**NCERT Solution For Class 10, Maths, Chapter 6 Triangles, Exercise 6.5** includes day to day examples which can be solved by practicing all the basic concepts of property of triangles. Exercise 6.5 class 10 maths, chapter 6 is formula based.

Table of Contents

Toggle## Class 10, Maths, Chapter 6, Exercise 6.5 Solutions

**Q.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**

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

Here the larger side is c = 25 cm (c^{2 }= 625)

We have, a^{2} + b^{2} = 7^{2} + 24^{2} = 49 + 576 = 625 = c^{2}

So, given sides are the sides of right triangle. Its hypotenuse is 25 cm.

**(ii) Let a = 3 cm, b = 8 cm and c = 6 cm**

Here the larger side is c= 8 cm

We have, a^{2} + c^{2} = 3^{2} + 6^{2} = 9 + 36 = 45 ≠ b^{2}

So, given sides are not the sides of right triangle.

**(iii) Let a = 50 cm, b = 80 cm and c = 100 cm**

Here the large side is c = 100 cm

We have, a^{2} + b^{2} = 50^{2} + 80^{2} = 2500 + 6400 = 8900 ≠ c^{2}

So, given sides are not the sides of right triangle.

**(iv) let a = 13 cm, b = 12 cm and c = 5 cm.**

Here the larger sides is a = 13 cm

We have, b^{2} + c^{2} = 12^{2} + 5^{2} = 144 + 25 = 169 = a^{2}

So, given sides are the sides of right triangle. Its hypotenuse is 13 cm.

**Q.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**

**Ans:**

Given: PQR is a triangle right angled at P.

PM⊥QR

**To prove:** PM^{2} = QM. MR

**Proof:** Since PM⊥QR

∴ ∆ PMQ ~ ∆RMP

⇒ $\frac{PM}{RM}=\frac{QR}{PM}$

⇒ PM^{2} = QM. MR **(Hence proved)**

**Q.3. In Figure, ABD is a triangle right angled at A and AC ****⊥**** BD. Show that**

**(i) AB ^{2} = BC . BD**

**(ii) AC ^{2} = BC . DC**

**(iii) AD ^{2} = BD . CD**

** **

**Ans:**

**Q.4. ABC is an isosceles triangle right angled at C. Prove that AB ^{2} = 2AC^{2}.**

**Ans:** Since, ABC is an isosceles right triangle, right angled at C.

∴ AB^{2} = AC^{2} + BC^{2}

⇒ AB^{2} = AC^{2} + AC^{2} [∵ BC = AC, Given]

⇒ AB^{2} = 2AC^{2}

**Q.5. ABC is an isosceles triangle with AC = BC. If AB ^{2} = 2AC^{2}, prove that ABC is a right triangle.**

**Ans:** Since, ABC is an isosceles right triangle with AC = BC and AB^{2} = 2AC^{2}

∴ AB^{2} = AC^{2} + BC^{2}

⇒ AB^{2} = AC^{2} + AC^{2} [∵ BC = AC, Given]

⇒ AB^{2} = AC^{2} + AC^{2}

⇒ AB^{2} = AC^{2} + BC^{2} [∵ AC = BC, Given]

Using converse of Pythagoras theorem

∴ ∆ABC is right angled at AC.

**Q.6. ABC is an equilateral triangle of side 2a. Find each of its altitudes.**

**Ans:**

**Q.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.**

**Ans:**

**Q.8. In Fig. 6.54, 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}.**

**Ans:** Join AO, BO and CO

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

By Pythagoras theorem, in right ∆s AFO, BOD and COE, we have

OA^{2} = AF^{2} + OF^{2}

OB^{2} = OD^{2 }+ BD^{2}

And, OC^{2} = CE^{2} + OE^{2}

Adding all these, we get

⇒ OA^{2} + OB^{2} + OC^{2} = AF^{2} + OF^{2} + OD^{2} + BD^{2 }+ CE^{2} + OE^{2}

⇒ OA^{2} + OB^{2} + OC^{2} – OD^{2} – OE^{2} – OF^{2} = AF^{2} + BD^{2} + CE^{2} **(Hence proved)**

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

We have proved that

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

By rearranging them, we get

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

⇒ AF^{2} + BD^{2} + CE^{2 }= AE^{2 }+ BF^{2 }+ CD^{2} or

⇒ AF^{2} + BD^{2} + CE^{2 }= AE^{2 }+ CD^{2}+ BF^{2 }**(Hence proved)**

**Q.9. A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.**

**Ans:**

**Q.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?**

**Ans:**

**Q.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 ** ** hours?**

**Ans:**

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

**Ans:**

**Q.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}.**

**Ans:**

**Q.14. The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD (see Fig. 6.55). Prove that 2AB ^{2} = 2AC^{2} + BC^{2}.**

**Ans:**

**Q.15. In an equilateral triangle ABC, D is a point on side BC such that BD = $\frac{1}{3}$BC. Prove that 9 AD ^{2} = 7 AB^{2}.**

**Ans:**

**Q.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.**

**Ans:**

**Q.17. Tick the correct answer and justify: In Δ ABC, AB = ** ** cm, AC = 12 cm and BC = 6 cm. The angle B is:**

**(A) 120° **

**(B) 60°**

**(C) 90°**

**(D) 45°**

**Ans:** **(C) is the correct answer.**