10th Maths 6.5

Exercise 6.5

Q1. Sides of triangles are given below. Determine which of them are right triangles In case of a triangle, write the length of its hypotenuse.

$7cm, 24cm, 25cm$

ii)

$3cm, 8cm, 6cm$

iii)

$50cm, 80cm, 100cm$

iv)

$13cm, 12cm, 5cm$ .

Sol. :

i) We use Pythagoras Theorem

$(25)^2=(24)^2+(7)^2$

$625=576+49$

$625=625$

So, it is a right triangle with hypotenuse is

$25$ .

Answer: Yes,

$25$ .

ii) We use Pythagoras Theorem

$(8)^2=(6)^2+(3)^2$

$64=36+9$

$64≠45$

No, it is not a right triangle.

Answer: No.

iii) We use Pythagoras Theorem

$(100)^2=(80)^2+(50)^2$

$10000=6400+2500$

$10000≠8900$

No, it is not a right triangle.

Answer: No.

iv) We use Pythagoras Theorem

$(13)^2=(12)^2+(5)^2$

$169=144+25$

$169=169$

Yes, it is a right triangle with hypotenuse is

$13$

Answer: Yes,

$13$ .

Q2.

$PQR$ is a triangle right angled at

$P$ and

$M$ is a point on

$QR$ such that

$PM⊥QR$ . Show that

$PM^2=QM.MR$ .

Sol. :

Given:

$PQR$ is a triangle right angled at

$p$ .

$PM⊥QR$

To Prove:

$PM^2=QM.MR$

Proof:

If a perpendicular is drawn from the vertex of the right angle of a right triangle to hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

$triangle PMR∼triangle QMR$

$(PM)/(QM)=(MR)/(PM)$

$PM^2=QM.MR$

Proved.

Q3. In figure,

$ABD$ is a triangle right angled at

$A$ and

$AC⊥BD$ . Show that

$AB^2=BC.BD$

ii)

$AC^2=BC.DC$

iii)

$AD^2=BD.CD$

Sol. :

Given:

$ABD$ is a triangle right angled at

$A$ and

$AC⊥BD$ .

To Prove:

$AB^2=BC.BD$

ii)

$AC^2=BC.DC$

iii)

$AD^2=BD.CD$

Proof:

If a perpendicular is drawn from the vertex of the right angle of a right triangle to hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

$triangleACB∼triangleBAD$ [ By Theorem 6.7]

$(AB)/(BD)=(BC)/(AB)$ [Sides are proportional]

$AB^2=BC.BD$

Proved.

ii)

$triangleACB∼triangleACD$ [ By Theorem 6.7]

$(AC)/(CD)=(BC)/(AC)$ [Sides are proportional]

$AC^2=BC.BD$

Proved

iii)

$triangleDCA∼triangleDAB$ [ By Theroem 6.7]

$(AD)/(BD)=(CD)/(AD)$ [ Side are proportional]

$AD^2=BD.CD$

Proved.

Q4.

$ABC$ is an isosceles triangle right angled at

$C$ . Prove that

$AB^2=2AC^2$ .

Sol. :

Given:

$ABC$ is an isosceles

$triangle$ right-angled at

$C$ .

To prove:

$AB^2=2AC^2$

Proof:

$triangle ABC$

By Pythagoras theorem

$AB^2=BC^2+AC^2$

$triangle ABC$ is an isosceles, So that

$AC=BC$

$AB^2=AC^2+AC^2$

$AB^2=2AC^2$

Proved.

Q5.

$ABC$ is an isosceles triangle with

$AC=BC$ . If

$AB^2=2AC^2$ , prove that

$ABC$ is a right triangle.

Sol. :

Given:

$triangle ABC, AC=BC$

$AB^2=2AC^2$

To Prove:

$ABC$ is a right.

Proof:

$triangle ABC$

$AB^2=2AC^2$

$AB^2=AC^2+AC^2$

$AB^2=BC^2+AC^2$ [

$AC=BC$ ]

$angleACB=90^circ$ [By converse of Pythagoras theorem]

This gives

The given

$triangle ABC$ is a triangle right angled at

$C$ .

Proved.

Q6.

$ABC$ is an equilateral triangle of side

$2a$ . Find each of its altitudes.

Sol. :

Given:

$ABC$ is an equilateral

$triangle$ .

$AB=BC=CA=2a$ .

To Find:

Each its altitudes.

Solve:

In equilateral

$triangle$

$AD=BE=CF$ ____(1) [Altitudes of equilateral

$triangle$ ]

$BD=1/2BC$ [Altitudes of equilateral

$triangle$ ]

$BD=1/2times2a=a$ ____(2)

$AD^2=AB^2-BD^2$ ___(3) [By Pythagoras theorem]

$AD^2=(2a)^2-(a)^2$ [From equations (1), (2) and (3)]

$AD^2=4a^2-a^2$

$AD^2=3a^2$ ________(4)

$AD=BE=CF=asqrt3$ [By equation (1) and (4)]

Answer:

The length of each altitude of equilateral

$triangle$ is

$asqrt3$ .

Q7. Prove that the sum of the squares of the side of a rhombus is equal to the sum of the squares of its diagonals.

Sol. :

Given:

rhombus, its diagonals.

To Prove:

$AB^2+BC^2+CD^2+DA^2=AC^2+BD^2$

Proof:

Let

$ABCD$ is rhombus with diagonal

$AC$ and

$BD$

$AB=BC=CD=DA=x$ ___(1) [Sides of rhombus]

$angleAOB=angleBOC=angleCOD=angleDOA=90^circ$

[Diagonal of rhombus bisect each other at right angle]

$OA=OC=1/2AC$

$OB=OD=1/2BD$

In right

$triangleAOB$

$AB^2=OA^2+OB^2$ [ By Pythagorus theorem]

$x^2=((AC)/2)^2+((BD)/2)^2$

$x^2=(AC)^2/4+(BD)^2/4$

$4x^2=AC^2+BD^2$ ________(2)

$AB^2+BC^2+CD^2+DA^2=4x^2$ _____(3)

[From equation (1)]

$AB^2+BC^2+CD^2+DA^2=AC^2+BD^2$

[By equation (2)&(3)]

Proved.

Q8. In figure,

$O$ is a point in the interior of a triangle

$ABC$ ,

$OD⊥BC,OE⊥AC$ and

$OF⊥AB$ . Show that

$OA^2+OB^2+OC^2-OD^2-OE^2-OF^2=AF^2+BD^2+CE^2$ ,

ii)

$AF^2+BD^2+CE^2=AE^2+CD^2+BF^2$ .

Sol. :

Given:

$triangle ABC$ ,

$O$ is interior

$OD⊥BC, OE⊥AC, OF⊥AB$

Construction:

Join

$OA, OB,$ and

$OC$

Proof:

i) In right

$triangle OFA$

$OA^2-OF^2=AF^2$ ____(1)

[By Pythagoras theorem]

In right

$triangle OBD$

$OB^2-OD^2=BD^2$ ____(2)

[By Pythagoras theorem]

In right

$triangleOEC$

$OC^2-OE^2=CE^2$ ____(3)

[By Pythagoras theorem]

By equation (1)+(2)+(3)

$OA^2-OF^2+OB^2-OD^2+OC^2-OE^2=AF^2+BD^2+CE^2$

$OA^2+OB^2+OC^2-OF^2-OD^2-OE^2=AF^2+BD^2+CE^2$ ___(4)

Proved.

ii)

In right

$triangle OEA$

$OA^2-OE^2=AE^2$ ____(5)

[By Pythagoras theorem]

In right

$triangle OFB$

$OC^2-OD^2=BF^2$ ____(6)

[By Pythagoras theorem]

In right

$triangle ODC$

$OC^2-OD^2=CD^2$ ____(7)

By equation (5)+(6)+(7)

$OA^2-OE^2+OB^2-OF^2+OC^2-OD^2=AE^2+BF^2+CD^2$ ___(8)

$AF^2+BD^2+CE^2=AE^2+CD^2+BF^2$

[By equation (4), (8)]

Proved.

Q9. A ladder

$10m$ long reaches a window

$8m$ above the ground. Find the distance of the foot of the ladder from base of the wall.

Sol. :

Given:

Length of ladder

$10 m$ , the height of the window

$8 m$

To find:

The distance of the foot of the ladder from the base of the wall.

Solve:

In right

$triangle ABC$

By Pythagoras theorem

$AC^2=AB^2-BC^2$

$AC^2=(10)^2-(8)^2$

$AC^2=100-64$

$AC^2=36$

$AC^2=(6)^2$

$AC=6$

Answer:

The distance of the foot of the ladder from the base of the wall is

$6 m$ .

Q10. A guy wire attached to a vertical pole of height

$18m$ is

$24m$ 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. :

Given:

Length of vertical pole

$18 m$

length of wire

$24 m$

To find:

The distance from the base of the pole should be the stake.

Solve:

In right

$triangle ABC$

By Pythagoras theorem

$AC^2= AB^2-BC^2$

$AC^2=(24)^2-(18)^2$

$AC^2=576-324$

$AC^2=252$

$AC^2=(6sqrt7)^2$

$AC=6sqrt7 m$

Answer:

The distance from the base of the pole should the stake is

$6sqrt7 m$ .

Q11. An aeroplane leaves an airport and flies due north at a speed of

$1000km$ per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of

$1200km$ per hour. How far apart will be the two planes after

$1(1)/2$ hours?

Sol. :

Given:

Speed of first aeroplane

$1000 km$ per hour

Speed of second aeroplane

$1200 km$ per hour

Time

$1(1)/2$ hour.

To find:

The distance of two planes after

$1(1)/2$ hour.

Solve:

First aeroplane cover after

$1(1)/2=3/2$ hour

distance

$=speed times time$

$=1000times3/2=1500 km$

Second aeroplane cover after

$1(1)/2=3/2$ hour

distance

$=speed times time$

$=1200times3/2=1800 km$

In right

$triangle ABC$

By Pythagoras theorem

$AB^2=AC^2+BC^2$

$AB^2=(1500)^2+(1800)^2$

$AB^2=2250000+3240000$

$AB^2=5490000$

$AB^2=(300sqrt61)^2$

$AB=300sqrt61 km$

Answer:

The distance of two plans after

$1(1)/2$ hour is

$300sqrt61 km$ .

Q12. Two poles of heights

$6m$ and

$11m$ stand on a plane ground. If the distance between the feet of the poles is

$12m$ , find the distance between their tops.

Sol. :

Given:

Height of poles

$6m, 11m$

Distance between poles

$12m$

To find:

Distance between their tops.

Construction:

Draw

$AE⊥CD$

$AE=BD=12m$

$ED=AB=6m$

$CE=5m$

Solve:

In right

$triangle AEC$

By Pythagoras theorem

$AC^2=AE^2+CE^2$

$AC^2=(12)^2+(5)^2$

$AC^2=144+25$

$AC^2=169$

$AC^2=(13)^2$

$AC=13 m$

Answer:

The distance between the top of the poles is

$13 m$ .

Q13.

$D$ and

$E$ are points on the sides

$CA$ and

$CB$ respectively of a triangle

$ABC$ right angled at

$C$ . Prove that

$AE^2+BD^2=AB^2+DE^2$ .

Sol. :

Given:

$triangle ABC, angleA=90^circ,$

$D,$ and

$E$ are points on

$AC$ and

$BC$ .

Construction:

Join

$DE, BD,$ and

$AE$ .

To Prove:

$AE^2+BD^2=AB^2+DE^2$

Proof:

In right

$triangle ACE$

By Pythagoras theorem

$AE^2=AC^2+EC^2$ ______(1)

In right

$triangle BCD$

By Pythagoras theorem

$BD^2=BC^2+DC^2$ ______(2)

We add equations (1) and (2)

$AE^2+BD^2=AC^2+EC^2+BC^2+DC^2$ ___(3)

In right

$triangle ACB$

By Pythagoras theorem

$AB^2=AC^2+BC^2$ _______(4)

In right

$triangle DCE$

By Pythagoras theorem

$DE^2=DC^2+EC^2$ _______(5)

We add equation (4) and (5)

$AB^2+DE^2=AC^2+BC^2+DC^2+EC^2$ ____(6)

from equations (3) and (6)

$AE^2+BD^2=AB^2+DE^2$

Proved.

Q14. The perpendicular from

$A$ on side

$BC$ of a

$triangleABC$ intersects

$BC$ at

$D$ such that

$DB=3CD$ (see figure). Prove that

$2AB^2=2AC^2+BC^2$ .

Sol. :

Given:

$triangle ABC,$

$AD⊥BC$

$DB=3CD$

$DB=3/4BC$

$CD=1/4BC$ ______(1)

To Prove:

$2AB^2=2AC^2+BC^2$

Proof:

In right

$triangle ABD$

By Pythagoras theorem

$AB^2 =AD^2+BD^2$ ___(2)

In right

$triangle ADC$

By Pythagoras theorem

$AC^2=AD^2+CD^2$ ____(3)

From equations (2) - (3)

$AB^2-AC^2=AD^2+BD^2-AD^2-CD^2$

$AB^2-AC^2=BD^2-CD^2$

$AB^2-AC^2=(BD+CD)(BD-CD)$

$AB^2-AC^2=(3/4BC+1/4BC)(3/4BC-1/4BC)$

$AB^2-AC^2=BCtimes1/2BC$

$AB^2-AC^2=1/2BC^2$

$2AB^2-2AC^2=BC^2$

$2AB^2=2AC^2+BC^2$ .

Proved.

Q15. In an equilateral triangle

$ABC, D$ is a point on side

$BC$ such that

$BD=1/3BC$ . Prove that

$9AD^2=7AB^2$ .

Sol. :

Given:

Equilateral

$triangle ABC$ ,

$D$ is on side

$BC$

such that

$BD=1/3BC$

Construction:

Draw

$AE⊥BC$

Proof:

$AB=BC=CA$ _____(1)

(Sides of equilateral

$triangle$ )

$BD=1/3BC$ ____(2) (Given)

$BE=EC=1/2BC=AB/2$ ___(3)

(Altitude

$AE$ of equilateral

$triangle$ bisects the

$BC$ )

$DE=BE-BD$

$DE=(BC)/2-(BC)/3$ __(4)[From eq. (2),(3)]

$DE=(3BC-2BC)/6$

$DE=(BC)/6=(AB)/6$ __(5)[From eq.(4), (1)]

In right

$triangle AEB$

By Pythagoras theorem

$AE^2=AB^2-BE^2$ _____(6)

In right

$triangle AED$

By Pythagoras theorem

$AE^2=AD^2-DE^2$ _____(7)

$AB^2-BE^2=AD^2-DE^2$ ___(8)[From eq.(6), (7)]

$AB^2-((AB)/2)^2=AD^2-((AB)/6)^2$ [From eq.(8), (3), (5)]

$AB^2-(AB)^2/4=AD^2-(AB)^2/36$

$36AD^2-9AB^2=36AD^2-AB^2$

$36AD^2=36AB^2+AB^2-9AB^2$

$36AD^2=28AB^2$

$9AD^2=7AB^2$

Proved.

Q16. 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. :

Given:

Let

$triangle ABC$ an equilateral,

$AD$ is altitude.

To Prove:

$3AB^2=4AD^2$

Proof:

$triangle ABC$

$BD=CD$ [

$D$ bisect

$BC$ ]

$1/2BC=1/2AB$ __(1) [

$AB=BC=CA$ ]

In right

$triangle ADB$

By Pythagoras theorem

$AB^2=AD^2+BD^2$ __(2)

$AB^2=AD^2+((AB)/2)^2$

$AB^2=AD^2+(AB)^2/4$

`4AB^2=AD^2+AB^2

$3AB^2=4AD^2$

Proved.

Q17. Tick the correct answer and justify: In

$triangleABC$ ,

$AB=6sqrt3cm$ and

$BC=6cm$ .

The angle

$B$ is:

$120^circ$

$60^circ$

$90^circ$

$45^circ$

Sol. :

$AB^2+BC^2=AC^2$ [By Pythagoras theorem]

$(6sqrt3)^2+(6)^2=(12)^2$

$108+36=144$

$144=144$

So that

$angleB=90^circ$ .

Answer:

(c)

$90^circ$

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10th Maths 6.5

NCERT Class 10th solution of Exercise 6.1

NCERT Class 10th solution of Exercise 6.2

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NCERT Class 10th solution of Exercise 7.1

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Exercise 6.5

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