The base and the height of a right-angled triangle are equal in magnitude and thelength of the third side is$$2\sqrt{2}$$​ cm. The area of the triangle is ?

Correct option is A

​We are given:

• It is a right-angled triangle

• Base = Height = same value, say x

• ​$$2\sqrt{2} cm$$​

  Hypotenuse (third side) =

Using the Pythagoras theorem:

Hypotenuse2=Base2+Height2=>(22)2=x2+x2=>8=2x2=>x2=4=>x=2\text{Hypotenuse}^2 = \text{Base}^2 + \text{Height}^2 \Rightarrow (2\sqrt{2})^2 = x^2 + x^2 \Rightarrow 8 = 2x^2 \Rightarrow x^2 = 4 \Rightarrow x = 2Hypotenuse2=Base2+Height2=>(22​)2=x2+x2=>8=2x2=>x2=4=>x=2​

Now, area of triangle:

Area=12×base×height=12×2×2=2 sq. cm\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 2 = 2 \text{ sq. cm}Area=21​×base×height=21​×2×2=2 sq. cm​

Answer: (A) 2 sq. cm