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Question
The length of the two sides of a right triangle containing the right angle differ by 2 cm. If the area of the triangle is 24xm^2, find the perimeter of the triangle.
Solution
Given: Area of triangle `24cm^2`
Let the sides be a and b, where a is the height and b is the base of triangle
`a-b=2cm`
`a=2+b` .............(1)
Area of triangle =`1/2xxbxxh`
⇒`24=1/2xxbxx(2+b)`
⇒`48=b+1/2b^2`
⇒`48=2b+b^2`
⇒`b^2+2b-48=0`
⇒`(b+8)(b-6)=0`
⇒`b=-8 or 6`
Side of a triangle cannot e negative.
Therefore, b=cm
Substituting the value of b=6 cm in equation (1), we get :
`a=2+6=8cm`
Now, `a=8cm, b=6cm`
In the given right triangle we have to find third side. Using the relation
`"(Hyp)"^2="(Oneside)"^2+"(otherside)"^2`
⇒`Hyp^2=8^2+6^2`
⇒`Hyp^2=64+36`
⇒`Hyp^2=100`
⇒`Hyp^2=10 cm`
So, the third side is 10 cm So,
perimeter of the triangle `a+b+c`
=`8+6+10`
=`24cm`
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