## Chapter 4 triangles R.D. Sharma Solutions for Class 10th Math MCQ's

**Multiple Choice Questions**

1. Sides of two similar triangles are in the ratio 4:9 . Areas of these triangles are in the ratio.

(a) 2:3

(b) 4:9

(c) 81:16

(d) 16:81

**Solution**

Given: Sides of two similar triangles are in the ration 4:9

To find: Ratio of these triangles

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of corresponding sides.

ar(triangle 1)/ar(triangle 2) = (side 1/side 2)

ar(triangle 1)/ar(triangle 2) = 16/81

Hence the correct answer is option d.

2. The areas of two similar triangles are in respectively 9 cm

(a) 3:4

(b) 4:3

(c) 2:3

(d) 4:5

To find: Ratio of these triangles

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of corresponding sides.

ar(triangle 1)/ar(triangle 2) = (side 1/side 2)

^{2}= (4/9)^{2}ar(triangle 1)/ar(triangle 2) = 16/81

Hence the correct answer is option d.

2. The areas of two similar triangles are in respectively 9 cm

^{2}and 16 cm^{2}. The ratio of their corresponding sides is(a) 3:4

(b) 4:3

(c) 2:3

(d) 4:5

**Solution**
Given: Areas of two similar triangles are 9 cm

^{2}and 16 cm^{2}.
To find : Ratio of their corresponding sides .

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides .

ar(triangle1)/ar(triangle2) = (side 1/side 2)

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides .

ar(triangle1)/ar(triangle2) = (side 1/side 2)

^{2}
9/16 = (side1/side2)

Taking square root on both sides, we get

side1/side2 = 3/4

So, the ratio of their corresponding sides is 3:4

Hence the correct answer is a.

3. The areas of two similar triangles â–³ABC and â–³DEF are 144 cm

(a) 20 cm

(b) 26 cm

(c) 27 cm

(d) 30 cm

Given : Areas of two similar triangles â–³ABC and â–³DEF are 144 cm

If the longest side of larger â–³ABC is 36 cm .

To find : the longest side of the smaller triangle â–³DEF

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of theirs corresponding sides.

ar(â–³ABC)ar(â–³DEF) = (longest side of larger â–³ABC/ longest side of smaller â–³DEF)

144/81 = (36/ longest side of smaller â–³DEF)

Taking square root on both sides, we get

12/9 = 36/longest side of smaller â–³DEF

Longest side of smaller â–³DEF = 36 Ã— 9/12 = 27 cm

Hence the correct answer is c .

4. â–³ABC and â–³BDE are two equilateral triangles such that D is the mid â€“ point of BC , The ratio of the areas of triangle ABC and BDE is

(a) 2:1

(b) 1:2

(c) 4:1

(d) 1:4

Given â–³ABC and â–³BDE are two equilateral triangles such that D is the midpoint of BC.

To find : ratio of areas of â–³ABC and â–³BDE .

^{2}Taking square root on both sides, we get

side1/side2 = 3/4

So, the ratio of their corresponding sides is 3:4

Hence the correct answer is a.

3. The areas of two similar triangles â–³ABC and â–³DEF are 144 cm

^{2}and 81 cm^{2}respectively . If longest sides of larger â–³ABC be 36 cm, then the longest side of the smallest triangle â–³DEF is(a) 20 cm

(b) 26 cm

(c) 27 cm

(d) 30 cm

**Solution**Given : Areas of two similar triangles â–³ABC and â–³DEF are 144 cm

^{2}and 81 cm^{2}.If the longest side of larger â–³ABC is 36 cm .

To find : the longest side of the smaller triangle â–³DEF

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of theirs corresponding sides.

ar(â–³ABC)ar(â–³DEF) = (longest side of larger â–³ABC/ longest side of smaller â–³DEF)

^{2}144/81 = (36/ longest side of smaller â–³DEF)

^{2}Taking square root on both sides, we get

12/9 = 36/longest side of smaller â–³DEF

Longest side of smaller â–³DEF = 36 Ã— 9/12 = 27 cm

Hence the correct answer is c .

4. â–³ABC and â–³BDE are two equilateral triangles such that D is the mid â€“ point of BC , The ratio of the areas of triangle ABC and BDE is

(a) 2:1

(b) 1:2

(c) 4:1

(d) 1:4

**Solution**To find : ratio of areas of â–³ABC and â–³BDE .

â–³ABC and â–³BDE are equilateral triangles ; hence they are similar triangles .

Since D is the midpoint of BC, BD = DC

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of the corresponding sides.

ar(â–³ABC)ar(â–³BDE) = (BC/BD)

^{2}
ar(â–³ABC)ar(â–³BDE) = (BD + DC/BD)

^{2}[D is the midpoint of BC]
ar(â–³ABC)ar(â–³BDE) = (BD + DC/BD)

^{2}
ar(â–³ABC)ar(â–³BDE) = (2BD/BD)

^{2}
ar(â–³ABC)ar(â–³BDE) = 4/1

Hence the correct answer is c .

5. If â–³ABC and â–³DEF are similar such that 2AB = DE and BC = 8 cm, then DF =

(a) 16 cm

(b) 12 cm

(c) 8 cm

(d) 4 cm

(a) 16 cm

(b) 12 cm

(c) 8 cm

(d) 4 cm

**Solution**

Given : â–³ABC and â–³DEF are similar such that 2AB = DE and BC = 8 cm .

To find : EF

We know that if two triangles are similar then there sides are proportional .

Hence, for similar triangles â–³ABC and â–³DEF

AB/DE = BC/EF = CA/FD

AB/DE = BC/EF

1/2 = 8/EF

EF = 16 cm

Hence the correct answer is a .

6. In â–³ABC and â–³DEF are two triangles such that AB/DE = BC/EF = CA/FD = 2/5, then Area (â–³ABC) : Area (â–³DEF) =

(a) 2:5

(b) 4: 25

(c) 4:15

(d) 8:125

(a) 2:5

(b) 4: 25

(c) 4:15

(d) 8:125

**Solution**

Given : â–³ABC and â–³DEF are two triangles such that AB/DE = BC/EF = CA/FD = 2/5 .

To find : ar(â–³ABC) : ar(â–³DEF)

We know that if the sides of two triangles are proportional, then the two triangles are similar.

Since AB/DE = BC/EF = CA/FD = 2/5, therefore, â–³ABC and â–³DEF are similar .

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

ar (â–³ABC)/ ar(â–³DEF) = AB

^{2}/DE^{2}
ar (â–³ABC)/ar(â–³DEF) = 2

^{2}/5^{2}
ar (â–³ABC)/ar(â–³DEF) = 4/25

Hence the correct answer is b .

7. XY is drawn parallel to the base BC of a â–³ABC Cutting AB at X and AC at Y. If AB = 4 BX and YC = 2 cm, then AY =

(a) 2 cm

(b) 4 cm

(c ) 6 cm

(d) 8 cm

(a) 2 cm

(b) 4 cm

(c ) 6 cm

(d) 8 cm

**Solution**

Given : XY is drawn parallel to the base BC of a â–³ABC Cutting AB at X and AC at Y. If AB = 4 BX and YC = 2 cm.

To find : AY

In â–³AXY and â–³ABC ,

âˆ AXY = âˆ B (Corresponding angles)

âˆ A = âˆ A (common)

âˆ´ â–³AXY ~ â–³ABC (AA similarity)

We know that if two triangles are similar , then their sides are proportional .

It is given that AB = 4BX = X.

Then, AX = 3X

AX/BX = AY/ YC

3x/1x = AY/2

AY = 3x Ã— 2/1x

AY = 6 cm

Hence the correct answer is c .

8. Two poles of height 6 m and 11 m stand vertically upright on a plane ground . If the distance between their foot is 12 m, the distance between their tops is

(a) 12 m

(b) 14 m

(c) 13 m

(d) 11 m

**Solution**

Given : Two poles of heights 6m and 11m stand vertically upright on a plane ground . Distance between their foot is 2 m.

To find : Distance between their tops.

Let CD be the pole with height 6m .

AB is the pole with the height 11 m, distance between their foot i.e. DB is 12 m .

Let us assume a point EE on the pole AB which is 6 m from the base of AB.

Hence

AE = AB â€“ 6 = 11 â€“ 6 = 5 m

Now in right triangle AEC, Applying Pythagoras theorem

AC

^{2}= AE^{2}+ EC^{2}
AC

^{2}= 5^{2}+ 12^{2}(Since CDEB forms a rectangle and opposite sides of rectangle are equal )
AC

^{2}= 25 + 144
AC

^{2}= 169
AC = 13 m

Thus, the distance between their tops is 13m.

Hence correct answer is c.

9. In â–³ABC, D and E are points on side AB and AC respectively such that DE | | BC and AD : DB = 3 : 1. If EA = 3.3 cm, then AC =

(a) 1.1 cm

(b) 4 cm

(c) 4.4 cm

(d) 5.5 cm

(a) 1.1 cm

(b) 4 cm

(c) 4.4 cm

(d) 5.5 cm

**Solution**

Given : In â–³ABC , D and E are points on the side AB and AC respectively such that DE | | BC and AD : DB = 3:1. Also , EA = 3.3 cm

To find : AC

In â–³ABC, DE | | BC,

Using corollary of basis proportionality theorem, we have

AD/AB = EA/AC

AD/AD+BD = 3.3/AC

AD/AD + 1/3 AD = 3.3/AC

EC = 4.4 cm

Hence the correct answer is C .

10. In triangle ABC and DEF , âˆ A = âˆ E = 40Â° , AB : ED = AC : EF and âˆ F = 65Â° , then âˆ B =

(a) 35Â°

(b) 65Â°

(c) 75Â°

(d) 85Â°

(a) 35Â°

(b) 65Â°

(c) 75Â°

(d) 85Â°

**Solution**

Given : In â–³ABC and â–³DEF

âˆ A = âˆ E = 40Â°

AB : ED = AC : EF

âˆ F = 65Â°

To find : Measure of angle B .

In â–³ABC and â–³DEF

âˆ A = âˆ E = 40Â°

AB : ED = AC : EF

â–³ABC ~ â–³DEF (S.A.S Similarity crieteria)

Hence in similar triangles â–³ABC and â–³DEF

âˆ A = âˆ E = 40Â°

âˆ C = âˆ F = 65Â°

âˆ B = âˆ D

We know that sum of all angles of a triangle is equal to 180Â° .

âˆ A + âˆ B + âˆ C = 180Â°

40Â° + âˆ B + 65Â° = 180Â°

âˆ B + 115Â° = 180Â°

âˆ B = 180Â° - 115Â°

âˆ B = 75Â°

Hence the correct answer is c .

11.If ABC and DEF are similar triangles such that âˆ A = 47Â° and âˆ E = 83Â° , then âˆ C =

(a)50Â°

(b)60Â°

(c)70Â°

(d)80Â°

(a)50Â°

(b)60Â°

(c)70Â°

(d)80Â°

**Solution**

Given: If â–³ABC and â–³DEF are similar triangles such that

âˆ A = 47Â°

âˆ E = 83Â°

To find : Measure of angle C

In similar â–³ABC and â–³DEF ,

âˆ A = âˆ D = 47Â°

âˆ B = âˆ E = 83Â°

âˆ C = âˆ F

We know that sum of all the angles of a triangle is equal 180Â° .

âˆ A + âˆ B + âˆ C = 180Â°

47Â° + 83Â° + âˆ C = 180Â°

âˆ C + 130Â° = 180Â°

âˆ C = 180Â° - 130Â°

âˆ C = 50Â°

Hence the correct answer is b .

âˆ C = âˆ F

We know that sum of all the angles of a triangle is equal 180Â° .

âˆ A + âˆ B + âˆ C = 180Â°

47Â° + 83Â° + âˆ C = 180Â°

âˆ C + 130Â° = 180Â°

âˆ C = 180Â° - 130Â°

âˆ C = 50Â°

Hence the correct answer is b .

(a) 1:4

(b) 1:2

(c) 2:3

(d) 4:5

**Solution**

Given : In â–³ABC, D, E and F are the midpoints of BC, CA, and AB respectively .

To find : Ratio of the areas of â–³DEF and â–³ABC

Since it is given that D and, E are the midpoints of BC, and AC respectively.

Therefore DE | | AB, DE | | FA â€¦(1)

Again it is given that D and , F are the midpoints of BC, and, AB respectively .

Therefore, DF | | CA, DF | | AE...(2)

From (1) and (2) we get AFDE is a parallelogram.

Similarly we can prove that BDEF is a parallelogram

Now, in â–³ADE and â–³ABC

âˆ FDE = âˆ A (Opposite angles of | |gm AFDE)

âˆ DEF = âˆ B (Opposite angles of | |gm BDEF)

â‡’ â–³ABC ~ â–³DEF (AA similarity criterion)

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides .

ar(â–³DEF)/ar(â–³ABC) = (DE/AB)

^{2}
ar(â–³DEF)/ar(â–³ABC) = [(1/2(AB))/AB] (Since DE = 1/2 AB)

ar(â–³DEF)/ar(â–³ABC) = (1/4)

Hence the correct option is a .

13. In a â–³ABC, âˆ A = 90Â°, AB = 5 cm and AC = 12 cm . If AD âŠ¥ BC, then AD =

(a) 13/2 cm

(b) 60/13 cm

(c) 13/60 cm

(d) 2âˆš15/13 cm

**Solution**

Given : In â–³ABC âˆ A = 90Â°, AD âŠ¥ BC, AC = 12 cm, and AB = 5cm

To find : AD

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides .

In âˆ ACB and âˆ ADC ,

âˆ C = âˆ C (Common)

âˆ A = âˆ ADC = 90Â°

âˆ´ â–³ACB ~ â–³ADC (AA Similarity)

AD/AB = AC/BC

AD = AB Ã— AC/BC

AD = 12Ã—5/13

AD = 60/13

We got the result as b .

14. If â–³ABC is an equilateral triangle such that AD âŠ¥ BC, then AD

^{2}=
(a) (3/2) DC

^{2}
(b) 2 DC

^{2}
(c) 3 CD

^{2}
(d) 4 DC

^{2}**Solution**

Given : In an equilateral â–³ABC, AD âŠ¥ BC .

Since AD âŠ¥ BC, BD = CD = BC/2

In â–³ADC

AC

^{2}= AD^{2}+ DC^{2}
BC

^{2}= AD^{2}+ DC^{2}(Since AC = BC)
(2DC)

^{2}= AD^{2}+ DC^{2}(Since BC = 2DC)
4DC

^{2}= AD^{2}+ DC^{2}
3DC

^{2}= AD^{2}
3CD

^{2}= AD^{2}
We got the result as c .

15. In a â–³ABC , AD is the bisect of âˆ BAC . If AB = 6 cm , AC = 5 cm and BD = 3 cm, then DC =

(a) 11.3 cm

(b) 2.5 cm

(c) 3:5 cm

(d) None of these

(a) 11.3 cm

(b) 2.5 cm

(c) 3:5 cm

(d) None of these

**Solution**

Given : In a â–³ABC, AD is the bisector of âˆ BAC . AB = 6 cm and AC = 5 cm and BD = 3 cm.

To find : DC

We know that the internal bisector of angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle ,

Hence ,

AB/AC = BD/DC

6/5 = 3/DC

DC = 5Ã—3/6

DC = 2.5 cm

Hence we got the result b.

16. In â–³ABC, AD is the bisector of âˆ BAC . If AB = 8 cm, BD = 6 cm and DC = 3 cm . Find AC

(a) 4 cm

(b) 6 cm

(c) 3 cm

(d) 8 cm

(a) 4 cm

(b) 6 cm

(c) 3 cm

(d) 8 cm

**Solution**

To find : AC

We know that the internal bisector of a triangle divides the opposite side internally in the ratio of the sides containing the angle .

AB/AC = BD/DC

8/AC = 6/3

AC = 8Ã—3/6

AC = 4 cm

Hence we got the result a.

(a) 7 cm

(b) 8 cm

(c) 9 cm

(d) 6 cm

**Solution**

Given : ABCD is a trapezium in which BC | | AD and AD = 4 cm

The diagonals AC and BD intersect at O such that AO/OC = DO/OB = 1/2

To find : DC

The diagonals AC and BD intersect at O such that AO/OC = DO/OB = 1/2

To find : DC

In â–³AOD and â–³COB

âˆ OAD = âˆ OCB (Alternate angles)

âˆ ODA = âˆ OBC (Alternate angles)

âˆ AOD = âˆ BOC (Vertically opposite angles)

So, â–³AOD ~ â–³COB (AAA similarity)

Now, corresponding sides of similar â–³â€™s are proportional .

AO/CO = DO/BO = AD/BC

â‡’ 1/2 = AD/BC

â‡’ 1/2 = 4/BC

â‡’ BC = 8 cm

Hence the correct answer is b .

18. If ABC is a right â€“ angled at B and M, N are the mid â€“ points of AB and BC respectively, then 4(AN

^{2}+ CM^{2}) =
(a) 4 AC

^{2}
(b) 5 AC

^{2}
(c ) 5/4 AC

^{2}
(d) AC

^{2}^{}

**Solution**

^{}

M is the mid â€“ point of AB.

âˆ´ BM = AB/2

N is the mid â€“ point of BC .

âˆ´ BN = BC/2

Now,

AN

^{2}+ CM^{2}= (AB^{2}+ (1/2 BC)^{2}) + ((1/2 AB)^{2}+ BC^{2}
= AB

^{2}+ 1/4 BC^{2}+ 1/4 AB^{2}+ BC^{2}
= 5/4 (AB

^{2}+ BC^{2})
â‡’ 4(AN

^{2}+ CM^{2}) = 5 AC^{2}
Hence option b is correct .

19. If in â–³ABC and â–³DEF , AB/DE = BC/FD , then â–³ABC ~ â–³DEF when

(a) âˆ A = âˆ F

(b) âˆ A = âˆ D

(c) âˆ B = âˆ D

(d) âˆ B = âˆ E

Given : In â–³ABC and â–³DEF , AB/DE = BC/FD .

We know that if in two triangles, one pair of corresponding sides are proportional and the included angles are equal , then the two triangles are similar .

Then ,

âˆ B = âˆ D

Hence, â–³ABC is similar to â–³DEF , we should have âˆ B = âˆ D

Hence the correct answer is c.

20. If in two triangles ABC and DEF , AB/DE = BC/FE = CA/FD , then

(a) â–³FDE ~ â–³CAB

(b) â–³FDE ~ â–³ABC

(c) â–³CBA ~ â–³FDE

(d) â–³BCA ~ â–³FDE

19. If in â–³ABC and â–³DEF , AB/DE = BC/FD , then â–³ABC ~ â–³DEF when

(a) âˆ A = âˆ F

(b) âˆ A = âˆ D

(c) âˆ B = âˆ D

(d) âˆ B = âˆ E

^{}**Solution**^{}Given : In â–³ABC and â–³DEF , AB/DE = BC/FD .

We know that if in two triangles, one pair of corresponding sides are proportional and the included angles are equal , then the two triangles are similar .

Then ,

âˆ B = âˆ D

Hence the correct answer is c.

20. If in two triangles ABC and DEF , AB/DE = BC/FE = CA/FD , then

(a) â–³FDE ~ â–³CAB

(b) â–³FDE ~ â–³ABC

(c) â–³CBA ~ â–³FDE

(d) â–³BCA ~ â–³FDE

**Solution**
We know that if two triangles are similar if their corresponding sides are proportional .

It is given that â–³ABC and â–³DEF are two triangles such that AB/DE = BC/EF = CA/FD .

âˆ A = âˆ D

âˆ B = âˆ E

âˆ C = âˆ F

âˆ´ â–³CAB ~ â–³FDE

Hence the correct answer is a .

21. â–³ABC ~ â–³DEF, ar(â–³ABC) = 9 cm

^{2}. If BC = 2.1 cm, then the measure of EF is
(a) 2.8 cm

(b) 4.2 cm

(c) 2.5 cm

(d) 4.1 cm

**Solution**

Given : Ar(â–³ABC) = 9 cm

^{2}, Ar(â–³DEF) = 16 cm

^{2}and BC = 2.1 cm

To find : measure of EF

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides .

Ar(â–³ABC)/Ar(â–³DEF) = BC2/EF2

9//16 = 2.12

3/4 = 2.1/EF

EF = 2.8 cm

Hence the correct answer is a .

22. The length of the hypotenuse of an isosceles right triangle whose one side is 4âˆš2 cm is

(a) 12 cm

(b) 8 cm

(c) 8âˆš2 cm

(d) 12âˆš2 cm

(a) 12 cm

(b) 8 cm

(c) 8âˆš2 cm

(d) 12âˆš2 cm

**Solution**

Given : One side of isosceles right triangle is 4âˆš2 cm

To find : Length of the hypotenuse .

We know that in isosceles triangle two sides are equal .

In isosceles right triangle ABC , let AB and AC be the two equal sides of measures 4âˆš2 cm .

Applying Pythagoras theorem, we get

BC

^{2}= AB^{2}+ AC^{2}
BC

^{2}= (4âˆš2)^{2}+ (4âˆš2)^{2}
BC

^{2}= 32 + 32
BC

^{2}= 64
BC = 8 cm

Hence correct answer is b .

23. A man goes 24 m due west and then 7 m due north. How far is he from the starting point ?

(a) 31 m

(b) 17 m

(c) 25 m

(d) 26 m

(a) 31 m

(b) 17 m

(c) 25 m

(d) 26 m

**Solution**

A man goes 24 m due to west and then 7 m due north.

Let the man starts from point B and goes 24 m due to west and reaches point A, then walked 7 m north and reaches point C.

Now we have to find the distance between the starting point and the end point i.e. BC.

In right triangle ABC , applying Pythagoras theorem we, get

BC

^{2},= AB^{2}+ AC^{2}
BC

^{2}= (24)^{2}+ (7)^{2}
BC

^{2}= 576 + 49
BC

^{2}= 625
BC = 25 m

Hence correct answer is c .

24. â–³ABC ~ â–³DEF . If BC = 3 cm , EF = 4 cm and ar(â–³ABC) = 54 cm

(b) 96 cm

(c) 48 cm

(d) 100 cm

^{2}, then ar(â–³DEF) = (a) 108 cm^{2}(b) 96 cm

^{2}(c) 48 cm

^{2}(d) 100 cm

^{2}**Solution**

Given : In â–³ABC and â–³DEF

â–³ABC ~â–³DEF

Ar(â–³ABC ) = 54 cm

^{2}BC = 3 cm and EF = 4 cm

To find : Ar(â–³DEF)

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides .

Ar(â–³ABC )/Ar(â–³DEF) = BC

^{2}/EF

^{2}

54/Ar(â–³DEF) = 3

^{2}/4

^{2}

54/ Ar(â–³DEF) = 9/16

Ar(â–³DEF) = 16Ã—54/9

Ar(â–³DEF) = 96 cm

^{2}

Hence the correct answer is b .

25. â–³ABC ~â–³PQR such that ar(â–³ABC) = 4 ar(â–³PQR). If BC = 12 cm, then QR =

(a) 9 cm

(b) 10 cm

(c) 6 cm

(d) 8 cm

**Solution**

Given : In â–³ABC and â–³PQR

â–³ABC ~â–³PQR

Ar(â–³ABC ) = 4Ar (â–³PQR)

BC = 12 cm

To find : Measure of QR

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides .

Ar(â–³ABC )/Ar(â–³PQR) = BC

^{2}/QR^{2}
4 Ar(â–³PQR)/ Ar(â–³PQR) = 12

^{2}/QR^{2}(Given Ar(â–³ABC ) = 4 Ar(â–³PQR))
4/1 = 12

^{2}/QR^{2}
2/1 = 12/QR

QR = 6 cm

Hence the correct answer is c .

26. The areas of two similar triangles are 121 cm

(a) 11 cm

(b) 8.8 cm

(c) 11.1 cm

(d) 8.1 cm

Given : The area of two similar triangles is 121 cm

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their medians .

ar(triangle1)/ar(triangle2) = (median1/median2)

121/64 = (12.1/median2)

Taking square root on both side, we get

11/8 = 12.1 cm/median2

â‡’ median

Hence the correct answer is b .

27. In an equilateral triangle ABC if AD âŠ¥ BC, then AD

(a) CD

(b)2CD

(c) 3CD

(d) 4CD

26. The areas of two similar triangles are 121 cm

^{2}and 64cm^{2}respectively .If the median of the first triangle is 12.1 cm, then the corresponding median of the other triangle is(a) 11 cm

(b) 8.8 cm

(c) 11.1 cm

(d) 8.1 cm

**Solution**Given : The area of two similar triangles is 121 cm

^{2}and 64 cm^{2}respectively . The median of the first triangle is 2.1 cm . To find : Corresponding medians of the other triangleWe know that the ratio of areas of two similar triangles is equal to the ratio of squares of their medians .

ar(triangle1)/ar(triangle2) = (median1/median2)

^{2}121/64 = (12.1/median2)

^{2}Taking square root on both side, we get

11/8 = 12.1 cm/median2

â‡’ median

^{2}= 8.8 cmHence the correct answer is b .

27. In an equilateral triangle ABC if AD âŠ¥ BC, then AD

^{2}=(a) CD

^{2}(b)2CD

^{2}(c) 3CD

^{2}(d) 4CD

^{2}**Solution**

In an equilateral â–³ABC , AD âŠ¥ BC .

In â–³ADC, applying Pythagoras theorem , we get

AC

^{2}= AD^{2}+ DC^{2}
BC

^{2}= AD^{2}+ DC^{2}(âˆµBC = 2DC)
4DC

^{2}= AD^{2}+ DC^{2}
3DC

^{2}= AD^{2}
Hence the correct option is c.

28. In an equilateral triangle ABC if AD âŠ¥ BC, then

(a) 5AB

(b) 3AB

(c) 4AB

^{2}= 4AD^{2}(b) 3AB

^{2}= 4AD^{2}(c) 4AB

^{2}= 3AD^{2}
(d) 2AB

â–³ABC is an equilateral triangle and AD âŠ¥ BC .

In â–³ABD, applying Pythagoras theorem, we get

AB

AB

AB

AB

3AB

We got the result as b .

29. â–³ABC is an isosceles triangle in which âˆ C = 90 . If AC = 6 cm, then AB =

(a) 6 âˆš2 cm

(b) 6 cm

(c) 2âˆš6 cm

^{2}= 3AD^{2}**Solution**â–³ABC is an equilateral triangle and AD âŠ¥ BC .

In â–³ABD, applying Pythagoras theorem, we get

AB

^{2}= AD^{2}+ BD^{2}AB

^{2}= AD^{2}+ (1/2BC)^{2}(âˆµ BD = 1/2 BC)AB

^{2}= AD^{2}+ (1/2 AB)^{2}(âˆµ AB = BC)AB

^{2}= AD^{2}+ 1/4 AB^{2}3AB

^{2}= 4AD^{2}We got the result as b .

(a) 6 âˆš2 cm

(b) 6 cm

(c) 2âˆš6 cm

(d) 4âˆš2 cm

Given : In an isosceles â–³ABC , âˆ C = 90Â° . If AC = 6 cm .

To find : AB

In an isosceles â–³ABC, âˆ C = 90Â° .

Therefore , BC = AC = 6 cm

**Solution**Given : In an isosceles â–³ABC , âˆ C = 90Â° . If AC = 6 cm .

To find : AB

In an isosceles â–³ABC, âˆ C = 90Â° .

Therefore , BC = AC = 6 cm

Applying Pythagoras theorem in â–³ABC , we get

AB

AB

AB2 = 36 + 36

AB2 = 72

AB = 6âˆš2 cm

We got the result as a.

30. If in two triangle ABC and DEF , âˆ A = âˆ E, âˆ B = âˆ F, then which of the following is not true?

(a) BC/DF = AC/DE

(b) AB/DE = BC/DF

(c) AB/EF = AC/DE

(d) BC/DF = AB/EF

In â–³ABC and â–³DEF

âˆ A = âˆ E

âˆ B = âˆ F

âˆ´ â–³ABC and â–³DEF are similar triangles.

Hence AB/EF = BC/FD = CA/DE

Hence the correct answer is b .

31 . In the given figure the measure of âˆ A and âˆ F are respectively

(a) 50Â°, 40Â°

(b) 20Â°, 30Â°

(c) 40Â°, 50Â°

(d) 30Â°, 20Â°

AB

^{2}= AC^{2}+ BC^{2}AB

^{2}= 62 + 62 (AC = BC, sides of isosceles triangle)AB2 = 36 + 36

AB2 = 72

AB = 6âˆš2 cm

We got the result as a.

30. If in two triangle ABC and DEF , âˆ A = âˆ E, âˆ B = âˆ F, then which of the following is not true?

(a) BC/DF = AC/DE

(b) AB/DE = BC/DF

(c) AB/EF = AC/DE

(d) BC/DF = AB/EF

**Solution**In â–³ABC and â–³DEF

âˆ A = âˆ E

âˆ B = âˆ F

âˆ´ â–³ABC and â–³DEF are similar triangles.

Hence AB/EF = BC/FD = CA/DE

Hence the correct answer is b .

31 . In the given figure the measure of âˆ A and âˆ F are respectively

(a) 50Â°, 40Â°

(b) 20Â°, 30Â°

(c) 40Â°, 50Â°

(d) 30Â°, 20Â°

**Solution**

â–³ABC and â–³DEF ,

AB/AC = EF/ED

âˆ A = âˆ E = 130Â°

â–³ABC ~ â–³EFD (SAS Similarity)

âˆ´ âˆ F = âˆ B = 30Â°

âˆ D = âˆ C = 20Â°

Hence the correct answer is b .

âˆ A = âˆ E = 130Â°

â–³ABC ~ â–³EFD (SAS Similarity)

âˆ´ âˆ F = âˆ B = 30Â°

âˆ D = âˆ C = 20Â°

Hence the correct answer is b .

32. In the given figure , the value of x for which DE | | AB is

(a) 4

(b) 1

(c) 3

(d) 2

(b) 1

(c) 3

(d) 2

**Solution**

Given : In â–³ABC , DE | | AB .

To find : the value x

According to basic proportionality theorem if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.

In â–³ABC , DE | | AB .

AD/DB = AE/EC

x+3/3x +19 = x/3x+4

(x+3)(3x+4) = (x)(3x+19)

3x

19x â€“ 13x = 12

6x = 12

x = 2

Hence we got the result d .

Given : RS | | DB | | PQ . CP = PD = 11 cm and DR = RA = 3 cm

To find : the value of x and y respectively .

In â–³ASR and â–³ABD ,

âˆ ASR = âˆ ABQ (Corresponding angles)

âˆ A = âˆ A (Common)

âˆ´ â–³ASR ~ â–³ABD (AA Similarity)

AR/AD = AS/AB = RS/DB

3/6 = RS/DB

1/2 = x/y

x = 2y

This relation is satisfied by option d.

Hence, x = 16 cm and y =8 cm

Hence the result is d .

To find : the value x

According to basic proportionality theorem if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.

In â–³ABC , DE | | AB .

AD/DB = AE/EC

x+3/3x +19 = x/3x+4

(x+3)(3x+4) = (x)(3x+19)

3x

^{2}+ 4x + 9x + 12 = 3x^{2}+ 19x19x â€“ 13x = 12

6x = 12

x = 2

Hence we got the result d .

33. In the given figure, if âˆ ADE = âˆ ABC, then CE =

(a) 2

(b) 5

(c) 9/2

(d) 3

Given: âˆ ADE = âˆ ABC

To find: The value of CE

Since âˆ ADE = âˆ ABC

âˆ´ DE | | BC (Two lines are parallel if the corresponding angles formed are equal)

According to basic proportionality theorem if a line is parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.

AD/DB = AE/EC

2/3 = 3/EC

EC = 3Ã—3/2

EC = 9/2

Hence we got the result c .

34. In the given figure, RS | | DB | | PQ. If CP = PD = 11 cm and DR = RA = 3 cm . Then the values of x and y are respectively .

(a) 2

(b) 5

(c) 9/2

(d) 3

**Solution**To find: The value of CE

Since âˆ ADE = âˆ ABC

âˆ´ DE | | BC (Two lines are parallel if the corresponding angles formed are equal)

According to basic proportionality theorem if a line is parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.

AD/DB = AE/EC

2/3 = 3/EC

EC = 3Ã—3/2

EC = 9/2

Hence we got the result c .

34. In the given figure, RS | | DB | | PQ. If CP = PD = 11 cm and DR = RA = 3 cm . Then the values of x and y are respectively .

(a) 12,10

(b) 14,6

(c) 10,7

(d) 16,8

(c) 10,7

(d) 16,8

**Solution**

Given : RS | | DB | | PQ . CP = PD = 11 cm and DR = RA = 3 cm

To find : the value of x and y respectively .

In â–³ASR and â–³ABD ,

âˆ ASR = âˆ ABQ (Corresponding angles)

âˆ A = âˆ A (Common)

âˆ´ â–³ASR ~ â–³ABD (AA Similarity)

AR/AD = AS/AB = RS/DB

3/6 = RS/DB

1/2 = x/y

x = 2y

This relation is satisfied by option d.

Hence, x = 16 cm and y =8 cm

Hence the result is d .

35. In the given figure, if PB | | CF and DP | | EF, then AD/ DE =

(a) 3/4

(b) 1/3

(c) 1/4

(d) 2/3

(a) 3/4

(b) 1/3

(c) 1/4

(d) 2/3

**Solution**

Given : PB | | CF and DP | | EF . AB = 2 cm and AC = 8 cm .

To find : AD:DE

According to basic proportionality theorem, if a line is drawn parallel to one side of a triangle intersecting the other two sides . then it divides the two sides in the same ratio .

AB/BC = AP/PF

AP/PF = 2/8-2

AP/PF = 2/6

AP/PF = 1/3 â€¦(1)

Again, DP | | EF .

AD/DE = AP/PF

AD/DE = 1/3

Hence we got the result option b .

36. A chord of a circle of radius 10 cm subtends a right at the centre. The length of the chord (in cm) is

(a) 5 âˆš2

(b) 10âˆš2

(c) 5 /âˆš2

(d) 105âˆš3

(a) 5 âˆš2

(b) 10âˆš2

(c) 5 /âˆš2

(d) 105âˆš3

**Solution**

AB

^{2}= OA

^{2}+ OB

^{2}(Pythagoras theorem)

â‡’ AB

^{2}= (10)

^{2}+ (10)

^{2}(OA = OB = 10 cm)

â‡’ AB

^{2}= 100 + 100 = 200

â‡’ AB = âˆš200 = 10âˆš2 cm

Thus, the length of the chord is 10âˆš2 cm

Hence the correct answer is option B.

37. A vertical stick 20 m long casts a shadow 10 m long on the ground . At the same time , a tower casts a shadow 50 m long on the ground . The height of the tower is

(a) 100 m

(b) 120 m

(c) 25 m

(d) 200 m

**Solution**

Given : Vertical stick 20 m long casts a shadow 10 m long on the ground . At the same time a tower casts the shadow 50 m long on the ground.

To determine : Height of the tower

Let AB be the vertical stick and AC be its shadow . Also , let DE be the vertical tower and DF be its shadow.

Join BC and EF.

In Î”ABC and Î”DEF , we have

âˆ A = âˆ D = 90Â°

âˆ C = âˆ F

Î”ABC ~ Î”DEF

We know that in any two similar triangles , the corresponding sides are proportional . Hence ,

AB/DE = AC/DF

20/DE = 10/50

DE = 100 m

Hence the correct answer is option a .

38. Two isosceles triangles have equal angles and their areas are in the ratio 16:25. The ratio of their corresponding heights is

(a) 4:5

(b) 5:4

(c) 3:2

(d) 5:7

(a) 4:5

(b) 5:4

(c) 3:2

(d) 5:7

**Solution**

Given two isosceles triangle have equal vertical angles and their areas are in the ratio of 16:25 .

To find : ratio of their corresponding heights .

Let Î”ABC and Î”PQR be two isosceles triangles such that âˆ A = âˆ P . Suppose AD âŠ¥ BC and PS âŠ¥ QR .

In Î”ABC and Î”PQR ,

AB/PQ = AC/PR

âˆ A = âˆ P

âˆ´ Î”ABC ~ Î”PQR (SAS similarity)

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes .

Hence,

Ar(Î”ABC)/Ar(Î”PQR) = (AD/PS)

^{2}
â‡’ 16/25 = (AD/PS)

^{2}
â‡’ AD/PS = 4/5

Hence we got the result as a .

39. Î”ABC is such that AB = 3 cm , BC = 2 cm and CA = 2.5 cm . If Î”DEF ~ Î”ABC and EF = 4 cm , then perimeter of Î”DEF is

(a) 7.5 cm

(b) 15 cm

(c) 22.5 cm

(d) 30 cm

(b) 15 cm

(c) 22.5 cm

(d) 30 cm

**Solution**

Given : In Î”ABC , AB = 3 cm, BC = 2 cm , CA = 2.5 cm . Î”DEF ~ Î”ABC and EF = 4 cm

To find : Perimeter of Î”DEF .

We know that if two triangles are similar , then their sides are proportional

Since Î”ABC and Î”DEF are similar ,

AB/DE = BC/EF = CA/FD

3/DE = 2/4 = 2.5/FD

3/DE = 2/4

DE = 6 cm â€¦(1)

2/4 = 2.5/FD

FD = 5 cm â€¦(2)

From (1) and (2) , we get

Perimeter of Î”DEF = DE + EF + FD = 6 + 4 + 5 = 15 cm

Hence the correct answer is b .

40. In Î”ABC , a line XY parallel to BC cuts AB at X AC at Y . If BY bisects âˆ XYC , then

(a) BC = CY

(b) BC = BY

(c) BC â‰ CY

(d) BC â‰ BY

Given XY | | BC and BY is bisector of âˆ XYC .

We know that if two triangles are similar , then their sides are proportional

Since Î”ABC and Î”DEF are similar ,

AB/DE = BC/EF = CA/FD

3/DE = 2/4 = 2.5/FD

3/DE = 2/4

DE = 6 cm â€¦(1)

2/4 = 2.5/FD

FD = 5 cm â€¦(2)

From (1) and (2) , we get

Perimeter of Î”DEF = DE + EF + FD = 6 + 4 + 5 = 15 cm

Hence the correct answer is b .

40. In Î”ABC , a line XY parallel to BC cuts AB at X AC at Y . If BY bisects âˆ XYC , then

(a) BC = CY

(b) BC = BY

(c) BC â‰ CY

(d) BC â‰ BY

**Solution**Given XY | | BC and BY is bisector of âˆ XYC .

Since XY | | BC

So âˆ YBC = âˆ BYC (Alternate angles)

Now, in triangles BYC two angles are equal . Therefore , the two corresponding sides will be equal

Hence, BC = CY

Hence option a is correct.

41. In a Î”ABC , âˆ A = 90Â°, AB = 5 cm and AC = 12 cm . If AD âŠ¥ BC , then AD =

(a) 13/2 cm

(b) 60/13 cm

(c) 13/60 cm

(d) 2âˆš15/13 cm

(a) 13/2 cm

(b) 60/13 cm

(c) 13/60 cm

(d) 2âˆš15/13 cm

**Solution**

Given : In Î”ABC âˆ A = 90Â°, AD âŠ¥ BC , AC = 12 cm, and AB = 5 cm .

To find : AD

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides .

In Î”ACB and Î”ADC ,

âˆ C = âˆ C (Common)

âˆ A = âˆ ADC = 90Â°

âˆ´ Î”ACB ~ Î”ADC (AA Similarity)

AD/AB = AC/BC

AD = AB Ã— AC/BC

AD = 12Ã—5/13

AD = 60/13

We got the result as b .

42. In a Î”ABC, perpendicular AD from A and BC meets BC at D . If BD = 8 cm, DC = 2 cm and AD = 4 cm, then

(a) Î”ABC is isosceles

(b) Î”ABC is equilateral

(c) AC = 2AB

(d) Î”ABC is right â€“ angled at A

(a) Î”ABC is isosceles

(b) Î”ABC is equilateral

(c) AC = 2AB

(d) Î”ABC is right â€“ angled at A

**Solution**

Given : In Î”ABC, AD âŠ¥ BC , BD = 8 cm, DC = 2 cm and AD = 4 cm .

In Î”ADC ,

AC

^{2}= AD^{2}+ DC^{2}
AC

^{2}= 42 + 22
AC

^{2}= 20 â€¦..(1)
Similarly , in Î”ADB

AB

^{2}= AD^{2}+ DC^{2}
AC

^{2}= 42 + 82
AB2 = 80 â€¦..(2)

Now, in Î”ABC

BC

^{2}= (CD + DB)^{2}= (2+8)^{2}= (10)^{2}= 100
and

AB

^{2}+ CA^{2}= 80 + 20 = 100
âˆ´ AB

^{2}+ CA^{2}= BC^{2}
Hence, triangle ABC is right angled at A .

We got the result as d .

43 . In a Î”ABC, point D is on side AB and point E is on side AC , such that BCED is a trapezium . If DE : BC = 3:5, then Area (Î”ADE) : Area(Î”BCED) =

(a) 3:4

(b) 9:16

(c) 3:5

(d) 9:25

(a) 3:4

(b) 9:16

(c) 3:5

(d) 9:25

**Solution**

Given : In Î”ABC,D is on side AB and point E is on side AC, such that BCED is a trapezium . DE : BC = 3:5.

To find : Calculate the ratio of the areas of Î”ADE and the trapezium BCED .

In Î”ADE and Î”ABC ,

âˆ ADE = âˆ B (Corresponding angles)

âˆ A = âˆ A (Common)

âˆ´ Î”ADE ~ Î”ABC (AA Similarity) .

We know that

Ar(Î”ADE)/Ar(Î”ABC) = DE

^{2}/BC^{2}
Ar(Î”ADE)/Ar(Î”ABC) = 3

^{2}/5^{2}
Ar(Î”ADE)/Ar(Î”ABC) = 9/25

Let Area of Î”ADE = 9x sq. units and Area of Î”ABC = 25x sq. units

Ar[trap BCED] = Ar(Î”ABC) â€“ Ar(Î”ADE)

= 25x â€“ 9x

= 16x sq units

Now,

Ar(Î”ADE)/Ar(trap BCED) = 9x/16x

Ar(Î”ADE)/Ar(trap BCED) = 9/16

Hence the correct answer is b .

44. If ABC is an isosceles triangle and D is a point of BC such that AD âŠ¥ BC , then

(a) AB

^{2}- AD^{2}= BD.DC
(b) AB

^{2}â€“ AD^{2}= BD^{2}â€“ DC^{2}
(c) AB

^{2}+ AD^{2}= BD.DC
(d) AB

^{2}+ AD^{2}= BD^{2}â€“ DC^{2}**Solution**

Given : Î”ABC is an isosceles triangle , D is a point on BC such that AD âŠ¥ BC

We know that in an isosceles triangle the perpendicular from the vertex bisects the Base .

âˆ´ BD = DC

Applying Pythagoras theorem in Î”ABD

AB

^{2}= AD^{2}+ BD^{2}
â‡’ AB

^{2}â€“ AD^{2}= BD Ã— BD
Since BD = DC

â‡’ AB

^{2}â€“ AD^{2}= BD Ã— BD
Since BD = DC

â‡’ AB

^{2}â€“ AD^{2}= BD Ã— BD
Hence correct answer is a .

45 . Î”ABC is a right triangle right â€“ angled at A and AD âŠ¥ BC . Then , BD/DC =

(a) (AB/AC)

^{2}
(b) AB/AC

(c) (AB/AD)

^{2}
(d) AB/AD

**Solution**

Given : In Î”ABC, âˆ A = 90Â° and AD âŠ¥ BC .

To find : BD : DC

âˆ CAD + âˆ BAD = 90Â° â€¦(1)

âˆ BAD + âˆ ABD = 90Â° â€¦(2) (âˆ ADB = 90Â°)

From (1) and (2),

âˆ CAD = âˆ ABD

In Î”ADB and Î”ADC,

âˆ ADB = âˆ ADC (90Â° each)

âˆ ABD = âˆ CAD (Proved)

âˆ´ Î”ADB ~ Î”ADC (AA Similarity)

â‡’ CD/AD = AC/AB = AD/BD (Corresponding sides are proportional)

Disclaimer : The question is not correct . The given ratio cannot be evaluated using he given conditions in the questions .

46. If E is a point on side CA of an equilateral triangle ABC such that BE âŠ¥ CA, then AB

(a)2 BE

(b) 3 BE

(c) 4 BE

(d) 6 BE

^{2}+ BC^{2}+ CA^{2}=(a)2 BE

^{2}(b) 3 BE

^{2}(c) 4 BE

^{2}(d) 6 BE

^{2}**Solution**In triangle ABC , E is a point on AC such that BE âŠ¥ AC .

We need to find AB

^{2}+ BC

^{2}+ AC

^{2}.

Since BE âŠ¥ AC , CE = AE = AC/2 (In a equilateral triangle, the perpendicular from the vertex bisects the base.

In triangle ABE, we haveAB

^{2}= BE

^{2 + AE2Since AB = BC = ACTherefore , AB2 = BC2 = AC2 = BE2 + AE2â‡’ AB2 + BC2 + AC2 = 3BE2 + 3AE2Since in triangle BE is an altitude, so BE = âˆš3/2 ABBE = âˆš3/2 AB }= âˆš3/2 Ã— AC

^{= âˆš3/2 Ã— 2AE = âˆš3AEâ‡’ AB2 + BC2 + AC2 = 3BE2 + 3(BE/âˆš3)2= 3BE2 + BE2 = 4BE2Hence option c is correct . }

^{}47. In a right triangle ABC right â€“ angled at B, if P and Q are points on the sides AB and AC respectively , then

(a) AQ

^{2}+ CP

^{2}= 2(AC

^{2}+ PQ

^{2})

(b) 2(AQ

^{2}+ CP

^{2}) = AC

^{2}+ PQ

^{2}

(c) AQ

^{2}+ CP

^{2}= AC

^{2}+ PQ

^{2}

(d) AQ + CP = 1/2 (AC + PQ)

**Solution**

Given : In the right â–³ABC , right angled at B. P and Q are points on the sides AB and BC respectively .

Applying Pythagoras theorem,

In â–³AQB,

AQ

^{2}= AB^{2}+ BQ^{2}â€¦(1)
In â–³PBC

CP

^{2}= PB^{2}+ BC^{2}â€¦(2)
Adding (1) and (2), we get

AQ

^{2}+ CP^{2}= AB^{2}+ BQ^{2}+ PB^{2}+ BC^{2}
â€¦(3)

In â–³ABC ,

AC

^{2}= AB^{2}+ BC^{2}â€¦(4)
In â–³PBQ ,

PQ

^{2}= PB^{2}+ BQ^{2}â€¦(5)
From (3), (4) and (5) , we get

AQ

^{2}+ CP^{2}= AC^{2}+ PQ^{2}
We got the result as c .

48. If â–³ABC ~ â–³DEF such that DE = 3 cm, EF = 2 cm, DF = 2.5 cm, BC = 4 cm, then perimeter of â–³ABC is

(a) 18 cm

(b) 20 cm

(c) 12 cm

(d) 15 cm

(a) 18 cm

(b) 20 cm

(c) 12 cm

(d) 15 cm

**Solution**

Given : â–³ABC and â–³DEF are similar triangles such that DE = 3 cm, EF = 2 cm, DF = 2.5 cm and BC = 4 cm .

To find : Perimeter of â–³ABC .

We know that if two triangles are similar then their corresponding sides are proportional .

Hence , AB/DE = BC/EF = CA/FD

Substituting the values we get

AB/BC = DE/EF

AB/4 = 3/2

AB = 6 cm â€¦(1)

Similarly,

CA/BC = DF/EF

CA/4 = 2.5/2

CA = 5 cm â€¦(2)

Perimeter of â–³ABC = AB + BC + CA

= 6+4+5

= 15 cm

Hence the correct option is d .

49. In â–³ABC ~ â–³DEF such that AB = 9.1 cm and DE = 6.5 cm . If the perimeter of â–³DEF is 25 cm , then the perimeter of â–³ABC is

(a) 36 cm

(b) 30 cm

(c) 34 cm

(d) 35 cm

(b) 30 cm

(c) 34 cm

(d) 35 cm

**Solution**
Given â–³ABC is similar to â–³DEF such that AB = 9.1 cm , DE = 6.5 cm . Perimeter of â–³DEF is 25 cm .To find : Perimeter of â–³ABC .

We know that the ratio of corresponding sides of similar triangles is equal to the ratio of their perimeters .

AB/DE = BC/EF = AC/DE = P1/P2

AB/DE = P(â–³ABC )/P(â–³DEF)

9.1/6.5 = P(â–³ABC )/25

P(â–³ABC ) = 9.1Ã—25/6.5

P(â–³ABC ) = 35 cm

Hence the correct answer is d .

50. In an isosceles triangle ABC if AC = BC and AB2 = 2AC2, then âˆ C =

(a) 30Â°

(b) 45Â°

(c) 90Â°

(d) 60Â°

Given : In Isosceles â–³ABC , AC = BC and AB

AB/DE = BC/EF = AC/DE = P1/P2

AB/DE = P(â–³ABC )/P(â–³DEF)

9.1/6.5 = P(â–³ABC )/25

P(â–³ABC ) = 9.1Ã—25/6.5

P(â–³ABC ) = 35 cm

Hence the correct answer is d .

50. In an isosceles triangle ABC if AC = BC and AB2 = 2AC2, then âˆ C =

(a) 30Â°

(b) 45Â°

(c) 90Â°

(d) 60Â°

**Solution**Given : In Isosceles â–³ABC , AC = BC and AB

^{2}= 2AC^{2}.
To find : Measure of angle C

In isosceles â–³ABC ,

AC = BC

âˆ B = âˆ A (Equal sides have equal angles opposite to them )

AB

^{2}= 2AC^{2}
AB

^{2}= AC^{2}+ AC^{2}
AB

^{2}= AC^{2}+ BC^{2}(AC = BC)
â‡’ â–³ABC is right angle triangle , with âˆ C = 90Â°

Hence the correct answer is c.