Question

The difference between the sides of right angle in right angle triangle is 14 cm. The area of the triangle is 120 cmsq. Find the perimeter of the triangle

Answer 1

Suppose one side of a right angle triangle is x
then another side is x+14 
Area of a triangle is 120 
1/2 * Altitude * Base =120 
1/2 * x * x+14 = 120 
x^2+ 14x = 120*2 
x^2+ 14x = 240 
x^2+ 14x-240 =0 
x^2+16x-15x-240=0 
x(x+16)-15 (x+16)=0 
(x-15) (x+16) =0 
x-15=0 OR x-16=0 
x=15 OR x=16 

Answer 2

Given,

• A right-angled triangle = 14cm
• Area of triangle = 120cm²

To find,

• Perimeter of the triangle

$\:\:\:\:\:\:$─────────────────────

Let the sides containing the right angle $\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$be X cm and (X - 14) cm.

$\:$

Then, its area,

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$ $\bigg[$ $\bf\dfrac{1}{2}$ $\sf × x×(x-14)$$\bigg]$ $\sf cm².$

$\:\:$

$\sf \:\:But, area = 120cm²\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:[given]$

$\:\:$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$$\therefore\:$ $\bf\dfrac{1}{2}$ $\sf x(x-14)=120$

$\:\:$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$ $\hookrightarrow$ $\sf \:x²-14x-240=0$

$\:\:$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$ $\hookrightarrow$ $\sf \:x²-24x+10x-240=0$

$\:\:$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$ $\hookrightarrow$ $\sf \:x(x-24)+10(x-24)=0$

$\:\:$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$$\hookrightarrow$ $\sf \:(x-24)(x+10)=0$

$\:\:$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$$\underline{\boxed{\sf{x=\frak{\pink{24}}}}}$

$\:\:$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$$\underline{\boxed{\sf{x=\frak{\purple{-10\:\:[neglecting]}}}}}$

$\:\:$

$\:$

${\bold { \underline{\normalsize{Now, }}}}$

$\:$

• $\sf One \:side=24cm,$
• $\sf Other\:side= (24-14)cm=10cm.$

$\:\:$

$\:\:\:\:\:\:\:\:\:\:\:\:\:$$\normalsize\mathtt {Hypotenuse=}$ $\displaystyle \sqrt{(24)²+(10)²}$

$\:\:$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$$=$ $\displaystyle \sqrt{576+100}$ $\: =\:$ $\displaystyle \sqrt{676}$

$\:\:$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$$\Large\mathtt {=26\:cm}$

$\:\:$

$\therefore$ $\sf perimeter\:of\: triangle=(24+10+26)cm$

$\:\:$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$ $\underline{\boxed{\sf{=\frak{\red{60\:cm.}}}}}$