Question

The sides of a scalene triangle are 11 cm, 12 cm and 13 cm. The length of the altitude corresponding to the side having length 12 cm is ________.

Answer 1

Given:
The sides of a triangle are 11 cm, 12 cm and 13 cm respectively.

Let ABC be a triangle with sides
AB = 11 cm
BC = 12 cm
AC = 13 cm

Let AD be the altitude corresponding to the side having length 12 cm.

Using Heron's formula:
If a, b and c are three sides of a triangle, then
area of the triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ , where $s=\frac{a+b+c}{2}$.


Here, $s=\frac{11+12+13}{2}=\frac{36}{2}=18$

$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}=\sqrt{18\left(18-11\right)\left(18-12\right)\left(18-13\right)} \\ =\sqrt{18\left(7\right)\left(6\right)\left(5\right)} \\ =\sqrt{\left(2\times 3\times 3\right)\left(7\right)\left(3\times 2\right)\left(5\right)} \\ =\left(2\times 3\right)\sqrt{105} \\ =6\sqrt{105}{\mathrm{cm}}^2 \end{gathered}$$


We know,
$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}=\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height} \\ \Rightarrow 6\sqrt{105}=\frac{1}{2}\times 12\times AD \\ \Rightarrow \sqrt{105}=AD \\ \Rightarrow AD=\sqrt{105}\mathrm{cm} \end{gathered}$$

Hence, the length of the altitude corresponding to the side having length 12 cm is $\underline{\sqrt{105}\mathrm{cm}}$.