NCERT Solution For Class 10, Maths, Chapter 6 Triangles, Exercise 6.5

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.

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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 (c²= 625)

We have, a² + b² = 7² + 24² = 49 + 576 = 625 = c²

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, a² + c² = 3² + 6² = 9 + 36 = 45 ≠ b²

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, a² + b² = 50² + 80² = 2500 + 6400 = 8900 ≠ c²

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, b² + c² = 12² + 5² = 144 + 25 = 169 = a²

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 PM² = QM. MR

Ans:

Given: PQR is a triangle right angled at P.

PM⊥QR

To prove: PM² = QM. MR

Proof: Since PM⊥QR

∴ ∆ PMQ ~ ∆RMP

⇒ $\frac{PM}{RM}=\frac{QR}{PM}$

⇒ PM² = QM. MR (Hence proved)

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

(i) AB² = BC . BD

(ii) AC² = BC . DC

(iii) AD² = BD . CD

Ans:

Q.4. ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC².

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

∴ AB² = AC² + BC²

⇒ AB² = AC² + AC² [∵ BC = AC, Given]

⇒ AB² = 2AC²

Q.5. ABC is an isosceles triangle with AC = BC. If AB² = 2AC², prove that ABC is a right triangle.

Ans: Since, ABC is an isosceles right triangle with AC = BC and AB² = 2AC²

∴ AB² = AC² + BC²

⇒ AB² = AC² + AC² [∵ BC = AC, Given]

⇒ AB² = AC² + AC²

⇒ AB² = AC² + BC² [∵ 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.

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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.

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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) OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE²,

(ii) AF² + BD² + CE² = AE² + CD² + BF².

Ans: Join AO, BO and CO

(i) OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE²

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

OA² = AF² + OF²

OB² = OD²+ BD²

And, OC² = CE² + OE²

Adding all these, we get

⇒ OA² + OB² + OC² = AF² + OF² + OD² + BD²+ CE² + OE²

⇒ OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE² (Hence proved)

(ii) AF² + BD² + CE² = AE² + CD² + BF²

We have proved that

AF² + BD² + CE²= OA² + OB² + OC² – OD² – OE² – OF²

By rearranging them, we get

⇒ AF² + BD² + CE²= OA² – OE²+ OB² – OF²+ OC² – OD²

⇒ AF² + BD² + CE²= AE²+ BF²+ CD² or

⇒ AF² + BD² + CE²= AE²+ CD²+ BF²(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.

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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?

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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?

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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.

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