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https://www.urbanpro.com/ncert-solutions/class-10/maths-triangles-exercise-6-5 # Learn Exercise 6.5 with Free Lessons & Tips

##### All Exercises - Chapter 6 - Triangles

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.

Let OA be the wall and AB be the ladder.

Therefore, by Pythagoras theorem,

OB=6m

Therefore, the distance of the foot of the ladder from the base of the wall is 6 m.

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Comments 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?

Let OB be the pole and AB be the wire.

By Pythagoras theorem,

Therefore, the distance from the base is m.

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Comments 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?

Distance travelled by the plane flying towards north in

Similarly, distance travelled by the plane flying towards west in

Let these distances be represented by OA and OB respectively.

Applying Pythagoras theorem,

Distance between these planes after, AB

= km

Therefore, the distance between these planes will be km after.

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Comments Tick the correct answer and justify :

In , AB = cm, AC = 12 cm and BC = 6 cm. The angle B is :

(A) (B) (C) (D)

Given : AB = 6√3cm

AC= 12 cm

BC=6 cm

Ans : check if (AC)²=(BC)²+(AB)²

12²=(6)²+(6√3)²

144= 36+108

144=144

Angle B is 90º

Option (C) is correct

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Comments

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

(i) The sides of the triangle are 7 cm, 24 cm, and 25 cm.

Squaring the lengths of these sides, we will obtain 49, 576, and 625.

49 + 576 = 625

Or,

Satisfies the pythagoras theorem

It is a right triangle.

The length of the hypotenuse is is 25 cm.

(ii) The sides of the triangle are 3 cm, 8 cm, and 6 cm.

Squaring the lengths of these sides, we will obtain 9, 64, and 36.

However, 9 + 36 ≠ 64

Or, 3^{2} + 6^{2} ≠ 8^{2}

The sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.

Therefore, the given triangle is not satisfying Pythagoras theorem.

Hence, it is not a right triangle.

(iii)The sides are 50 cm, 80 cm, and 100 cm.

Squaring the lengths of these sides, we will obtain 2500, 6400, and 10000.

However, 2500 + 6400 ≠ 10000

Or, 50^{2} + 80^{2} ≠ 100^{2}

The sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.

Therefore, the given triangle is not satisfying Pythagoras theorem.

Hence, it is not a right triangle.

(iv)Thesides are 13 cm, 12 cm, and 5 cm.

Squaring the lengths of these sides, we will obtain 169, 144, and 25.

144 +25 = 169

Or,

The sides of the given triangle are satisfying Pythagoras theorem.

Therefore, it is a right triangle.

The length of the hypotenuse of this triangle is 13 cm.

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Comments PQR is a triangle right angled at P and M is a point on QR such that . Show that .

Ler

In

Similarly, in

In ,

( By AAA similarity)

Therefore,

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Comments In the figure, ABD is a triangle right angled at A and .

Show that:

(i)

(ii)

(iii)

(Each 90 degress)

(Common angle)

(AA similarity criteria)

(Each 90 degress)

( By AAA similarity)

∠DCA = ∠ DAB (Each 90º)

∠CDA = ∠ ADB (Common angle)

( By AAA similarity)

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

Given that ΔABC is an isosceles triangle.

∴ AC = CB

Applying Pythagoras theorem in ΔABC (i.e., right-angled at point C), we obtain

(AC=CB)

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

(As AC =BC)

The triangle satisfies Pythagoras theorem

Hence it is a right angled triangle

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

Let AD be the altitude in the given equilateral triangle, ΔABC.

We know that altitude bisects the opposite side.

∴ BD = DC = *a*

*In *

*Applying pythagoras theorem, we get*

In an equilateral triangle, all the altitudes are equal in length.

Therefore, the length of each altitude will be.

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

In ΔAOB, ΔBOC, ΔCOD, ΔAOD,

Applying Pythagoras theorem, we obtain

On addition of the above equations, we get

(diagonals bisect each other)

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Comments In the figure, O is a point in the interior of a triangle ABC, and .

Show that:

(i) ,

(ii)

Join OA, OB, and OC.

(i) Applying Pythagoras theorem in ΔAOF, we obtain

Similarly, in ΔBOD,

Similarly, in ΔCOE,

On addition of these equations,

(ii) From the above result,

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

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Comments 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.

Given : distance btween two poles = 12 m

Small pole height = 6 m

Big pole height = 11 m

To find : distnace btween two poles

Small to top pole distance can be find by pythgores therom top (hypotenuse)²=(base)²+(height)²

Base =12 m

Height = 11-6= 5 m

(Hypotenuse)²=(12)²+(5)²

Hypotenuse =√(12)²+(5)²m

Hypotenuse=√144+25= √169m

distance between their tops is13 m.

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

On applying the Pythagoras theorem in

Using (i) and (ii) we obtain

Applying the pythagoras theorem in we get

Applying the pythagoras theorem in we get

Placing the values in (iii)

AE^{2 }+ BD^{2} = AB^{2} + DE^{2}

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

Applying Pythagoras theorem for ΔACD, we obtain

Applying Pythagoras theorem in ΔABD, we obtain

From (i) and (ii)

It is given that 3DC=DB

and

Placing these values in (iii)

On solving, it is proved that 2 AB^{2} = 2 AC^{2} + BC^{2}

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

Let the side of the equilateral triangle be *a*, and AE be the altitude of ΔABC.

∴ BE = EC = =

And, AE =

Given that, BD = BC

∴ BD =

DE = BE − BD =

Applying Pythagoras theorem in ΔADE, we obtain

AD^{2} = AE^{2} + DE^{2}

^{}

^{}

^{}

⇒ 9 AD^{2} = 7 AB^{2}

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Comments 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.

Let the side of the equilateral triangle be *a*, and AE be the altitude of ΔABC.

∴ BE = EC = =

Applying Pythagoras theorem in ΔABE, we obtain

AB^{2} = AE^{2} + BE^{2}

4AE^{2} = 3*a*^{2}

⇒ 4 × (Square of altitude) = 3 × (Square of one side)

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Comments Exercise 6.5

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