NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 6 |

Chapter Name |
Triangles |

Exercise |
Ex 6.5 |

Number of Questions Solved |
17 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5

**Question 1.**

**Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.**

(i) 7 cm, 24 cm, 25 cm (ii) 3 cm, 8 cm, 6 cm

(iii) 50 cm, 80 cm, 100 cm (iv) 13 cm, 12 cm, 5 cm

**Solution:**

**(i) 7 cm, 24 cm,-25 cm**

(7)^{2} + (24)^{2} = 49 + 576 = 625 = (25)^{2} = 25

∴ The given sides make a right angled triangle with hypotenuse 25 cm

(ii) 3 cm, 8 cm, 6 cm

(8)^{2} = 64

(3)^{2} + (6)^{2} = 9 + 36 = 45

64 ≠ 45

The square of larger side is not equal to the sum of squares of other two sides.

∴ The given triangle is not a right angled.

**
(iii)** 50 cm, 80 cm, 100 cm

(100)

^{2}= 10000

(80)

^{2}+ (50)

^{2}= 6400 + 2500

= 8900

The square of larger side is not equal to the sum of squares of other two sides.

∴The given triangle is not a right angled.

**13 cm, 12 cm, 5 cm**

(iv)

(iv)

(13)

^{2}= 169

(12)

^{2}+ (5)

^{2}= 144 + 25 = 169

= (13)

^{2}= 13

Sides make a right angled triangle with hypotenuse 13 cm.

**Question 2.**

**PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM ^{2} = QM • MR.**

**Solution:**

In right angled ∆QPR,

∠P = 90°, PM ⊥ QR

∴ ∆PMQ ~ ∆RMP

[If ⊥ is drawn from the vertex of right angle to the hypotenuse then triangles on both sides of perpendicular are similar to each other, and to whole triangle]

⇒ [Corresponding sides of similar

⇒ PM x MP = RM x MQ ⇒ PM2 = QM.MR

**Question 3.**

**In the given figure, ABD is a triangle right angled at A and AC i. BD. Show that**

**(i) AB ^{2} = BC.BD
(ii) AC^{2} = BC.DC
(iii) AD^{2} = BD.CD**

**Solution:**

**Question 4.**

**ABC is an isosceles triangle right angled at C. Prove that AB ^{2} = 2AC^{2}.**

**Solution:**

**Given:**In ∆ABC, ∠C = 90° and AC = BC

**To Prove:**AB

^{2}= 2AC

^{2}

**Proof:**In ∆ABC,

AB

^{2}= BC

^{2}+ AC

^{2}

AB

^{2}= AC

^{2}+ AC

^{2}[Pythagoras theorem]

= 2AC

^{2}

**Question 5.**

**ABC is an isosceles triangle with AC = BC. If AB ^{2} = 2AC^{2} , Prove that ABC is a right triangle.**

**Solution:**

**Question 6.**

**ABC is an equilateral triangle of side la. Find each of its altitudes.**

**Solution:**

**Given:** In ∆ABC, AB = BC = AC = 2a

We have to find length of AD

In ∆ABC,

AB = BC = AC = 2a

and AD ⊥ BC

BD = \(\frac { 1 }{ 2 }\) x 2 a = a

In right angled triangle ADB,

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

⇒ AD^{2} = AB^{2} – BD^{2}= (2a)^{2} – (a)^{2} = 4a^{2}– a^{2}= 3a^{2}

AD = √3a

**Question 7.**

**Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.**

**Solution:**

**Given:** ABCD is a rhombus. Diagonals AC and BD intersect at O.

**To Prove:** AB^{2}+ BC^{2}+ CD^{2}+ DA^{2 }= AC^{2}+ BD^{2}

**Question 8.**

**In the given figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that**

(i) OA^{2} + OB^{2} + OC^{2} – OD^{2} – OE^{2} – OF^{2} = AF^{2} + BD^{2} + CE^{2}

(ii) AF^{2} + BD^{2} + CE^{2} = AE^{2} + CD^{2} + BF^{2}.

**Solution:**

**Question 9.**

**A ladder 10 m long reaches a window 8 m above the ground. ind the distance of the foot of the ladder from base of the wall.**

**Solution:**

Let AC be the ladder of length 10 m and AB = 8 m

In ∆ABC, BC^{2} + AB^{2} = AC^{2}

⇒ BC^{2}= AC^{2} – AB^{2}= (10)^{2} – (8)^{2}

BC^{2} = 100-64 – 36 BC = √36 = 6 m

Hence distance of foot of the ladder from base of the wall is 6 m.

**Question 10.**

**A guy wire attached to a vertical pole of height 18 m is 24 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:**

**Question 11.**

**An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after 1\(\frac { 1 }{ 2 }\) hours?**

**Solution:**

**Question 12.**

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

**Solution:**

Length of poles is 6 m and 11m.

DE = DC – EC = 11m-6m = 5m

In ∆DAE,

AD^{2} = AE^{2} + DE^{2} [ ∵AE = BC]

= (12)^{2} + (5)^{2} =144 + 25 = 169

AD = √l69 = 13

**Question 13.**

**D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE ^{2} + BD^{2} = AB^{2} + DE^{2}.**

**Solution:**

**Question 14.**

**The perpendicular from A on side BC of a ∆ABC intersects BC at D such that DB = 3CD (see the figure). Prove that 2AB ^{2 }= 2AC^{2} + BC^{2}.**

**Solution:**

**Question 15.**

**In an equilateral triangle ABC, D is a point on side BC, such that BD = \(\frac { 1 }{ 3 }\)BC. Prove that 9AD ^{2} = 7AB^{2}.**

**Solution:**

**Question 16.**

**In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.**

**Solution:**

**Question 17.**

**Tick the correct answer and justify : In ∆ABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm. The angle B is:**

**(a) 120°
(b) 60°
(c) 90°
(d) 45**

**Solution:**

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