Question

Find the area of the triangle whose sides are 42 cm, 34 cm and 20 cm. Also, find the height corresponding to the longest side.

• A. Area = 336 cm${}^2$ and Height = 10 cm
• B. Area = 300 cm${}^2$ and Height = 10 cm
• C. Area = 336 cm${}^2$ and Height = 16 cm
• D. Area = 300 cm${}^2$ and Height = 16 cm

Answer 1

The correct option is C Area = 336 cm${}^2$ and Height = 16 cm

$\text{Semiperimeter (s)}=\frac{(42+34+20)}{2}$
$=\frac{96}{2}$
$=48cm$

Using heron's formula:
Area of the triangle
$=\sqrt{s(s-a)(s-b)(s-c)}$

By substituting s and a, b , c i.e the given sides, we get

$\text{Area of the triangle}$
$=\sqrt{48\times 6\times 14\times 28}$
$=\sqrt{(4\times 6\times 2)\times 6\times (7\times 2)\times (7\times 4)}$
$=4\times 6\times 2\times 7$
$=336{\text{cm}}^2$

Hence area of the triangle is $336{\text{cm}}^2$.

Here, longest side is 42 cm. Let's consider this side as base of the triangle.
Let height corresponding to this longest side be h.

Area of triangle = $\frac{1}{2}\times \text{base}\times \text{height}$

$\Rightarrow 336=\frac{1}{2}\times 42\times h$

$\Rightarrow$ h = 16 cm

So height corresponding to the longest side is 16 cm.