Question

2. Find the area of a triangle two sides of which are 18 cm
and 10 cm and the perimeter is 42 cm.
​

Answer 1

Let sides be 18cm, 10cm and x

$18 + 10 + x = 42$

$x = 14cm$

Using Heron's formula

S= Perimeter/2= 42/2= 21

Area=

$\sqrt{s(s - a )(s - b)(s - c) }$

$= \sqrt{21 \times 3 \times 11 \times 7}$

$= \sqrt{4851}$

$= 69.64c {m}^{2}$

Answer 2

Answer:

Following the the given image drawn by me:-$AC=18cm\\AB=10cm\\perimeter=42cm\\10+18+BC=42\\=>BC=42-28\\=>BC=14cm\\\\s=\dfrac{p}{2} \\=\dfrac{42}{2} \\=21\\area=\sqrt{s(s-a)(s-b)(s-c)} \\\implies \sqrt{21*3*7*11}\\\implies 21\sqrt{11}\\\implies 21*3.317\\\implies 69.657 cm^{2}$

Here formula used:

Heron's formula = $\sqrt{s(s-a)(s-b)(s-c)$

plzz mrk it as the brainliest one