Question

a floor design is made on a floor of a room by joining 4 triangular tiles of dimensions 12cm, 20cm and 24 cm each. Find the cost of the four tiles at the rate of Rs √14 per cm^2.​​

Answer 1

Answer:

Rs 1796

Step-by-step explanation:

Area of one tile, by herons formula = 32√14 cm²

Area of 4 tiles = 4×32√14 = 128√14 cm²

Cost of 1cm²= √14

Cost of 128√14cm² = 128√14 × √14 = 1796

Answer 2

$\huge \fbox \red {\sf{Answer:- }}$

Given:

• Dimension of triangular tiles is 12cm×20cm×24cm.
• Cost of 4 tiles at the rate of ₹√14 cm².

To find:

• Area of triangular tiles.
• Cost of 4 tiles.

Formula required:

Area of triangle by using heron's formula:

$\boxed{\bold{\sqrt{(s-a)(s-b)(s-c)}}}$

Where,

• s=semi-perimeter of triangle.
• a,b and c is sides of triangle.

According to Question:

$\bold {</u><u>A</u><u>rea \: of \: triangle = \sqrt{(s)(s - a)(s - b)(s - c)}}$

Where,

$\bold{s =\frac{12 + 20 + 24}{2}=\frac{56}{2}=28}$

Then,

Area of triangular tile=

$\sqrt{(28)(28 - 12)(28 - 20)(28 - 24)}$

Area of triangular tile=

$\sqrt{28 \times 16 \times 8 \times 4}= \sqrt{14336}$

Area of 1 triangular tiles=

$\bold{ \sqrt{14336} = 119.73}$

Area of 4 triangular tiles=(119.73×4)cm²=478.92cm²

$\bold{∵cost \: of \: 1 \: tile = \sqrt{14}}$

$\bold{∴cost \: of \: 4 \: tiles = 478.92 \times \sqrt{14}}$

$\bold{∴cost \: of \: 4 \: triangular \: tiles =₹ 1791.95}$

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