6

Q1.

• 1) 50^o
• 2) 120^o
• 3) 60^o
• 4) 70^o

Solution

Q2. In similar triangles ABC and MNP, AG and MH are the angle bisectors. What is the value of  given that AG = 4, MH = 7?

• 1) None
• 2)  
• 3)  
• 4)  

Solution

  We know that  for two similar triangles the ratio of areas is equal to the ratio of squares of their corresponding angle bisectors. So 

Q3. Given two triangles PQR and MNS with .  Are the two triangles RPQ and MNS similar ? If yes, by which rule? Explain your Answer..

Solution

If we match the information given we see that the sides of the two triangles are proportional, but to have the required triangles to be similar we must have the corresponding sides of RPQ and MNS Proportional. If we check the given information, We have . But if we must have   triangles   RPQ and MNS similar, then we must have which is not the  given ratio. So we can conclude that even though the sides of the given triangles are in proportion, the corresponding sides of   triangles , RPQ and MNS are not proportional Hence the triangles RPQ  and MNS are not similar.

Q4. M is the mid-point of BC and N is the mid-point of QR. The area of ABC = 100 sq. cm and the area of PQR = 144 sq. cm If AM = 4 cm then PN is:

• 1) 12 cm
• 2) 4 cm
• 3) 5.6 cm
• 4) 4.8 cm

Solution

_{AM is median of triangle ABC and PN is median of triangle PQR. We know that ratio of areas of two similar triangles is equal to the ratio of the squares of their corresponding medians.} PN = 4.8 cm

Q5. Calculate the area of right triangle PQR where OP = 6 cm, OR = 8 cm and QR = 26 cm.

Solution

In POR, PR² = PO² + OR² (Pythagoras Theorem) PR² = 6² + 8² PR² = 10² PR = 10 cm In PQR, QR² = PQ² + PR² (Pythagoras Theorem) PQ² = 26² - 10² = 676 - 100 = 576 PQ = 24 cm

Q6. ABC PQR. If area (ABC) = 2.25 m², area (PQR) = 6.25 m², PQ = 0.5 m, then length of AB is:

• 1) 50 m
• 2) 0.5 m
• 3) 3 m
• 4) 0.3 m

Solution

Q7. Perimeters of two   similar triangles PQR   and MNS are in the ratio 13:15 The sides of Triangle MNP   are MN=8, NS=10  MS =12 find the sides of triangle  PQR.

Solution

According to question = =(corresponding sides of similar triangles are proportional) , , , ,  

Q8. In an equilateral triangle ABC, D is a point on side BC such that BC. Prove that 9AD² = 7AB².

Solution

Draw AP BC. BD = Since ABC is equilateral and AP BC, BP = PC = BC = AB Using Pythagoras theorem in ABP, AP² + BP² = AB² AP² = AB² - BP² = AB² - = … (1) Using Pythagoras theorem in ADP, AD² = AP² + DP² = AB² + (BP - BD)² = AB² + = 36AD² = 28AB² 9AD² = 7AB² Hence, proved.

Q9. In the figure, DE || BC, then x is equal to:

• 1) 1.4 cm
• 2) 4 cm
• 3) 2 cm
• 4) 2.5 cm

Solution

Since, DE || BC, by Basic proportionality theorem,

Q10. The areas of two similar triangles ABC and PQR are in the ratio 4 : 9. If the sum of the perimeters is 30 m, find the perimeter of each triangle.

Solution

Let x and y, s and t, u and v, be the corresponding sides of the triangles. Then according to question, So = ratio of perimeters of triangles ABC and PQR Let 2z and 3z be the respective perimeters. So according to question 2z + 3z = 30 So z = 6 Hence the required perimeters are 12 cm and 18 cm.

Q11.

Solution

Q12. A carpenter wants to make a stool with a square top. He takes the length of the diagonal as 28.2 cm What should be the length of side of the square?

Solution

Q13. In given figure, DE||BC, if AB = 7.6 cm, AD = 1.9 cm, then AE : EC is :

• 1) 1:4
• 2) 3:1
• 3) 1:3
• 4) 4:1

Solution

Since, DE || BC, by Basic Proportionality theorem, Thus, AE: EC = 1:3.

Q14. In figure, O is a point inside PQR such that POR = 90^o, OP = 6 cm and OR = 8 cm. If PQ = 24 cm, QR = 26 cm, prove that QPR is a right angled triangle.

Solution

In POR, PR² = PO² + OR² = 6² + 8² = 36 + 64 = 100 PR = 10 cm Now, PQ² = 24² = 576 QR² = 26² = 676 So, PQ² + PR² = 676 = QR² By the converse of Pythagoras theorem, Hence, QPR is a right angled triangle. 

Q15. PQRS is a trapezium with PQ||SR. Diagonals PR and SQ intersect at M. PMS QMR. Prove that PS = QR.

Solution

In MSR and MQP, 1 = 2 3 = 4 (Alternate angles) [AA similarity] Also, From (1) and (2), QM² = PM² QM = PM … (3) From (1), (3), … (4) From (2), (3), (4), we have:

Q16. Two right triangles ABC and DBC are drawn on the same hypotenuses BC and on the same side of BC. If AC and BD intersect at P, prove that: AP PC = BP DP

Solution

In BPA and CPD, BAP = CDP = 90^o APB = DPC (Vertically opposite angles) _{Hence, proved.}

Q17.   An equilateral triangle is based on the side of a square of area 64 sq. cm and another on the diagonal of the same square. Find the ratio of their areas.

Solution

  We see that the side of the square = 8 cm So the diagonal of the square =  cm Since both are equilateral triangles, so they are equiangular, each of their angles is . So the two triangles are similar. Hence the ratio of their areas = ratio of squares of their corresponding sides.   Let A be the area of the equilateral triangle based on the side 8 cm. Let A' be the area of the equilateral triangle based on the diagonal  cm.

Q18. In figure, , then y + z is:

• 1)
• 2)
• 3)
• 4)

Solution

Given,

Q19. In a trapezium, show that any line drawn parallel to the parallel sides of the trapezium, divides the non-parallel sides proportionally.

Solution

ABCD is a trapezium and EF||AB||CD Join AC to cut EF at G. In ADC, EG||DC, so by Basic Proportionality theorem, we have: ...(1) In ACB, GF||AB, so by Basic Proportionality theorem, we have: ...(2) From (1) and (2), we get Hence proved.

Q20. In the figure given below, if DE || BC, then x equals :

• 1) 3 cm
• 2) 6.7 cm
• 3) 2 cm
• 4) 4 cm

Solution

Since, DE || BC, ADE = ABC AED = ACB (by AA similarity) Hence, BC = x = 6.7 cm

Q21. In the given figure, ABCD is a trapezium in which AB || DC. The diagonals AC and DB intersect at O. Prove that:

Solution

In AOB and COD, we have: AOB = COD(Vertically opposite angles) OAB = OCD(Alternate interior angles, as AB || CD) AOB COD (By AA similarity)

Q22. In the given figure, PQRS is a rhombus such that RQ is produced to E and SE is joined. SE and PR intersect at D. Prove that SD. RE = SR. DE.

Solution

In PDS and RDE 1 = 2 [Alternate interior angles] 3 = 4 [Vertically opposite angles] PDS RDE [AA similarity] [ PS = SR] DS. RE = SR. DE

Q23. In the given figure, ABC is an isosceles right angled triangle with B = 90^o such that PQ AC, ST AC, where P lies on AB and S lies on BC. Then, prove that AQP CTS.

Solution

Since, ABC is an isosceles right angled triangle, AB = BC In AQP and CTS, QAP = TCS [In a triangle, angles opposite to equal sides are equal] AQP = CTS = 90^o AQP CTS [AA similarity]

Q24. A vertical pole of 10 m casts a shadow of 6 m long on the ground. At the same time, a tree casts a shadow 60 m long on the ground. The height of the tree is

• 1) 100 m
• 2) 120 m
• 3) 200 m
• 4) 25 m

Solution

Let AB be the pole and AC be its shadow. Let DE be the tree and DF be its shadow. ∆ABC ∼∆DEF… (AA test) AB/DE=AC/DF 10/6 = AC/60AC = 100 m 

Q25. In the given figure YXZ = XPZ, then is equal to:

• 1)
• 2) PZ²
• 3) ZY ZP
• 4) XZ²

Solution

Q26. In given figure, is a right triangle PQR, right angled Q. X and Y are the point on PQ and QR such that PX : XQ = 1:2 and QY : YR = 2:1. Prove that 9 (PY² + XR²) = 13PR².

Solution

X divides PQ in ratio 1: 2. ...(1) Also, Y divides QR in ratio 2:1 ...(2) In right PQY, PY² = PQ² + QY² PY² = PQ² + PY² = PQ² + QR² 9PY² = 9PQ² + 4QR² ...(3) In right XQR, XR² = XQ² + QR² 9XR² = 4PQ² + 9QR² ...(4) Adding (3) and (4), we get, 9PY² + 9XR² = 9PQ² + 4QR² + 4 PQ² + 9QR² 9 (PY² + XR²) = 13 PQ² + 13 QR² 9(PY² + XR²) = 13 (PQ² + QR²) = 13 PR²

Q27. In figure BAC = 90°, AD BC. Prove that: AB² + CD² = BD² + AC².

Solution

Q28. If LM II CB and LN II CD, prove that           

Solution

In ACB, _{LM || CB} Therefore, by                                 In ADC, NL II DC, hence by BPT                                From (I) and (II)       Or                      (INVERTENDO)                                                                                                                       Or               (INVERTENDO)

Q29. In the given figure, AD BC. Prove that AB² + CD² = BD² + AC².

Solution

In ADB, AB² = BD² + AD² (Pythagoras theorem) ...(1) Also in ADC, AC² = CD² + AD² (Pythagoras theorem) AD² = AC² - CD² ...(2) From (1) and (2), we get, AB² - BD² = AC² - CD² AB² + CD² = BD² + AC²

Q30. Triangle ABC is right angled at B and D is mid-point of BC. Prove that: AC² = 4AD² - 3AB²

Solution

In right ABC, using Pythagoras theorem, AC² = AB² + BC² …(1) In right ABD, using Pythagoras theorem, AD² = AB² + BD² AD² = AB² + AD² = AB² +BC² 4AD² = 4AB² + BC² BC² = 4AD² − 4AB² …(2) Substituting (2) in (1), we get AC² = AB² + 4AD² − 4AB² = 4AD² - 3AB² AC² = 4AD² - 3AB²

Q31. In Figure  DE II OQ and DF II OR show that EF II QR.       

Solution

      In PQO;  DE II  OQ                                     (BPT)                    Also, in POR; DF II  OR                     (BPT)                    From (1) and (2) we have                      From converse of BPT, we have EF II QR.  Hence proved

Q32. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O.  Show that

Solution

In Figure ABCD is a trapezium, AB II DC through O draw PQ II AB           In ADB, PO II AB            (By BPT)             Or                              Again in ACD, PO II DC ( AB II DC II PO )                     (By BPT)       (2)                    From (1) and (2)                                                  (by alternando)        Hence proved

Q33. In a trapezium ABCD, E and F are points on AB and AD respectively. Also G and H are points on CB and CD respectively such that AE=3, AB=9, AF=2, AD=6, CG=1, CB=4, CH=3, CD=4. Check if EF || GH.

Solution

Q34. From the given figure, find 

Solution

From the given figure, Therefore, ABC is similar to LMN (SSS similarity) MLN = BAC = 180 - (50^o + 70^o) = 60^o

Q35. In the given figure, PQR is right-angled triangle right-angled at Q. DE PR. Prove PQR PED and find the lengths of PE and DE if PD = 3 cm, QD = 2 cm and QR = 12 cm.

Solution

In right PQR, PR² = PQ² + QR² = 25 + 144 = 169 PR = 13 cm Let PE = x, then ER = 13 - x In PQR and PED, PQR = PED QPR = EPD PQR PED [AA similarity]

Q36. P and Q are points on sides AB and AC, respectively, of ∆ABC. If AP = 4 cm, PB = 12 cm, AQ = 2 cm and QC = 6 cm, then BC = 

• 1) 2PQ
• 2) 4PQ
• 3) PQ
• 4) 3PQ

Solution

AB = AP + PB = 16 cm AC = AQ + QC = 8 cm  

Q37. In two triangles ABC and DEF, BC = EF, AB = DE , If AG and DH are the medians of triangles ABC  and DEF respectively. Find = ?

Solution

In triangles ABC  and DEF, AB=DE BC=EF So (SAS  rule) Now consider  triangles ABG and DEH, (as AG and DH are medians) So BG = EH        AB=DE       Thus we have  (SAS rule) So (  corresponding sides of similar triangles are proportional)   = Hence, =Ratio   of the corresponding sides. =

Q38.

Solution

Q39. Which of the following cannot be the sides a right triangle?

• 1) 400 mm, 300 mm, 500 mm
• 2) 2 cm, 1 cm, cm
• 3) 9 cm, 5 cm, 7cm
• 4) 9 cm, 15 cm, 12 cm

Solution

9 cm, 5 cm, 7 cm cannot form the sides of a right triangle as the Pythagoras theorem is not satisfied in this case.

Q40. In the given figure, . Find the value of x.

Solution

ABC XYZ Or x = 4.8

Q41. Three squares are based on the sides of a right angled triangle. The area of the two smaller ones are 144 sq. cm and 256 sq. cm. What is the area of the third one?

• 1) 400 sq. cm
• 2) 625 sq. cm
• 3) 361 sq. cm
• 4) 900 sq. cm

Solution

Area of square = (side)² The smaller sides of the triangle are 12 and 16 cm. respectively  This essentially is the Pythagorean principle, so the hypotenuse =  Largest square"s area will be = 20² =  400 sq.cm

Q42. In given figure, in ABC, PQ||BC and BC = 3PQ. Find the ratio of the area of PQR and area of CRB where PC and BQ intersect at R.

Solution

Since PQ||BC, 1 = 2 [Alternate interior angles] 3 = 4 [Alternate interior angles] PQR CBR [By AA similarity] ar (PQR) : ar(CRB) = 1: 9

Q43. In quadrilateral ABCD,  If AD² = AB² + BC² + CD². Prove that 

Solution

Q44. In figure, ABC PQR, then y + z is

• 1) 4 + 3
• 2) 2 +
• 3) 4 +
• 4) 3+ 4

Solution

Since triangles ABC and PQR are similar, their corresponding sides are proportional.

Q45. A vertical stick 30 m long casts a shadow 15 m long on the ground. At the same time, a tower casts a shadow 75 m long on the ground. The height of the tower is:

• 1) 25 m
• 2) 100 m
• 3) 200 m
• 4) 150 m

Solution

In the figure, DE represents the stick and EF is its shadow; AB represents the tower and BC is its shadow. ABC = DEF = 90^o ACB = DFE = Thus, the height of the tower is 150 m.

Q46. In a quadrilateral ABCD, given that A + D = 90^o. Prove that AC² + BD² = AD² + BC².

Solution

We have, A + D = 90^o In APD, by angle sum property, A + D + P = 180^o 90^o + P = 180^o P = 180^o - 90^o = 90^o In APC, by Pythagoras theorem, AC² = AP² + PC²...(1) In BPD, by Pythagoras theorem, BD² = BP² + DP²...(2) Adding equations (1) and (2), AC² + BD² = AP² + PC² + BP² + DP² AC² + BD² = (AP² + DP²) + (PC² + BP²) AC² + BD² = AD² + BC²[By Pythagoras theorem]

Q47.

Solution

Q48. In the given figure, ABC is an isosceles triangle with AB = AC. D is mid-point of BC. If DE AB, DF AC, prove that DE = DF.

Solution

In DEB and DFC, E = F = 90^o EBD = FCD    [In an isosceles triangle angles opposite to equal sides are equal] DB = DC [D is mid-point of BC] Therefore, by Angle-Angle-Side criterion of congruence, the triangles, DEB and DFC are congruent. [AAS congruence] The corresponding parts of the congruent triangles are congruent. DE = DF [cpct]

Q49. ABCD is a parallelogram and E,G,H and F are the mid points of the sides AB, BC, CD, DA. Prove that =

Solution

We see that  AB = CD (opposite sides of a parallelogram are equal)   Similarly AF = GC Also (opposite angles of a parallelogram are equal)  (SAS rule) This is also a special case of similarity where the ratio of corresponding sides is 1:1 Hence area of triangles AEF and GCH are equal.

Q50. In fig. ABD is a triangle in which ÐDAB = 90° and AC BD. Prove that AD² = BD CD.

Solution

In DCA and DAB D is common DCA = DAB = 90^o Therefore, DCA DAB, by AA similarity Therefore, corresponding sides are proportional. DA² = DC x DB

Q51. In figure, ABC is right angled at C. DE AB. If BC = 12 cm. AD = 3 cm and DC = 2 cm, then prove that and hence find the lengths of AE and DE.

Solution

ABC ADE (by AAS) In right AB² = BC² + AC² (by Pythagoras theorem) AB² = 5² + 12² = 25 + 144 = 169 AB = 13 Substituting AB = 13 cm, BC = 12 cm and AC = 5 cm in (1), DE = and AE =

Q52. In given figure, ABCD is a rectangle, in which BC = 2AB. A point E lies on CD produced such that CE = 2BC. Find AC: BE.

Solution

In ABC and BCE, and B = C = 90^o

Q53. The diagonals AC and BD of a rhombus ABCD are of length 6 cm and 8 cm respectively. What is the perimeter of the rhombus?

Solution

Q54. ABC is right angled at B. AD and CE are the two medians drawn from A and C, respectively. If AC = 5 cm, AD = cm, find the length of CE.

Solution

By Pythagoras theorem, we have, AB² + BD² = AD² AC² - BC² + BD² = AD² AC² - AD² = BC² - BD²

Q55. In fig, PA QB and RC are perpendiculars to AC. Prove that

Solution

In _{...(1) Similarly, ...(2)} From (1) and (2) xy = xz + yz Dividing by xyz, we get,

Q56. Find value of x from given figure of DE II  BC    

Solution

         

Q57. What is the value of a if a, a-1,a+8 are the lengths of the sides of a right triangle where a is a natural number?

Solution

Q58. In given figure ABC, X and Y are two points lying on the side BC such that BX = CY. If DX||AC and YE||AB, then prove that DE||BC.

Solution

BX = CY (Given) _{… (1)} Now in ABC, DX||AC. Using BPT, … (2) [Using (1)] In ABC, YE||AB. Using BPT, _{… (3)} From (2) and (3), DE||BC [ By converse of BPT]

Q59. In the given figure, 1 = 2 and 3 = 4. Show that AE.BC = AC.DE.

Solution

In ADE and ABC, As, 1 = 2 Or DAE = BAC And 3 = 4 or AE.BC = AC.DE

Q60. In given figure, in ABC, DE||BC such that ar (ADE) = ar (BCED). Then, prove that .

Solution

Since, DE||BC So, 1 = 2 [Corresponding angles] 3 = 4 [Corresponding angles] [AA similarity] … (1) Also, given that ar(ADE) = ar(BCED) ar(ADE) = ar(ABC) - ar(ADE) 2ar (ADE) = ar(ABC) … (2) From (1) and (2), we have: Hence, proved.

Q61. Two triangles ABC and PQR  are such that  and . If the ratio of their perimeters is 3 : 5, Find the ratio of their areas.

Solution

From the given information, We see that (AA  rule) So,  So according to question , ( ratio of their areas is the same as the ratio of the squares of their corresponding sides.)

Q62. The length of an altitude of an equilateral triangle of side a is:

• 1)
• 2)
• 3)
• 4)

Solution

ABC is the equilateral triangle. Let AD be the altitude. Using Pythagoras theorem in ABC, we have: AD² = AB² - BD² Hence, AD =

Q63. In the given figure, if and PQR = PRQ. Prove that PQS TQR.

Solution

Given : ...(1) PQ = PR From (1), In PQS and TQR, Q is common and By SAS similarity criterion, PQS TQR

Q64. There is a triangular park ABC in a colony (see figure). The resident Welfare Association of the colony wishes to divide this park intwo two parts - one for planting trees and raising a lawn and the other for providing place for children to play. They decided to divide the park by creating a boundary along XY which is parallel to AC (as shown in fgure) such that ar(BXY) = 2ar(ACYX). State how this boundary XY can be drawn so that X and Y lie on AB and BC respectively.  

Solution

ar (BXY) = 2 ar(ACYX) ar (BXY) = 2 [ar (BAC) - ar (BXY)] 3 ar(BXY) = 2 ar(BAC) ...(1) In BXY and BAC, B = B (Common) BXY = BAC (Corresponding angles) BXY BAC [AA similarity]  

Q65. In the given figure, AC BD and CE AD. Prove that AC² = DA. AE.

Solution

In ACD and AEC, ACD = AEC = 90^o CAD = EAC [common] ACD AEC [AA similarity] AC² = AE.AD

Q66. X and Y are points on the sides PQ and PR respectively of a PQR. If the length of PX, QX, PY and YR (in centimeters) are 4, 4.5, 8 and 9 respectively. Then show XY||QR

Solution

In PQR, XY||QR [By converse of BPT]

Q67. In fig. if A = B and AD = BE show that DE || AB in ABC

Solution

Since A = B, AC = BC Also AD = BE (Given) DC = EC DE || AB (By converse of BPT)

Q68.

Solution

Q69.

Solution

Q70. In given figure XY||BC, the lengths of AX,AY,XY and BC in centimeters are 2.4, 3.2, 2 and 5 respectively. Find XB and YC.

Solution

XY||BC 1 = 2 3 = 4 [corresponding angles] AXY ABC [AA similarity] AB = 6 cm and AC = 8 cm BX = AB - AX = 6 - 2.4 = 3.6 cm CY = AC - AY = 8 - 3.2 = 4.8 cm

Q71.

Solution

Q72. The sides of a right triangle that contain the right angle are in the ratio 3:4. If the area of the triangle is given to be 150 sq.cm. Find the sides.

Solution

Q73. Triangle ABC is right angled at B and D is mid point of BC. Prove that : AC² = 4AD² - 3AB².

Solution

In right ABC, using Pythagoras theorem, we have: AC² = AB² + BC² ...(1) In right ABD, using Pythagoras theorem, we have: AD² = AB² + BD² BC² = 4AD² - 4AB² ...(2) Substituting (2) in (1), we get, AC² = 4AD² - 3AB² Hence, proved.

Q74. In the figure, D is a point on the side BC of ABC such that BAC = ADC. Prove that CA² = CB CD.

Solution

In ADC and ABC, C is common ADC = BAC (Given) By AA similarity, ADC BAC Or, AC² = BC DC

Q75. In figure below, QA and PB are perpendicular to AB. If AO = 10 cm, BO = 6 cm and PB = 9 cm and then find AQ.

Solution

AOQ = POB(vertically opposite angles) QAO = PBO (Each 90^o) By AA similarity criterion, Given AO = 10 cm, BO = 6 cm, PB = 9 cm AQ = 15 cm

Q76. If one diagonal of a trapezium divides the other diagonal in the ratio 1: 2. Prove that one of the parallel sides is double the other.

Solution

  Given: ABCD is a trapezium. AB||CD. Diagonal BC divides the diagonal AD in the ratio 1: 2 at O. That is, OA: OD = 1: 2. To prove: CD = 2 AB Proof: In AOB and COD, AOB = COD (Vertically opposite angles) ABO = OCD (Alternate angles) Hence, proved.

Q77. In PQR, PD QR such that D lies on QR. If PQ = a, PR = b, QD = c and DR = d and a, b, c, d are positive units, prove that (a + b) (a - b) = (c + d) (c - d).

Solution

In PQD, PDQ = 90^o Using Pythagoras theorem, PD² = a² - c² ...(1) Similarly, in PDR, PDR = 90^o PD² = b² - d² ...(2) From (1) and (2), a² - c² = b² - d² a² - b² = c² - d² (a + b) (a - b) = (c + d) (c - d)

Q78. In the given figure, T and B are right angles. If the lengths of AT, BC and AS (in centimeters) are 15, 16 and 17 respectively, then the length of TC (in centimeters) is:

• 1) 18
• 2) 12
• 3) 19
• 4) 16

Solution

Using Pythagoras theorem in , TS =

Q79. The areas of two similar triangles are 49 cm² and 64 cm² respectively. If the difference of the corresponding altitudes is 10 cm, then find the length of altitudes (in centimetres).

Solution

  Given, ABC PQR ... (1) According to the question, ar(ABC) = 49 cm², ar(PQR) = 64 cm² Let AM = x, then PN = 10 + x In ABM and PQN, B = Q M = N = 90^o [AA similarity] Or ... (2) From (1) and (2),

Q80. Two poles of height p and q metres are standing vertically on a level ground, a metres apart. Prove that the height of the point of intersection of the lines joining the top of each pole to the foot of the opposite pole is given by metres.

Solution

CD and AB are the two poles of height p and q respectively. Let AQ = x, so AQ = a - x (Since, CA = a, given) In CQP and CAB, C = C CQP = CAB = 90^o Adding (1) and (2),

Q81. If AD and PM are medians of ABC and PQR respectively where ABC PQR. Prove that

Solution

_{(Since, AD and PM are medians)} In

Q82. A pole of length 10 m casts a shadow 2 m long on the ground. At the same time a tower casts a shadow of length 50 m on the ground. Find the height of the tower.

Solution

In the figure, DE represents the pole and EF is its shadow; AB represents the tower and BC is its shadow. ABC = DEF = 90^o ACB = DFE =

Q83. In ABC, D and E the points on the sides AB and AC respectively such that DE||BC. If AD = 6x - 7, DB = 4x - 3, AE = 3x - 3 and EC = 2x - 1, then find the value of x.

Solution

AD = 6x - 7, DB = 4x - 3 AE = 3x - 3, EC = 2x - 1 DE||BC By B.P.T

Q84. In the figure given below, ABC and DBC are two triangles on the same base BC. If AD intersect BC at O then show that : .

Solution

Construction: Draw AM BC and DN BC _{Now, in AMO and DNO, we have} AOM = DON (Vertically opposite angles) AMO = DNO = 90^o (By construction) AMO DNO (By AA similarity)

Q85. P and Q are points on sides AB and AC respectively of ABC. If AP = 3 cm, PB = 6 cm, AQ = 5 cm and QC = 10 cm, show that BC = 3PQ.

Solution

_{Given, AP = 3 cm and PB = 6 cm AB = 9 cm Also, given AQ = 5 cm and QC = 10 cm AC = 15 cm}

Q86. In given figure, in ABC; D, E and F are points on AB, BC and AC respectively such that ADEF is a parallelogram, then prove that .

Solution

ADEF is a ||gm AD||EF and AF||DE AB||BF and AC||DE Using Basic Proportionality theorem, In ABC, AB||EF ...(1) Also, AC||DE ...(2) From (1) and (2), Hence proved.

Q87.

Solution

Q88. If the areas of two similar triangles are equal, prove that they are congruent.

Solution

Given, ar(ABC) = ar(PQR) and _{We know that:} 1 = AB = PQ Similarly BC = QR and AC = PR Thus, ABC PQR (SSS congruency criterion)

Q89. In PQR, PD QR such that D lies on QR. If PQ = a, PR = b, QD = c and DR = d, where a, b, c, d are positive units. Prove that (a + b) (a - b)=(c + d)(c - d)

Solution

In PQD, Using Pythagoras theorem. PD² = a² - c² …(1) Similarly in PD² = b² - d² …(2) From (1) and (2), we get, a² - c² = b² - d² a² - b² = c² - d² (a + b) (a - b) = (c + d) (c - d)

Q90.

Solution

Q91. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at O. Prove that.

Solution

Draw OE parallel to AB and CD. In ABD, OE || AB By Basic Proportionality theorem, ... (i) In CDA, OE || CD By Basic Proportionality theorem, ... (ii) From (i) and (ii), we get, Or,

Q92. If two scalene triangles are equiangular, prove that the ratio of the corresponding sides is same as the ratio of the corresponding angle bisector segments.

Solution

Let AD be the bisector of A and PS be the bisector of P. In ABC and PQR, A = P, B = Q, C = R (Given) Also, In ABD and PQS, 1 = 2 B = Q From (1) and (2), Hence, the ratio of the corresponding sides is same as the ratio of the corresponding angle bisector segments.

Q93. In the given figure, ABC is an isosceles triangle with AB = AC. D and E are points on AB and AC respectively such that AD = AE. Show that C + EDB = 180^o.

Solution

In ABC and ADE, (SAS congruence criterion) Hence, C + EDB = 180^o.

Q94.

Solution

Q95. In figure, DE||OQ and DF||OR. Show that EF||QR.

Solution

In PQO, (BPT) In POR, (BPT) Therefore, By converse of BPT, in PQR, we get, EF||QR

Q96.

Solution

Q97.

Solution

Q98. ∆ABC and ∆XYZ are similar, and if AB =11 cm, XY= 7 cm and BC = 22 cm, then YZ = 

• 1) 15
• 2) 16
• 3) 14
• 4) none of these

Solution

If two triangles are similar, then their corresponding sides are proportional.

Q99. In figure if AD BC, then prove that AB² + CD² =AC²+ BD²

Solution

We will apply Pythagoras theorem. In ADC, AD² = AC² - CD² In ABD, AD² = AB² - DB² Therefore, AB² - DB² = AC² - CD² or AB² + CD² = AC² + DB²

Q100. In fig. PQR and SQR are two triangles on the same base QR. If PS intersect QR at O, then show that

Solution

Construction :- Draw PA QR and SB QR _{From (1) and (2),}

Q101. In figure, ACB = 90^o and CD AB. Prove that .

Solution

In ACD and ABC, ADC = ACB = 90^o DAC = BAC (Common) Similarly, BC² = BD AB Therefore,

Q102. Triangle ABC is an isosceles right triangle right-angled at B. An equilateral triangle ADC based on the hypotenuse and another equilateral triangle BEC are to be compared for their areas. Find the ratio of those two areas.      

Solution

We know that the area of an equilateral triangle whose side is a  is  equal to Now we are given that ABC is a right angled triangle with BC as the hypotenuse. As it is isosceles, we assume the sides AB= BC to be x cm then the hypotenuse AC =  From the above information we see that And So, ==2

Q103. Prove that the equilateral triangles described on the two sides of a right-angled triangle are together equal to the equilateral triangle described on the hypotenuse in terms of their areas.

Solution

To prove:

Q104. In the given figure, ST and UV are intersecting at O such that the lengths of OS, OT, OU and OV (in centimeters) are 4.2, 6.3, 1.2 and 1.8. Prove SU||VT.

Solution

OS = 4.2 cm, OT = 6.3 cm, OU = 1.2 cm, OV = 1.8 cm Also, In SOU and TOV, and 1 = 2 [Vertically opposite angles] SOU TOV [SAS similarity] But they form alternate interior angles when SU and VT are cut by transversal ST.

Q105. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. Prove it.

Solution

Given: ΔPQR, in which XY || QR, XY intersects PQ and PR at X and Y respectively. To prove: Construction: Join RX and QY and draw YN perpendicular to PQ and XM perpendicular to PR. Proof: Since, Area of a triangle = Therefore, ar (ΔPXY) = …(1) Also, ar (ΔPXY) = …(2) Similarly, ar (ΔQXY) = …(3) And, ar (ΔRXY) = …(4) Dividing (1) by (3), we get Again, dividing (2) by (4), we get Since the area of triangles with same base and between same parallel lines are equal, so (as and are on same base XY and between same parallel lines XY and QR) Therefore, from (5), (6) and (7), we get Hence, proved.

Q106. In fig., (a) and (b) sides AB, BC and median AD of ABC are respectively proportional to sides PQ, QR and median PM of PQR. Prove that ABC PQR.

Solution

Given: Since, D and M are mid points of BC, QR respectively. In ABC and PQR, we have,

Q107. In figure, XN || CA and XM || BA. T is a point on CB produced. Prove the TX² = TB.TC.

Solution

In TXM, BN || XM …(1) In TCM, XN || CM From (1) and (2) we get

Q108. Given  triangle ABC  with BE  and DC  as perpendiculars  to AC and AB respectively. Prove that  .    

Solution

In triangles ABE and ACD, (each is )                                                is common                                      So (AA rule)                                      Hence (corresponding sides of similar triangles are proportional) Hence proved.

Q109. In trapezium ABCD, AB||DC and DC = 2AB. EF||AB, where E and F lie on BC and AD respectively such that. Diagonal DB intersects EF at G. Prove that 7EF = 11AB.

Solution

In trapezium ABCD, AB||DC and DC = 2AB. Also, In trapezium ABCD, EF||AB||CD, using BPT, In BGE and BDC, B = B(Common) BEG = BCD(Corresponding angles) As, (1) Similarly, DGF DBA FG = AB(2) Adding (1) and (2), EG + FG = CD + AB EF = (2AB) + AB = AB + AB = AB 7EF = 11 AB

Q110. In trapezium ABCD, AB is parallel to CD and AB = 2 CD. If area of AOB = 84 cm², find the area of COD.

Solution

_{Give, AB || CD}

Q111. E and F are points on the sides PQ and PR respectively of a PQR.  State whether EF || QR. PE = 3.9cm, EQ = 3cm, PF =3.6cm and FR = 2.4cm.

Solution

          ,   Hence EF is not parallel to QR

Q112. In the given figure, PQR is an isosceles triangle with PQ = PR. S is a point on QR and T is a point on QP produced such that , prove that .

Solution

or Now in PQS = TQR [common] PQS TQR [SAS similarity]

Q113. D, E and F are respectively the mid-points sides PQ, QR and PR of PQR. Find the ratio of the areas of DEF and RPQ.

Solution

D, E and F are mid-points of PQ, QR and PR DE||PR and DF||QR DE||FR and DF||ER DERF is a || gm Similarly, DEFP is a || gm Now, in DEF and RPQ, DEF = P FDE = R [Opposite angle of a || gm are equal] DEF RPQ [AA similarity] Required ratio is 1 : 4

Q114. The area of a rectangular field is 200 sq m and its perimeter is 60 m. What is the length of its diagonal?

Solution

Q115. In ABC, AB = AC and D is a point on side AC such that BC² = AC.CD. Prove that BD = BC.

Solution

BC² = AC CD ... (1) In DBC and BAC, [From (1)] BCD = ACB [Common] DBC BAC [By SAS similarity criterion]

Q116. In figure, P, Q and R are respectively the mid-points of sides AB, BC and CA of ABC. Show that ar (PBQR) = ar(ABC)

Solution

Here, PR = BC, PQ = AC, QR = AB [ P, Q, R are mid-points of AB, BC, CA] Now, PR||BC and QR||AB PR||BQ and QR||PB PBQR is a parallelogram ar(PBQR) = 2 ar (PQR)                                             ...(2) From (1) and (2), ar (PBQR) = 2 ar (CAB) = ar (CAB) Hence proved.

Q117.

Solution

Q118. Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Solution

Given: ABCD is a rhombus To prove: AB² + BC² + CD² + AD² = AC² + BD² Proof: We know, diagonals of a rhombus bisect at right angles. Therefore, from triangle AOB, AB² = AO² + OB² 4AB² = AC² + BD² Thus, AB² + BC² + CD² + DA² = 4AB² = AC² + BD² [As AB = BC = CD = DA]

Q119. In and D is a point on side AC such that BC² = AC.CD. Show that BD = BC

Solution

BC² = AC CD BD = BC

Q120. If the areas of two similar triangles are equal, then show that triangles are congruent.

Solution

Let Given, ar(ABC) = ar(PQR) _{We know that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.} AB = PQ; BC = QR ; AC = PR ABC PQR [By SSS congruency criterion]

Q121. In given figure, AB || DE and BD||EF. Prove that DC² = CF AC.

Solution

In CDB, EF||BD, (by BPT) ...(1) In CAB, DE||AB, (by BPT) ...(2) From (1) and (2), DC² = CF AC

Q122. Two similar triangles ABC and DEF are such that their perimeters are 30 cm and 18 cm respectively.If AB=6,then find DE.

Solution

Q123. If the areas of two similar triangles are equal, prove that they are congruent.

Solution

Let ABC DEF We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides. Given, ar (ABC) = ar (DEF) AB = DE, BC = EF, AC = DF ABC DEF by Side-Side-Side criterion of congruency.

Q124. In figure below, DE||BC and AD:DB = 5:4. Find

Solution

In ABC, DE||BC ABC = ADE, ACB = AED [corresponding s] ABC ADE (AA similarity) 1 + = 1 + _{Now, in triangles DEF and CBF, we have:} FDE = BCF DEF = FBC [Alternate s] DEF CBF [AA similarity]

Q125. Prove that the line segment joining the mid points of any two sides of a triangle is parallel to the third side.

Solution

In ABC, D and E are mid points of AB and AC respectively  to prove DE II BC.  as D is mid point of AB (1) Similarly      (2)                     Hence from (1) and (2)  by converse of BPT DE II BC.  Hence proved.

Q126.

Solution

Q127. In the given figure, in ABC, XY||AC and XY divides the ABC into two regions such that ar(BXY) = 2ar (ACYX). Determine .

Solution

ar(BXY) = 2ar (ACYX) ar(BXY) = 2 [ar(BAC) - ar(BXY)] ar(BXY) = 2ar (BAC) - 2 ar(BXY) 3 ar (BXY) = 2ar (BAC) ...(1) In BXY and BAC, B = B (Common) BXY = BAC (Alternate angles, as XY || AC) BXY BAC (AA similarity) _{[Using (1)]}