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Find the area of a triangle, whose sides are 0.24 m, 28 cm and 32 cm.

A84\sqrt{15}$ $cm^2

B82\sqrt{15}$ $cm^2

C86\sqrt{15}$ $cm^2

D88\sqrt{15}$ $cm^2

Answer:

84\sqrt{15}$ $cm^2

Read Explanation:

the area of a triangle with sides a, b, and c is given by:

Area = (s×(s−a)×(s−b)×(s−c))\sqrt{(s × (s - a) × (s - b) × (s - c))}

where s is the semi-perimeter of the triangle:

s=(a+b+c)/2s = (a + b + c) / 2

Inserting our values:

s=(24+28+32)/2=42s = (24 + 28 + 32) / 2 = 42

Area=(42×(42−24)×(42−28)×(42−32))Area = \sqrt{(42 × (42 - 24) × (42 - 28) × (42 - 32))}

Area=(42×18×14×10)Area = \sqrt{(42 × 18 × 14 × 10)}

Area=105840Area = \sqrt{105840}

We recognize that the expression can be rewritten to include a perfect square:

Area=(15×7056)Area = \sqrt{(15 × 7056)}

Area=8415Area = 84\sqrt{15} cm²


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In the figure ABCD is a square. The length of its diagonal is 4√2 centimetres. The area of the square is :

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