The base of a right angled triangle is 8 m and its hypotenuse is 10 m. Then its area is

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The base of a right angled triangle is 8 m and its hypotenuse is 10 m. Then its area is

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m^{2}

m^{2}

m^{2}

m^{2}

answer is D.

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Detailed Solution

Concept-
The area of a right-angled triangle is calculated using the formula A=

\frac{1}{2}

×b×h, where b is the triangle's base and h is its height.
A right-angled triangle's base measures 8 metres, and its hypotenuse is 10 metres.

Let x represent the right angle triangle's height.
We shall now apply the pythagoras theorem to determine the triangle's height. As a result, we have

{AB}^{2}

{AC}^{2}

{BC}^{2}

.
When the equation is rewritten, we get

{AC}^{2}

{AB}^{2}

{BC}^{2}

.
We obtain

x^{2}

10^{2}

8^{2}

by replacing AB=10 and BC=8 in the equation above.
When the exponent is applied to the terms, we obtain

x^{2}

=100-64.
By removing the words, we obtain

x^{2}

=36.
Square roots on both sides give us x =

\sqrt

36 and x=6.
The triangle's height can never be negative. So,

\Rightarrow x = 6 m

Consequently, a triangle has a height of 6 m.
We shall now calculate the triangle's area.
We obtain

A = \frac{1}{2} \times 8 \times 6

by changing b=8 and h=6 in the formula

A = \frac{1}{2} \times b \times h .

When we divide the terms, we get

A = 4 \times 6 .

When we multiply the terms, we obtain

A = 24 m^{2} .

Hence, option 4 is correct.

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