KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.5
KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.5
Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangles Exercise 2.5
Question 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.
Solution:
In ⊥∆ABC, ∠B = 90°.
Let AB = a, BC = b, Hypotenuse AC = c then
AC2 = AB2 + BC2
c2 = a2 + b2
∴ Here diagonal is the greatest side.
i) a, b,c
7 cm, 24 cm, 25 cm,
c2 = a2 + b2
252 = (7)2 + (24)2
625 = 49 + 576
625 = 625
625 = 49 + 576 625 = 625
∴ This is right angled triangle.
Measurement of Hypotenuse, c = 25 cm.
ii) a c b
3 cm, 8 cm, 6 cm.
c2 = a2 + b2
82 = (3)2 + (6)2
64 = 9 + 36
64 ≠ 45
∴ These are not sides of right angled triangle.
iii) a b c
50 cm, 80 cm, 100 cm.
c2 = a2 + b2
1002= (50)2 + (80)2
10000 = 2500 + 6400
10000 ≠ 8900
∴ These are not sides of right angled triangle.
iv) a b c
12 cm, 5 cm, 13 cm,
c2 = a2 + b2
132 = (12)2 + (5)2
169 = 144 + 25
169 = 169
∴ These are sides of right angled triangle.
Measurement of Hypotenuse =13 cm.
Question 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.
Solution:
Data: PQR is a triangle 9 right angled at P and M is a point on QR such that PM ⊥ QR.
To Prove: PM2 = QM.MR
Proof: In ⊥D QPR, ∠P = 90°,
QR2 = QP2 + PR2 ……………. (i)
In ⊥D PMQ, ∠M = 90°
QP2 = PM2 + QM2 ………. (ii)
In ⊥D PMR, ∠M = 90°
PR2 = PM2 + MR2 ……………. (iii)
By Adding eqn. (ii) and (iii)
QP2 + PR2 = PM2 + QM2 + PM2 + MR2
QR2 = 2PM2 + QM2 + MR2 (. Eqn. (i))
(QM + MR)2 = 2PM2 + QM2 + MR2
QM2 + MR2 + 2QM.MR = 2PM2 + QM2 + MR2
QM2 – QM2 + MR2 – MR2 + 2QM.MR = 2PM2
2QM.MR = 2PM2
2PM2 = 2QM.MR
∴ PM2 = QM.MR
Question 3.
In the following 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
Solution:
Data: In ∆ABD, ∠A = 90°,
AC ⊥ BD.
To Proved: AB2 = BC.BD
ii) AC2 = BC.DC
iii) AD2 = BD.CD
Question 4.
ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2.
Solution:
Data: ABC is an isosceles triangle right angled at C.
To Prove: AB2 = 2AC2.
In ⊥∆ACB, ∠C = 90°
∴ AB2= AC2 + BC2
(∵ Pythagoras Theorem)
AB2 = AC2 + AC2 (∵ AC = BC)
AB2 = 2 AC2.
Question 5.
ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that ABC is a right-angled triangle.
Solution:
Data: ABC is an isosceles triangle with AC = BC.
AB2 = 2 AC2.
To Prove: ∆ABC is a right angled triangle
AB2 = 2AC2 (Data)
AB2 = AC2 + AC2
AB2 = AC2 + BC2 (∵ AC = BC)
Now, in ∆ABC, square of one side is equal to squares of other two sides.
∆ABC is a right angled triangle, Opposite angle to AB, i.e., ∠C is 90°.
Question 6.
ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Solution:
Data: ABC is an equilateral triangle of side 2a.
Question 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.
Solution:
Data: ABCD is a rhombus.
Here, AB = BC = CD = DA.
Diagonals AC and BD intersects at ’O’.
To Prove: AB2 + BC2 + CD2 + DA2 = AC2 + BD2.
In rhombus diagonals bisects perpendicularly.
∴ ∠AOB = ∠AOD = 90°.
Question 8.
In the following figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC, OF ⊥ AB. Show that
i) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2
ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2.
Solution:
Data: O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC, OF⊥ AB.
To Proved: i) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2
ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2
(i) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2
OA2 = AF2 + OF2 → (i)
OB2 = BD2 + OD2 → (ii)
OC2 = OE2 + EC2 → (iii)
By Adding equations (i) + (ii) + (iii),
OA2 + OB2+ OC2 = AF2 + OF2 + BD2 + OD2 + OE2 +EC2
∴ OA2 + OB2 + OC2 – OE2 – OF2 – OD2 = AF2 + BD2 + CE2
(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2.
OA2 = AF2 + OF2
∴ AF2 = OA2 – OF2 → (i)
OB2 = BD2 + OD2
∴BD2 = OB2 – OD2 → (ii)
OC2 = OE2 + EC2
∴ CE2 = OC2 – OE2 → (iii)
From adding equations (i) + (ii) + (iii),
AF2 + BD2 + CE2 = OA2 – OF2 + OB2 – OD2 + OC2 – OE2
AF2 + BD2 + CE2 = OA2 – OE2 + OB2 – OF2 + OC2 – OE2
∴ AF2 + BD2 + CE2 = AE2 + FB2 + CD2 .
Question 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.
Solution:
In ⊥∆ACB, ∠C = 90°, BC = ?
AC2 + CB2 = AB2
(8)2 + CB2 = (10)2
64 + CB2 = 100
CB2 = 100 – 64
CB2 = 36
∴ CB = 6
∴ Ladder is at a distance of 6m from the base of the wall.
Question 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?
Solution:
In ⊥∆PQR, ∠Q = 90°, QR = ?
∴ PQ2 + QR2 = PR2
(18)2 + QR2 = (24)2
324 + QR2 = 576
QR2 = 576 – 324
QR2 = 252
∴ QR = 15.8 m.
Question 11.
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 1 1/2 hours?
Question 12.
Two poles of heights 6 m and 11m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.
Solution:
Pole AB = 6 m.
Pole CD = 11 m.
Distance between poles BD = 12 m.
Distance between feet of the poles, AC = ?
ABDM is a rectangle, AB = MD = 6 m.
BD = AM = 12 m.
In ⊥∆AMC, ∠AMC = 90°
AM = 12 m, CM = 5 m. AC = ?
AC2 = AM2 + CM2
= (12)2 + (5)2
= 144 + 25
AC2 = 169
∴ AC = 13 m.
Question 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.
Solution:
Data: In ∆ABC, ∠C = 90°, D and E are points on the sides CA and CB respectively
Top Prove: AE2 + BD2 = AB2 + DE2
In ⊥∆ACE, ∠C = 90°
∴ AE2 = AC2 + CE2 ………. (i)
In ⊥∆DEB, ∠C = 90°
∴ BD2 = DC2 + CB2 ………… (ii)
From adding equations (i) + (ii)
AE2 + BD2 = AC2 + CE2 + DC2 + CB2
= AC2 + CB2 + DC2 + CE2
∴ AE2 + BD2 = AB2 + DE2 ( . Theorem 8).
Question 14.
The perpendicular from A on side BC of a ∆ABC intersects BC at D such that DB = 3CD (see the following figure) Prove that 2AB2 = 2AC2 + BC2.
Question 15.
In an equilateral triangle ABC, D is a point on side BC such that BD=1/3 BC, Prove that 9AD2 = 7AB2.
Solution:
∆ABC is an equilateral triangle.
Question 17.
Tick the correct answer and justify:
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