Question

Find the area of a triangle whose sides are 28cm, 21cm and 35cm.
Also, find the length of the altitude corresponding to the largest side
of the triangle.​

Answer 1

Answer:

Area = 294cm²

Altitude = 16.8cm

$☣ \large \rm\bf{Given}$

Sides of a triangle = 28 cm, 21 cm and 35 cm

$☣ \large\rm \bf{Find}$

1. Area of a triangle
2. Length of the altitude corresponding to the largest side of a triangle

$☣ \large\rm\bf{Instructions}$

Let a = 28cm , b = 21cm and c = 35cm

Then,

s = $\frac{1}{2}(28 + 21 + 35)$ = 42

(s - a) = (42 - 28) = 14

(s - b) = (42 - 21) = 21

(s - c) = (42 - 35) = 7

$☣ \large\rm\bf{Explanation}$

$\small\rm{ Area \: of \: a \: triangle = \sqrt{s(s - a)(s - b)(s - c)}}$

$\small\rm{ ⤍ \sqrt{42 \times 14 \times 21 \times 7 } {cm}^{2}}$

$\small\rm{ ⤍ \sqrt{21 \times 2 \times 7 \times 2 \times 21 \times 7} \: {cm}^{2}}$

$\small\rm{ ⤍(21 \times 2 \times 7) {cm}^{2}}$

$\small\rm {⤍294 \: {cm}^{2}}$

Area = 294cm²

Largest side = 35cm

Let the altitude corresponding to largest side be h cm

Then,

$\small\bf Area \: of \: the \: traingle = ( \frac{1}{2} \times 35 \times h) {cm}^{2}$

$\small\rm{ ⤍ \frac{1}{2} \times 35 \times h = 294 }$

$\small\rm{⤍ \: 35 \times h =294 \times 2 }$

$\small\rm{ h = \frac{294 \times 2}{35} cm}$

$\small\rm {h = 16.8cm}$

Hence, the required altitude is 16.8cm