# 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:
In ∆QMP and ∆QPR,
∠QMP = ∠QPR [Each 90°]
∠Q = ∠Q [Common]
⇒ ∆QMP ~ ∆QPR …. (1) [AA similarity]
Again, in ∆PMR and ∆QPR,
∠PMR = ∠QPR [Each = 90°]
∠R = ∠R [Common]
⇒ ∆PMR ~ ∆QPR …… (2) [AA similarity]
From (1) and (2), we have

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

Solution:
Data: In ∆ABD, ∠A = 90°,
AC ⊥ BD.
To Proved: AB2 = BC.BD
ii) AC2 = BC.DC

i) AB2 = BC.BD
$$\frac{\mathrm{AB}}{\mathrm{BD}}=\frac{\mathrm{BC}}{\mathrm{AB}}$$
∴ 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

$$\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{CD}}{\mathrm{AD}}=\frac{\mathrm{AC}}{\mathrm{AB}}$$
∴ AD2 = BD × DC

Question 4.
ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2.
Solution:
We have, right ∆ABC such that ∠C = 90° and AC = BC.

∴ By Pythagoras theorem, we have AB2 = AC2 + BC2 = AC2 + AC2 = 2AC2
[∵ BC = AC (given)]
Thus, AB2 = 2AC2

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.

To Prove: Altitude of ∆ABC,
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 = 4a2 – a2
$$\sqrt{A D^{2}}=\sqrt{3 a^{2}}$$
∴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.
Solution:
Let us have a rhombus ABCD.

∵ Diagonal of a rhombus bisect each other at right angles.
∴ OA = OC and OB = OD
Also, ∠AOB = ∠BOC [Each = 90°]
And ∠COD = ∠DOA [Each = 90°]
In right ∆AOB, we have,
AB2 = OA2 + OB2 …… (1)
[Using Pythagoras theorem]
Similarly, in right ∆BOC,
BC2 = OB2 + OC2 …… (2)
In right ∆COD,
CD2 = OC2 + OD2 …… (3)
In right ∆AOD,
DA2 = OD2 + OA2 ……. (4)
Adding (1), (2), (3) and (4)
AB2 + BC2 + CD2 + DA2
= [OA2 + OB2] + [OB2 + OC2] + [OC2 + OD2] + [OD2 + OA2]
= 2OA2 + 2OB2 + 2 OC2 + 2OD2 = 2[OA2 + OB2 + OC2 + OD2]
= 2[OA2 + OB2 + OA2 + OB2]

Thus, sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

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:
Let AB is the wire and BC is the vertical pole. The point A is the stake.

Now, in the right AABC, using Pythagoras Theorem, we have

Thus, the stake is required to be taken at $$6 \sqrt{7}$$m from the base of the pole to make the wire taut.

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?

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 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:
Let the two poles AB and CD are such that the distance between their feet AC = 12m.
∵ Height of pole-1, AB = 11 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.

Solution:
In ∆ACD,
AD2 = AC2 – DC2……….. (1)
In ∆ABD,
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.

AB = BC = AC = a
In ∆ABC, AE is perpendicular line.
$$B E=E C=\frac{B C}{2}$$
Altitude, $$\mathrm{AE}=\frac{\mathrm{a} \sqrt{3}}{2}$$
$$\mathrm{BD}=\frac{1}{3} \mathrm{BC}(\mathrm{Data})$$
$$B D=\frac{a}{3}$$
$$=\quad \frac{a}{2}-\frac{a}{3}=\frac{a}{6}$$
$$=\left(\frac{\mathrm{a} \sqrt{3}}{2}\right)^{2}+\left(\frac{\mathrm{a}}{6}\right)^{6}$$
$$=\frac{3 a^{2}}{4}+\frac{a^{2}}{36}=\frac{27 a^{2}+a^{2}}{36}$$
$$=\frac{28 \mathrm{a}^{2}}{36}$$
$$=\frac{7}{9} \mathrm{a}^{2}$$
$$\mathrm{AD}^{2}=\frac{7}{9} \mathrm{AB}^{2} \quad 9 \mathrm{AD}^{2}=7 \mathrm{AB}^{2}$$

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.

Solution:
Data: ABC is an equilateral triangle.
Here, AB = BC = CD.
∴ AC2 = AD2 + DC2
$$=A D^{2}+\left(\frac{1}{2} A C\right)^{2}$$
$$\left( \begin{array}{l}{\mathrm{DC}=\frac{1}{2} \mathrm{BC}} \\ {\mathrm{DC}=\frac{1}{2} \mathrm{AC}}\end{array}\right)$$
$$\mathrm{AC}^{2}=\mathrm{AD}^{2}+\frac{1}{4} \mathrm{AC}^{2}$$
$$\frac{\mathrm{AC}^{2}}{1}-\frac{1}{4} \mathrm{AC}^{2}=\mathrm{AD}^{2}$$
$$\frac{4 \mathrm{AC}^{2}-1 \mathrm{AC}^{2}}{4}=\mathrm{AD}^{2}$$
Tick the correct answer and justify : In ∆ABC, AB = $$6 \sqrt{3}$$ cm, AC = 12 cm and BC = 6 cm. The angle B is
(C): We have, AB = $$6 \sqrt{3}$$ cm, AC = 12 cm, and BC = 6 cm