What is the area of a trapezoid with base lengths of 12 and 40, and side lengths of 17 and 25?

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Answer 1 · 2015-11-26T09:26:21

$A = 390 {units}^{2}$ $A = 390 " units"^2$

Please take a look at my drawing:

enter image source here

To compute the area of the trapezoid, we need the two base lengths (which we have) and the height $h$ $h$ .

If we draw the height $h$ $h$ as I did in my drawing, you see that it builds two right angle triangles with the side and the parts of the long base.

About $a$ $a$ and $b$ $b$ , we know that $a + b + 12 = 40$ $a + b + 12 = 40$ holds which means that $a + b = 28$ $a + b = 28$ .

Further, on the two right angle triangles we can apply the theorem of Pythagoras:

${ (17 ^2 = a ^2 + h^2), (25^2 = b^2 + h^2) :}$

Let's transform $a + b = 28$ into $b = 28 - a$ and plug it into the second equation:

${ (17 ^2 = color(white)(xxxx)a ^2 + h^2), (25^2 = (28-a)^2 + h^2) :}$

${ (17 ^2 = color(white)(xxxxxxxx)a ^2 + h^2), (25^2 = 28^2 - 56a + a^2 + h^2) :}$

Subtracting one of the equations from the other gives us:

$25^2 - 17^2 = 28^2 - 56a$

The solution of this equation is $a = 8$ , so we conclude that $b = 20$ .

With this information, we can compute $h$ if we plug either $a$ in the first equation or $b$ in the second one:

$h = 15$ .

Now that we have $h$ , we can compute the area of the trapezoid:

$A = (12 + 40 )/2 * 15 = 390 " units"^2$