# The sides of a triangle are 35 cm,

Question:

The sides of a triangle are 35 cm, 54 cm and 61 cm, respectively. The length of its longest altitude

(a) 1675 cm

(b) 1o75 cm

(c) 2475 cm

(d) 28 cm

Thinking Process

(i) First, determine the semi-perimeter, s and then determine the area of triangle by using Heron’s formula.

(ii) For the longest altitude, take base as the smallest side. Apply the formula,

Area = ½ x Base x Altitude

(iii) Equate the area obtained using the two formula’s and obtain the required height.

Solution:

(c) Let $A B C$ be a triangle in which sides $A B=35 \mathrm{~cm}, B C=54 \mathrm{~cm}$ and $C A=61 \mathrm{~cm}$

Now, semi-perimeter of a triangle,

$s=\frac{a+b+c}{2}=\frac{35+54+61}{2}=\frac{150}{2}=75 \mathrm{~cm}$

$\left[\because\right.$ semi-perimeter, $\left.s=\frac{a+b+c}{2}\right]$

$\because \quad$ Area of $\triangle A B C=\sqrt{s(s-a)(s-b)(s-c)}$ [by Heron's formula]

$=\sqrt{75(75-35)(75-54)(75-61)}$

$=\sqrt{75 \times 40 \times 21 \times 14}$

$=\sqrt{25 \times 3 \times 4 \times 2 \times 5 \times 7 \times 3 \times 7 \times 2}$

$=5 \times 2 \times 2 \times 3 \times 7 \sqrt{5}=420 \sqrt{5} \mathrm{~cm}^{2}$

Aso, Area of $\triangle A B C=\frac{1}{2} \times A B \times$ Altitude

$\Rightarrow \quad \frac{1}{2} \times 35 \times C D=420 \sqrt{5}$

$\Rightarrow \quad C D=\frac{420 \times 2 \sqrt{5}}{35}$

$\therefore \quad C D=24 \sqrt{5}$

Hence, the length of altitude is $24 \sqrt{5} \mathrm{~cm}$.