**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

AC^{2} = AB^{2} + BC^{2}

c^{2} = a^{2} + b^{2}

∴ Here diagonal is the greatest side.

i) a, b,c

7 cm, 24 cm, 25 cm,

c^{2} = a^{2} + b^{2}

25^{2} = (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.

c^{2} = a^{2} + b^{2}

8^{2} = (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.

c^{2} = a^{2} + b^{2}

100^{2}= (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,

c^{2} = a^{2} + b^{2}

13^{2} = (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 PM^{2} = 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) AB^{2}= BC.BD

ii) AC^{2} = BC.DC

iii) AD^{2} = BD.CD

Solution:

Data: In ∆ABD, ∠A = 90°,

AC ⊥ BD.

To Proved: AB^{2} = BC.BD

ii) AC^{2} = BC.DC

iii) AD^{2} = BD.CD

i) AB^{2} = BC.BD

∆ACB ~ ∆BAD (. Theorem7)

\(\frac{\mathrm{AB}}{\mathrm{BD}}=\frac{\mathrm{BC}}{\mathrm{AB}}\)

∴ AB^{2}= BC × BD.

ii) AC^{2} = BC.DC

∆BCA ~ ∆ACD

\(\frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{AC}}{\mathrm{CD}}=\frac{\mathrm{BC}}{\mathrm{AC}}\)

∴ AC × AC = BC × CD

∴ AC^{2} = BC × CD

iii) AD^{2} = BD.CD

∆ACD ~ ∆BAD

\(\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{CD}}{\mathrm{AD}}=\frac{\mathrm{AC}}{\mathrm{AB}}\)

∴ AD × AD = BD × DC

∴ AD^{2} = BD × DC

Question 4.

ABC is an isosceles triangle right angled at C. Prove that AB^{2} = 2AC^{2}.

Solution:

We have, right ∆ABC such that ∠C = 90° and AC = BC.

∴ By Pythagoras theorem, we have AB^{2} = AC^{2} + BC^{2} = AC^{2} + AC^{2} = 2AC^{2}

[∵ BC = AC (given)]

Thus, AB^{2} = 2AC^{2}

Question 5.

ABC is an isosceles triangle with AC = BC. If AB^{2} = 2AC^{2}, prove that ABC is a right-angled triangle.

Solution:

Data: ABC is an isosceles triangle with AC = BC.

AB^{2} = 2 AC^{2}.

To Prove: ∆ABC is a right angled triangle

AB^{2} = 2AC^{2} (Data)

AB^{2} = AC^{2} + AC^{2}

AB^{2} = AC^{2} + BC^{2} (∵ 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,

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°

∴ AD^{2} + BD^{2} = AB^{2}

AD^{2} + a^{2} = (2a)^{2}

AD^{2} + a^{2} = 4a^{2}

∴ AD^{2} = 4a^{2} – a^{2}

AD^{2} = 3a^{2}.

\(\sqrt{A D^{2}}=\sqrt{3 a^{2}}\)

∴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,

AB^{2} = OA^{2} + OB^{2} …… (1)

[Using Pythagoras theorem]

Similarly, in right ∆BOC,

BC^{2} = OB^{2} + OC^{2} …… (2)

In right ∆COD,

CD^{2} = OC^{2} + OD^{2} …… (3)

In right ∆AOD,

DA^{2} = OD^{2} + OA^{2} ……. (4)

Adding (1), (2), (3) and (4)

AB^{2} + BC^{2} + CD^{2} + DA^{2}

= [OA^{2} + OB^{2}] + [OB^{2} + OC^{2}] + [OC^{2} + OD^{2}] + [OD^{2} + OA^{2}]

= 2OA^{2} + 2OB^{2} + 2 OC^{2} + 2OD^{2} = 2[OA^{2} + OB^{2} + OC^{2} + OD^{2}]

= 2[OA^{2} + OB^{2} + OA^{2} + OB^{2}]

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) OA^{2} + OB^{2} + OC^{2} – OD^{2} – OE^{2} – OF^{2} = AF^{2} + BD^{2} + CE^{2}

ii) AF^{2} + BD^{2} + CE^{2} = AE^{2} + CD^{2} + BF^{2}.

Solution:

Data: O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC, OF⊥ AB.

To Proved: i) OA^{2} + OB^{2} + OC^{2} – OD^{2} – OE^{2} – OF^{2} = AF^{2} + BD^{2} + CE^{2}

ii) AF^{2} + BD^{2} + CE^{2} = AE^{2} + CD^{2} + BF^{2}

(i) OA^{2} + OB^{2} + OC^{2} – OD^{2} – OE^{2} – OF^{2} = AF^{2} + BD^{2} + CE^{2}

OA^{2} = AF^{2} + OF^{2} → (i)

OB^{2} = BD^{2} + OD^{2} → (ii)

OC^{2} = OE^{2} + EC^{2} → (iii)

By Adding equations (i) + (ii) + (iii),

OA^{2} + OB^{2}+ OC^{2} = AF^{2} + OF^{2} + BD^{2} + OD^{2} + OE^{2} +EC^{2}

∴ OA^{2} + OB^{2} + OC^{2} – OE^{2} – OF^{2} – OD^{2} = AF^{2} + BD^{2} + CE^{2}

(ii) AF^{2} + BD^{2} + CE^{2} = AE^{2} + CD^{2} + BF^{2}.

OA^{2} = AF^{2} + OF^{2}

∴ AF^{2} = OA^{2} – OF^{2} → (i)

OB^{2} = BD^{2} + OD^{2}

∴BD^{2} = OB^{2} – OD^{2} → (ii)

OC^{2} = OE^{2} + EC^{2}

∴ CE^{2} = OC^{2} – OE^{2} → (iii)

From adding equations (i) + (ii) + (iii),

AF^{2} + BD^{2} + CE^{2} = OA^{2} – OF^{2} + OB^{2} – OD^{2} + OC^{2} – OE^{2}

AF^{2 }+ BD^{2} + CE^{2} = OA^{2} – OE^{2} + OB^{2} – OF^{2} + OC^{2} – OE^{2}

∴ AF^{2} + BD^{2} + CE^{2} = AE^{2} + FB^{2} + CD^{2} .

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 = ?

AC^{2} + CB^{2} = AB^{2}

(8)^{2} + CB^{2} = (10)^{2}

64 + CB^{2} = 100

CB^{2} = 100 – 64

CB^{2} = 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,

AB^{2} = OA^{2} + OB^{2}

= (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 AE^{2} + BD^{2} = AB^{2} + DE^{2}.

Solution:

Data: In ∆ABC, ∠C = 90°, D and E are points on the sides CA and CB respectively

Top Prove: AE^{2} + BD^{2} = AB^{2} + DE^{2}

In ⊥∆ACE, ∠C = 90°

∴ AE^{2} = AC^{2} + CE^{2} ………. (i)

In ⊥∆DEB, ∠C = 90°

∴ BD^{2} = DC^{2} + CB^{2} ………… (ii)

From adding equations (i) + (ii)

AE^{2} + BD^{2} = AC^{2} + CE^{2} + DC^{2} + CB^{2}

= AC^{2} + CB^{2} + DC^{2} + CE^{2}

∴ AE^{2} + BD^{2} = AB^{2} + DE^{2} ( . 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 2AB^{2} = 2AC^{2} + BC^{2}.

Solution:

In ∆ACD,

AC^{2} = AD^{2} + DC^{2}

AD^{2} = AC^{2} – DC^{2}……….. (1)

In ∆ABD,

AB^{2} = AD^{2} + DB^{2}

AD^{2} = AB^{2} – DB^{2}………. (2)

From equations (1) and (2),

AC^{2} – DC^{2} = AB^{2} – DB^{2} ………… (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),

\(\mathrm{AC}^{2}-\left(\frac{\mathrm{BC}}{4}\right)^{2}=\mathrm{AB}^{2}-\left(\frac{3 \mathrm{BC}}{4}\right)^{2}\)

\(A C^{2}-\frac{B C^{2}}{16}=A B^{2}-\frac{9 B C^{2}}{16}\)

\(\frac{16 \mathrm{AC}^{2}-\mathrm{BC}^{2}}{16}=\frac{16 \mathrm{AB}^{2}-9 \mathrm{BC}^{2}}{16}\)

16AC^{2} – BC^{2} = 16 AB^{2} – 9BC^{2}

16AB^{2} – 16AC^{2} = 9 BC^{2} – BC^{2}

16AB^{2} – 16AC^{2} = 8BC^{2}

8(2AB^{2} – 2AC^{2} = BC^{2})

2AB^{2} – 2AC^{2} = BC^{2}

2AB^{2} = 2AC^{2} ≠ BC^{2}

^{}

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 9AD^{2} = 7AB^{2}.

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, \(\mathrm{AE}=\frac{\mathrm{a} \sqrt{3}}{2}\)

\(\mathrm{BD}=\frac{1}{3} \mathrm{BC}(\mathrm{Data})\)

\(B D=\frac{a}{3}\)

DE = BE – AD

\(=\quad \frac{a}{2}-\frac{a}{3}=\frac{a}{6}\)

In ∆ADE,

AD^{2} = AE^{2} + DE^{2}

\(=\left(\frac{\mathrm{a} \sqrt{3}}{2}\right)^{2}+\left(\frac{\mathrm{a}}{6}\right)^{6}\)

\(=\frac{3 a^{2}}{4}+\frac{a^{2}}{36}=\frac{27 a^{2}+a^{2}}{36}\)

\(=\frac{28 \mathrm{a}^{2}}{36}\)

\(=\frac{7}{9} \mathrm{a}^{2}\)

\(\mathrm{AD}^{2}=\frac{7}{9} \mathrm{AB}^{2} \quad 9 \mathrm{AD}^{2}=7 \mathrm{AB}^{2}\)

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.

AD ⊥ BC.

To Prove: 3AC^{2} = 4AD^{2}

In ⊥∆ADC, ∠ADC = 90°

∴ AC^{2} = AD^{2} + DC^{2}

\(=A D^{2}+\left(\frac{1}{2} A C\right)^{2}\)

\(\left( \begin{array}{l}{\mathrm{DC}=\frac{1}{2} \mathrm{BC}} \\ {\mathrm{DC}=\frac{1}{2} \mathrm{AC}}\end{array}\right)\)

\(\mathrm{AC}^{2}=\mathrm{AD}^{2}+\frac{1}{4} \mathrm{AC}^{2}\)

\(\frac{\mathrm{AC}^{2}}{1}-\frac{1}{4} \mathrm{AC}^{2}=\mathrm{AD}^{2}\)

\(\frac{4 \mathrm{AC}^{2}-1 \mathrm{AC}^{2}}{4}=\mathrm{AD}^{2}\)

∴ 3AC^{2} = 4AD^{2}.

Question 17.

Tick the correct answer and justify : In ∆ABC, AB = \(6 \sqrt{3}\) cm, AC = 12 cm and BC = 6 cm. The angle B is

(A) 120°

(B) 60°

(C) 90°

(D) 45°

Solution:

(C): We have, AB = \(6 \sqrt{3}\) cm, AC = 12 cm, and BC = 6 cm

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