Area of triangle $A B C$ whose sides are $24 m, 40 m$ and $32 m$ . is :

96 m^{2}

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384 m^{2}

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43 m^{2}

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192 m^{2}

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Solution

The correct option is B $384 m^{2}$
A] a=24
b = 40
c = 32
$S = \frac{a + b + c}{2} = \frac{24 + 40 + 32}{2} = \frac{96}{2} = 48$

Hurons formula
$= \sqrt{S (S - a) (S - b) (S - c)}$
$= \sqrt{48 (48 - 24) (48 - 40) (48 - 32)}$
$= \sqrt{48 \times 24 \times 8 \times 16}$
$= \sqrt{4 \times 3 \times 2 \times 2 \times 4 \times 3 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2}$
$= \sqrt{(4 \times 4) \times (3 \times 3) \times (2^{10})}$
$= 4 \times 3 \times 2^{5}$
= $12 \times 32$
=384 $M^{2}$

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Area of triangle ABC whose sides are 24 m,40 m and 32 m. is :

The correct option is B 384 m2A] a=24 b = 40c = 32S=a+b+c2=24+40+322=962=48Hurons formula =√S(S−a)(S−b)(S−c)=√48(48−24)(48−40)(48−32)=√48×24×8×16=√4×3×2×2×4×3×2×2×2×2×2×2×2×2=√(4×4)×(3×3)×(210)=4×3×25= 12×32=384 M2

Area of triangle $A B C$ whose sides are $24 m, 40 m$ and $32 m$ . is :