Question

If the perimeter of a right angled triangle is 60 cm and its hypotenuse is 25 cm, find
its area.
Please help me to do this sum ​

Answer 1

Given Information:

• The Perimeter of Right-angled triangle is 60cm.
• Hypotenuse of the triangle is 25cm.

Need to Find:

• The Area of triangle.

Solution:

Let ∆ABC be the given Right-angled triangle, with Perpendicular, AB; Base, BC and Hypotenuse, AC. Let the base of the triangle be 'x'

Here,

$\: \: \: \: \: \longmapsto\tt{Perimeter=60\: cm}$

$\: \: \: \: \: \longmapsto\tt{Hypotenuse, CA=25\: cm}$

We know, Sum of all sides of triangle is equal to the perimeter of triangle. So,

$\: \: \: \: \: \longrightarrow\sf{Sum\: of\: all\: sides=Perimeter}$

$\: \: \: \: \: \longrightarrow\sf{AB+BC+CA=60}$

$\: \: \: \: \: \longrightarrow\sf{AB+x+25=60}$

$\: \: \: \: \: \longrightarrow\sf{AB+x=60-25}$

$\: \: \: \: \: \longrightarrow\sf{AB+x=35}$

$\: \: \: \: \: \longrightarrow\bold{AB=35-x}$

We know, In a Right-angled triangle, the sum of two side is equal to it's Hypotenuse. That is,

$\: \: \: \: \: \implies\sf{{AC}^{2}={AB}^{2}+{BC}^{2}}$

$\: \: \: \: \: \implies\sf{{25}^{2}={(35-x)}^{2}+{x}^{2}}$

From, (a-b)²=a²+b²-2ab.

$\: \: \: \: \: \implies\sf{625={35}^{2}+{x}^{2}-2(35\times x)+{x}^{2}}$

$\: \: \: \: \: \implies\sf{625=1225+{x}^{2}-70x+{x}^{2}}$

$\: \: \: \: \: \implies\sf{1225-625+2{x}^{2}-70x=0}$

$\: \: \: \: \: \implies\sf{600+2{x}^{2}-70x=0}$

$\: \: \: \: \: \implies\sf{2{x}^{2}-70x+600=0}$

• Taking 2 as common,

$\: \: \: \: \: \implies\sf{2({x}^{2}-35x+300=0}$

$\: \: \: \: \: \implies\sf{{x}^{2}-35x+300=0}$

~By Middle term splitting,

$\: \: \: \: \: \implies\sf{{x}^{2}-(15+20)x+300=0}$

$\: \: \: \: \: \implies\sf{{x}^{2}-15x-20x+300=0}$

$\: \: \: \: \: \implies\sf{x(x-15)-20(x-15)=0}$

$\: \: \: \: \: \implies\sf{(x-15)(x-20)=0}$

$\: \: \: \: \: \implies\bold{x=15\: \: \: \: or\: \: \: \: x=20}$

• Let value of 'x' be 20 i.e, x=20. Then,

>> AB= 35-x= 35-20= 15

>> BC= x= 20

We know, Area of triangle is 1/2(b×h). So,

$\: \: \: \: \: \longrightarrow\sf{Area\: of\: Triangle= \dfrac{1}{2}\times (b\times h)}$

$\: \: \: \: \: \longrightarrow\sf{Area\: of\: Triangle=\dfrac{1}{2}\times (BC\times AB)}$

$\: \: \: \: \: \longrightarrow\sf{Area\: of\ Triangle=\dfrac{1}{2}\times (20\times 15)}$

$\: \: \: \: \: \longrightarrow\sf{Area\: of\: Triangle=\dfrac{1}{2}\times 300}$

$\: \: \: \: \: \longrightarrow\bold{Area\: of\: Triangle=\color{red}{150\: {cm}^{2}}}$

Now, Let the Value of 'x' be 15

>> AB= 35-x= 35-15= 20

>> BC= x= 15

So,

$\: \: \: \: \: \implies\sf{Area\: of\: Triangle=\dfrac{1}{2}\times (b\times h)}$ $\\ \: \: \: \: \: \implies\sf{Area\: of\: Triangle=\dfrac{1}{2}\times (BC\times AB)}$

$\: \: \: \: \: \implies\sf{Area\: of\: Triangle=\dfrac{1}{2}\times (15\times 20)}$

$\: \: \: \: \: \implies\sf{Area\: of\: Triangle=\dfrac{1}{2}\times 300}$

$\: \: \: \: \: \implies\sf{Area\: of\: Triangle=\color{red}{150\: {cm}^{2}}}$

Required Answer:

• Henceforth, Area of Right-angled triangle is 150cm²

Answer 2

Answer:

150cm ° this is a correct answer