Question

In the given figure, AB = 8 cm, BC = 6 cm, AC = 10 cm, then the area of $\Delta ABC$ is equal to

• A. $(\sqrt{3}+20)c{m}^2$
• B. $(20\sqrt{3}+5)c{m}^2$
• C. $(24-\sqrt{3})c{m}^2$
• D. $24c{m}^2$

Answer 1

The correct option is D $24c{m}^2$



Given: $AB=8cm,BC=6cm,AC=10cm$
Consider: $A{B}^2+B{C}^2=8^2+6^2$
$=64+36$
$=100$
$={10}^2$
$=A{C}^2$
$\Rightarrow A{B}^2+B{C}^2=A{C}^2$
Therefore, $\Delta ABC$ is a right angle triangle, right-angled at B.
$\therefore$ Area of $\Delta ABC=\frac{1}{2}\times BC\times AB$
$=\frac{1}{2}\times 6\times 8$
$=3\times 8$
$=24c{m}^2$
Hence, the correct answer is option (d).