Class 10 RD Sharma Solutions - Chapter 4 Triangles - Exercise 4.6 | Set 1

This exercise contains questions regarding the similarity of triangles. Questions in this exercise are based on the concepts of properties of two similar triangles. The RD Sharma Solutions Class 10 provides all solutions required for a quick preparation for exams. The following questions and their solutions are prepared to help and boost the morale of students as many find mathematics a tough subject.

Question 1. Triangles ABC and DEF are similar.

(i) If area (△ABC) = 16cm², area (△DEF) = 25 cm² and BC = 2.3 cm, find EF.

Solution:

Since, △ABC ∼ △DEF,

So, 

ar.(△ACB)/ar.(△DEF) = (BC)²/(EF)²

16/25 = (2.3)²/EF²

(4)²/(5)² = (2.3)²/EF² 

4/5 = 2.3/EF

EF = (2.3 × 5)/4

EF = 11.5/4 = 2.875 cm

Hence, EF = 2.875 cm

(ii) If area (△ABC) = 9cm², area (△DEF) = 64 cm² and DE = 5.1 cm, find AB.

Solution:

Since, △ABC ∼ △DEF

So, 

ar.(△ACB)/ar.(△DEF) = (AB)²/(DE)²

9/5 = AB²/(5.1)²

(3)²/(8)² = AB²/(5.1)²

3/8 = AB/5.1

AB = (3 × 5.1)/8 = 15.3/8

Hence, AB = 1.9125 cm

(iii) If AC = 19cm and DF = 8 cm, find the ratio of the area of two triangles.

Solution:

Since, △ABC ∼ △DEF

So, 

ar.(△ACB)/ar.(△DEF) = (AB)²/(DE)²

= (19)/(8)² = 361/64

Hence, ar.(△ABC):ar.(△DEF) = 361:64

(iv) If area (△ABC) = 36cm², area (△DEF) = 64 cm² and DE = 6.2 cm, find AB.

Solution:

Since, △ABC ∼ △DEF

So, 

ar.(△ACB)/ar.(△DEF) = (AB)²/(DE)²

⇒ 36/64 = AB²/(6.2)²

⇒ 6/8= AB/(6.2)

⇒ AB/6.2 = 6/8

⇒ AB = (6 × 6.2)/8 = 37.2/8  

AB = 4.65

Hence AB = 4.65 cm

(v) If AB = 1.2 cm and DE = 1.4 cm, find the ratio of the areas of △ABC and △DEF.

Solution:

Since, △ABC ∼ △DEF

So, 

ar.(△ACB)/ar.(△DEF) = (AB)²/(DE)²

(1.2)²/(1.4)² = 1.44/1.96 = 144/196 = 36/49

Hence, ar.(△ABC):ar.(△DEF) = 36:49

Question 2. In fig. below ∆ACB ~ ∆APQ. If BC = 10 cm, PQ = 5 cm, BA = 6.5 cm and AP = 2.8 cm, find CA and AQ. Also, find the area (∆ACB): area (∆APQ).

Solution:

According to the question

It is given that, BC = 10 cm, PQ = 5 cm, BA = 6.5 cm and AP = 2.8 cm

Also, △ACB ∼ △APQ

So, BC/PQ = AB/AQ = AC/AP

10/5 = 6.5/AQ = AC/2.8

6.5/AQ = 10/5

⇒ AQ = (6.5 × 5)/10 = 3.25 

and AC/2.8 = 10/5

⇒ (2.8 × 10)/5 = 5.6

AC = 5.6m, AQ = 3.25cm

As we know that △ACB ∼ △APQ, 

So, ar.(△ACB)/ar.(△APQ) = (BC)²/(PQ)²

ar.(△ACB)/ar.(△APQ) = (10)²/(5)² = 100/25 = 4/1

Hence, ar(△ACB):ar(△APQ) = 4:1

Question 3. The areas of two similar triangles are 81 cm² and 49 cm² respectively. Find the ratio of their corresponding heights. What is the ratio of their corresponding medians?

Solution:

Let us consider two similar triangles, ABC and PQR whose altitudes are AD and PO

It is given that the areas of two similar triangles are 81 cm² and 49 cm²

So, △ABC = 81 cm²  and △PQR = 49 cm²

So, ar.(△ABC)/ar.(△PQR) = (AB)²/(PQ)²

81/49 = (AB/PQ)²

9/7 = AB/PQ

Now, in △ABD and △PQO

∠B = ∠Q

∠ADB = ∠POQ = 90°

Hence, △ABD ~ △PQO

So, AB/PQ = AD/PO

Hence, AD/PO = 9/7 

Or

AD:PO = 9:7

As we know, that the ratio of the areas of two similar triangles are proportional to the square of their corresponding altitudes and also the squares of their corresponding medians.

So, the ratio in their medians = 9 : 7

Question 4. The areas of two similar triangles are 169 cm² and 121 cm² respectively. If the longest side of the larger triangle is 26 cm, find the longest side of the smaller triangle.

Solution:

Given that both the triangles are similar 

So, the area of larger triangle(ABC) = 169 cm²

and area of the smaller triangle(PQR) = 121 cm²

The length of the longest sides of the larger triangles(AC) = 26 cm

Let us assume the length of longest side of the smaller triangle(PR) = x 

So, the ar.(△ABC)/ar.(△PQR) = (AC)²/(PR)²

 169/121 = (26)²/(x)²

 13/11 = 26/x

x = (13 × 26)/11

x = 22 

Hence, the length of the longest side of the smaller triangle is 22 cm.

Question 5. The areas of two similar triangles are 25 cm² and 36 cm² respectively. If the altitude of the first triangle is 2.4 cm, find the corresponding altitude of the other.

Solution:

Given that, the area of the first triangle = 25 cm²

and the area of second = 36 cm²

Altitude of the first triangle = 2.4 cm

Let us consider the altitude of the second triangle = x

It is given that both the triangles are similar, so

ar.(first triangle)/ar.(second triangle) = (Altitude of the first triangle)²/(Altitude of the second triangle)²

⇒ 25/36 = (2.4)²/x²

⇒ 2.4/x = 5/6

⇒ x = (2.4 × 6)/5 = 14.4/5 = 2.88

Hence, the altitude of the second triangle is 2.88cm

Question 6. The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively. Find the ratio of their areas. 

Solution:

Given that the length of the corresponding altitude of two triangles are 6 cm and 9 cm

Also, both the triangles are similar

So, 

ar.(first triangle)/ar.(second triangle) = (6)²/(9)²

= 36/81

= 4/9

Hence, the ratio of the areas of triangles is 4:9

Question 7. ABC is a triangle in which ∠A =90°, AN⊥ BC, BC = 12 cm and AC = 5cm. Find the ratio of the areas of ∆ANC and ∆ABC.  

Solution:

Given that, 

In ∆ABC, ∠A = 90°

AN ⊥ BC

BC = 12 cm, AC = 5 cm

So, in ∆ANC and ∆ ABC,

∠ANC = ∠BAC = 90°       

∠C =∠C              [Common]

So, by AA,

∆ANC ∼ ∆ ABC           

ar.(∆ANC)/ar.(∆ABC) = (AC)²/(BC)² = (5)²/(12)² = 25/144 

Hence, the ratio of the areas of ∆ANC and ∆ABC is 25:144

Question 8. In Fig., DE || BC  

(i) If DE = 4 cm, BC = 6 cm and Area (∆ADE) = 16 cm², find the area of ∆ABC.

(ii) If DE = 4cm, BC = 8 cm and Area (∆ADE) = 25 cm², find the area of ∆ABC.

(iii)If DE : BC = 3 : 5. Calculate the ratio of the areas of ∆ADE and the trapezium BCED.

Solution:

Given that, DE || BC  

So, In ∆ADE and ∆ABC

∠ADE = ∠B

∠BAC = ∠DAE

So, by AA 

∆ADE ~ ∆ABC

(i) Given that DE = 4 cm, BC = 6 cm and ar(∆ADE) = 16 cm²

As we know that ∆ADE ∼ ∆ABC

So, ar(∆ADE)/ar(∆ABC) = DE²/BC²

16/ar(∆ABC) = 4²/6² = 16/36

So,  16 × area ∆ABC = 16 × 36

⇒ ar.(∆ABC) = 36cm²

(ii) Given that DE = 4cm, BC = 8 cm and ar(∆ADE) = 25 cm²

As we know that ∆ADE ∼ ∆ABC

So, ar(∆ADE)/ar(∆ABC) = DE²/BC²

25/area(∆ABC) = (4)²/(8)² = 16/64

area(∆ABC) = (25 × 64)/16 = 100 cm²

(iii) Given that, DE : BC = 3 : 5

As we know that ∆ADE ∼ ∆ABC

area(∆ADE)/area(∆ABC) = DE²/BC² = (3/5)² = 9/25 

25 (area(∆ADE)) = 9 (area ∆ABC)

25 (area(∆ADE)) = 9(area ∆ADE + area trapezium BCED)

area(∆ADE)/area of trapezium BCED = 9/16

Hence, the ratio of the areas of ∆ADE and the trapezium BCED is 9:16

Question 9. In ∆ABC, D and E are the mid-points of AB and AC respectively. Find the ratio of the areas of ∆ADE and ∆ABC. 

Solution:

Given that, in ∆ABC, D and E are the mid points of AB and AC

So, DE||BC and DE = 1/2BC

In ∆ADE and ∆ABC,

∠ADE = ∠B 

∠DAE = ∠BAC   [Common]

By AA

∆ADE ∼ ∆ABC 

So, ar(∆ADE)/ar(∆ABC) = (DE)²/(BC)²

Hence, the ratio of the areas of ∆ADE and ∆ABC is 1:4

Question 10. The areas of two similar triangles are 100 cm² and 49 cm² respectively. If the altitude of the bigger triangle is 5 cm, find the corresponding altitude of the other. 

Solution:

Let us consider ∆ABC and ∆DEF

It is given that, area ∆ABC = 100 cm²

and area ∆DEF = 49 cm²

AL perpendicular BC  and DM/EF

AL = 5cm,

Let DM = x cm

It is given that ∆ABC ~ ∆DEF    

So, ar(∆ABC)/ar(∆DEF) = AL²/DM²

100/49 = (5)²/(x)²

100/49 = 25/x²

x² = (25 × 49)/100 = 49/4

x = 

Hence, the length of altitude of second triangle is 3.5cm

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