Practice the AP 10th Class Maths Bits with Answers Chapter 8 Similar Triangles on a regular basis so that you can attempt exams with utmost confidence.

## AP State Syllabus 10th Class Maths Bits 8th Lesson Similar Triangles with Answers

Question 1.

Write maximum possible tangents that can drawn to a circle.

Answer:

Infinity

Question 2.

If ΔABC ~ ΔDEF and area (ΔABC): area (ΔDEF) = 49 :100. Then find DE : AB.

Answer:

10:7

Question 3.

Name the theorem applied to divide the line segment in the given ratio.

Answer:

Thales theorem.

Question 4.

Write an example for the side of a right angled triangle.

Answer:

5, 12, 13.

Question 5.

If ΔPQR ~ ΔXYZ and ∠X = 30°, ∠Q = 50°, then find ∠Z.

Answer:

∠Z = ∠R = 100°

Explanation:

∠X = ∠P = 30°, ∠Q = ∠Y = 50°, then ∠Z = ∠R = 180° – (30° + 50°) = 100°

Question 6.

The perimeters of two similar triangles are in 4 : 9 ratio, find the ratio of their’ corresponding sides.

Answer:

4:9.

Explanation:

Ratio of the perimeters of two similar triangles is equal to the ratio of their corresponding sides.

∴ 4 : 9.

Question 7.

ΔABC ~ ΔDEF and area of ΔABC, ΔDEF are 64 cm2 and 121 cm2, then find the ratio of corresponding sides.

Answer:

8:11

Explanation:

Ratio of corresponding sides

= \(\sqrt{64: 121}\) = 8 : 11

Question 8.

In ΔABC, AC = 12 cm, AB = 5 cm and ∠BAC = 30°, find the area of ΔABC,

Answer:

15 cm^{2}

Question 9.

From the given figure, find ‘x’.

Answer:

3

Explanation:

\(\frac{3}{x}=\frac{5}{5} \Rightarrow \frac{3}{x}=1 \Rightarrow x=3\)

Question 10.

If a man walks 6 m to east and 8 m to north. Now he is at a distance of from origin point.

Answer:

10m

Explanation:

Question 11.

If ΔPQR ~ ΔXYZ, QR = 3 cm, YZ = 4 cm, ΔPQR area = 54 cm2. Then find ΔXYZ areAnswer:

Answer:

96 cm^{2}

Question 12.

Write height of an equilateral triangle whose sides is ‘a’ cm.

Answer:

\(\frac{\sqrt{3}}{2} \mathrm{a}\)

Question 13.

Find are of a regular hexagon whose side is ‘a’ cm.

Answer:

\(6\left(\frac{\sqrt{3}}{4} \mathrm{a}^{2}\right)\)

Question 14.

In the figure, find ∠BDE.

Answer:

45°

Explanation:

From figure, ∠B = 60°,

∠E = ∠C = 75°, then ∠BDE = 180°- 135° = 45°

Question 15.

ΔABC ~ ΔPQR and ∠A + ∠B = 115°, then find ∠R.

Answer:

65°

Question 16.

In ΔABC, E and F are the points on the sides AB and AC respectively. If AE = 2 cm, EB = 2.5 cm, AF = 4 cm and FC = 5 cm, then write the relation.

Answer:

EF || BC

Explanation:

Question 17.

In the figure ΔABC, DE || BC, AD = 1.5 cm, DB = 6 cm, AE = x cm, EC = 8 cm, then find ‘x’.

Answer:

2 cm

Explanation:

Question 18.

Find ∠CAD in the given figure.

Answer:

50°

Question 19.

In a right angled triangle with inte¬gral sides at least one of its measure¬ments must be

Answer:

Multiple of 3 and multiple of 2.

Question 20.

ΔABC ~ ΔXYZ, ∠C = 60°, ∠B = 70°, then find ∠X.

Answer:

∠X = 50°

Question 21.

Express ‘x’ in terms of a, b and c in the following figure.

Answer:

x = \(\frac{a c}{b+c}\)

Question 22.

Write the altitude of an equilateral tri¬angle of side ‘x’ cm.

Answer:

\(\frac{\sqrt{3}}{2} \mathrm{x}\) cm

Question 23.

When we construct a triangle similar to a given triangle as per given scale factor, on which similarity we con-struct.

Answer:

Basic proportionality theorem.

Question 24.

In Heron’s formula, area of triangle =\(\sqrt{s(s-a)(s-b)(s-c)}\), write ‘s’ is of the triangle.

Answer:

Half of perimeter (or) s = \(\frac{a+b+c}{2}\)

Question 25.

In ΔABC, ∠C = 90°, BC = a, AB = c, AC = b and ‘p’ is length of height drawn from ‘C’ to AB, then write the correct relation between a, b, c, p.

Answer:

\(\frac{1}{\mathrm{p}^{2}}=\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}\)

Question 26.

ΔABC ~ ΔDEF and their areas are re-spectively 64 cm^{2} and 121 cm^{2}, then \(\frac{\mathbf{B C}}{\mathbf{E F}}\)

Answer:

\(\frac{8}{11}\)

Explanation:

\(\frac{\mathrm{BC}}{\mathrm{EF}}=\sqrt{\frac{\mathrm{ar} \Delta \mathrm{ABC}}{\mathrm{ar} \Delta \mathrm{DEF}}}=\sqrt{\frac{64}{121}}=\frac{8}{11}\)

Question 27.

From the given figure, find ar (ΔADE) : ar (ΔABC)

Answer:

9.25

Explanation:

\(\frac{{ar} \Delta \mathrm{ADE}}{{ar} \Delta \mathrm{ABC}}=\left(\frac{3}{5}\right)^{2}=\frac{9}{25}=9: 25\)

Question 28.

ΔABC ~ ΔDEF is given, then draw rough diagram.

Answer:

Question 29.

Areas of 2 similar triangles are 100cm^{2} and 64 cm^{2}. If the median of bigger

triangle is 10 cm, then find the me dian of the smaller triangle.

Answer:

8cm

Question 30.

From the given figure, find the value represented by x.

Answer:

x = 13 cm

Explanation:

From figure, AC = 5 cm

(by Pythagoras triplet)

In ∆ABC, AB = x = 13 cm

(by Pythagoras triplet)

Question 31.

In the figure, D, E are mid-points of AB and AC, then find ΔADE: 10 BCED.

Answer:

1:3

Question 32.

In a right triangle, write the relation between sides.

Answer:

Hypotenuse^{2} = (Adj. side)^{2} + (Opp. side)^{2}

Question 33.

In an isosceles ΔPQR, PR = QR and PQ^{2} = 2PR^{2}, then find ∠R.

Answer:

90°

Question 34.

In ΔPQR; PQ = 6\(\sqrt{3}\) cm, PR = 12 cm and QR = 6 cm, then find ∠Q.

Answer:

90°

Explanation:

Phthagoras theorem is satisfied.

∴ ∠Q = 90°.

Question 35.

If ΔABC ~ ΔXYZ; ∠C = 60°, ∠B = 75°, then find ∠Z.

Answer:

60°

Question 36.

If ΔABC ~ ΔDEF, BC = 4 cm, EF = 5 cm and ΔABC = 80 cm^{2}, then find ΔDEF.

Answer:

125 cm^{2}

Question 37.

The sides PQ and PR of right triangle PQR are such that PQ = 5 cm, PR = 13 cm. If ∠Q = 90°, then find QR.

Answer:

12 cm

Question 38.

From the figure, find ∠DAC.

Answer:

35°

Question 39.

If in two triangles, corresponding sides are in the same ratio, then the two triangles are similar, which crite¬rion it relates ?

Answer:

S.S.S. criterion.

Question 40.

In the figure DE // BC and AD : DB =1:2, then find ΔADE: ΔABC.

Answer:

1 : 9

Explanation:

AD : DB = 1 : 2 ⇒ AD : AB = 1 : 3

AE : EC = 1 : 2 ⇒ AE : AC = 1 : 3

ar ∆ADE : ar ∆ABC = 1^{2} : 3^{2} = 1 : 9

Question 41.

ΔABC and ΔBDE are two equilateral triangles such that ‘D’ is the midpoint of BC. Find the ratio of the areas of triangles AABC and ABDE.

Answer:

4:1.

Question 42.

From the figure, find ‘x’.

Answer:

x = 15 cm

Question 43.

In the following figure, DE // BC, then find x’.

Answer:

± \(\sqrt{7}\)

Explanation:

\(\frac{A D}{D B}=\frac{A E}{E C}\)

\(\frac{x+4}{x+3}=\frac{2 x-1}{x+1}\)

(x + 4) (x + 1) = (2x – 1) (x + 3)

x^{2} + x + 4x + 4 = 2x^{2} + 6x – x – 3

x^{2} + 5x + 4 = 2x^{2} + 5x – 3

=> x^{2} – 2x^{2} + 4 + 3 = 0

x^{2} – 7 = 0 x = \(\pm \sqrt{7}\)

Question 44.

The ratio of the areas of two similar triangles is 1 : 4, then write the ratio of their corresponding sides.

Answer:

1 : 2

Explanation:

Ratio of corresponding sides

= \(\sqrt{1}: \sqrt{4}\) =1:2

Question 45.

The perimeter of ΔABC ~ ΔLMN are 60 cm and 48 cm of LM = 8 cm, then find AB.

Answer:

10 cm

Explanation:

Question 46.

ΔABC ~ ΔPQR, ∠A = 32°, ∠R = 65°, then find ∠B.

Answer:

83°

Question 47.

All circles are …………….

Answer:

Similar

Question 48.

All squares are …………..

Answer:

Similar

Question 49.

All……………triangles are similar.

Answer:

Equilateral.

Question 50.

Write the longest side in a right tri-angle.

Answer:

Hypotenuse.

Question 51.

In the figure, AB = 2.5 cm, A AC = 3.5 cm. If AD is the bisector of ∠BAC, then find BD : DC. 8

Answer:

5 : 7

Explanation:

According to the angle bisector theorem

\(\frac{B D}{D C}=\frac{A B}{A C} \Rightarrow \frac{B D}{D C}=\frac{25}{35}=\frac{5}{7}\)

∴ BD : DC = 5 : 7

Question 52.

The ratio of corresponding sides of two similar triangles is 3 : 2, then find the ratio of their corresponding heights.

Answer:

3 : 2

Question 53.

The diagonals of a rhombus are 24 cm and 32 cm, then find its perimeter.

Answer:

80 cm

Question 54.

In ΔABC, BC^{2} + AB^{2} = AC^{2}, then where is the right angle ?

Answer:

∠B

Question 55.

Write each angle in an equilateral tri-angle.

Answer:

60°

Question 56.

ΔABC ~ ΔDEF if DE : AB = 2 : 3 and ar ΔDEF = 44 sq. units, then find ar (ΔABC).

Answer:

99

Explanation:

Question 57.

In an equilateral triangle ABC, AD⊥BC meeting BC in D, then find AD^{2}.

Answer:

3 BD^{2}

Question 58.

ΔABC ~ ΔDEF and 2AB = DE and BC = 8 cm, then find EF = ……………..cm.

Answer:

16

Question 59.

From the figure, find AD.

Answer:

2.4 cm

Question 60.

Side of rhombus is 4 cm, then find its perimeter.

Answer:

16 cm

Question 61.

Write any one example for the sides of, a right triangle.

Answer:

10 cm, 8 cm, 6 cm.

Question 62.

In the figure, ΔABC, DE//BC and \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{3}{5}\) AC = 5.6, then find AE.

Answer:

2.1 cm

Question 63.

If 82 + 152 = k2, then find k.

Answer:

k = 17.

Question 64.

In the figure PQR, ∠QPR = 90°, PQ = 24 cm and QR = 26 cm and in ΔPKR, ∠PKR = 90° and KR = 8 cm, then find PK.

Answer:

PK 6 cm.

Explanation:

By Pythagoras theorem,

In APQR,

QR^{2} = QP^{2} + PR^{2}

⇒ PR = 10 cm

In ∆PKR, PK = 6 cm.

Question 65.

In the figure, DE // BC if AD = x, AE = x + 2, DB = x – 2 and CE = x – 1, then find ‘x’.

Answer:

4

Question 66.

The ratio of areas of two similar triangles is equal to the ratio of the squares of corresponding ……….

Answer:

Sides

Question 67.

If the diagonal of a square is \(7 \sqrt{2}\) cm, then find its areAnswer:

Answer:

49 cm^{2}

Question 68.

In the figure, AC = ……………..cm.

Answer:

12 cm

Question 69.

Write the height of an equilateral tri-angle whose side is ‘a’ units.

Answer:

\(\frac{\sqrt{3}}{2} \mathrm{a}\)

Question 70.

The ratio of the corresponding sides of two similar triangles is 5 : 3. Then find the ratio of their areas.

Answer:

25:9

Question 71.

The areas of two similar triangles are 36 cm2 and 64 cm^{2}. If one side of the first triangle is 6 cm, then find the cor-responding side of the later triangle.

Answer:

8 cm

Explanation:

Ratio of area of two Similar triangles is equal to the ratio of their corre¬sponding sides.

\(\frac{6}{x}=\sqrt{\frac{36}{64}} \Rightarrow \frac{6}{x}=\frac{6}{8}\)

Question 72.

The lengths of diagonals of a rhombus are 24 cm and 32 cm, then find the pe-rimeter of the rhombus.

Answer:

80 cm

Question 73.

In the given figure, DE // BC and AD : DB = 5 : 4, then find \(\frac{\Delta \text { DEF }}{\Delta \text { CFB }}\)

Answer:

\(\frac{25}{81}\)

Question 74.

Write each exterior angle of an equi-lateral triangle.

Answer:

120°

Question 75.

ΔABC ~ ΔPQR, ∠A = 50°, then find ∠Q + ∠R.

Answer:

130°

Question 76.

ΔABC ~ ΔPQR; M is the midpoint of BC and N is the midpoint of QR. If ΔABC = 100 cm2 and ΔPQR = 144 cm^{2} and AM = 4 cm, then find PN.

Answer:

PN 4.8 cm

Question 77.

In the figure, PQ // MN, \(\frac{\mathbf{K P}}{\mathbf{P M}}=\frac{4}{13}\) and

KN = 20.4cm, them find KQ.

Answer:

4.8 cm.

Explanation:

PQ // MN, by Thales theorem KP

⇒ 4(20.4-KQ) = 13 KQ

⇒ 81.6-4 KQ – 13 KQ .

⇒ 13 KQ + 4 KQ = 81.6

⇒ KQ = 4.8 cm.

Question 78.

The diagonal of a trapezium ABCD in which AB // CD intersect at ‘O’. If AB = 2CD, then find the ratio of ar¬eas of triangles AOB and COD.

Answer:

4:1

Question 79.

In a trapezium how diagonals divide each other ?

Answer:

Proportionally to each other.

Question 80.

In the figure, \(\frac{\mathbf{P S}}{\mathrm{SQ}}=\frac{\mathbf{P T}}{\mathbf{T R}}\)

∠PST = ∠PRQ, then find ∆PQR is which type of triangle ?

Answer:

Isosceles

Question 81.

From the figure, y = …………….. cm.

y = 15 cm

Question 82.

∆ABC is an isosceles right triangle, ∠C= 90°, then find AB^{2}.

Answer:

AC^{2} + BC^{2}

Question 83.

In a square, the diagonal is ……………… time of its side.

Answer:

\(\sqrt{2}\)

Question 84.

In the figure, CD =……………….cm.

Answer:

\(6 \sqrt{3}\) cm.

Question 85.

In the figure, AABC is an isosceles tri¬angle right angled at B. Two equilat¬eral triangles are constructed with sides AC and BC. Then find ∆BCD.

I’m 67

Answer:

∆ABC

Question 86.

The areas of two similar triangles are 25 m2 and 36 m2. If the median of smaller triangle is 10 m, then find the median of the larger triangle.

Answer:

12 m

Explanation:

(The ratio of sides)^{2} = (ratio of medians)^{2}

= Ratio of areas

∴ Ratio of medians = \(\sqrt{\text { Ratio of areas }}\)

\(\frac{x}{10}=\sqrt{\frac{25}{36}} \Rightarrow \frac{x \cdot}{10}=\frac{5}{6}\)

⇒ x = 12 m

Question 87.

∆ABC ~ ∆PQR, then find AB : PQ.

Answer:

AC : PR

Question 88.

In ∆LMN, ∠L = 60°, ∠M = 50° and ∆LMN ~ ∆PQR, then find ∠R.

Answer:

∠R = 70°

Question 89.

In the figure, ∠BAD = ∠CAD; AB = 3.4 cm, BD = 4 cm, BC = 10 cm, then find AC.

Answer:

5.1 cm

Question 90.

Two sides of a right triangle are 3 cm and 4 cm, then find the third side.

Answer:

5 cm.

Question 91.

∆ABC ~ ∆DEF, BC = 4 cm, EF = 5 cm and area of ∆ABC = 80 cm2, then find area of ∆DEF.

Answer:

125 cm^{2}

Question 92.

In the figure, find x.

Answer:

x = 8 cm

Question 93.

In the figure, if AB // CD, then find x.

Answer:

7 cm

Question 94.

When two polygons are similar ? Write similarity conditions.

Answer:

If corresponding angles are equal & corresponding sides are equal.

Question 95.

The side of an equilateral triangle is ‘a’ units, then write its height in terms of side.

Answer:

\(\frac{\sqrt{3} a}{2}\)

Question 96.

The angles of a triangle are in the ra¬tio 1 :2 : 3, then find the largest angle.

Answer:

90°

Question 97.

∆ABC ~ ∆DEF and their areas are re-spectively 64 cm^{2} and 121 cm2 if EF = 15.4 cm, then find BC.

Answer:

11.2 cm

Question 98.

In the figure, DE // AB and FE // DB, then find DC^{2}.

Answer:

CF x AC

Question 99.

If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are pro-portional, the two triangles are simi-lar. This property is

Answer:

S.Answer:S. criterion.

Question 100.

In ∆ABC, the midpoints are D, E and F of the sides, AB, BC and CA, then find ∆DEF : ∆ABC.

Answer:

1:4

Question 101.

In ∆ABC, XY || BC, AX : XB = 2 : 1, then find ∆AXY : ∆ABC.

Answer:

4 : 9

Question 102.

In the figure, ∠ABC.

Answer:

60°.

Question 103.

In the figure, in ∆PQR, QR // ST, \(\frac{\mathbf{P S}}{\mathbf{S Q}}=\frac{3}{5}\) and PR = 28 cm, then find PT.

Answer:

10.5 cm

Question 104.

∆ABC ~ ∆PQR, AB : PQ = 3 : 4, then find ar (∆ABC) far (∆PQR).

Answer:

9 : 16

Question 105.

The bisector of ∠A of ∆ABC intersects BC at D. If BD : DC = 4 : 7 and AC = 3.5. Then find AB.

Answer:

2 cm

Explanation:

Question 106.

A perpendicular is drawn from the vertex of a right angle to the hypot-enuse, then the triangles on each side of the perpendicular are

Answer:

Similar.

Question 107.

In the figure, D, E are the midpoints of the sides AB and AC. If DE = 4 cm, then find BC.

Answer:

BC = 8 cm.

Explanation:

By B.P.T. BC = 2 x DE

⇒ BC = 2 x 4 = 8 cm.

Question 108.

In ∆ABC, AD bisects ∠Answer: AB = 6 cm, BD = 8 cm and DC = 6 cm, then find AC.

Answer:

4.5 cm

Question 109.

If ∆ABC ~ ∆PQR, then find y + z.

Answer:

4 + 3\(\sqrt{3}\)

Question 110.

In the figure, find ‘x’.

Answer:

∠x = 135°.

Question 111.

In ∆ABC, AB = BC = AC, then find ∠A = ∠B = ∠C.

Answer:

60°

Question 112.

If the diagonals in a quadrilateral di-vide each other proportionally then it is which type of polygon ?

Answer:

Trapezium.

Question 113.

∆ABC ~ ∆PQR, ∠A + ∠B = 100°, then find ∠R.

Answer:

80°

Question 114.

If the sides of two similar triangles are in the ratio 7 : 2, then find the ratio of their areas.

Answer:

49 : 4

Question 115.

In the figure, QA⊥AB and PB ⊥ AB, if AO = 20 cm, BO = 12 cm, PB = 18 cm, then find AQ.

Answer:

AQ = 30 cm

Question 116.

D, E and F are the mid-points of the sides BC, CA and AB respectively of ∆ABC, then find the ratio of the areas of ∆DEFand ∆ABC.

Answer:

1 : 4

Question 117.

In the figure, DE divides AB and AC in the ratio 1 : 3. If DE = 2.4 cm, then find BC.

Answer:

7.2 cm

Question 118.

In the figure, APQR and ASQR are two triangles on the same base QR. If PS intersects QR at ‘O’, then find ∆PQR : ∆SQR.

Answer:

PO : SO.

Question 119.

In the figure, LM // CB and LN // CD, then find \(\frac{\mathbf{A M}}{\mathbf{A B}}\)

Answer:

\(\frac{AN}{AD}\)

Question 120.

In the figure, two triangles are siml Iar, then find ‘x’.

Answer:

x = 7.5 cm

Question 121.

In the figure, ∠A = ∠B and AD = BE, then write the relation.

Answer:

DE//AB

Choose the correct answer satisfying the following statements.

Question 122.

Statement (A) : ∆ABC ~ ∆DEF such that ar(∆ABC) = 36 cm^{2} and ar (∆DEF) = 49 cm^{2}, then AB : DE = 6 : 7.

Statement (B): If ∆ABC ~ ∆DEF, then \(\frac{{ar}(\Delta \mathrm{ABC})}{{ar}(\Delta \mathrm{DEF})}=\frac{\mathrm{AB}^{2}}{\mathrm{DE}^{2}}=\frac{\mathrm{BC}^{2}}{\mathrm{EF}^{2}}=\frac{\mathrm{AC}^{2}}{\mathrm{DF}^{2}}\)

i) Both A and B are true.

ii) A is true, B is false.

iii) A is false, B is true.

iv) Both A and B are false.

Answer:

(i)

Explanation:

So, both A and B are correct and B explains Answer:

Hence, (i) is the correct option.

Question 123.

Statement (A) : In AABC, DE // BC such that AD — (7x – 4) cm,

AE = (5x – 2) cm, DB = (3x + 4) cm and EC = 3x cm, then x equal to 5.

Statement (B): If a line is drawn par¬allel to one side of a triangle to inter¬sect the other two sides in distant point, then the other two sides are divided in the same ratio.

i) Both A and B are true.

ii) A is true, B is false.

iii) A is false, B is true.

iv) Both A and B are false.

Answer:

(iii)

Explanation:

We have, \(\frac{A D}{D B}=\frac{A E}{E C} \Rightarrow \frac{7 x-4}{3 x+4}=\frac{5 x-2}{3 x}\)

⇒ 21x^{2} – 12x = 15x^{2} + 20x – 6x – 8

⇒ 6x^{2} – 26x + 8 = 0

⇒ 3x^{2} – 13x + 4 = 0

⇒ 3x^{2} – 12x – x + 4 = 0

⇒ 3x (x – 4) – 1 (x – 4) = 0

⇒ (x-4) (3x – 1) = 0

x = 4, \(\frac { 1 }{ 3 }\)

So, A is incorrect but B is correct. Hence, (iii) is the correct option.

Question 124.

Statement (A) : ∆ABC is an isosceles triangle right angled of C, then

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

Statement (B) : In right∆ABC, right angled at B, AC^{2} = AB^{2} + BC^{2}.

i) Both A and B are true.

ii) A is true, B is false.

iii) A is false, B is true.

iv) Both A and B are false.

Answer:

(i)

Explanation:

In an isosceles AABC, right angled at C is AB^{2} = AC^{2} + BC^{2}

⇒ AB^{2} = AC^{2} + AC^{2}

⇒ AB^{2} = 2AC^{2}

So, both A and B are correct and B explains Answer:

Hence, (i) is the correct option.

Question 125.

Statement (A): In the ∆ABC, AB = 24 cm, BC = 10 cm and AC = 26 cm, then ∆ABC is a right angle triangle.

Statement (B) : If in two triangles, their corresponding angles are equal, then the triangles are similar.

i) Both A and B are true.

ii) A is true, B is false.

iii) A is false, B is true.

iv) Both A and B are false.

Answer:

(i)

Explanation:

We have, AB^{2} + BC^{2} = (24)^{2} + (10)^{2}

= 576 + 100 = 676 = AC^{2}

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

∴ ABC is a right angled triangle. Also, two triangles are similar, if their corresponding angles are equal.

So, both A and B are correct.

Hence, (i) is the correct option.

Question 126.

Statement (A) : Two similar triangles are always congruent.

Statement (B): If the areas of two simi-lar triangles are equal, then the tri¬angles are congruent.

i) Both A and B are true.

ii) A is true, B is false.

iii) A is false, B is true.

iv) Both A and B are false.

Answer:

(iii)

Explanation:

Two similar triangles are not congru¬ent generally.

So, A is incorrect, but B is correct. Hence, (iii) is the correct option.

Question 127.

Statement (A) : If in a AABC, a line DE // BC, intersects, AB in D and AC in E, then \(\frac{AD}{DB}=\frac{AE}{EC}\)

Statement (B): If a line is drawn paral¬lel to one side of a triangle intersecting the other two sides, then the other two sides ore divided in the same ratio.

i) Both A and B are true.

ii) A is true, B is false.

iii) A is false, B is true.

iv) Both A and B are false.

Answer:

(i)

Explanation:

Statement (B) is true [This is Thale’s Theorem]

For statement (A),

Since DE // BC

∴ By Thales’ Theorem

\(\frac{AD}{DB}=\frac{AE}{EC} \Rightarrow \frac{DB}{AD}=\frac{EC}{AE}\)

∴ Statement (A) is true.

Since statement (B) gives statement (A).

∴ Option (i) is correct.

Question 128.

Statement (A) : ABC is an isosceles, right triangle, right angled at C. Then AB^{2} = 3 AC^{2}.

Statement (B): In an isosceles triangle ABC if AC = BC and AB^{2} = 2AC^{2}, then ∠C .= 90°.

i) Both A and B are true.

ii) A is true, B is false.

iii) A is false, B is true.

iv) Both A and B are false.

Answer:

(iii)

Explanation:

In right angled ΔABC,

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

(By Pythagoras Theorem)

AB^{2} = AC^{2} + AC^{2} (∵ BC = AC)

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

Statement (A) is false.

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

= AC^{2} + BC^{2} (∵ AC = BC given)

∠C = 90° (By converse of Pythagoras

Theorem)

∴ Statement (B) is true.

Hence, (iii) is the correct option.

Question 129.

Statement (A): ABC and DEF are two similar-triangles such that BC = 4 cm, EF = 5 cm and area of AABC 64 cm2, then area of ∆DEF =100 cm^{2}. Statement (B): The areas of two simi¬lar triangles are in the ratio of the squares of the corresponding altitudes.

i) Both A and B are true.

ii) A is true, B is false.

iii) A is false, B is true.

iv) Both A and B are false.

Answer:

(i)

Explanation:

Statement (B) is true.

[∵ It is Standard result]

For statement (A),

since ΔABC ~ ΔDEF

( ∵ Ratio of areas of two similar tri¬angles is equal to the ratio of squares of corresponding sides)

∴ Statement (A) is true. But statement (B) is not the correct explanation for statement (A).

∴ Option (i) is correct.

In figure, AD is a median of a triangle ABC and AM ⊥ BC.

Question 130.

AD^{2} + BC . DM = \(\left(\frac{B C}{2}\right)^{2}\) is equal to

Answer:

AC^{2}

Explanation:

AD is the median, so D is the mid point of BC.

So, BD = DC = \(\frac { 1 }{ 2 }\)BC …………. (1)

In right angled AAMC,

AC^{2} = AM^{2} + MC^{2} ………..(2)

In right angled AAMD,

AM^{2} = AD^{2} – MD^{2} ……………….(3)

Putting AM^{2} from (3) in (2),

We get AC^{2} = AD^{2} – MD^{2} + MC^{2}

= AD^{2} – MD^{2} + (MD + DC)^{2}

= AD^{2} + 2DM + \(\frac{BC}{2}+\left(\frac{BC}{2}\right)^{2}\)

So, AC^{2} = AD^{2} + BC . DM + \(\left(\frac{BC}{2}\right)^{2}\)

Question 131.

AD^{2} – BC . DM = \(\left(\frac{B C}{2}\right)^{2}\) is equal to

Answer:

AB^{2}

Explanation:

In right angled AABM,

AB^{2} = AM^{2} + BM^{2}

From AAMD, AM^{2} = AD^{2} – MD^{2}

So, AB^{2} = AD^{2} – MD^{2} + BM^{2}

= AD^{2} – MD^{2} + (BD – MD)^{2}

= AD^{2} – MD^{2} + BD^{2} – 2BD.MD+MD^{2}

=» AB^{2} = AD^{2} – BC . DM + \(\left(\frac{BC}{2}\right)^{2}\) proved.

Question 132.

2AD^{2} + \(\frac { 1 }{ 2 }\) BC^{2} is equal to

Answer:

AC^{2} + AB^{2}

Explanation:

From the solution of above two ques-tions

AC^{2} = AD^{2} + BC . DM + \(\left(\frac{BC}{2}\right)^{2}\) …. (i)

and AB^{2} – AD^{2} – BC . DM + \(\frac { 1 }{ 2 }\) BC^{2} …. (ii)

Adding results of (i) and (ii) we get

AC^{2} + AB^{2} = AD^{2} + BC . DM + \(\left(\frac{BC}{2}\right)^{2}\) + AD^{2} – BC . DM + \(\left(\frac{BC}{2}\right)^{2}\)

AC^{2} + AB^{2} = 2 AD^{2} + \(\frac { 1 }{ 2 }\) (BC)^{2}

Read the below passages and an-swer to the following questions.

Raju said that the triangle with sides 25 cm, 5 cm and 24 cm a right tri¬angle.

Question 133.

Are you support with Raju ? (or) not ?

Answer:

No, I didn’t.

Question 134.

Why you support / does not support with him ? Give reason.

Answer:

Given measurements are not support¬ing the Pythagoras theorem.

Question 135.

Which mathematical concept was used to check your answer from textbook ?

Answer:

Similar Triangles.

If ∆ABC ~ ∆DEF, AB = 4 cm, DE = 6 cm, EF = 9 cm and FD = 12 cm.

Question 136.

Find AC.

Answer:

8 cm

Question 137.

Find perimeter of ∆ABC.

Answer:

Perimeter of ∆ABC =18 cm.

∆ABC ~ ∆EDF such that AB = 5 cm, AC = 7 cm, DF = 15 cm and DE = 12 cm.

Question 138.

Which property was used to solve the given problem ?

Answer:

Similar triangles.

Question 139.

Find “EF”.

Answer:

EF = 16.8 cm.

Question 140.

Find “BC”.

Answer:

BC = 6.25 cm

Write the correct matching options.

Question 141.

If in ∆ABC, DE // BC and intersects AB in D and AC in E, then

Answer:

A – (ii), B – (i).

Question 142.

If in ∆ABC, DE // BC and intersects AB in D and AC in E, then

Answer:

A – (iv), B – (iii).

Question 143.

Match the correct option.

Answer:

A-(ii), B – (i)

Question 144.

Match the correct option

Answer:

A – (iii), B – (iv).

Question 145.

Write correct matching

Answer:

A – (ii), B – (i).

Question 146.

Write correct matching.

Answer: A – (iii), B – (iv).

Question 147.

In figure, the line segment XY // AC of side ∆ABC and it divides the triangle into two parts of equal areas, then

Answer:

A – (i), B – (ii).

Question 148.

In figure, the line segment XY // AC of side AABC and it divides the triangle into two parts of equal areas, then

Answer:

A – (iii), B – (iv).

Question 149.

The perimeters of two similar triangles are 24 cm and 18 cm respectively. If one side of the first triangle is 8 cm then what is the corresponding side of second triangle ?

Answer:

6.

Question 150.

In ∆ABC, ∠B = 90°, BD ⊥ AC.

If AD = 8 cm and BD = 4 cm, then what is the length of CD ?

Answer:

2 cm.

Question 151.

Among “Circles”, “Squares” and “Tri¬angles”, which are always similar ?

Answer:

Circles and squares are always similar.

Question 152.

In AABC and ADEF,

if ∠B = ∠E, ∠C = ∠F, then which of the following is a true statement ?

Answer:

(c)

Question 153.

Given DE || BC, AD = 4.5 cm, BD = 9 cm, and EC = 8 cm. What is the length of AE?

Sol.

AD = 4.5 cm, BD = 9 cm, EC = 8 cm

\(\frac{AD}{DB}=\frac{AE}{EC} \Rightarrow \frac{4.5}{9}=\frac{AE}{8}\)

⇒ 9 AE = 36

∴ AE = 4 cm

Question 154.

The sides of two equilateral triangles ABC and DEF are 4 cm. and 5 cm. respectively. Find \(\frac{{ar}(\Delta D E F)}{{ar}(\Delta A B C)}\)

Solution:

Ratio of areas of Δ = (Ratio of corresponding sides)^{2}

\(\frac{(\Delta {DEF})}{(\Delta \mathrm{ABC})}=\frac{5^{2}}{4^{2}}=\frac{25}{16}\)

Question 155.

ΔABC ~ ΔDEF and ∠A = ∠D = 90°, find the measure of ∠B + ∠F.

Solution:

ΔABC ~ ΔDEF

⇒ ∠A = ∠D = 90°

∠B = ∠E;∠C = ∠F

Now in ΔABC, ∠A + ∠B + ∠C = 180°

Now putting ∠A = ∠D = 90°

⇒ 90° + ∠B + ∠C = 180°

⇒ ∠B + ∠C = 90°

Now putting ∠C = ∠F ⇒ ∠B + ∠F = 90°

Question 156.

Given AB || CD, identify the pair of similar triangles in this image and justify.

Solution:

∠AEB = ∠CED (Vertically opp. angle)

∴ ∠BEA and ∠CDE are similar by

AAA – Similarity rule

ΔBEA ~ ΔCDE