Question

Match the following triangles with the measure of their respective areas (given in sq cm).

• A. $4\sqrt{3}$
• B. $84$
• C. $6\sqrt{6}$

Answer 1

Case (1):
In $△$ ABC,
since sides AB, BC and CA form a pythagorean triplet,
$△$ ABC is a right angled triangle.
So, area of $△$ ABC = $\frac{1}{2}\times base\times height$
= $\frac{1}{2}\times 24\times 7$
= $84c{m}^2$

Case (2):
It is mentioned that all the sides of $△$ PQR are equal.
So, $△$ PQR is an equilateral triangle
Area of $△$ PQR = $\frac{\sqrt{3}}{4}\times {side}^2$
= $\frac{\sqrt{3}}{4}\times 4^2$
= $4\sqrt{3}c{m}^2$

Case (3):
The given triangle XYZ is a scalene triangle.
So, applying Heron's formula
area of $△$XYZ = $\sqrt{s\times (s-a)\times (s-b)\times (s-c)}$
where, s= $\frac{a+b+c}{2}$
So, s = $\frac{5+6+7}{2}$ = 9cm
Now, area of $△$ XYZ = $\sqrt{9\times (9-6)\times (9-5)\times (9-7)}$
= $\sqrt{9\times 3\times 4\times 2}$
= $6\sqrt{6}c{m}^2$