Question

If it is given that the base of an isosceles right triangle is 30 cm then the area of the triangle is:
(a) $450c{{m}^{2}}$
(b) $225\sqrt{3}c{{m}^{2}}$
(c)$5\sqrt{2}c{{m}^{2}}$
(d) None of these.

Answer 1

Hint: Isosceles right-angle triangle is a triangle with equal base and perpendicular. Area of a triangle is given by the relation
Area of a triangle = $\dfrac{1}{2}\times Base\times Height$
Apply the above definition of isosceles right-angle triangle and the relation of area with base and perpendicular.

Complete step-by-step solution -
As we know isosceles triangle is a triangle with two equal sides and a right angle triangle is a triangle with one angle as ${{90}^{\circ }}$ .It means an isosceles right angle triangle will be a triangle with two equal sides and one angle of the triangle should be ${{90}^{\circ }}$ . as hypotenuse of a right-angled triangle is of greatest length than the lengths of other two sides of a right-angle triangle. It means the other two sides (Base and Perpendicular) of the given triangle will be equal to each other and both will have a length of 30cm as given in the question. Now we can draw diagram with the given information as

Where AB is perpendicular to BC and both should be equal to each other, with length of 30cm.
Now, we know the area of any triangle is given as
Area of a triangle = $\dfrac{1}{2}\times Base\times Height............\left( i \right)$
Now the base of the given triangle is 30cm and height i.e. perpendicular length to the base is 30cm as well. Hence, area of the given triangle can be calculated with the help of equation(i) as
Area of triangle $\begin{align}
  & =\dfrac{1}{2}\times 30\times 30 \\
 & =\dfrac{900}{2}=450c{{m}^{2}} \\
\end{align}$
Hence, the area of the given triangle is $450c{{m}^{2}}$ .
So, option(a) is the correct answer.

Note: There may be another approach to get the area of the triangle that we can get the length AC by using Pythagoras theorem and hence, by applying Heron’s formula, we can get the area of the triangle as well.
Pythagoras theorem is given as
${{\left( \text{Hypotenuse} \right)}^{\text{2}}}\text{=}{{\left( \text{Base} \right)}^{\text{2}}}\text{+}{{\left( \text{Perpendicular} \right)}^{\text{2}}}$
And Heron’s formula is given as
$Area=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$
Where a, b, c are the sides of the triangle and the value of ‘s’ is $\dfrac{a+b+c}{2}$ . This approach will be longer for this question but will help in other questions like for the area of a general triangle i.e. $a\ne b\ne c$ and any angle of this triangle is not ${{90}^{\circ }}$ . So, it will help with these kinds of questions.
One may go wrong if he/she takes two equal sides as base and Hypotenuse or perpendicular and hypotenuse will become 30cm, then one side of the triangle will be 0cm by Pythagoras theorem. So, don’t confuse with this part of the question. Hypotenuse of any right-angle triangle cannot be equal to base or perpendicular.