Cross Sectional Area

Geometry Level 1

The trapezoid shown in the given figure represents a cross section of the rudder of a ship.

If the distance from A A to B B is 13 feet, what is the area of the cross section of the rudder in square feet?


The answer is 42.

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Yahia El Haw by Yahia El Haw , 18, Egypt

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5 solutions

Yahia El Haw
Nov 8, 2016

5 Helpful 0 Interesting 0 Brilliant 0 Confused
Joe Potillor
Nov 10, 2016

3 Helpful 0 Interesting 0 Brilliant 0 Confused

By pythagorean theorem, we have A C = 1 3 2 − 5 2 = 12 f t . AC=\sqrt{13^2-5^2}=12~ft.

The figure is a trapezoid, so the area is 1 2 ( 2 + 5 ) ( 12 ) = 42 f t . 2 \dfrac{1}{2}(2+5)(12)=42~ft.^2

0 Helpful 0 Interesting 0 Brilliant 0 Confused

A C = 1 3 2 − 5 2 = 12 AC=\sqrt{13^2-5^2}=12

A = 1 2 ( 2 + 5 ) ( 12 ) = 42 feet 2 A=\dfrac{1}{2}(2+5)(12)=42~\text{feet}^2

0 Helpful 0 Interesting 0 Brilliant 0 Confused

From the 5 − 12 − 13 5-12-13 right triangle, the base of the trapezoid is 12 12 feet. So the area is

A = 1 2 ( 2 + 5 ) ( 12 ) = 42 A=\dfrac{1}{2}(2+5)(12)=42

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