A floor design is made on a floor of a room by joining $4$ triangular tile of dimensions $12 c m, 20 c m$ and $24 c m$ each. Find the cost of the tile at the rate of $R s . 14$ per $s q . c m$ .

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Solution

The triangle tiles are of dimension $12 c m, 20 c m$ and $24 c m$ .
We know that,
Heron's formula,
$S = \frac{1}{2} (12 + 20 + 24) = 28$
Area $= \sqrt{28 (28 - 12) (28 - 20) (28 - 24)}$
$\begin{matrix} = \sqrt{28 \times 16 \times 8 \times 4} = \sqrt{14336} = 119.73 \end{matrix}$
Area $= 119.73$
The cost of each tile $= 119.73 \times \sqrt{14}$
$= 448$
Then,
We get the cost of each tile is $448$ .

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A floor design is made on a floor of a room by joining 4 triangular tile of dimensions 12 cm,20 cm and 24 cm each. Find the cost of the tile at the rate of Rs.14 per sq.cm.

The triangle tiles are of dimension 12cm,20cm and 24cm.We know that,Heron's formula,S=12(12+20+24)=28Area =√28(28−12)(28−20)(28−24)=√28×16×8×4=√14336=119.73Area =119.73The cost of each tile =119.73×√14=448Then,We get the cost of each tile is 448.

A floor design is made on a floor of a room by joining $4$ triangular tile of dimensions $12 c m, 20 c m$ and $24 c m$ each. Find the cost of the tile at the rate of $R s . 14$ per $s q . c m$ .