These TS 10th Class Maths Chapter Wise Important Questions Chapter 8 Similar Triangles given here will help you to solve different types of questions.

## TS 10th Class Maths Important Questions Chapter 8 Similar Triangles

Previous Exams Questions

Question 1.

Construct a triangle of sides 4.2 cm, 5.1 cm and 6 cm. Then construct a triangle similar to it, whose sides are \(\frac{2}{3}\) of corresponding sides of the first triangle. (A.P. Mar. ’15)

Solution:

- Draw a triangle ABC, with sides AB = 4.2 cm, BC = 5.1 cm, CA = 6 cm.
- Draw a ray BX making an acute angle with BC on the side opposite to vertex A.
- Locate 3 points B
_{1}, B_{2}, B_{3}on BX so that BB_{1}= B_{1}B_{2}= B_{2}B_{3} - Join B
_{3}, C and draw a line through B_{2}parallel to B_{3}C intersecting BC and C^{1}. - Draw a line through C
^{1}parallel to CA intersect AB at A^{1}. - ∆A
^{1}BC^{1}is required triangle.

Question 2.

Is a square similar to a rectangle ? Justify your answer. (A.P. Mar.’15)

Solution:

In a square and rectangle, the corresponding angles are equal. But the corresponding sides are not proportional.

∴ A square and a rectangle are not similar.

Question 3.

In a ∆DEF, A, B and C are mid points of EF, FD and DE respectively. If the area of ∆DEF is 14.4 cm^{2} then find the area of ∆ABC. (T.S. Mar. ’15)

Solution:

Area of ∆ABC = 1/4 of area of ∆DEF

= 1/4 (14.4) = 3.6 cm^{2}

Question 4.

Observe the below diagram and find the values of x and y. (T.S. Mar. ’15)

Solution:

∆ABC ~ AEFC

\(\frac{\mathrm{x}}{6}\) = \(\frac{9}{15}\)

x = \(\frac{9 \times 6}{15}\) = \(\frac{18}{5}\) = 3.6 cm.

∆BDC ~ ABEF

\(\frac{\mathrm{x}}{9.6}\) = \(\frac{\mathrm{y}}{15}\)

⇒ y = \(\frac{3.6 \times 15}{9.6}\) = \(\frac{45}{8}\) = 5.625 cm

Question 5.

Observe the figure given below in ∆FQR if XY // PQ, \(\frac{\mathrm{PX}}{\mathrm{XR}}\) = \(\frac{5}{3}\) and QR = 7.2. Then find the length of RY. (T.S. Mar. ’15)

Solution:

XY || PQ

⇒ \(\frac{\mathrm{PX}}{\mathrm{XR}}\) = \(\frac{\mathrm{QY}}{\mathrm{YR}}\)

⇒ \(\frac{5}{3}\) + 1 = \(\frac{\mathrm{QY}}{\mathrm{RY}}\) + 1

⇒ \(\frac{8}{3}\) = \(\frac{\mathrm{QR}}{\mathrm{RY}}\)

⇒ RY = 7.2 × \(\frac{3}{8}\) = 2.7 cm.

Question 6.

Find the value of ‘x’ in the given figure where ∆ ABC ~ ∆ ADE. (T.S. Mar. ’15)

Solution:

∆ABC ~ ∆ADE

∴ \(\frac{\mathrm{AB}}{\mathrm{AD}}\) = \(\frac{\mathrm{BC}}{\mathrm{DE}}\) = \(\frac{\mathrm{AC}}{\mathrm{AE}}\) ;

BC = x, DE = 5, AE = 3, AC = 9

By substituting \(\frac{\mathrm{BC}}{\mathrm{DE}}\) = \(\frac{\mathrm{AC}}{\mathrm{AE}}\)

∴ \(\frac{\mathrm{x}}{5}\) = \(\frac{9}{3}\) ⇒ x = \(\frac{9 \times 5}{3}\) = 15

∴ x = 15 cm

Question 7.

Construct a triangle of sides 5 cm, 6 cm, and 7 cm then construct a triangle similar to it, whose sides are \(\frac{2}{3}\) of the corresponding sides of the triangle. (T.S. Mar. ’15)

Solution:

Construction steps :

- Draw a triangle ∆ABC with sides AB = 5 cm, BC = 6 cm and CA = 7 cm.
- Draw a ray \(\overrightarrow{\mathrm{BX}}\) making an acute angle with BC on the side opposite to vertex A.
- Locate 3 points B
_{1}, B_{2}, B_{3}on BX. So that BB_{1}= B_{1}B_{2}= B_{2}B_{3}. - Join B
_{3}, C and draw a line through B_{2}parallel to B_{3}C intersecting BC at C’. - Draw a line through C’ parallel to CA intersect AB at A’.
- ∆A’B’C’ is required triangle.

Additional Questions

Question 1.

In a ∆ ABC, DE || BC and AD = \(\frac{1}{3}\) BD. If BC = 5.6 cm, find DE.

Solution:

Given DE || BC, AD = \(\frac{1}{3}\) BD, BC = 5.6 cm

AD = \(\frac{1}{3}\) BD ⇒ BD = 3 AD …………… (1)

From similar triangles, ABC and ADE

We have \(\frac{\mathrm{AD}}{\mathrm{AB}}\) = \(\frac{\mathrm{DE}}{\mathrm{BC}}\)

⇒ \(\frac{\mathrm{AD}}{\mathrm{AD}+\mathrm{DB}}\) = \(\frac{\mathrm{DE}}{5.6}\)

⇒ \(\frac{\mathrm{AD}}{\mathrm{AD}+3\mathrm{DB}}=\) = \(\frac{\mathrm{DE}}{5.6}\)

[∵ from (1) DB = BD = 3AD]

⇒ \(\frac{\mathrm{AD}}{4 \mathrm{AD}}\) = \(\frac{\mathrm{DE}}{5.6}\)

⇒ \(\frac{1}{4}\) = \(\frac{\mathrm{DE}}{5.6}\)

⇒ DE = \(\frac{5.6}}{4\)

∴ DE = 1.4 cm

Question 2.

In the adjacent figure ∆ABC ~ ∆AHK. If AK = 8 cm, BC = 4.5 cm and HK = 9 cm find AC.

Solution:

Given AK = 8 cm, BC 4.5 cm, HK from similar ∆les, ABC and AHK

We have \(\frac{\mathrm{AC}}{\mathrm{AK}}\) = \(\frac{\mathrm{BC}}{\mathrm{HK}}\)

AC = \(\frac{\mathrm{AK} \times \mathrm{BC}}{\mathrm{HK}}\)

= \(\frac{8 \times 4.5}{9}\)

= 8 × 0.5 = 4

∴ AC = 4 cm

Question 3.

In the below given figure P and Q are points on the sides AB and AC respectively of ∆ABC such that AQ = 3 cm, QC = 5 cm and PQ || BC. Find the ratio of areas of ∆APQ and ∆ABC.

Solution:

Given AQ = 3 cm, QC = 5 cm and PQ || BC.

We know that by theorem,

Ratio of the areas of two similar triangles = Ratio of the squares of their corresponding sides

Now AC = AQ + QC = 3 + 5 = 8 cm

By Theorem \(\frac{\text { area of } \triangle \mathrm{APQ}}{\text { area of } \triangle \mathrm{ABC}}\) = \(\frac{(\mathrm{AQ})^2}{(\mathrm{AC})^2}\)

= \(\frac{3^2}{8^2}\)

= \(\frac{9}{64}\)

∴ Ratio of areas of ∆APQ and ∆ABC = 9 : 64

Question 4.

What value of (5) of x will make DE || AB, in the given figure ?

AD = 5x + 5, CD = x + 2 BE = 3x + 3, CE = x

Solution:

Given in ∆ ABC, DE || AB

AD = 5x + 5, CD = x + 2

BE = 3x + 3, CE = x

y Basic proportional theorem,

If DE || AB, then we have

\(\frac{\mathrm{CD}}{\mathrm{DA}}\) = \(\frac{\mathrm{CE}}{\mathrm{EB}}\)

⇒ \(\frac{x+2}{5 x+5}\) = \(\frac{x}{3 x+3}\)

⇒ (5x + 5) x = (x +2) (3x + 3)

⇒ 5x^{2} + 5x = 3x^{2} + 3x + 6x + 6

⇒ 5x^{2} – 3x^{2} + 5x – 3x – 6x – 6 = 0

⇒ 2x^{2} – 4x – 6 = 0

⇒ 2x^{2} – 6x + 2x – 6 = 0

⇒ 2x (x – 3) + 2(x – 3) = 0

⇒ (x – 3) + (2x + 2) = 0

⇒ x-3 = 0 or 2x + 2 = 0

⇒ x = 3 or 2x = -2

⇒ x = –\(\frac{2}{2}\) = -1

∴ The value of x = 3 will make DE || AB.

Question 5.

∆ABC ~ ∆PQR and their areas are respectively 81 cm^{2} and 144 cm^{2}. If QR = 16cm then find BC.

Solution:

\(\frac{\text { area of } \triangle \mathrm{ABC}}{\text { area of } \triangle \mathrm{DEF}}\) = \(\left(\frac{\mathrm{BC}}{\mathrm{QR}}\right)^2\)

⇒ \(\frac{81}{144}\) = \(\left(\frac{\mathrm{BC}}{16}\right)^2\)

⇒ \(\frac{9}{12}\) = \(\frac{\mathrm{BC}}{16}\)

⇒ BC = \(\frac{9}{12}\) × 16 = \(\frac{144}{12}\) = 12

∴ BC = 12 cm.

Question 6.

∆ABC ~ ∆DEF, BC = 5 cm, EF = 6 cm and area of ∆DEF = 72 cm^{2}. Determine the area of ∆ABC.

Solution:

Given ∆ABC ~ ∆DEF

and BC = 5cm, EF = 6cm

Area of ∆DEF = 72 cm^{2}

We know that areas of two similar triangles are in the ratios of the squares of the corresponding sides.

Question 7.

A ladder 13m long reaches a window of building 12cm above the ground. Determine the distance of the foot of the ladder from the building.

Solution:

In ∆ ABC, B = ∠90°

⇒ AC^{2} = AB^{2} + BC^{2} (By Pythagoras theorem)

Let AC = 13m, AB = 12m, BC = ?

⇒ 13^{2} = 12^{2} + BC^{2}

⇒ 169 = 144 + BC^{2}

⇒ BC^{2} = 169 – 144 = 25

⇒ BC = \(\sqrt{25}\) = 5m

Hence, the foot of the ladder from the building is at a distance of 5m.

Question 8.

The hypotenuse of a right angled triangle is 3 m more than twice of the shortest side. If the third side is 1 m less than the hypotenuse find the sides of the triangle.

Solution:

Let the shortest side = x m = BC

Then hypotenuse = 2x + 3 m = AC

The third side = (2x + 3) – 1

= (2x + 2) m = AB

By phythagoras theorem, we have

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

⇒ (2x + 3)^{2} = (2x + 2)^{2} + x^{2}

⇒ 4x^{2} + 9 + 12x = 4x^{2} + 4 + 8x + x^{2}

⇒ x^{2} + 8x – 12x + 4 – 9 = 0

⇒ x^{2} – 4x – 5 = 0

⇒ x^{2} – 5x + x – 5 = 0

⇒ x (x – 5) + 1 (x – 5) = 0

⇒ (x – 5) (x + 1) = 0

⇒ x – 5 = 0 or x + 1 = 0

⇒ x = 5 or x = -1

But x can’t be negative x = 5

Hence, the sides of the triangle are

x, 2x + 3, 2x + 2

⇒ x = 5, 2 × 5 + 3, 2 × 5 + 2

= 5, 10 + 3, 10 + 2

i.e. 5, 13, 12

Question 9.

A ladder 25 m long reaches a window which is 15 m above the ground an one side of the road. Keeping its foot at the same point, the ladder is turned to other side of the road to reach a window 20 m high. Find the width of the road.

Solution:

Let A and D be the windows on the either sides of the road

From phythagoras theorem,

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

⇒ 25^{2} = 15^{2} + BC^{2}

⇒ BC^{2} = 25^{2} – 15^{2} = 625 – 225 = 400

BC = \(\sqrt{400}\) = 20m ……………. (1)

Also CD^{2} = CE^{2} + DE^{2}

25^{2} = CE^{2} + 20^{2}

CE^{2} = 25^{2} – 20^{2} = 625 – 400 = 225

CE = \(\sqrt{225}\) = 15m

Width of the road = BE = BC + CE

= 20 + 15

= 35 m

Question 10.

In the given figure below, If AD ⊥ BC, prove that AB^{2} – BD^{2} = AC^{2} – CD^{2}.

Solution:

Given in ∆ ABC, AD ⊥ BC

R.T.P : AB^{2} – BD^{2} = AC^{2} – CD^{2}

Proof : ∆ ABD is a right angled triangle.

We have AB^{2} = AD^{2} + BD^{2} (By pythagaros theorem)

⇒ AB^{2} – BD^{2} = AD^{2} ……………… (1)

∆ ACD is a right angled triangle.

We have AC^{2} = AD^{2} + CD^{2}

⇒ AC^{2} – CD^{2} = AD^{2} ………………. (2)

From (1) and (2), we have

AB^{2} – BD^{2} = AC^{2} – CD^{2}

Question 11.

In an equilateral triangle ABC, if AD is the altitude prove that 3AB^{2} = 4 AD^{2}.

Solution:

Given ABC is an equilateral triangle.

Here, AD ⊥ BC and AB = BC = CA

In triangles ADB and ADC

∠ADB = ∠ADC = 90°

Hypotenuse AB = Hypotenuse AC

AD is common

∴ ∆ ADB ≅ ∆ ADC

∴ BD = DC

In ∆ ADB, ∠ADB = 90°(By Phythagoras theorem)

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

= AD^{2} + \(\left(\frac{\mathrm{BC}}{2}\right)^2\) (∵ BD = CD)

= AD^{2} + \(\frac{\mathrm{BC}^2}{4}\)

AB^{2} = \(\frac{4 A D^2+B C^2}{4}\)

⇒ 4AB^{2} = 4AD^{2} + BC^{2}

⇒ 4AB^{2} – BC^{2} = 4AD^{2}

⇒ 4AB^{2} – AB^{2} = 4AD^{2} (∵ BC = AB)

∴ 3AB^{2} = 4AD^{2}

Question 12.

A wire attached to a vertical pole of height 15 m is 25 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut ?

Solution:

Let AB = Vertical pole

AC = Wire

B = Base

C = Stake

By Pythagoras theorem

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

25^{2} = 15^{2} + BC^{2}

BC^{2} = 25^{2} – 15^{2}

BC^{2} = 62^{2} – 225 = 400

BC = \(\sqrt{400}\) = 20 m

∴ The stake should be driven 20 m for the base of the pole so that the wire will be taut.

Question 13.

The larger of two complimentary angles is double the smaller. Find the angles.

Solution:

Let first complementary angle = x

Second complementary angle = 2x

Sum of the two complementary angles = 90°

∴ x + 2x = 90

3x = 90

x = \(\frac{90}{3}\)

x = 30

First angle = 30°, second angle = 60°.

Question 14.

In ∆ ABC, DE || BC and \(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{3}{5}\). If AE = 2.1 cm, then find AC.

Solution:

Here, Given DE || BC and \(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{3}{5}\)

AE = 2.1

∴ \(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{\mathrm{AE}}{\mathrm{CE}}\)

\(\frac{3}{5}\) = \(\frac{2.1}{\mathrm{CE}}\)

CE = \(\frac{5 \times 2.1}{3}\)

CE = \(\frac{10.5}{3}\) = 3.5

CE = 3.5

Then AC = AE + CE

AC = 2.1 + 3.5

AC = 5.6

Question 15.

What can you say about the ratio of areas of two similar triangles ?

Solution:

The ratio of areas of two similar triangles is( equal to the ratio of the squares of their corresponding sides.

Here

Question 16.

Construct an isosceles triangle whose base is 8 cm and altitude is 4 cm. Then, draw another similar triangle whose side are 1 \(\frac{1}{2}\) times the corresponding sides of the isosceles triangle.

Solution:

- Draw a lines segment BC = 8 cm
- Draw the perpendicular bisector PQ of BC intersecting BC at ‘O’
- Mark a point ‘A’ on PQ such that \(\overline{\mathrm{OA}}\) = 4 cm
- Join AB and AC to the isosceles triangle ABC
- Extend BC on either side O’C’ = 1 \(\frac{1}{2}\) times; BC = 12 times.
- Similarly extend OA So that OA’ = 1 \(\frac{1}{2}\) times A = \(\frac{12}{2}\) = 6
- Join A’ B’ and A’ C’
- Now A’ B’ C’ – The required triangle

Question 17.

Give two different examples of pair of 3 similar figures and non similar figures.

Solution:

Similar figures :

- Any two circles
- Any two squares

Non-similar figures :

- A square and a rhombus
- A square and a rectangle

Question 18.

∆ ABC ~ ∆ DEF and their areas are respectively 64 cm^{2} and 121 cm^{2}. If EF = 15.4 cm then find BC.

Solution:

Given area of ∆ABC = 64 cm^{2}

area of ∆DEF = 121 cm^{2}

EF = 15.4 cm, BC = ?

We know that \(\frac{\text { area of } \triangle \mathrm{ABC}}{\text { area of } \triangle \mathrm{DEF}}\)=\(\frac{\mathrm{BC}^2}{\mathrm{EF}^2}\)

⇒ \(\frac{64}{121}\) = \(\frac{\mathrm{BC}^2}{(15.4)^2}\)

⇒ BC^{2} = \(\frac{64}{121}\) × (15.4)^{2}

= \(\frac{64}{121}\) × 15.4 × 15.4

BC^{2} = 125.44

∴ BC = \(\sqrt{125.44}\) = 11.2 cm.

Question 19.

In D ABC, DE || BC and \(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{3}{5}\) ; AC = 5.6 find AE.

Solution:

But \(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{3}{5}\) , So \(\frac{\mathrm{AE}}{\mathrm{EC}}\) = \(\frac{3}{5}\)

given AC = 5.6

from (1) \(\frac{3}{5}\) = \(\frac{\mathrm{AE}}{\mathrm{AC – AE}}\)

\(\frac{3}{5}\) = \(\frac{\mathrm{AE}}{5.6-\mathrm{AE}}\) (By cross-multiplication)

5 AE = 3 (5.6 – AE)

5 AE = 3 × 5.6 – 3AE

5AE + 3 AE = 3 × 5.6

8AE = 16.8

AE = \(\frac{16.8}{8}\) = 2.1 cm

Question 20.

State and prove basic proportionality theorem.

Solution:

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

Given : A triangle ABC in which DE || BC and DE intersects AB in D and AC in E.

To Prove : \(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{\mathrm{AE}}{\mathrm{EC}}\)

Construction : Join BE, CD and draw EF ⊥ AB. DG ⊥ AC.

Proof: In ∆ EAD and ∆ EDB, Here as EF ⊥ AB

therefore EF is the height for both of triangles EAD and EDB.

Now, Area of ∆ EAD = \(\frac{1}{2}\) (base × height)

= \(\frac{1}{2}\) × AD × EF

Area of EDB = \(\frac{1}{2}\) (base × height)

= \(\frac{1}{2}\) × DB × EF

∴ ∆ DBF, ECD are on the same base DE and between the same parallels DE || BC,

We have area of ∆DBE = area of ∆ECD

Hence (1) = (2)

i.e, \(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{\mathrm{AE}}{\mathrm{EC}}\) (Q.E.D)

Question 21.

Construct a triangle of sides 4 cm, 5 cm and 6 cm then construct a triangle. Similar to it. Whose sides are \(\frac{2}{3}\) of the corresponding sides of the first triangle.

Solution:

Steps of construction :

- Draw ∆ ABC with AB = 4 cm, BC = 5 cm and CA = 6 cm.
- Draw a Ray \(\overrightarrow{\mathrm{BX}}\) making an acute angle with BC on the side opposite to vertex A.
- Mark three points B
_{1}, B_{2}and B_{3}on \(\overrightarrow{\mathrm{BX}}\) Such that BB_{1}= B_{1}B_{2}= B_{2}B_{3}. - Join B
_{3}, C. - Draw a line parallel B
_{3}, C through B_{2}meeting BA at A’. - ∆ A’ BC’ is required triangle.