Question

_ Given:

AB = 9 cm, AC = 9 cm, BC = 10 cm

To Find:

Area of ∆ ABC

Answer 1

Given:

AB = 9 cm, AC = 9 cm, BC = 10 cm

To Find:

Area of ∆ ABC

Solution:

$\boxed{Area of ∆ ABC = \large \frac{ height × base}{2}}$

So, we will draw a perpendicularly line from the vertex A of ∆ ABC, namely AM.

Since, AB = AC, AM will divide BC into 2 equal parts, i.e., BM = CM.

→ BM = CM = BC/2 = 10/2 cm

→ BM = CM = 5 cm

So, in ∆ ABM,

$\angle AMB = 90°$

AB = 9 cm, BM = 5 cm

Using Pythagoras Theorem,

AB² = AM² + BM²

9² = AM² + 5²

81 = AM² + 25

AM² = 81 - 25

AM² = 56

AM = $\sqrt{56} \:$ cm

AM = 7.48 cm

Therefore, height (h) = AM = 7.48 cm

and, base (b) = BC = 10 cm

Area of ∆ ABC = $\Large \frac{height × base}{2}$

$= \Large \frac{7.48×10}{2}$

$= \Large \frac {\: ^{^{3.74}}\: \cancel {7.48} × 10}{\cancel 2}$

$= \boxed{37.4 \: cm^2}$

Hence, Area of $\triangle$ABC is 37.4 cm².

Answer 2

Area of ∆ ABC=1/2×10√(9)²+(10/2)²cm²=5√81+25 cm²=5√106 cm²