If the sides of a triangle are 5 cm- 4 cm and -41- then the area of the triangle is

The correct option is C 10cm^2 Since #160; 5^2+4^2= ( √(41) )^2, the given agnle is a #160;right angled #160; Δ . ∴ Area of #160; ...

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Standard IX

Mathematics

Area of a Triangle

If the sides ...

Question

If the sides of a triangle are $5 c m, 4 c m$ and $\sqrt{41}$ , then the area of the triangle is

20 c m^{2}

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(5 + 4 + \sqrt{41}) c m^{2}

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\frac{5 + 4 + \sqrt{41}}{2} c m^{2}

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10 c m^{2}

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Solution

The correct option is C $10 c m^{2}$
Since $5^{2} + 4^{2} = {(\sqrt{41})}^{2},$ the given agnle is a right angled $Δ$ .

$∴$ Area of $Δ = \frac{1}{2} \times 5 \times 4 = 10 c m^{2}$

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If the sides of a triangle are 5cm,4cm and √41, then the area of the triangle is

The correct option is C 10cm2Since 52+42=(√41)2, the given agnle is a right angled Δ.∴ Area of Δ=12×5×4=10cm2

If the sides of a triangle are $5 c m, 4 c m$ and $\sqrt{41}$ , then the area of the triangle is

The correct option is C $10 c m^{2}$
Since $5^{2} + 4^{2} = {(\sqrt{41})}^{2},$ the given agnle is a right angled $Δ$ .

$∴$ Area of $Δ = \frac{1}{2} \times 5 \times 4 = 10 c m^{2}$