The sides of a triangle are $5 c m, 12 c m$ and $13 c m$ , what is its area?

0.0024

m^{2}

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0.0026

m^{2}

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0.003

m^{2}

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0.0015

m^{2}

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Solution

The correct option is A $0.003$ $m^{2}$
Area of $△ = \sqrt{s (s - a) (s - b) (s - c)}, s = \frac{a + b + c}{2} = \frac{13 + 12 + 5}{2} = 15 c m$

$∴$ ar of $△ = \sqrt{15 (15 - 13) (15 - 12) (15 - 5)} = \sqrt{15 \times 2 \times 3 \times 10} = 30 {c m}^{2}$

Therefore area of triangle $= {30 c m}^{2} = 0.003 m^{2} [∵ 1 m^{2} = 10000 {c m}^{2}]$

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Similar questions

Find the third side of a triangle whose perimeter is $8 m^{2} - 9 m + 4$ and two of its sides are $3 m^{2} - 5 m$ and $4 m^{2} + 12$ .

Find the third side of a triangle whose two sides are $(3 m^{2} - 5 m$ ) and $(4 m^{2} + 12$ ) and perimeter is $(8 m^{2} - 9 m + 4$ ).

5 c m, 12 c m

13 c m

336

m^{2}

26

30

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