NCERT Solutions class 10 Maths Exercise 6.5 Ch 6 Triangles

NCERT Solutions for Class 10 Maths Exercise 6.5 Chapter 6 Triangles – FREE PDF Download

NCERT Class 10 Maths Ch 6 is one of the most important ones in the NCERT syllabus. Duly following NCERT Solutions for Class 10 Maths Chapter
6 ensures you that there will be no hindrance when you opt for more advanced branches of Maths. This is where CoolGyan comes in. Our free Class 10 Triangles solutions will help you understand the chapter thoroughly.

NCERT Solutions for Class 10 Maths Chapter 6 – Triangles

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 = 7 cm, = 24 cm and = 25 cm

Here the larger side is = 25 cm.

We have, 

So, the triangle with the given sides is a right triangle. Its hypotenuse = 25 cm

(ii) Let = 3 cm, = 8 cm and = 6 cm

Here the larger side is = 8 cm.

We have, 

So, the triangle with the given sides is not a right triangle.

(iii) Let = 50 cm, = 80 cm and = 100 cm

Here the larger side is= 100 cm.

We have, 

So, the triangle with the given sides is not a right triangle.

(iv) Let = 13 cm, = 12 cm and = 5 cm

Here the larger side is = 13 cm.

We have, 

So, the triangle with the given sides is a right triangle. Its hypotenuse = 13 cm

2. PQR is a triangle right angled at P and M is a point on QR such that PM  QR. Show that PM2 = QM x MR.

Ans. Given: PQR is a triangle right angles at P and PMQR

To Prove: PM2 = QM.MR

Proof: Since PMQR
 QMPPMR
 QMPM=PMRMQMPM=PMRM
 

3. In the given figure, ABD is a triangle right angled at A and ACBD. Show that:

(i) AB2 = BC.BD

(ii) AC2 = BC.DC

(iii) AD2 = BD.CD

Ans. Given: ABD is a triangle right angled at A and ACBD.

To Prove: (i) AB2 = BC.BD, (ii) (iii) 

Proof:(i) Since ACBD

 CBACAD and each triangle is similar to ABD

 ABDCBA

 

 

(ii) Since ABCDAC

 

 

(iii) Since CADABD

 

 

4. ABC is an isosceles triangle right angled at C. Prove that.

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

 
 [BC = AC, given]
 

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

Ans. Since ABC is an isosceles right triangle with AC = BC and 
 
  [BC = AC, given] ABC is right angled at C. 

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

Ans. Let ABC be an equilateral triangle of side  units.
Draw ADBC. Then, D is the midpoint of BC. 

 BD = BC = 

Since, ABD is a right triangle, right triangle at D.

 

 = AD2 + 

 = 

 Each of its altitude = 

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

Ans. Let the diagonals AC and BD of rhombus ABCD intersect each other at O. Since the diagonals of a rhombus bisect each other at right angles.

  and AO = CO, BO = OD

Since AOB is a right triangle, right angled at O.

 

 = 

[OA = OC and OB = OD]

 ……….(1)

Similarly, we have ……….(2)

……….(3)

……….(4)

Adding all these results, we get

= 

 = 

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) = 

(ii) = 

Ans. Join AO, BO and CO.

(i) In right s OFA, ODB and OEC, we have

, and

Adding all these, we get

= 

 = 

(ii) In right s ODB and ODC, we have

 and 

 ……….(1)

Similarly, we have ……….(2)

and = ……….(3)

Adding equations (1), (2) and (3), we get

=

= 

–= 0

 = 

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

Ans. Let AB be the ladder, B be the window and CB be the wall. Then, ABC is a

right triangle, right angled at C.

 

 

 = 100 – 64

 = 36

 AC = 6

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

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 hand. How far from the base of the pole should the stake be driven so that the wire will be taut?

Ans. Let AB (= 24m) be a guy wire attached to a vertical pole. BC of height 18 m. To keep the wire taut, let it be fixed to a stake at A. Then, ABC is a right triangle, right angled at C.

 

 

 = 576 – 324

 = 252

 AC = 6

Hence, the stake may be placed at a distance of 6m from the base of the pole.

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. Let the first aeroplane starts from O and goes upto A towards north where

OA = km = 1500 km

Let the second aeroplane starts from O at the same time and goes upto

B towards west where

OB = km = 1800 km

According to the question the required distance = BA

In right angled triangle ABC, by Pythagoras theorem, we have,

= 

= 2250000 + 3240000

= 5490000 = 

 AB = km

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. Let AB = 11 m and CD = 6 m be the two poles such that BD = 12 m

Draw CEAB and join AC.

 CE = DB = 12 m

AE = AB – BE = AB – CD = (11 – 6)m = 5 m

In right angled triangle ACE, by Pythagoras theorem, we have

= 

= 144 + 25 = 169

 AC = 13 m

Hence, the distance between the tops of the two poles is 13 m.

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

Ans. In right angled s ACE and DCB, we have

and 

 = 

 = 

 = 

[By Pythagoras theorem,  and ]

14. The perpendicular from A on side BC of a ABC intersects BC at D such that DB = 3CD (see figure). Prove that.

Ans. We have, DB = 3CD

Now, BC = DB + CD

 BC = 3CD + CD

 BC = 4CD

 CD = BC and DB = 3CD = BC ……….(1)

Since, ∆ ABD is a right triangle, right angled at D. Therefore by Pythagoras theorem, we have,

……….(2)

Similarly, from ACD, we have, ……….(3)

From eq. (2) and (3)

 =  [Using eq.(1)]

 

 = 

 AB2 – AC2 = 

 

 

15. In an equilateral triangle ABC, D is a point on side BC such that BD = BC. Prove that.

Ans. Let ABC be an equilateral triangle and let D be a point on BC such that BD = BC

Draw AEBC, Join AD.

In s AEB and AEC, we have,

AB = AC[ABC is equilateral]

AEB = AEC [ each ]

And AE = AE

 By SAS-criterion of similarity, we have

AEB  AEC

 BE = EC

Thus, we have, BD = BC, DC = BC and BE = EC = BC ………(1)

Since, C = 

 ADC is an acute angle triangle.

 

= [using eq.(1)]

 

=  [ AB = BC = AC]

 

 =

 

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. Let ABC be an equilateral triangle and let ADBC. In s ADB and ADC, we have,

AB = AC [Given]

B = C = [Given]

And ADB = ADC[Each = ]

 ADB  ADC[By RHS criterion of congruence]

 BD = DC

 BD = DC = BC

Since ADB is a right triangle, right angled at D, by Pythagoras theorem, we have,

 

 

 [ BC = AB]

 

 

17. Tick the correct answer and justify: In ABC, AB =  cm, AC = 12 cm and BC = 6 cm. the angles A and B are respectively:

(A)  and 

(B)  and 

(C)  and 

(D)  and 

Ans. (C) In ABC, we have, AB = cm, AC = 12 cm and BC = 6 cm.
Now, = = 144 = 
Thus, ABC is a right triangle, right angled at B. B = 

Let D be the midpoint of AC. We know that the midpoint of the hypotenuse of a right triangle is equidistant from the vertices. 

AD = BD = CD

 CD = BD = 6 cm [ CD = AC]

Also, BC = 6 cm

 In BDC, we have, BD = CD = BC

 BDC is equilateral

 ACB = 

 A = 

Thus, A =  and B =