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Find the area of an isosceles trapezoid with legs 16 and bases 15 and 29

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The diagram represents the interpretation of the question for proper clarity.

To be able to obtain the are of the Isosceles trapezoid, we have to find the height first.

To find the height, we are going to use the Pythagoras theorem. Thus, we have:


\begin{gathered} (\text{Hypotenuse)}^2=(\text{Opposite)}^2+(\text{Adjacent)}^2 \\ 16^2=h^2+7^2 \\ 256=h^2+49 \\ 256-49=h^2 \\ 207=h^2 \\ h=\sqrt[]{207} \\ h=14.3875 \end{gathered}

The area of the Isosceles Trapezoid is given as:


A=(1)/(2)(a+b)h

Thus, we have:


\begin{gathered} A=(1)/(2)(15+29)14.3875 \\ A=(1)/(2)*44*14.3875 \\ A=316.53\text{ square unit} \end{gathered}

Hence, the area of the Isosceles trapezoid is 316.53 square unit

Find the area of an isosceles trapezoid with legs 16 and bases 15 and 29-example-1
User Tom Verelst
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