Question

what is the area of triangle whose sides are AB=8 cm, BC=11 cm and AC=4cm.

please explain with the answer I need it urgently!​

Answer 1

Answer:-

Given:-

• AB = 8 cm
• BC = 11 cm
• AC = 4 cm

To Find:-

Area occupied by the triangle.

__________...

∴ △ABC is a scalene triangle since all sides are not same.

We know,

Area of △ :- (Heron's Formula)

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

where,

• s = Semi perimeter,
• a, b, c = Sides of the ∆.

∴ Semi perimeter = (a + b + c) × ½

= [(8 + 11 + 4) cm] × ½

= 23 cm × ½

= 11.5 cm

∴ ar(△ABC) =

√11.5(11.5 - 8)(11.5 - 11.0)(11.5 - 4)

= √11.5 × 3.5 × 0.5 × 7.5

= √40.25 × 3.75

≈ √151 cm²

≈ 12.28 cm² (Approx. val.)

∴ Area of the triangle was 12.28 cm² approximately.

Answer 2

Answer:

The area of triangle = 12.2 $cm^{2}$

Step-by-step explanation:

Given:

AB = a =8 cm

BC= b = 11 cm and

AC= c = 4 cm.

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

where s = a+b+c/2

             = 8+11+4/2 =23/2 = 11.5 cm

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

                          =$\sqrt{11.5(11.5-8) (11.5-11) (11.5-4)}$

                          =$\sqrt{11.5 * 3.5 * 0.5 * 7.5}$

                          =$\sqrt{151}$ (approx...)

                         =12.2 $cm^{2}$

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