Question

Find the area of a triangle whose sides are 18cm ,10cm and 14cm .

Answer 1

Answer:

This is a question of area of triangle.

Answer 2

Answer:-

$\sf{Area \ of \ the \ triangle \ is \ 21\sqrt11 \ cm^{2}}$

Given:

• The sides of the triangle are 18 cm, 10 cm and 14 cm.

To find:

• The area of the triangle.

Solution:

By heron's formula

$\sf{s=\frac{a+b+c}{2}}$

$\sf{s=\frac{18+10+14}{2}}$

$\sf{\therefore{s=\frac{42}{2}}}$

$\boxed{\sf{\therefore{s=21}}}$

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

$\sf{\therefore{Area \ of \ triangle=\sqrt{21(21-18)(21-10)(21-14)}}}$

$\sf{\therefore{Area \ of \ triangle=\sqrt{21\times3\times11\times7}}}$

$\sf{\therefore{Area \ of \ triangle=21\sqrt11 \ cm^{2}}}$

$\sf\purple{\tt{\therefore{Area \ of \ the \ triangle \ is \ 21\sqrt11 \ cm^{2}}}}$