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

### Question 1. Triangles ABC and DEF are similar.

### (i) If area (â–³ABC) = 16cm^{2}, area (â–³DEF) = 25 cm^{2} and BC = 2.3 cm, find EF.

**Solution:**

Since, â–³ABC âˆ¼ â–³DEF,

So,

ar.(â–³ACB)/ar.(â–³DEF) = (BC)

^{2}/(EF)^{2}16/25 = (2.3)

^{2}/EF^{2}(4)

^{2}/(5)^{2 }= (2.3)^{2}/EF^{2 }4/5 = 2.3/EF

EF = (2.3 Ã— 5)/4

EF = 11.5/4 = 2.875 cm

Hence, EF = 2.875 cm

### (ii) If area (â–³ABC) = 9cm^{2}, area (â–³DEF) = 64 cm^{2} and DE = 5.1 cm, find AB.

**Solution:**

Since, â–³ABC âˆ¼ â–³DEF

So,

ar.(â–³ACB)/ar.(â–³DEF) = (AB)

^{2}/(DE)^{2}9/5 = AB

^{2}/(5.1)^{2}(3)

^{2}/(8)^{2 }= AB^{2}/(5.1)^{2}3/8 = AB/5.1

AB = (3 Ã— 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, â–³ABC âˆ¼ â–³DEF

So,

ar.(â–³ACB)/ar.(â–³DEF) = (AB)

^{2}/(DE)^{2}= (19)/(8)

^{2 }= 361/64Hence, ar.(â–³ABC):ar.(â–³DEF) = 361:64

### (iv) If area (â–³ABC) = 36cm^{2}, area (â–³DEF) = 64 cm^{2} and DE = 6.2 cm, find AB.

**Solution:**

Since, â–³ABC âˆ¼ â–³DEF

So,

ar.(â–³ACB)/ar.(â–³DEF) = (AB)

^{2}/(DE)^{2}â‡’ 36/64 = AB

^{2}/(6.2)^{2}â‡’ 6/8

^{ }= AB/(6.2)â‡’ AB/6.2 = 6/8

â‡’ AB = (6 Ã— 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 â–³ABC and â–³DEF.

**Solution:**

Since, â–³ABC âˆ¼ â–³DEF

So,

ar.(â–³ACB)/ar.(â–³DEF) = (AB)

^{2}/(DE)^{2}(1.2)

^{2}/(1.4)^{2 }= 1.44/1.96 = 144/196 = 36/49Hence, ar.(â–³ABC):ar.(â–³DEF) = 36:49

### Question 2. In fig. below âˆ†ACB ~ âˆ†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 (âˆ†ACB): area (âˆ†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, â–³ACB âˆ¼ â–³APQ

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

10/5 = 6.5/AQ = AC/2.8

6.5/AQ = 10/5

â‡’ AQ = (6.5 Ã— 5)/10 = 3.25

and AC/2.8 = 10/5

â‡’ (2.8 Ã— 10)/5 = 5.6

AC = 5.6m, AQ = 3.25cm

As we know that â–³ACB âˆ¼ â–³APQ,

So, ar.(â–³ACB)/ar.(â–³APQ) = (BC)

^{2}/(PQ)^{2}

^{ }ar.(â–³ACB)/ar.(â–³APQ) = (10)^{2}/(5)^{2 }= 100/25 = 4/1Hence, ar(â–³ACB):ar(â–³APQ) = 4:1

### Question 3. The areas of two similar triangles are 81 cm^{2} and 49 cm^{2} 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 cm

^{2}and 49 cm^{2}So, â–³ABC = 81 cm

^{2}and â–³PQR = 49 cm^{2}So, ar.(â–³ABC)/ar.(â–³PQR) = (AB)

^{2}/(PQ)^{2}81/49 = (AB/PQ)

^{2}9/7 = AB/PQ

Now, in â–³ABD and â–³PQO

âˆ B = âˆ Q

âˆ ADB = âˆ POQ = 90Â°

Hence, â–³ABD ~ â–³PQO

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 cm^{2 }and 121 cm^{2} 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 cm

^{2}and area of the smaller triangle(PQR) = 121 cm

^{2}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.(â–³ABC)/ar.(â–³PQR) = (AC)

^{2}/(PR)^{2}169/121 = (26)

^{2}/(x)^{2}13/11 = 26/x

x = (13 Ã— 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 cm^{2} and 36 cm^{2} 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 cm

^{2}and the area of second = 36 cm

^{2}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}â‡’ 25/36 = (2.4)

^{2}/x^{2}â‡’ 2.4/x = 5/6

â‡’ x = (2.4 Ã— 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 âˆ A =90Â°, ANâŠ¥ BC, BC = 12 cm and AC = 5cm. Find the ratio of the areas of âˆ†ANC and âˆ†ABC.

**Solution:**

Given that,

In âˆ†ABC, âˆ A = 90Â°

AN âŠ¥ BC

BC = 12 cm, AC = 5 cm

So, in âˆ†ANC and âˆ† ABC,

âˆ ANC = âˆ BAC = 90Â°

âˆ C =âˆ C [Common]

So, by AA,

âˆ†ANC âˆ¼ âˆ† ABC

ar.(âˆ†ANC)/ar.(âˆ†ABC) = (AC)

^{2}/(BC)^{2}= (5)^{2}/(12)^{2}= 25/144Hence, the ratio of the areas of âˆ†ANC and âˆ†ABC is 25:144

### Question 8. In Fig., DE || BC

**(i) If DE = 4 cm, BC = 6 cm and Area (âˆ†ADE) = 16 cm ^{2}, find the area of âˆ†ABC.**

**(ii) If DE = 4cm, BC = 8 cm and Area (âˆ†ADE) = 25 cm ^{2}, find the area of âˆ†ABC.**

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

**Solution:**

Given that, DE || BC

So, In âˆ†ADE and âˆ†ABC

âˆ ADE = âˆ B

âˆ BAC = âˆ DAE

So, by AA

âˆ†ADE ~ âˆ†ABC

(i)Given that DE = 4 cm, BC = 6 cm and ar(âˆ†ADE) = 16 cm^{2}As we know that âˆ†ADE âˆ¼ âˆ†ABC

So, ar(âˆ†ADE)/ar(âˆ†ABC) = DE

^{2}/BC^{2}16/ar(âˆ†ABC) = 4

^{2}/6^{2}= 16/36So, 16 Ã— area âˆ†ABC = 16 Ã— 36

â‡’ ar.(âˆ†ABC) = 36cm

^{2}

(ii)Given that DE = 4cm, BC = 8 cm and ar(âˆ†ADE) = 25 cm^{2}As we know that âˆ†ADE âˆ¼ âˆ†ABC

So, ar(âˆ†ADE)/ar(âˆ†ABC) = DE

^{2}/BC^{2}25/area(âˆ†ABC) = (4)

^{2}/(8)^{2 }= 16/64area(âˆ†ABC) = (25 Ã— 64)/16 = 100 cm

^{2}

(iii)Given that, DE : BC = 3 : 5As we know that âˆ†ADE âˆ¼ âˆ†ABC

area(âˆ†ADE)/area(âˆ†ABC) = DE

^{2}/BC^{2 }= (3/5)^{2}= 9/2525 (area(âˆ†ADE)) = 9 (area âˆ†ABC)

25 (area(âˆ†ADE)) = 9(area âˆ†ADE + area trapezium BCED)

area(âˆ†ADE)/area of trapezium BCED = 9/16

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

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

**Solution:**

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

So, DE||BC and DE = 1/2BC

In âˆ†ADE and âˆ†ABC,

âˆ ADE = âˆ B

âˆ DAE = âˆ BAC [Common]

By AA

âˆ†ADE âˆ¼ âˆ†ABC

So, ar(âˆ†ADE)/ar(âˆ†ABC) = (DE)

^{2}/(BC)^{2}Hence, the ratio of the areas of âˆ†ADE and âˆ†ABC is 1:4

### Question 10. The areas of two similar triangles are 100 cm^{2} and 49 cm^{2} respectively. If the altitude of the bigger triangle is 5 cm, find the corresponding altitude of the other.

**Solution:**

Let us consider âˆ†ABC and âˆ†DEF

It is given that, area âˆ†ABC = 100 cm

^{2}and area âˆ†DEF = 49 cm

^{2}AL perpendicular BC and DM/EF

AL = 5cm,

Let DM = x cm

It is given that âˆ†ABC ~ âˆ†DEF

So, ar(âˆ†ABC)/ar(âˆ†DEF) = AL

^{2}/DM^{2}100/49 = (5)

^{2}/(x)^{2}100/49 = 25/x

^{2}x

^{2 }= (25 Ã— 49)/100 = 49/4x =

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

## Please

Loginto comment...