Question

In AABC, AB = 3 cm, BC = 4 cm and AC = 5 cm, find area of AABC​

Answer 1

Answer:

Given: ∠A=90

∘

⇒AB

2

+AC

2

=BC

2

⇒AC

2

=5

2

−4

2

=3

2

⇒AC=3cm

⇒ Area =

2

1

×AB×AC

⇒ Area =

2

1

×4×3=6 cm

2

Answer 2

$\huge\bold{\gray{\sf{Answer:}}}$

$\bold{Explanation:}$

Above figure is a scalene triangle .

Given :

AB=3cm ,BC=4cm and AC=5cm

To Find :

Area of $\triangle{ABC}$

There are 3 sides given :

So we first find 'S'

$\sf S= \dfrac{AB + BC + AC}{2}$

$\sf S= \dfrac{3 + 4 + 5}{2} = \dfrac{12}{2} = 6$

Now,we will find area of triangle

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

$\sf= > \sqrt{6(6 - 3)(6 - 4)(6 - 5)}$

$\sf= > \sqrt{6(3)(2)(1)} = \sqrt{6 \times 6} = \sqrt{36} = 6$

$\therefore$ Area of triangle is $\bold{\red{6 {cm}^{2}}}$