KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.5 

KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.5 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangles Exercise 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:
KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.5 1
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
KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.5 2
Solution:
Data: In ∆ABD, ∠A = 90°,
AC ⊥ BD.
To Proved: AB2 = BC.BD
ii) AC2 = BC.DC
iii) AD2 = BD.CD

i) AB2 = BC.BD
∆ACB ~ ∆BAD (. Theorem7)
\(\frac{\mathrm{AB}}{\mathrm{BD}}=\frac{\mathrm{BC}}{\mathrm{AB}}\)
∴ AB2= BC × BD.

ii) AC2 = BC.DC
∆BCA ~ ∆ACD
\(\frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{AC}}{\mathrm{CD}}=\frac{\mathrm{BC}}{\mathrm{AC}}\)
∴ AC × AC = BC × CD
∴ AC2 = BC × CD

iii) AD2 = BD.CD
∆ACD ~ ∆BAD
\(\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{CD}}{\mathrm{AD}}=\frac{\mathrm{AC}}{\mathrm{AB}}\)
∴ AD × AD = BD × DC
∴ AD2 = BD × DC

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.
KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.5 3
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.
KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.5 4
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.
KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.5 5
To Prove: Altitude of ∆ABC,
AD =?
In equilateral triangle
D bisect base.
∴ AB = BC = CA = 2a. 3 D a
If BC = 2a,
\(\frac{1}{2} \mathrm{BC}=\mathrm{a}\) unit
∴ BD = DC = a.
Now, in ⊥∆ADB, ∠D = 90°
∴ AD2 + BD2 = AB2
AD2 + a2 = (2a)2
AD2 + a2 = 4a2
∴ AD2 = 4a2 – a2
AD2 = 3a2.
\(\sqrt{A D^{2}}=\sqrt{3 a^{2}}\)
∴Altitude,\(\mathrm{AD}=\sqrt{3} \mathrm{a}\) unit.

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.
KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.5 6
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°.
In ⊥∆AOB,
AB2 = AO2 + BO2 ……..(i)
In ⊥∆BOC,
AC2 = BO2 + CO2 ……..(ii)
In ⊥∆COD,
CD2 = OC2 + OD2 ……….(iii)
In ⊥∆AOD,
AD2 = AO2 + OD2…………(iv)
By Adding equations (i) + (ii) + (iii) + (iv)
AB2 + BC2 + CD2 + DA2 =
= AO2 + BO2 + BO2 + CO2 + CO2 + DO2 + AO2 + OD2
= 2AO2 + 2BO2 + 2CO2 + 2DO2
= 2AO2 + 2CO2 + 2BO2 + 2DO2
Now, RHS = 2AO2 + 2CO2 + 2BO2 + 2DO2
\(=2 \times\left(\frac{1}{2} \mathrm{AC}\right)^{2}+2 \times\left(\frac{1}{2} \mathrm{AC}\right)^{2}+2 \times\left(\frac{1}{2} \mathrm{BD}\right)^{2}+2 \times\left(\frac{1}{2} \mathrm{BD}\right)^{2}\)
\(=2 \times \frac{1}{4} \mathrm{AC}^{2}+2 \times \frac{1}{4} \mathrm{AC}^{2}+2 \times \frac{1}{4} \mathrm{BD}^{2}+2 \times \frac{1}{4} \mathrm{BD}^{2}\)
\(=\frac{1}{2} A C^{2}+\frac{1}{2} A C^{2}+\frac{1}{2} B D^{2}+\frac{1}{2} B D^{2}\)
RHS = AC2 + BD2
∴ LHS = RHS
∴ AB2 + BC2 + CD2 + DA2 = AC2 + BD2.

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.
KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.5 7
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
KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.5 8
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.
KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.5 9
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?
KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.5 10
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 \(1 \frac{1}{2}\) hours?
KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.5 11
Solution:
Distance travelled by aeroplace trowards North is \(1 \frac{1}{2}\) hours :
\(=1000 \times 1 \frac{1}{2}\)
\(=1000 \times \frac{3}{2}\)
= 1500 km.
Diatance travelled by aeroplane towards West in \(1 \frac{1}{2}\) Hours :
\(=1200 \times 1 \frac{1}{2}\)
\(=1200 \times \frac{3}{2}\)
= 1800 km.
In ⊥∆AOB,
AB2 = OA2 + OB2
= (1500)2 + (1800)2
= 2250000 + 3240000 = 5490000
\(\mathrm{AB}=\sqrt{5490000}\)
\(A B=\sqrt{90000 \times 61}\)
\(A B=300 \sqrt{61} \mathrm{km}\) km
∴ Two planes are 300V6T km. apart after 14 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.
KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.5 12
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.
KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.5 13
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.
KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.5 14
Solution:
In ∆ACD,
AC2 = AD2 + DC2
AD2 = AC2 – DC2……….. (1)
In ∆ABD,
AB2 = AD2 + DB2
AD2 = AB2 – DB2………. (2)
From equations (1) and (2),
AC2 – DC2 = AB2 – DB2 ………… (3)
3DC = DB (data given)
\(\mathrm{DC}=\frac{\mathrm{BC}}{4}, \text { and } \mathrm{DB}=\frac{3 \mathrm{BC}}{4}\) ……….. (3)
Substituting eqn. (4) in eqn. (3),
\(\mathrm{AC}^{2}-\left(\frac{\mathrm{BC}}{4}\right)^{2}=\mathrm{AB}^{2}-\left(\frac{3 \mathrm{BC}}{4}\right)^{2}\)
\(A C^{2}-\frac{B C^{2}}{16}=A B^{2}-\frac{9 B C^{2}}{16}\)
\(\frac{16 \mathrm{AC}^{2}-\mathrm{BC}^{2}}{16}=\frac{16 \mathrm{AB}^{2}-9 \mathrm{BC}^{2}}{16}\)
16AC2 – BC2 = 16 AB2 – 9BC2
16AB2 – 16AC2 = 9 BC2 – BC2
16AB2 – 16AC2 = 8BC2
8(2AB2 – 2AC2 = BC2)
2AB2 – 2AC2 = BC2
2AB2 = 2AC2 ≠ BC2

Question 15.
In an equilateral triangle ABC, D is a point on side BC such that \(B D=\frac{1}{3} B C\), Prove that 9AD2 = 7AB2.
Solution:
∆ABC is an equilateral triangle.
KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.5 15
AB = BC = AC = a
In ∆ABC, AE is perpendicular line.
\(B E=E C=\frac{B C}{2}\)
Altitude, \(\mathrm{AE}=\frac{\mathrm{a} \sqrt{3}}{2}\)
\(\mathrm{BD}=\frac{1}{3} \mathrm{BC}(\mathrm{Data})\)
\(B D=\frac{a}{3}\)
DE = BE – AD
\(=\quad \frac{a}{2}-\frac{a}{3}=\frac{a}{6}\)
In ∆ADE,
AD2 = AE2 + DE2
\(=\left(\frac{\mathrm{a} \sqrt{3}}{2}\right)^{2}+\left(\frac{\mathrm{a}}{6}\right)^{6}\)
\(=\frac{3 a^{2}}{4}+\frac{a^{2}}{36}=\frac{27 a^{2}+a^{2}}{36}\)
\(=\frac{28 \mathrm{a}^{2}}{36}\)
\(=\frac{7}{9} \mathrm{a}^{2}\)
\(\mathrm{AD}^{2}=\frac{7}{9} \mathrm{AB}^{2} \quad 9 \mathrm{AD}^{2}=7 \mathrm{AB}^{2}\)

Question 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.
KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.5 16
Solution:
Data: ABC is an equilateral triangle.
Here, AB = BC = CD.
AD ⊥ BC.
To Prove: 3AC2 = 4AD2
In ⊥∆ADC, ∠ADC = 90°
∴ AC2 = AD2 + DC2
\(=A D^{2}+\left(\frac{1}{2} A C\right)^{2}\)
\(\left( \begin{array}{l}{\mathrm{DC}=\frac{1}{2} \mathrm{BC}} \\ {\mathrm{DC}=\frac{1}{2} \mathrm{AC}}\end{array}\right)\)
\(\mathrm{AC}^{2}=\mathrm{AD}^{2}+\frac{1}{4} \mathrm{AC}^{2}\)
\(\frac{\mathrm{AC}^{2}}{1}-\frac{1}{4} \mathrm{AC}^{2}=\mathrm{AD}^{2}\)
\(\frac{4 \mathrm{AC}^{2}-1 \mathrm{AC}^{2}}{4}=\mathrm{AD}^{2}\)
∴ 3AC2 = 4AD2.

Question 17.
Tick the correct answer and justify:
In ∆ABC, \(A B=6 \sqrt{3} \mathrm{cm}\). AC = 12 cm. and BC = 6 cm. The angle B is
A) 120°
b) 60°
C) 90°
D) 45°
Solution:
C) 90°
KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.5 17
Justification:
a             b           c
\(6 \sqrt{3}\)        6             12
Here largest side is 12 cm.
If square of the hypotenuse is equal to square of other two sides, then it is a right angled triangle.
∴ c2 = a2 + b2
AC2 = AB2 + BC2
\((12)^{2}=(6 \sqrt{3})^{2}+(6)^{2}\)
144 = 36 × 3 + 36
144 = 108 + 36
144 = 144
∴ ∆ABC is a right angled triangle and angle opposite to hypotenuse, ∠B = 90°.

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