The three sides of a triangle are $3 c m, 4 c m$ and $5 c m$ respectively; then its area is:

6 {c m}^{2}

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\sqrt{23} {c m}^{2}

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\sqrt{12} {c m}^{2}

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\sqrt{32} {c m}^{2}

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Solution

The correct option is A $6 {c m}^{2}$
Given sides of triangle
$a = 3, b = 4, c = 5$

Using heron's formula.
$S = \frac{3 + 4 + 5}{2} = 6$

Area $A = \sqrt{S (S - a) (S - b) (S - c)} = \sqrt{6 \times 3 \times 2 \times 1} = 6 c m^{2}$

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The three sides of a triangle are 3cm,4cm and 5cm respectively; then its area is:

The correct option is A 6cm2Given sides of trianglea=3,b=4,c=5 Using heron's formula. S=3+4+52=6Area A=√S(S−a)(S−b)(S−c)=√6×3×2×1=6cm2

The three sides of a triangle are $3 c m, 4 c m$ and $5 c m$ respectively; then its area is:

The correct option is A $6 {c m}^{2}$
Given sides of triangle
$a = 3, b = 4, c = 5$

Using heron's formula.
$S = \frac{3 + 4 + 5}{2} = 6$

Area $A = \sqrt{S (S - a) (S - b) (S - c)} = \sqrt{6 \times 3 \times 2 \times 1} = 6 c m^{2}$