The area of a right angled triangle is 80 sq. cm. The ratio of the bas

1.	The area of a right angled triangle is 80 sq. cm. The ratio of the base and the height of the triangle is 4 : 5. Find the length of hypotenuse.1). 82cm2). 2 82cm3). 2 41cm4). 3 82cm
Answer» Solution Let base and height be $4x$ and $5X$ respectively. Area of TRIANGLE = $\frac{1}{2} \times$ base $\times$ height = 80 => $\frac{1}{2} \times 4x \times 5x = 80$ => $x^2 = \frac{80}{10} = 8$ => $x = \sqrt{8} = 2 \sqrt{2}$ => Base = $8 \sqrt{2}$ and Height = $10 \sqrt{2}$ $\therefore$ Hypotenuse = $\sqrt{(8 \sqrt{2})^2 + (10 \sqrt{2})^2}$ = $\sqrt{128 + 200} = \sqrt{328}$ $\approx 18$ cm

The area of a right angled triangle is 80 sq. cm. The ratio of the base and the height of the triangle is 4 : 5. Find the length of hypotenuse.1). 82cm2). 2 82cm3). 2 41cm4). 3 82cm

Answer»

Solution

Let base and height be $4x$ and $5X$ respectively.

Area of TRIANGLE = $\frac{1}{2} \times$ base $\times$ height = 80

=> $\frac{1}{2} \times 4x \times 5x = 80$

=> $x^2 = \frac{80}{10} = 8$

=> $x = \sqrt{8} = 2 \sqrt{2}$

=> Base = $8 \sqrt{2}$ and Height = $10 \sqrt{2}$

$\therefore$ Hypotenuse = $\sqrt{(8 \sqrt{2})^2 + (10 \sqrt{2})^2}$

= $\sqrt{128 + 200} = \sqrt{328}$

$\approx 18$ cm