NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5

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

• Triangles Class 10 Ex 6.1
• Triangles Class 10 Ex 6.2
• Triangles Class 10 Ex 6.3
• Triangles Class 10 Ex 6.4
• Triangles Class 10 Ex 6.6

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5

NCERT Solutions for Class 10 Maths

Page No: 150

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). Given that sides are 7 cm, 24 cm, and 25 cm. Squaring the lengths of these sides we get 49, 576, and 625.
Clearly, 49 + 576 = 625 or 7² + 24² = 25² .
Therefore, given triangle is satisfying Pythagoras theorem. So, it is a right triangle. The longest side in a right angled triangle is the hypotenuse.
Therefore length of hypotenuse of this triangle = 25 cm.

(ii).Given that sides are 3 cm, 8 cm, and 6 cm. Squaring the lengths of these sides we may get 9, 64, and 36. Clearly, sum of squares of lengths of two sides is not equal to square of length of third side. Therefore given triangle is not satisfying Pythagoras theorem. So, it is not a right triangle

(iii).Given that sides are 50 cm, 80 cm, and 100 cm. Squaring the lengths of these sides we may get 2500, 6400, and 10000. Clearly, sum of squares of lengths of two sides is not equal to square of length of third side. Therefore given triangle is not satisfying Pythagoras theorem. So, it is not a right triangle.

(iv).Given that sides are 13 cm, 12 cm, and 5 cm. Squaring the lengths of these sides we may get 169, 144, and 25. Clearly, 144 +25 = 169 Or, 12² + 5² = 13².
Therefore given triangle is satisfying Pythagoras theorem. So, it is a right triangle.
The longest side in a right angled triangle is the hypotenuse.
Therefore length of hypotenuse of this triangle = 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 PM² = QM × MR.

Solution:

Question 3. 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

Solution:

iii.     In  ∆DCA & ∆DAB
∠DCA = ∠DAB = 90º
∠CDA = ∠ADB            (common angle)
∠DAC = ∠DBA            (remaining angle)

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

Solution:

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

Solution:

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

Solution:

Question 7. Prove that the sum of the squares of the sides of rhombus is equal to the sum of the squares of its diagonals.

Solution:

In ∆AOB, ∆BOC, ∆COD, ∆AOD
Applying Pythagoras theorem

Page No: 151

Question 8. 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.

Solution:

Question 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.

Solution:

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 1,000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km per hour. How far apart will be the two planes after hours?

Solution:

Question 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.

Solution:

Question 13. D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE² + BD² = AB² + DE².

Solution:

Question 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 2AB² = 2AC²+ BC².

Solution:

Question 15. In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3BC. Prove that 9AD² = 7AB².

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 ΔABC, 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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