Let us discuss with the following questions in solving the right angled triangle.

### Solved Examples

**Question 1: **Solve the right angled triangle for b, when a = 5 cm and c = 13 cm.

** Solution: **

We have a = 5 cm and c = 13 cm

The given data will satisfy the equation,

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

=> 5^{2} + b^{2} = 13^{2}

=> 25 + b^{2} = 169

=> b^{2} = 169 - 25

= 144

= 12^{2}

=> ** b = 12 cm**

**Question 2: **In the given figure, $\angle PSR$ = 90

^{o} , PQ = 10 cm, QS = 6 cm and RQ= 9 cm, calculate the length of PR.

** Solution: **

In Right triangle PQS, $\angle PSQ$ = 90^{o}.

Therefore applying Pythagoras theorem, we get

PS^{2} + QS^{2} = PQ^{2}

=> PS^{2} + 6^{2} = 10^{2}

=> PS^{2} + 36 = 100

=> PS^{2} = 100 - 36

= 64

= 8^{2}

=> PS = 8 cm

In triangle PSR, $\angle PSR$ = 90^{o}

Therefore we have RS = RQ + QS

= 9 + 6

= 15 cm

PS = 8 cm (proved)

Applying Pythagoras Theorem, we get

PR^{2} = PS^{2} + RS^{2}

=> PR^{2} = 8^{2} + 15^{2}

= 64 + 225

= 289

= 17^{2}

=> PR = 17 cm

Therefore, **the length of PR = 17 cm.**

**Question 3: **In an isosceles triangle the equal sides measure 15 cm. If the unequal side measure 18 cm, find the area of the triangle.

** Solution: **

In the above triangle we have AB = AC = 15 cm [ since the triangle is isosceles ]

It is given that the unequal side = 18 cm

=> BC = 18 cm

Since the perpendicular from the vertex bisects the third side, we have BD = DC = 9 cm

In $\Delta ABC$, $\angle ADC$ = 90

^{o} Therefore, applying Pythagoras Theorem, we get

AD

^{2} + BD

^{2} = AB

^{2} => AD

^{2} + 9

^{2} = 15

^{2} => AD

^{2} + 81 = 225

=> AD

^{2} = 225 - 81

= 144

=> AD = $\sqrt 144$

= 12 cm

Therefore,

**Height of the triangle = 12 cm** Area of the triangle =

$\frac{1}{2}$ x Base x Height

=

$\frac{1}{2}$ x 18 x 12

=

**108 cm**^{2}
**Question 4: **Find the height of an equilateral triangle whose equal sides measure 12 cm.

** Solution: **

In the above triangle KL = LM = MN = 12 cm

The perpendicular from the vertex bisects the base.

( i. e ) KD $\perp$ LM.

Therefore, LD = DM = 6 cm

In $\Delta KLD$, $\angle KDL$ = 90

^{o} Therefore, applying Pythagoras Theorem, we get,

KD

^{2} + LD

^{2} = KL

^{2} => KD

^{2} + 6

^{2} = 12

^{2} => KD

^{2} + 36 = 144

=> KD

^{2} = 144 - 36

= 108

=> KD = $\sqrt108$

= 10.392 cm

Therefore,

**Height of the Equilateral triangle of side 12 cm is 10.392 cm**
**Question 5: **Find the length of the sides of a filed of rhombus in shape, whose diagonals measure 30 m and 16 m.

** Solution: **
The above figure shows the Rhombus PQRS, whose diagonals measure 30 cm and 16 cm respectively.

Since the diagonals of a rhombus bisect each other at right angles we have,

PO = OR = 8 cm and

QO = OS = 15 cm

In $\Delta POQ$, $\angle POQ$ = 90

^{o} Applying Pythagoras Theorem, we have

PQ

^{2} = PO

^{2} + QO

^{2} = 8

^{2} + 15

^{2} = 64 + 225

= 289

=> PQ = $\sqrt 289$

= 17

Since the sides of a rhombus measure same, we have

** PQ = QR = RS = SP = 17 cm**