Question

A triangle has sides 35 cm, 54cm, and 61 cm long find it's area also find the smallest altitude

Answer 1

Perimeter of the triangle = 150cm
SemiPerimeter of the triangle = 75cm

By Heron's Formula
Area = $\sqrt{s(s-a)(s-b)(s-c)}$
Area = $\sqrt{75*40*21*14}$
Area = 939.14 $cm^2$

Answer 2

Answer:

The area of the triangle is $420 \sqrt{5} \mathrm{cm}^{2}$ and the smallest altitude is $24 \sqrt{5} \mathrm{cm}$

Step-by-step explanation:

To find the area of the triangle we first find it by using Heron’s Formula:

Heron’s Formula $\bold{=\sqrt{s(s-a)(s-b)(s-c)}}$

Now a = 35 cm, b = 54cm, c = 61cm. To find S, we first find the average of the three sides

The averages of the three sides are $\frac{(35+54+61)}{3}=\frac{150}{2}=75$

Hence the value of S is 75

Therefore putting the value of a, b, c, s; we get  

$\begin{array}{l}{\text { Area }=\sqrt{s(s-a)(s-b)(s-c)}} \\ {\text { Area }=\sqrt{75(75-35)(75-54)(75-61)}} \\ {\text { Area }=\sqrt{(75 \times 40 \times 21 \times 14)}=420 \sqrt{5}}\end{array}$

Now as we have found the whole area of the triangle.

Let us find the area of the triangle in which the smallest side is the altitude.

The smallest width or breadth (B) of the triangle is 35 cm.

To find the “height” of the triangle let us form$=\frac{1}{2} \times B \times H=\frac{1}{2} \times 35 \times H=420 \sqrt{5} \mathrm{cm}^{2}$

$H=\frac{2}{35} \times 420 \sqrt{5} \mathrm{cm}^{2}=24 \sqrt{5} \mathrm{cm}$

Therefore the height of the smallest altitude with area $420 \sqrt{5} \mathrm{cm}^{2} \mathrm{is} \bold{24 \sqrt{5} \mathrm{cm}}$