 360° = 2 × 180° = (n – 2) × 180° A D C B Geometry: Quadrilateral and Regular Polygon

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 235 Vedanta Excel in Mathematics - Book 8 ∴ The sum of ‘n’ number of interior angles = (n – 2) × 180° ∴ The size of each interior angle = (n – 2) × 180° n Thus, each interior angle of a regular polygon = (n – 2) × 180° n 15.8 Sum of the exterior angles of a polygon The figure alongside is a quadrilateral. The sides of the quadrilateral are produced only to one direction. Let, a, b, c, and d are the interior angles of the quadrilateral. Also, w, x, y, and z are the exterior angles of the quadrilateral. Here, the sum of the interior angles of the quadrilateral is a + b + c + d = (4 – 2) × 180° = 2 × 180° Also,a + z = 180°, b + w = 180°, c + x = 180° and d + y = 180° ∴(a + z) + (b + w) + (c + x) + (d + y) = 4 × 180° or, (a + b + c + d) + (w + x + y + z) = 4 × 180° or, 2 × 180° + (w + x + y + z) = 4 × 180° or, w + x + y + z = 4 × 180° – 2 × 180° = 2 × 180° = 360° Thus, the sum of the exterior angle of any polygon = 360° In the case of a regular polygon, its n number of exterior angles are equal. ∴ The sum of n number of exterior angles = 360° ∴ The size of each exterior angle = 360° n Thus, each exterior angle of a regular polygon = 360° n . Worked-out Examples Example 1: Find the sum of interior angles of an octagon. Solution: In an octagon, number of sides (n) = 8 Now, the sum of interior angles of an octagon = (n – 2) × 180° = (8 – 2) × 180° = 6 × 180° = 1080° Example 2: Find the interior and exterior angle of a regular hexagon. Solution: In a regular hexagon, number of sides (n) = 6 Now, interior angle of the regular hexagon = (n – 2) × 180° n = (6 – 2) × 180° 6 = 4 × 30° = 120° Again, the exterior angle of the regular hexagon = 360° n = 360° 6 = 60° c b d a z y x w Geometry: Quadrilateral and Regular Polygon

Vedanta Excel in Mathematics - Book 8 236 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Creative Section - A 2. Find the sum of interior angles of the following polygons by using formula. a) Quadrilateral b) Pentagon c) Hexagon d) Heptagon e) Octagon f) Nonagon g) Decagon h) Dodecagon 3. Find the size of interior angles of the following regular polygons by using formula. a) Quadrilateral b) Pentagon c) Hexagon d) Heptagon e) Octagon f) Nonagon g) Decagon h) Dodecagon 4. Find the size of exterior angles of the following regular polygons by using formula. a) Quadrilateral b) Pentagon c) Hexagon d) Heptagon e) Octagon f) Nonagon g) Decagon h) Dodecagon 5. Find the number of sides of polygons whose sum of interior angles are given below. a) 540° b) 720° c) 900° d) 1080° e) 1440° 6. The size each interior angle of regular polygons are given below. Find the number of sides of the polygons. a) 90° b) 108° c) 120° d) 135° e) 140° EXERCISE 15.3 General Section - Classwork 1. Let’s say and write the correct answer in the blank spaces. a) The sum of the interior angle of a regular polygon with n number of side is obtained by the formula ...................................... b) The sum of the exterior angle of any regular polygon is ................................ c) Each interior angle of a regular polygon with n number of sides is obtained by the formula .................................................... d) Each exterior angle of a regular polygon with n number of sides is obtained by the formula .................................................... e) The number of sides (n) of the following polygons are (i) pentagon, n = .................. (ii) hexagon, n = .................. (iii) heptagon, n = .................. (iv) octagon, n = .................. (v) nonagon, n = .................. (vi) decagon, n = .................. Example 3: In the adjoining hexagon, find the size of x°. Solution: Here, the sum of the interior angles of hexagon= (n – 2) × 180° x° + 130° + (180° – 70°) + 105° + (180° – 70°) + 130° = (6–2)×180° or, x° + 130° + 110° + 105° + 110° + 130° = 4 × 180° or, x° + 585° = 720° or, x° = 720° – 585° = 135° 70° x° 70° 105° 130° 130° Geometry: Quadrilateral and Regular Polygon

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 237 Vedanta Excel in Mathematics - Book 8 8. Find the unknown sizes of angles of the following polygons. Creative Section - B 9. Find the value of x° in each case. a) b) It’s your time - Project work! 10. a) Let’s draw six equilateral triangles of the same measurement of length in a chart paper. Cut out each triangle and arrange them taking different number of triangles to form the following polygons. (i) Rhombus (ii) Trapezium (iii) Hexagon b) How many rhombus did you form? c) How many trapezium did you form? d) How many hexagon did you form? 105° 75° 80° 95° 100° 60° a) b) x x x c) 105° 87° 95° 70° 25° 20° d) 80° 130° 65° 30° x e) x 64° 75° 110° f) 130° x h) x x 165° 160° 160° (x+10)° 50° 60° i) 150° 150° 72° x 140° 135° g) x c) d) 7. Find the number of sides of regular polygon whose measure of each exterior angles are given below. a) 72° b) 60° c) 45° d) 40° e) 36° x x x x Geometry: Quadrilateral and Regular Polygon

Vedanta Excel in Mathematics - Book 8 238 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 16.1 Coordinates - Looking back Classroom - Exercise 1. Let's study the adjoining graph, tell and write the answers as quickly as possible. a) The straight line XOX' is called .................................... b) The straight line YOY' is called .................................... c) The point O is called .................................... d) The coordinates of the points A, B, C and D are ........................, ........................., ........................ and ......................... e) The coordinates of the origin are ......................... In the graph given alongside, XOX' and YOY' are known as the coordinate axes. Here, XOX' is called the x-axis (abscissa) and YOY' is called the y-axis (ordinate). The point O is called the origin. In the graph, the coordinates of the point A is (5, 3), B is (–4, 2), C is (–3, –5), and D is (2, –3). Furthermore, the intersected x-axis and y-axis divide the coordinate plane into 4 regions. These regions are called the quadrants. The region XOY is called the first quadrant. The region X’OY is called the second quadrant. The region X’OY’ is called the third quadrant. The region XOY’ is called the fourth quadrant. Y Y' O X A D B C X' Y' X' X Y O P (0, 0) Q R S Unit 16 Coordinates

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 239 Vedanta Excel in Mathematics - Book 8 16.2 Pythagoras Theorem The adjoining triangle is a right angled triangle ABC right angled at B. The side AC opposite to the right angle is called its hypotenuse. The sides AB and BC are called its perpendicular and base respectively. Pythagoras was a Greek Mathematician who lived around 2500 years ago. He discovered a significant fact about right-angled triangles known as Pythagoras Theorem. Pythagoras Theorem states that in a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of perpendicular and base. ∴ (hypotenuse)2 = (perpendicular)2 + (base)2 h2 = p2 + b2 Study the following illustration and try to understand the Pythagoras Theorem. In the given figure, DABC is a right angled triangle, right angled at B. Here AC = 5 cm is the hypotenuse, AB = 4 cm is perpendicular and BC= 3 cm is base. Here, area of the square ACDE = 25 sq. cm. Area of square BCHI = 9 sq. cm. Area of square ABFG = 16 sq. cm. Thus, area of square ACDE = area of square BCHI + area of square ABFG i.e. square of hypotenuse = square of perpendicular + square of base h2 = p2 + b2 . 16.3 Pythagorean Triples Suppose the hypotenuse (h) of a right angled triangle be 5, perpendicular (p) be 4, and base (b) be 3. Then, according to Pythagoras theorem, h2 = p2 + b2 or, 52 = 42 + 32 or, 25 = 16 + 9 or, 25 = 25 Thus, the relation h2 = p2 + b2 is satisfied by the particular values of h, p and b. Here, 5, 4, and 3 are called the Pythagorean Triples. When a set of Pythagorean triple is known, we can find other sets of triples multiplying each number of triple by the same natural number. For example, A B C Base Hypotenuse Perpendicular www.geogebra.org/classroom/reuczgru Classroom code: REUC ZGRU Vedanta ICT Corner Please! Scan this QR code or browse the link given below: F G B I H C E D A Coordinates

Vedanta Excel in Mathematics - Book 8 240 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 3, 4 and 5 is a triple then, 3 × 2 = 6, 4 × 2 = 8 and 5 × 2 = 10 So, the other set of triples is 6, 8, and 10. Similarly, 15 (3 × 5), 20 (4 × 5), and 25 (5 × 5) are also the triples. Experiment 19: Experimental verification of Pythagoras Theorem. Step 1: Draw three right angled triangles ABC of different sizes and right angled at B. A B C A B A B C C Fig (i) Fig (ii) Fig (iii) Step 2: Measure the lengths of the sides of each triangle and write the measurements in the table. Fig No. AB BC CA AB2 BC2 CA2 AB2 + BC2 Results (i) CA2 = AB2 + BC2 (ii) CA2 = AB2 + BC2 (iii) CA2 = AB2 + BC2 Conclusion: The square of the hypotenuse of a right-angled triangle is equal to the sum of the squares of perpendicular and base. Worked-out Examples Example 1: In the adjoining right angled triangle ABC, calculate the length of AC. Solution Here, base (b) = BC = 8 cm perpendicular (p) = AB = 6 cm hypotenuse (h) = AC = ? Now, by using the Pythagoras theorem, h2 = p2 + b2 or, h2 = 62 + 82 or, h2 = 36 + 64 or, h2 = 100 or, h = 100 = 10 cm ∴ The required length of AC is 10 cm. A B C 8 cm 6 cm Coordinates

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 241 Vedanta Excel in Mathematics - Book 8 Example 2: In the given figure, a ladder 13 m long rest against a vertical wall. If the height of the wall at which the upper end of the ladder is supported is 12 m, find the distance of the foot of the ladder from the wall. Solution: Let PR be the length of the ladder, PQ be the height of the wall, and QR be the distance between the wall and foot of the ladder. Here, hypotenuse (h) = PR = 13 m perpendicular (p)= PQ = 12 m base (b) = QR Now, by using Pythagoras theorem, h2 = p2 + b2 or, 132 = 122 + b2 or, 169 = 144 + b2 or, b2 = 169 – 144 = 25 or, b = 25 = 5 m. So, the required distance is 5 m. 13 m P R Q 12 m EXERCISE 16.1 General Section – Classwork 1. Let's say and write the answers as quickly as possible. a) p = 1, b = 2, h = .................. b) p = 2, b = 3, h = ...................... c) p = 3, b = 4, h = ................. d) h= 3, p = 1, b = ........................ e) h = 5, b = 3, p = ................. f) h = 5, p = 4, b = ........................ 2. Let's say and write the distance between origin and the given points. a) (0,0) and (1, 2) distance = ................ b) (0,0) and (2, 3) distance = ................. c) (0,0) and (3, 1) distance = .................d) (0,0) and (2, 1) distance = ................. Creative Section - A 3. Find the unknown lengths of sides of the following right angled triangles. 4. Using Pythagoras theorem, find whether the following sides of triangles are the sides of right angled triangles. a) 3 cm, 4 cm, 5 cm b) 6 cm, 8 cm, 10 cm c) 9 cm, 12 cm, 16 cm d) 7 cm, 10 cm, 14 cm e) 5 cm, 5 3 cm, 10 cm f) 3 3 cm, 4 cm, 8 cm A B C a) 4 cm 3 cm Q P R 6 cm 10 cm b) X Y Z 13 cm 12 cm c) K M L 9 cm 12 cm d) Coordinates

Vedanta Excel in Mathematics - Book 8 242 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 8. a) A ladder 10 m long rests against a vertical wall 6 m above the ground. At what distance does the ladder touch the ground from the bottom of the wall? A B C 10 m 6 m 5. Test and identify whether the following sets of numbers are pythagorean tripples or not. a) 6, 8, 10 b) 9, 15, 20 c) 7, 24, 25 6. a) In the adjoining figure, ABC is a triangle. If BD ⊥ AC, AB = BC = 15 cm, BD = 9 cm, find the length of AC. b) Find the length of diagonal of the adjoining rectangle. c) If the length and breadth of a rectangle are 12 cm and 5 cm respectively, what will be the length of its diagonal? A C B D 9 cm 15 cm 15 cm 12 cm 16 cm b) The vertical height of a tree is 18 m. The uppermost part of the tree is broken by wind 5 m above the ground. At what distance does the uppermost part of the tree touch the ground from the foot of the tree? c) In the figure alongside, ABCD is a rectangular ground of length 40 m and breadth 30 m. Ramesh reached at C walking from A to B and B to C. But Shyam reached to C walking from A to C directly. Who walked the shorter distance and by how much? A B C 5 m B A C D 40 m 30 m Creative Section -B 7. a) In the adjoining figure, find the values of x and y. b) Find the length of x and y in the adjoining kite. P Q R S 13 cm 4 cm x cm 3 cm y cm A C B E D 12 cm 5 3cm y cm 13 cm x cm Coordinates

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 243 Vedanta Excel in Mathematics - Book 8 Example 1: Find the distance between the point A (4, 1) and B (10, 9) . Solution : In A (4, 1), x1 = 4 and y1 = 1 In B (10, 9), x2 = 10 and y2 = 9 Now, by using the distance formula, AB = (x2 – x1 ) 2 + (y2 – y1 ) 2 = (10 – 4)2 + (9 – 1)2 = 62 + 82 = 36 + 64 = 100 = 10 units . (4, 1) A B (10, 9) 16.4 Distance between two points Let the coordinates of the point A be (x1 , y1 ) and that of B be (x2 , y2 ). Draw perpendiculars AP from A and BQ from B to the x - axis. Also, draw perpendicular AR from A to BQ. Here, OP = x1 PA = QR = y1 OQ =x2 QB = y2 PQ = AR = OQ – OP = x2 – x1 BR = QB – QR = y2 – y1 Thus, AR = x2 – x1 and BR = y2 – y1 Now, from right angled-triangle ARB, by Pythagoras theorem, h2 = p2 + b2 AB2 = BR2 + AR2 AB2 = (y2 – y1 ) 2 + (x2 – x1 ) 2 AB = (x2 – x1 ) 2 + (y2 – y1 ) 2 is the distance between the points A and B. Y Y' O X' X x2 -x1 y2 -y1 R P Q B(x2 , y2 ) A(x1 ,y1 ) d) In the figure alongside, AB is an electric pole. When an electric wire falls, its one end touches the ground at P, 5 m away from the base of the pole. If the length of the wire from the top of the pole to the ground is 13 m, complete the following problems. (i) Find the height of the pole. (ii) If the other end of the wire touches the grand at Q, 9 m away from the base to the opposite side of the pole, find the length of the falling wire. e) The length of each side of the given equilateral triangle is x cm. When the length of 1 cm, 2 cm and 3 cm are reduced from the sides of the triangle respectively, a right angled triangle is formed. (i) Find the length of each side of the equilateral triangle. (ii) Find the length of sides of the newly formed right angled triangle. A Q P 5 m B 9 m 13 m Coordinates

Vedanta Excel in Mathematics - Book 8 244 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Example 2: A point G on the x-axis is 8 units right from the origin. Another point H on the y-axis is 6 units above the origin. Find the distance between the points G and H. Solution: Here, the coordinates of the point G is (8, 0) and that of H is (0, 6) So, x1 = 8, x2 = 0, y1 = 0, y2 = 6 Now, by using distance formula, GH = (x2 – x1 ) 2 + (y2 – y1 ) 2 = (0 – 8)2 + (6 – 0)2 = 64 + 36 = 100 = 10 units ∴ The distance between the point G and H is 10 units. Example 3: The centre of a circle is O (5, 8) and A (12, 9) is any point in its circumference. Find the diameter of the circle. Does another point B (9, 4) lie in the circumference of the circle? Solution: Here, O (5, 8) is the centre of the circle. A (12, 9) is a point in its circumference. So, x1 = 5, x2 = 12, y1 = 8 and y2 = 9. Now, by using distance formula, OA = (x2 – x1 ) 2 + (y2 – y1 ) 2 = (12 – 5)2 + (9 – 8)2 = 72 + 12 = 50 = 25 × 2 = 5 2 units ∴ The radius of circle (OA) = 5 2 units ∴ The diameter of the circle = 2(OA) = 2 × 5 2 = 10 2 units. Again, to find the distance between O (5, 8) and B (9, 4) OB = (9 – 5)2 + (4 – 8)2 = 42 + (–4)2 = 32 = 16 × 2 = 4 2 units ∴ OA ≠ OB. So, OB is not the radius of the given circle. The point B (9, 4) does not lie in the circumference of the circle. O (5, 8) A (12, 9) Y' X' X Y O(0, 0) H G Coordinates

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 245 Vedanta Excel in Mathematics - Book 8 Example 4: Show that the points A (5, 3), B (1, 2), C (2, –2), and D (6, –1) are the vertices of a square. Solution: Let, A (5, 3), B (1, 2), C (2, –2), and D (6, –1) are the vertices of a quadrilateral ABCD. Now, by using the distance formula, AB = (x2 – x1 ) 2 + (y2 – y1 ) 2 = (1 – 5)2 + (2 – 3)2 = (–4)2 + (–1)2 = 17 units BC = (x2 – x1 ) 2 + (y2 – y1 ) 2 = (2 – 1)2 + (–2 – 2)2 = 12 + (–4)2 = 17 units CD = (x2 – x1 ) 2 + (y2 – y1 ) 2 = (6 – 2)2 + (–1+ 2)2 = 42 + 12 = 17 units DA = (x2 – x1 ) 2 + (y2 – y1 ) 2 = (5 – 6)2 + (3+ 1)2 = (–1)2 + 42 = 17 units Also, diagonal AC = (x2 – x1 ) 2 + (y2 – y1 ) 2 = (2– 5)2 + (–2 – 3)2 = 34 units And, diagonal BD = (x2 – x1 ) 2 + (y2 – y1 ) 2 = (6– 1)2 + (–1 – 2)2 = 34 units Thus, in quadrilateral ABCD, The sides AB = BC = CD = DA = 17 units Also, the diagonals AC = BD = 34 units Here, all sides of the quadrilateral ABCD are equal and it's diagonals are also equal. So, the quadrilateral ABCD is a square and the given points are the vertices of the square. Example 5: Show that the points (4, 3), (3, 2), and C (2, 1) are collinear. Solution: Let the points be A (4, 3), B (3, 2), and C (2, 1). Now, by using distance formula, AB = (x2 – x1 ) 2 + (y2 – y1 ) 2 = (3 – 4)2 + (2 – 3)2 = (–1)2 + (–1)2 = 2 units BC = (x2 – x1 ) 2 + (y2 – y1 ) 2 = (2 – 3)2 + (1 – 2)2 = (–1)2 + (–1)2 = 2 units AC = (x2 – x1 ) 2 + (y2 – y1 ) 2 = (2 – 4)2 + (1 – 3)2 = (–2)2 + (–2)2 = 8 = 2 2 units Here, AB + BC = 2 + 2 = 2 2 = AC Hence, the given points are collinear. D(6, –1) C(2, –2) A(5, 3) B(1, 2) A (4, 3) B (3, 2) C (2, 1) Coordinates

Vedanta Excel in Mathematics - Book 8 246 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Example 6: If the distance between the points (a, 0) and (0, 3) is 5 units, find the value of a. Solution: Let P (a, 0) and Q (0, 3) be the two given points. Here, PQ = 5 units. Now, by using distance formula, PQ = (x2 – x1 ) 2 + (y2 – y1 ) 2 or, 5 = (0 – a)2 + (3 – 0)2 or, 5 = (–a)2 + 32 or, 5 = a2 + 9 EXERCISE 16.2 General Section 1. Say and write the distance between the given points. a) (m1 , n1 ) and (m2 , n2 ) = .......................... b) (0,0) and (3, 4) distance = .......................... c) (0,0) and (4, 3) distance = .......................... d) (0,0) and (1, 3) distance = .......................... e) (0,0) and (2, 3) distance = .......................... Creative Section - A 2. Find the distance between the following points. a) (1, 2) and (3, 4) b) (5, 3) and (1, 6) c) (–2, 1) and (4, 3) d) (4, 5) and (–3, 6) e) (–5, –4) and (–13, 2) f) (5 + 3, 2 – 3) and (7 + 3, 2 + 3) 3. a) A point P is on the x-axis, 6 units right from the origin and another point Q is on the y-axis, 8 units above the origin. Find the distance between P and Q. b) A point A lies on the x-axis 9 units left from the origin and another point B lies on the y-axis 12 units below the origin. Find the distance between A and B. c) Find the distance between origin and a point P (–5, 5). Creative Section - B 4. a) When the map of Nepal is presented in a coordinate plane, the coordinates of Kathamndu is found to be (5, 3) and that of Pokhara (4, 2 3). Find the map distance between these places. If 1 unit represents 100 km, find the actual distance of Pokhara from Kathmandu. b) Biratnagar lies at the point (1, 5) and Dharan lies at (3, 9) when the map of Nepal is placed on the graph of coordinate plane. If 1 unit represents 4.5 5 km, find the actual distance between these two places. Squaring both sides 52 = ( a2 + 9) 2 or, 25 = a2 + 9 or, a2 = 25 – 9 or, a2 = 16 or, a = 16 = 4 So, the required value of a is 4. P (a, 0) Q (0, 3) Coordinates

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 247 Vedanta Excel in Mathematics - Book 8 c) The adjoining figure is the map of Nepal presented in the coordinate axes of the graph. Find the distance between Tikapur and Bhairahawa. If 1 unit represents 18 5 km, find the actual distance of Tikapur to Bhairahawa. d) From the given graph of map of Nepal, find the actual distance between Hetauda and Damak, if 1 unit represents 4.1 85 km. 5. a) The centre of a circle lies at O (4, 6) and A (–5, 18) is any point in its circumference. (i) Find the radius of the circle. (ii) Find the diameter of the circle. (iii) Show that B (13, –6) also lies in the circumference of the circle. b) P (1, 3) is the centre of a circle and a point Q (7, 11) is any point in its circumference. Find the diameter of the circle. Does another point (–5, 10) lie in the circumference of the circle? 6. a) Show that the points A (3, 4), B (7, 8), and C (11, 4) are the vertices of an Isosceles triangle. b) Show that the points P (–6, 2), Q (1, 7), and R (6, 3) are the vertices of a scalene triangle. c) Show that the points A (2, –2), B (8, 4), C (5, 7), and D (–1, 1) are the vertices of a rectangle ABCD. d) Show that the points P (1, 1), Q (4, 4), R (4, 8), and S (1, 5) are the vertices of a parallelogram. Tikapur O X Y Bhairahawa O X Y Hetauda Damak Coordinates

Vedanta Excel in Mathematics - Book 8 248 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur e) Show that the points A (3, 1), B (8, 1), C (8, 6), and D (3, 6) are the vertices of a square. f) Show that the points (4, 6), (–1, 5), (–2, 0), and (3, 1) are the vertices of a rhombus. g) Show that the points P (1, 1), Q (–1, –1), and R (– 3, 3) are the vertices of an equilateral triangle. 7. Show that the following points are collinear. a) (1, 2), (4, 5), (8, 9) b) (–3, 7), (0, 4), (2, 2) c) (–1, –1), (2, 3), (8, 11) d) (4, –3), (2, 1), (–1, 7) 8. a) If the distance between the points (a, 3) and (8, 3) is 7 units, find the value of a. b) If the distance between the points (0, 5) and (a, 0) is 13 units, find the value of a. It's your time - Project Work and Activity Section 9. a) Let's draw three straight lines of length in whole number of centimetres (say 3 cm, 4 cm, 5 cm, ... ) in a squared graph paper by using a ruler. Now, write the coordinates of the joining points of each straight line and find the lengths by using distance formula. Are the lengths given by distance formula and the lengths measured by a ruler equal? 10. a) Let's draw the following plane shapes in the separate squared graph papers. (i) An equilateral triangle (ii) An isosceles triangle (iii) A square (iv) A rectangle Then write the coordinates of the vertices of these shapes and find the length of sides of each plane shape by using distance formula. Also find the length of diagonals in the case of square and rectangle. b) Let's draw a circle with centre at origin in a squared graph paper by using a pencil compass. Then find the length of it's any 3 radii and a diameter. Show that radii of the same circle are equal and diameter is two times the radius of the circle. c) Let's draw two right angled triangles in a squared graph paper by using a ruler. Write the coordinates of the vertices of each triangle. Find the length of perpendicular, base and hypotenuse by using distance formula. Show that h2 = p2 + b2 in each of the right-angled triangles. Y' X' X Y C D O A B P Q Coordinates

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 249 Vedanta Excel in Mathematics - Book 8 17.1 Area of plane figures - Looking back Classroom - Exercise 1. Let the area of each square room in the graph be 1 cm2 . Say and write the area of each plane figure. (Area of half room = 1 2 cm2 , area of more than half room = 1 cm2 and ignore the room covered less than half.) A E S R K L N M P Q F G W X Y H B D C Area of square ABCD = ................... Area of rectangle EFGH = ................... Area of triangle WXY = ................... Area of parm. PQRS = .................. Area of trapezium KLMN = ................... 2. Let’s say and write the answers as quickly as possible. a) If p cm is the length of a square, its area is ................... b) If the length of a rectangle is x cm and breadth is y cm, its area is ............ c) If b is the base of a triangle with height h, its area is ........... d) If b is the base of a parallelogram with height h, its area is ........... The plane surface enclosed by the boundary line of a plane closed figure is known as its area. Area is measured in square units. For example, sq. centimeters (cm2 ), sq. metres (m2 ), etc. Unit 17 Area and Volume

Vedanta Excel in Mathematics - Book 8 250 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 17.2 Area of triangle Let's cut a cardboard paper to form a triangular shape. Say the triangle is ABC. First fold a perpendicular from the vertex A to the base BC. Open and mark the perpendicular with a pen. It is the height of the triangle and mark it as h. Fold the height in half so that vertex A sits on the base BC. Mark half of the height as h 2. B D C A h B P Q X D C A h 2 h 2 B F E D C A 1 2 Now, cut the folded edges AX, PX, and QX. Place the edge AP along PB and the edge AQ along QC. Thus, a rectangle BCEF is formed by this arrangement. Here, area of DABC = area of rectangle BCEF = length × breadth = BC × CE = base × height = b × 1 2 h = 1 2 bh Thus, area of triangle = 1 2 base × height = 1 2 b × h. (i) Area of a right-angled triangle In the given right-angled D ABC right angled at B, the perpendicular AB is the height of the triangle and BC is its base. Area of right-angled D ABC = 1 2 base × height = 1 2 base × perpendicular = 1 2 BC × AB (ii) Area of an equilateral triangle In the given equilateral triangle ABC, AB = BC = CA = a (suppose). AD perpendicular to BC is drawn. In an equilateral triangle, perpendicular drawn from a vertex to its opposite side bisects the side. www.geogebra.org/classroom/vnavdzkh Classroom code: VNAV DZKH Vedanta ICT Corner Please! Scan this QR code or browse the link given below: A B D C A B C p b www.geogebra.org/classroom/smsb7t2j Classroom code: SMSB 7T2J Vedanta ICT Corner Please! Scan this QR code or browse the link given below: Area and Volume