In this chapter, we provide RD Sharma Class 10 Ex 4.7 Solutions Chapter 4 Triangles for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 10 Ex 4.7 Solutions Chapter 4 Triangles pdf, Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 4 |

Chapter Name | Triangles |

Exercise | 4.7 |

Category | RD Sharma Solutions |

**RD Sharma Solutions for Class 10 Chapter** **4** **Triangles Variables Ex 4.7 Download PDF**

**Chapter 4: Triangles Exercise – 4.7**

**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:**

We have,

Sides of triangle

AB = 3 cm

BC = 4 cm

AC = 6 cm

∴ AB^{2} = 3^{2} = 9

BC^{2 }= 4^{2} = 16

AC^{2} = 6^{2} = 36

Since, AB^{2} + BC^{2} ≠ AC^{2}

Then, 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:**

We have,

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

∴ a^{2} = 49, b^{2} = 576 and c^{2} = 625

Since, a^{2} + b^{2} = 49 + 576 = 625 = C^{2}

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

We have,

a = 9 cm, b = 16 cm and c = 18 cm

∴ a^{2} = 81, b^{2} = 256 and c^{2} = 324

Since, a^{2} + b^{2} = 81 + 256 = 337 ≠ c^{2}

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

We have,

a = 1.6 cm, b = 3.8 cm and C = 4 cm

∴ a^{2} = 64, b^{2} = 100 and c^{2} = 36

Since, a^{2} + c^{2} = 64 + 36 = 100 = b^{2}

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

**Question: 3**

A man goes 15 metres due west and then 8 metres 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.

∴ In ∆ABO, by Pythagoras theorem AO^{2} = AB^{2} + BO^{2}

⟹ AO^{2} = 8^{2} + 15^{2}

⟹ AO^{2} = 64 + 225 = 289

⟹ AO = √289 = 17m

∴ He 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 Pythagoras theorem

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

⟹ 15^{2} + BC^{2} = 17^{2}

⟹ 225 + BC^{2} = 17^{2}

⟹ BC^{2} = 289 – 225

⟹ BC^{2} = 64

⟹ BC = 8 m

∴ 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 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 triangle APC, by applying Pythagoras theorem

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

12^{2} + 5^{2} = AC^{2}

AC^{2} = 144 + 25 = 169

AC = 13

Therefore 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:**

We have

AB = AC = 25 cm and BC = 14

In ∆ABD and ∆ACD

∠ADB = ∠ADC [Each = 90°]

AB = AC [Each = 25 cm]

AD = AD [Common]

Then, ∆ABD ≅ ∆ACD [By RHS condition]

∴ BD = CD = 7 cm [By c.p.c.t]

In ∆ADB, by 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 Pythagoras theorem

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

⟹ l^{2} = 8^{2} + 6^{2} … (i)

In ∆BCE, by pythagoras theorem

BE^{2} = BC^{2} + CE^{2}

⟹ l^{2} = BC^{2} + 8^{2} … (ii)

Compare (i) and (ii)

BC2 + 8^{2} = 8^{2} + 6^{2}

⟹ BC^{2} + 6^{2}

⟹ BC = 6 m

**Question: 8**

Two poles of height 9 in and 14 m stand on a plane ground. If the distance between their feet is 12 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 Pythagoras theorem,

AE^{2} = AB^{2} + BE^{2}

⟹ AE^{2} = 5^{2} + 12^{2}

⟹ AE^{2} = 25 + 144 = 169

⟹ AE= √169 = 13 m

∴ Distance between their tops = 13 m

**Question: 9**

Using Pythagoras theorem determine the length of AD in terms of b and c shown in Fig. 4.219

**Solution:**

We have,

In ∆BAC, by Pythagoras theorem

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

⟹ BC^{2} = c^{2} + b^{2}

In ∆ABD and ∆CBA

∠B = ∠B [Common]

∠ADB = ∠BAC [Each 90°]

Then, ∆ABD ͏~ ∆CBA [By AA similarity]

[Corresponding parts of similar ∆ are proportional]

**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.

= 30 cm^{2}

**All Chapter RD Sharma Solutions For Class10 Maths**

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