NCERT Solutions For Class 10 Math Chapter – 6 Exercise – 6.5

NCERT Solutions For Class 10 Math Chapter – 6 Exercise – 6.5

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

7 cm , 24 cm , 25 cm
3 cm , 8 cm , 6 cm
50 cm , 80 cm , 100 cm
13 cm , 12 cm , 5 cm

Solution:-

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

49 + 576 = 652

(7)² + (24)² = (25)²

Length of hypotenuse = 25 cm.

Sides of triangle are 3 cm , 8 cm and 6 cm .

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

3²+ 6² ≠ 8²

Hence the given triangles does not satisfies pythagoras theorem.

Squaring the lengths of these sides we wll get 2500 , 6400 and 10000.
2500 + 6400 ≠10000

50² + 80² ≠ 100²

Hence it is not a right triangle.

(iv ) Squaring the lengths of these sides , we will get 169 , 144 and 25.

144 + 25 = 169

12³+ 5²= 13²

Therefore it is a right triangle.

Hence , length of the hypotenuse of this triangle is 13 cm.

Q2. PQR is a triangle right angled at P and M a point on QR such that PM ⊥ QR. Show that PM² = QM X MR.

Solution:-

PM² = PQ² – QM² (I)

In ∆PMR by Pythagoras theorem

PNR²= PM² + MR²

PM² = PR²– MR² (II)

Adding equation (I) and (ii) we get,

2PM² = (PQ² + PM²) – (QM² + MR²)

= QR² – QM²– MR²

= (QM + MR)²– QM² – MR²

= 20M X MR

PM² = QM X MR

Q3. In figure ABD is a triangle right angled at A and AC ⊥ BD. Show that

AB²= BC X BD
AC²= BC X DC
AD²= BD X CD

Solution:-

In ∆ADB and ∆CAB,

∟DAB = ∟ACB (Each 90⁰)

∟ABD = ∟CBA ( Common angle)

∆ADB = ∆CAB ( AA similarity criterion)

AB² = CB X BD

(II) In ∆CBA ,

∟CBA = 180⁰ – 90⁰ – X

∟CBA = 90⁰ – X

Similarity in ∆CAD

∟CAD = 90⁰ – ∟CBA

= 90⁰– X

∟CDA = 180⁰ – 90⁰ – (90⁰ – X)

∟CDA = X

In ∆CBA and ∆CAD , we have

∟CBA = ∟CAD

∟CAB = ∟CDA

∟ACB = ∟DCA

∆CBA = ∆CAD [ AAA similarity criterion]

AC/DC = BC/AC

AC² = DC X BC

(iii ) In ∆DCA and ∆DAB

∟DCA = ∟DAB

∟CDA = ∟ADB

∆DCA = ∆DAB [ AA similarity criterion]

DC/DA = DA/DA

AD² = BD X CD

Q4. ABC is an isosceles triangle right angled at C. Prove that AB²= 2AC².

Solution:-

In ∆ACB , ∟C = 90⁰

AC = BC (By isosceles triangle)

AB² = AC² + BC² [ By pythagoras theorem]

= AC² + AC² [ Since , AC = BC]

AB² = 2AC² .

Q5. ABC is an isosceles triangle with AC = BC If AB² = 2AC² , Prove that ABC is a right triangle.

Solution:-

Triangles Exercise 6.5 Answer 5

In ∆ACB,

AC = BC

AB² = 2AC²

AB² = AC² + AC²

= AC² + BC²

∆ ABC is a right triangle.

Q6. ABC is an equilateral triangle of side 2a Find each of its altitudes.

Solution :-

In ∆ADB and ∆ADC ,

AB = AC

AD = AD

∟ADB = ∟ADC

Therefore , ∆ADB = ∆ADC by RHS congruence.

Hence , BD = DC

In right angled ∆ADB ,

AB² = AD²+ BD²

(2a)² = AD² + A²

AD²= 4A² – A²

AD = √3a.

Q7. Prove that the sum of the square of the sides of rhombus is equal to the sum of the squares of its diagonals.

Solution:-

In ∆AOB ,

∟AOB = 90⁰

AB² = AO² + BO² (I)

AD² = AO² + DO² (II)

DC² = DO² + CO² (III)

BC² = CO²+ BO² (IV)

Adding equations (I) + (ii) + (iii) + (iv) we get

Hence proved.

Q8. In fig 6.54 O is a point in the interior of a triangle

ABC , OD ⊥ BC , OE ⊥ AC and OF ⊥ AB. Show that

OA²+ OB² + OC²– OD²– 0E² – OF² = AF²+ BD² + CE²
AF²+ BD² + CE² = AE² + CD² + BF²

Solution:-

By pythagoras theorem in ∆AOF , we have

OA² = OF² + AF²

Similarly , in ∆BOD

OB² = OD² + BD²

Similarly in ∆COE

OC² = OE² + EC²

Adding these equations,

OA² + OB² + OC²= 0F² + AF² + OD² + BD² + OE² + EC²

= AF² + BD² + CE² .

Triangles Exercise 6.5 Answer 8

(II) AE² + CD² + BF².

Q9. A ladder 10 m long reaches a window 8m above the ground. Find the distance of the foot of the ladder from base of the wall.

Solution:-

AC² = AB² + BC²

10² = 8² + BC²

BC² Triangles Exercise 6.5 Answer 10 100 – 64

BC² = 36

BC = 6 m

Therefore 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 1 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:-

AC² = AB² + BC²

24² = 18² + BC²

BC² = 576 – 324

BC²=252

BC = 6√7 cm.

Q11. An aeroplane leaves an airport and flies sue 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 3/2 hours ?

Solution:-

Distance covered by first aeroplane flying due north in 3/2 hours (OA) = 100 X 3/2 = 1500 km.

Speed of second aeroplane = 1200 km/hr

Distance covered by second aeroplane flying due west in 3/2 = 1200 x 3/2 = 1800 km

AB² = (1500)²+ (1800)²

AB = (2250000 + 3240000)

= 5490000

AB = 300√61 km

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:

Given, Two poles of heights 6 m and 11 m stand on a plane ground.

And distance between the feet of the poles is 12 m.

Let AB and CD be the poles of height 6m and 11m.

Therefore, CP = 11 – 6 = 5m

From the figure, it can be observed that AP = 12m

By Pythagoras theorem for ΔAPC, we get,

AP2 = PC2 + AC2

(12m)2 + (5m)2 = (AC)2

AC2 = (144+25) m2 = 169 m2

AC = 13m

Therefore, the distance between their tops is 13 m.

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:

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

By Pythagoras theorem in ΔACE, we get

AC2 + CE2 = AE2 ………………………………………….(i)

In ΔBCD, by Pythagoras theorem, we get

BC2 + CD2 = BD2 ………………………………..(ii)

From equations (i) and (ii), we get,

AC2 + CE2 + BC2 + CD2 = AE2 + BD2 …………..(iii)

In ΔCDE, by Pythagoras theorem, we get

DE2 = CD2 + CE2

In ΔABC, by Pythagoras theorem, we get

AB2 = AC2 + CB2

Putting the above two values in equation (iii), we get

DE2 + AB2 = AE2 + BD2.

The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD (see Figure). Prove that 2AB2= 2AC2+ BC2.

Solution:

Given, the perpendicular from A on side BC of a Δ ABC intersects BC at D such that;

DB = 3CD.

In Δ ABC,

AD ⊥BC and BD = 3CD

In right angle triangle, ADB and ADC, by Pythagoras theorem,

AB2 = AD2 + BD2 ……………………….(i)

AC2 = AD2 + DC2 ……………………………..(ii)

Subtracting equation (ii) from equation (i), we get

AB2 – AC2 = BD2 – DC2

= 9CD2 – CD2 [Since, BD = 3CD]

= 8CD2

= 8(BC/4)2 [Since, BC = DB + CD = 3CD + CD = 4CD]

Therefore, AB2 – AC2 = BC2/2

⇒ 2(AB2 – AC2) = BC2

⇒ 2AB2 – 2AC2 = BC2

∴ 2AB2 = 2AC2 + BC2.

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

Solution:

Given, ABC is an equilateral triangle.

And D is a point on side BC such that BD = 1/3BC

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

∴ BE = EC = BC/2 = a/2

And, AE = a√3/2

Given, BD = 1/3BC

∴ BD = a/3

DE = BE – BD = a/2 – a/3 = a/6

In ΔADE, by Pythagoras theorem,

AD2 = AE2 + DE2

⇒ 9 AD2 = 7 AB2

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:

Given, an equilateral triangle say ABC,

Let the sides of the equilateral triangle be of length a, and AE be the altitude of ΔABC.

∴ BE = EC = BC/2 = a/2

In ΔABE, by Pythagoras Theorem, we get

AB2 = AE2 + BE2

4AE2 = 3a2

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

Hence, proved.

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:

Given, in ΔABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm.

We can observe that,

AB2 = 108

AC2 = 144

And, BC2 = 36

AB2 + BC2 = AC2

The given triangle, ΔABC, is satisfying Pythagoras theorem.

Therefore, the triangle is a right triangle, right-angled at B.

∴ ∠B = 90°

Hence, the correct answer is (C).