# Triangles : Exercise - 6.5 (Mathematics NCERT Class 10th)

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**Q.1Â Â Â 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.**

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

Â Â Â Â Â Â Â Â Â Â Â **(ii) 3 cm, 8 cm, 6 cm**

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

Â Â Â Â Â Â Â Â Â Â Â **(iv) 13 cm, 12 cm, 5 cm**

**Sol.**

(i)Â Â Â Â Â Let a = 7 cm, b = 24 cm and c = 25 cm

Â Â Â Â Â Â Â Â Â Here the larger side is c = 25 cm

Â Â Â Â Â Â Â Â Â We have,

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

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

Â Â Â Â Â Â Â Â Â Â Here the larger side is b = 8 cm

Â Â Â Â Â Â Â Â Â We have,

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

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

Â Â Â Â Â Â Â Â Â Â Here the larger side is c = 100 cm

Â Â Â Â Â Â Â Â Â Â We have, = 8900

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

(iv)Â Â Â Â Let a = 13 cm, b = 12 cm and c = 5 cm

Â Â Â Â Â Â Â Â Â Â Here the larger side is a = 13 cm

Â Â Â Â Â Â Â Â Â Â We have,

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

**Q.2Â Â Â Â PQR is a triangle right angled at P and M is a point on QR such that . Show that .**

**Sol. **

Given : PQR is a triangle right angled at P and .

To prove :

Proof : Since

Therefore

**Q.3Â Â Â In figure, ABD is a triangle right angled at A and . SHow that**

Â Â Â Â Â Â Â Â Â Â Â Â **(i) **

Â Â Â Â Â Â Â Â Â Â Â Â **(ii) **

Â Â Â Â Â Â Â Â Â Â Â Â **(iii) **

** ****Sol.**

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

To prove :

(i)

(ii)

(iii)

Proof :

(i) Since

Therefore and each triangle is similar to

Because

Therefore

(ii) Since,

Therefore

(iii) Since,

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

* Sol.*Â Â Â Â Â

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

Therefore

[Because BC = AC, given]

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

**Sol.**

Since, ABC is an isosceles triangle with AC = BC and

[Because AC = BC, given]

Therefore ABC is right angled at AC.

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

*Sol.*

Let ABC be an equilateral triangle of side 2a units.

Draw . Then, D is the mid-point of BC.

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

Therefore

Therefore each of its altitude = .

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

**Sol. **

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

Therefore =

and, AO = CO, BO = OD

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

Therefore

[Because OA = OC and OB = OD]

................ (1)

Siumilarly, we have

............... (2)

............... (3)

and ................(4)

Adding all these results, we get

Â **Q.8Â Â Â Â In figure, O is a point in the interior of a triangle ABC, , and . Show that**

Â Â Â Â Â Â Â Â Â Â Â **(i) Â Â Â Â Â Â Â Â **

**(ii)**

**Sol.**Join AO, BO and CO

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

Â Â Â Â Â Â Â

Â Â Â Â Â Â Â

Â Â Â Â Â Â Â and,

Â Â Â Â Â Â Â Â Â Â Â Adding all these, we get

Â Â Â Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â Â Â Â =

(ii)Â Â Â Â Â In right ODB and ODC, we have

Â Â Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â Â Â and,

Â Â Â Â Â Â Â Â Â Â ............... (1)

Â Â Â Â Â Â Â Â Â Â Suimilarly, we have

Â Â Â Â Â Â Â Â Â Â ................ (2)

Â Â Â Â Â Â Â Â Â Â and, ................. (3)

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

Â Â Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â Â Â Â =

Â Â Â Â Â Â Â Â Â Â Â = 0

Â Â Â Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â Â Â Â Â = .

**Q.9Â Â Â 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.**

**Sol. **

Let AB be the ladder, B be the window and CB be the wall. Then, ABC is a right triangle, right angled at C.

Therefore

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

**Q.10Â Â Â 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 ?**

* Sol.*Â Â Â Â Â Â

Let AB (= 24 m) 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.

Because

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

**Q.11Â Â Â 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 1Â½ hours ?**

Sol.

Let the first aeroplane starts from O and goes upto A towards north where km = 1500 km.

Let the second aeroplane starts from O at the same time and goes upto B towards west where km = 1800 km.

According to the problem the required distane = BA.

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

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =

km.

**Q.12Â Â 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.**

**Sol. **

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

Draw and join AC

Therefore CE = DB = 12 m,

AE = AB â€“ BE = AB â€“ CD = (11â€“6) m = 5 m.

In right angled , by Pythagoras theorem, we have

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =

Â Â Â Â Â Â Â Â Â Â Â Â =

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

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

** Sol.**

In right angled ACE and DCB, we have

and,

[By Pythagoras theorem, and ]

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

** ****Sol.**

We have, DB = 3CD

Now, BC = DB + CD

BC = 3CD + CD [Because BD = 3CD]

BC = 4CD

Therefore and,

............. (1)

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

............... (1)

Similarly, from , we have

........... (2)

(1) â€“ (2) gives,

[Using (1)]

Which is the required result.

**Q.15 Â In an equilateral triangle ABC, D is a point on side BC such that . Prove that **

**Sol.**

Let ABC be an equilateral triangle and let D be a point on BC such that

Draw . Join AD.

In AEB and AEC, we have

AB = AC [Because is equilateral]

[Because each = 90Â°]

and, AE = AE

Therefore by SAS-criterion of similarity, we have

BE = EC

Thus, we have

and, ................. (1)

Since,

Therefore is an acute triangle.

Therefore

= [Using (1)]

=

= [Because AB = BC = AC]

=

, which is the required result.

**Q.16Â Â Â 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.**

**Sol.**

Let ABC be an equilateral triangle and let .

In and , we have

AB = AC [Given]

[Each = 60Â°]

and, [Each = 90Â°]

By RHS criterion of congruence, we have

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

[Because BC = AB]

Hence, the result.

**Q.17Â Â Â Tick the correct answer and justify : In , cm, AC = 12 cm and BC = 6 cm. The angles A and B are respectively :**

Â Â Â Â Â Â Â Â Â Â Â Â Â **(a) 90Â° and 30Â°**

Â Â Â Â Â Â Â Â Â Â Â Â Â **(b) 90Â° and 60Â°**

Â Â Â Â Â Â Â Â Â Â Â Â Â **(c) 30Â° and 90Â°**

Â Â Â Â Â Â Â Â Â Â Â Â Â **(d) 60Â° and 90Â°**

**Sol.**

In , we have

AB = cm,

AC = 12 cm

and BC = 6 cm ............... (1)

Now,

= 36 Ã— 3 + 36 = 108 + 36

=

Thus, is a right triangle, right angled at B.

Therefore

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

AD = BD = CD

CD = BD = 6 cm [Because ]

Also BC = 6 cm

Therefore in , we have

BD = CD = BC

is equilateral

Therefore

= 180Â° â€“ (90Â° + 60Â°) = 30Â°

Thus, and

Therefore (C) is the correct choice.