Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev

Mathematics (Maths) Class 10

Class 10 : Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev

The document Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev is a part of the Class 10 Course Mathematics (Maths) Class 10.
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Q.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
Sol. (i) 7 cm, 24 cm, 25 cm
(7)2 + (24)2 = 49 + 576
= 625 = (25)2 = 25
∴ The given sides make a right angled triangle with hypotenuse 25 cm
(ii) 3 cm, 8 cm, 6 cm
(8)2 = 64
(3)2 + (6)2 = 9 + 36 = 45
64 ≠ 45
The square of larger side is not equal to the sum of square of other two sides.
∴ The given triangle is not a right angled.
(iii) 50 cm, 80 cm, 100 cm
(100)2 = 10000
(80)2 + (50)2 = 6400 + 2500 = 8900
The square of larger side is not equal to the sum of squares of other two sides.
∴ The given triangle is not a right angled.
(iv) 13 cm, 12 cm, 5 cm
(13)2 = 169
(12)2 + (5)2 = 144 + 25 - 169
Sides make a right angled triangle with hypotenuse 13 cm.

Q.2. PQR is a triangle, right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM· MR.
Sol. In right angled ΔQPR,
∠P = 90°, PM QR
∴ ΔPMQ ~ ΔRMP
Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev
[If is drawn from the vertex of right angle to the hypotenuse then triangles on both sides of perpendicular are similar to each other, and to whole triangle]
Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev
[Corresponding sides of similar triangles]
Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev

Q3. In the figure, ABD is a triangle, right angled at A and AC ⊥ BD. Show that
(i) AB2 = BC.BD 
(ii) AC2 = BC.DC 
(iii) AD2 = BD.CD 
Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev
Sol. (i) In ΔABC, ∠BAD = 90°
AC BD
Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev
Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev

Q.4. ABC is an isosceles triangle, right angled at C. Prove that AB2 = 2AC2.
Sol. Given: In ΔABC, ∠C = 90° and AC = BC
Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev
To Prove: AB2 = 2AC2
Proof: In ΔABC, AB2 = BC2 + AC2
AB2 = AC2 + AC2 [Pythagoras theorem]
= 2AC2

Q.5. ABC is an isosceles triangle with AC = BC. If AB2 = 2 AC2, prove that ABC is a right triangle.
Sol. Given: In ΔABC, AC = BC and AB2 = 2 AC2
Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev
To Prove: ACB is a right angled triangle
Proof: AB2 = 2AC2
AB2 = AC2 + AC2
AB2 = AC2 + BC2         [∵ AC = BC]
⇒ ∠ACB = 90°
[By converse of Pythagoras theorem]
∴ ΔABC is right angled triangle

Q.6. ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Sol.  Given: In Δ ABC, AB = BC = AC = 2a
Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev
To Prove: Length of AD
Proof: In Δ ABC,
AB = BC = AC = 2a
and AD ⊥BC
Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev
In right angled triangle ADB,
AD2 + BD2 = AB2 
⇒ AD2 = AB2 - BD2
= (2a)2 - (a)2 = 4a2 - a2 = 3a2
 Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev

Q.7. 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. Given: ABCD is a rhombus. Diagonals AC and BD intersect at O.
To Prove: AB2 + BC+ CD+ DA2 = AC2 + BD2 
Proof: In ΔAOB, AB2 = AO2 + OB2 [Pythagoras theorem]
Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev
Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev
AB2 + AB2 + AB2 + AB2 = AC2 + BD2 
AB2 + BC2 + CD2 + DA2 = AC2 + BD2
[∵ AB = BC = CD = DA]

Q.8. In the figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that
(i) OA2 + OB2 + OC2 - OD2 - OE2 - OF2 = AF+ BD2 + CE2,
(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2.
Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev
Sol. Given: A triangle ABC in which OD ⊥ BC, OE ⊥ AC and OF ⊥ AB
To Prove: (i) OA2 + OB2 + OC2 - OD2 - OE2 - OF2 = AF2 + BD2 + CE2 
Construction: Join OA,.OB and OC
Proof: (i) In ΔAOF,
OA2 = OF2 + AF2 [Pythagoras theorem]
AF2 = OA2 - OF2 
In ΔBDO, OB2 = BD2 + OD2 
BD2 = OB2 - OD2 
In ΔCEO, OC2 = CE2 + OE2 
CE2 = OC2 - OE2

Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev
Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev
Equating (i) and (ii)
AF2 + BD2 + CE2 = AE2 + CD2 + BF2

Q.9. 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 AC be the ladder of length 10 m and AB = 8 m
In ΔABC,
BC2 + AB2 = AC2 
⇒ BC2 = AC2 - AB2
= (10)2 - (8)2 
BC2 = 100 - 64 = 36
Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev
Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev
Hence distance of foot of the ladder from base of the wall is 6 m.

Q.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 end. How far from the base of the pole should the stake be driven so that the wire will be taut?
Sol. In Δ ABC, AC2 = AB2 + BC2 
Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev
⇒ (24)2 = (18)2 + BC2
⇒ 576 = 324 + BC2
⇒ 576 - 324 = BC2
BC2 = 252
Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev
Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev

Q.11. An aeroplane leaves an airport and flies due north 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 Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev hours ?
Sol. Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev
Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev
Distance travelled by the aeroplane due west in
Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev

Q.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.
Sol. Length of poles is 6 m and 11m.
Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev
DE = DC - EC = 11 m - 6 m = 5 m
In ΔDAE, AD2 = AE2 + DE2    [∴ AE - BC]
= (12)2 + (5)2 = 144 + 25 = 169
Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev

Q.13. 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.
Sol.  In ΔABC, ∠C = 90°
Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev
AB2 = AC2 + BC2 [Pythagoras theorem]
In ΔDCE, DE2 = DC2+CE2 
⇒ AB2 + DE2 = AC2 + BC2 + DC2 + CE2
= AC2 + CE2 + DC2 + BC2
= AE2 + BD2

Q.14. The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3 CD (see figure). Prove that 2 AB2 = 2 AC2 + BC2.
Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev
Sol. In ΔADB,
Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev
Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev

Q.15. In an equilateral triangle ABC, D is a point on side BC such that Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRevProve that 9 AD2 = 7 AB2.
Sol. Given: In Δ ABC, AB = BC = AC
Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev
Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev
Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev
Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev
Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev
Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev
On dividing both sides by 4, we get,
7AB2 = 9 AD2

Q.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.
Sol. Given: In ΔABC,
AB = BC = AC
AD ⊥ BC
Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev

To Prove: 3AB2 = 4 AD2
Proof: In ΔADB,
AB2 = AD2 + BD2 
AD2 = AB2 - BD2
Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev
⇒ 4AD2 = 4AB2 - BC2
= 4AB2 - AB2 [∴BC = AB = AC]
4AD2 = 3AB2

Q.17. Tick the correct answer and justify: In Δ ABC, AB =  6√3 cm, AC = 12 cm and BC = 6 cm. The angle B is:
(A) 120° 
(B) 60° 
(C) 90° 
(D) 45°
Sol. In ΔABC,
AB = 6√3 cm, AC = 12 cm and BC = 6 cm
Ex 6.5 NCERT Solutions- Triangles Class 10 Notes | EduRev
ΔABC is right angled at B.
∴ ∠B = 90°
Correct answer is (c).

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