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

Question

Ella Armstrong · Accepted Answer

#A = 390 " units"^2#

Please take a look at my drawing:

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

If we draw the height #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# and #b#, we know that #a + b + 12 = 40# holds which means that #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#

Parker Allen · Answer

undefined The area of a trapezoid can be calculated using the formula:

$$ \text{Area} = \frac{1}{2} \times (\text{sum of the lengths of the bases}) \times (\text{height}) $$

In this case, the sum of the lengths of the bases is $12 + 40 = 52$. To find the height, we can use the Pythagorean theorem because the side lengths of 17 and 25 form a right triangle. So,

$$ \text{height} = \sqrt{(25^2 - 17^2)} = \sqrt{(625 - 289)} = \sqrt{336} $$

Thus, the area of the trapezoid is:

$$ \text{Area} = \frac{1}{2} \times 52 \times \sqrt{336} $$

$$ \text{Area} = 26 \times \sqrt{336} $$

$$ \text{Area} \approx 26 \times 18.33 $$

$$ \text{Area} \approx 477.58 $$

Therefore, the area of the trapezoid is approximately $477.58$ square units.