# Class 10 RD Sharma Solutions – Chapter 4 Triangles – Exercise 4.7 | Set 1

### Question 1. If the sides of a triangle are 3 cm, 4 cm, and 6 cm long, determine whether the triangle is a right-angled triangle.

**Solution:**

According to the question

The sides of triangle are:

AB = 3 cm

BC = 4 cm

AC = 6 cm

According to Pythagoras Theorem:

AB

^{2}= 3^{2}= 9BC

^{2}= 4^{2}= 16AC

^{2}= 6^{2}= 36Since, AB

^{2}+ BC^{2}≠ AC^{2}Hence, by converse of Pythagoras theorem, triangle is not a right triangle

### Question 2. The sides of certain triangles are given below. Determine which of them right triangles are.

**(i) a = 7 cm, b = 24 cm and c = 25 cm**

**(ii) a = 9 cm, b = 16 cm and c = 18 cm**

**(iii) a = 1.6 cm, b = 3.8 cm and c = 4 cm**

**(iv) a = 8 cm, b = 10 cm and c = 6 cm**

**Solution:**

(i)According to the questionThe sides of triangle are:

a = 7 cm, b = 24 cm and c = 25 cm

According to Pythagoras Theorem:

The hypotenuse or longest side of a right angle triangle is equal to the sum of

squares of other two sides of the triangle

∴ a

^{2}= 49, b^{2}= 576 and c^{2}= 625(Longest side)

^{2}= c^{2 }= 625Sum of squares of shorter sides = (a)

^{2 }+ (b)^{2}= 49 + 576 = 625∴ 625 = 625

Hence, by converse of Pythagoras theorem, given triangle is a right triangle.

(ii)According to the questionThe sides of triangle are: a = 9 cm, b = 16 cm and c = 18 cm

According to Pythagoras Theorem:

The hypotenuse or longest side of a right angle triangle is equal to the sum of

squares of other two sides of the triangle

∴ a

^{2}= 81, b^{2}= 256 and c^{2}= 324(Longest side)

^{2}= c^{2 }= 324Sum of squares of shorter sides = (a)

^{2 }+ (b)^{2 }= 81 + 256 = 337∴324 ≠ 337

Hence, by converse of Pythagoras theorem, given triangle is not a right triangle.

(iii)According to the questionThe sides of triangle are: a = 1.6 cm, b = 3.8 cm and C = 4 cm

According to Pythagoras Theorem:

The hypotenuse or longest side of a right angle triangle is equal to the sum of

squares of other two sides of the triangle

(Longest side)

^{2}= 16Sum of squares of shorter sides = (1.6)

^{2 }+ (3.8)^{2 }= 2.56 + 14.44 = 17∴ 16 ≠ 17

Hence, by converse of Pythagoras theorem, given triangle is not a light triangle.

(iv)According to the questionThe sides of triangle are: a = 8 cm, b = 10 cm, c = 6 cm

According to Pythagoras Theorem:

The hypotenuse or longest side of a right angle triangle is equal to the sum of

squares of other two sides of the triangle

(Longest side)

^{2}= (10)^{2}= 100Sum of squares of shorter sides = (8)

^{2}+ (6)^{2}= 64 + 36 = 100∴ 100 = 100

Hence, by converse of Pythagoras theorem, given triangle is a light triangle.

### Question 3. A man goes 15m due west and then 8m due north. How far is lie from the starting point?

**Solution:**

Let the starting point of the man be O and final point be A.

So, In ∆ABO,

By using Pythagoras theorem

AO

^{2 }= AB^{2}+ BO^{2}⟹ AO2 = 8

^{2}+ 15^{2}⟹ AO2 = 64 + 225 = 289

⟹ AO = √289 = 17m

Hence, the man is 17m far from the starting point.

### Question 4. A ladder 17 m long reaches a window of a building 15 m above the ground. Find the distance of the foot of the ladder from the building.

**Solution:**

In ∆ABC,

By using Pythagoras theorem

AB

^{2}+ BC^{2}= AC^{2}⟹ 152 + BC

^{2}= 172⟹ 225 + BC

^{2}= 172⟹ BC

^{2}= 289 – 225⟹ BC

^{2}= 64⟹ BC = 8 m

Hence, the distance of the foot of the ladder from building = 8 m

### Question 5. Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.

**Solution:**

Let us considered CD and AB be the poles of height 11 and 6 in.

Therefore, CP = 11 – 6 = 5 m

From the figure we may observe that AP = 12m

In ∆APC,

By using Pythagoras theorem

AP

^{2}+ PC^{2}= AC^{2}12

^{2}+ 5^{2}= AC^{2}AC

^{2}= 144 + 25 = 169 mAC = 13 m

Hence, the distance between their tops = 13 m.

### Question 6. In an isosceles triangle ABC. AB = AC = 25 cm, BC = 14 cm. Calculate the altitude from A on BC.

**Solution:**

According to the question

AB = AC = 25 cm and BC = 14

In ∆ABD and ∆ACD

∠ADB = ∠ADC [Each = 90°]

AB = AC [Each = 25 cm]

AD = AD [Common]

So, by RHS condition

∆ABD ≅ ∆ACD

Hence, by corresponding parts of congruent triangles

∴ BD = CD = 7 cm

In ∆ADB,

By using Pythagoras theorem

AD

^{2}+ BD^{2}= AB^{2}⟹ AD

^{2}+ 7^{2}= 25^{2}⟹ AD

^{2}= 625 – 49 = 576⟹ AD = √576 = 24 cm

### Question 7. The foot of a ladder is 6 m away from a wall and its top reaches a window 8 m above the ground. If the ladder is shifted in such a way that its foot is 8 m away from the wall, to what height does its tip reach?

**Solution:**

Let length of ladder be AD = BE = l m

In ∆ACD,

By using Pythagoras theorem

AD

^{2}= AC^{2}+ CD^{2 }⟹ l

^{2}= 8^{2}+ 6^{2}… (i)In ∆BCE,

By using Pythagoras theorem

BE

^{2}= BC^{2}+ CE^{2 }⟹ l

^{2}= BC^{2}+ 8^{2}… (ii)From eq(i) and (ii), we get

BC

^{2}+ 8^{2}= 8^{2}+ 6^{2}⟹ BC

^{2}+ 6^{2}⟹ BC = 6 m

### Question 8. Two poles of height 9 m in and 14 m stand on a plane ground. If the distance between their feet is 12 m in, find the distance between their tops.

**Solution: **

We have,

AC = 14 m. DC = 12m and ED = BC = 9 m

Construction: Draw EB ⊥ AC

AB = AC – BC = 14 — 9 = 5 m

And, EB = DC = 12m

In ∆ABE,

By using Pythagoras theorem

AE

^{2}= AB^{2}+ BE^{2}⟹ AE

^{2}= 5^{2}+ 12^{2}⟹ AE

^{2}= 25 + 144 = 169⟹ AE= √169 = 13 m

Hence, the distance between their tops = 13 m

### Question 9. Using Pythagoras theorem determine the length of AD in terms of b and c shown in the below fig.

**Solution:**

We have,

In ∆BAC,

By using Pythagoras theorem

BC

^{2}= AB^{2}+ AC^{2}⟹ BC

^{2}= c^{2}+ b^{2}….(1)

In ∆ABD and ∆CBA

∠B = ∠B [Common]

∠ADB = ∠BAC [Each 90°]

So, by AA both the triangles are similar

∆ABD ͏~ ∆CBA

Hence, by corresponding parts of similar ∆ are proportional

Hence, the length of AD in terms of b and c is

### Question 10. A triangle has sides 5 cm, 12 cm, and 13 cm. Find the length to one decimal place, of the perpendicular from the opposite vertex to the side whose length is 13 cm.

**Solution:**

Let, AB = 5cm, BC = 12 cm and AC = 13 cm. Then, AC

^{2}= AB^{2}+ BC^{2}.This proves that ∆ABC is a fight triangle. right angles at B.

Let BD be the length of perpendicular from B on AC.

Now, Area of ∆ABC = 1/2(BC x BA)

= 1/2(12 x 5)

= 30 cm

Also, Area of ∆ABC = 1/2(AC x BD) = 1/2(13 x BD) = 1/2(13 x 2)

BD = 60/13 cm

### Question 11. ABCD is a square, F is the mid-point of AB. BE is one third of BC. If the area of ∆FBE = 108 cm^{2} find the length of AC.

**Solution:**

Given,

ABCD is a square.

F is the mid-point of AB.

BE is one third of BC.

Area of ∆ FBE = 108cm

^{2}To find: length of AC

Let’s the sides of the square to be y.

AB = BC = CD = DA = y cm

So, AF = FB = x/2 cm

and, BE = x/3 cm

Now,

Area of ∆ FBE = 1/2 x BE x FB

⇒ 108 = (1/2) x (x/3) x (x/2)

⇒ x

^{2}= 108 x 2 x 3 x 2 = 1296⇒ x = √(1296)

x = 36cm

In ∆ ABC,

By using Pythagoras theorem

AC

^{2}= AB^{2}+ BC^{2 }⇒ AC

^{2}= x^{2}+ x^{2}= 2x^{2}⇒ AC

^{2}= 2 x (36)^{2}⇒ AC = 36√2 = 36 x 1.414 = 50.904 cm

Hence, the length of AC is 50.904 cm.

### Question 12. In an isosceles triangle ABC, if AB = AC = 13 cm and the altitude from A on BC is 5 cm, find BC.

**Solution:**

Given that,

An isosceles triangle ABC, AB = AC = 13cm, AD = 5cm

To find: BC

Now In ∆ ADB,

By using Pythagoras theorem

AD

^{2}+ BD^{2}= 13^{2}5

^{2}+ BD^{2}= 169BD

^{2}= 169 – 25 = 144BD = √144 = 12 cm

Similarly, In ∆ADC,

By using Pythagoras theorem

AC

^{2 }= AD^{2}+ DC^{2}13

^{2}= 5^{2}+ DC^{2}169 – 25 = DC

^{2}DC = √144 = 12 cm

So, BC = BD + DC = 12 + 12 = 24 cm

Hence, in isosceles triangle ABC side BC is 24 cm

### Question 13. In a ∆ABC, AB = BC = CA = 2a and AD ⊥ BC. Prove that

**(i) AD = a √3**

**(ii) area (∆ABC) = √3 a ^{2} **

**Solution:**

(i)In ∆ABD and ∆ACD,∠ADB = ∠ADC = 90° [Given]

AB = AC [Given]

AD = AD [Common]

So, By RHS condition

∆ABD ≅ ∆ACD

Hence, by corresponding parts of congruent triangles

∴ BD = CD = a

Now In ∆ABD,

By using Pythagoras theorem

AD

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

^{2}+ a^{2}= 2a^{2}AD

^{2}= 4a^{2}– a^{2}= 3a^{2}AD = a√3

(ii)Area (∆ABC) = 1/2 x BC x AD= 1/2 x (2a) x (a√3)

= √3 a

^{2}Hence Proved

### Question 14. The lengths of the diagonals of a rhombus are 24 cm and 10 cm. Find each side of the rhombus.

**Solution:**

To find: Each side of the rhombus.

Let ABCD be a rhombus with diagonals AC and BD.

∴ AC = 24cm and BD = 10cm

We know that,

AO = OC = 12cm and BO = OD = 3cm [Diagonals of a rhombus bisect each other at right angle]

Now In ∆AOB,

By using Pythagoras theorem

AB

^{2}= AO^{2}+ BO^{2 }= 12

^{2}+ 5^{2}= 144 + 25

= 169

∴ AB = √(169) = 13cm

∴ AB = BC = CD = AD = 13cm. [The sides of rhombus are all equal.]

Hence, the sides of the rhombus are as follows AB = BC = CD = AD = 13cm.