Question

The area of a right triangle is 54cm^2. If one of its legs is 12cm long. Find the perimeter.​

Answer 1

Answer:

36cm

Explanation:

Area of triangle = ½ × b × h

Given that,

Area of triangle = 54cm²

b = 12cm

=> 54 = ½ × 12 × h

=> 54 = 6 × h

=> h = 54/6

=> h = 9cm

Now,

Base = 12cm, Height = 9cm

Using Pythagoras Theorem,

(Hypotenuse)² = (Base)² + (Height)²

=> (Hypotenuse)² = (12)² + (9)²

=> (Hypotenuse)² = 144 + 81

=> (Hypotenuse)² = 225

=> Hypotenuse = 15cm

Hence, perimeter = (12+9+15) cm = 36cm

Answer 2

AnsWer :

36 Cm.

Solution :

We have,

$\tt \dagger \: \: \: \: \: Area \: of_\triangle = \frac{1}{2} \times B \times H.$

Let long side be Base.

• B = 12 Cm.

$\tt\longmapsto 54 = \frac{1}{2} \times 12 \times H.$

$\tt\longmapsto 54 = \frac{1}{\cancel2} \times \cancel{12} \times H.$

$\tt\longmapsto 54 = 6H.$

$\tt\longmapsto \dfrac{\cancel{54}}{\cancel6} = H.$

$\tt\longmapsto H = 9 \: cm.$

$\rule{90}1$

Now,

We need to Find Other Sides of triangle,

* Let Apply Pythagoras theorem,

$\tt \dagger \: \: \: \: \: {H }^{2} = {P }^{2} + {B }^{2}$

Where as,

• P = 9 Cm ( Height )
• B = 12 Cm ( Base )
• H = ?

$\tt\longmapsto {H }^{2} = {9}^{2} + {12}^{2} .$

$\tt\longmapsto {H }^{2} = 81 + 144.$

$\tt\longmapsto H = \sqrt{225}$

$\tt\longmapsto H = 15 \: cm.$

Now, We have all Sides of triangle,

• a = 15 Cm.
• b = 9 Cm.
• c = 12 Cm.

$\rule{90}1$

$\tt \dagger \: \: \: \: \: Perimeter \: of_\triangle = a + b + c.$

$\tt \longmapsto 15 + 9 + 12.$

$\tt \longmapsto 36 \: cm.$

Therefore, the perimeter of given triangle be 36 Cm.

$\rule{200}3$

Note : Diagram Provide above attachment.