I was given the following assignment:

A trapezoid's one angle is right and other is 45°. The length of the shorter parallel side is a. The length of the longer diagonal is 4a. Define the area of the trapezoid.

(Please mind the possible mistakes in the text. I translated the assignment from Finnish and I'm not too familiar with English math vocabulary)

I have usually been able to solve problems like this, but now it feels like I've hit a wall. Could someone explain step by step how to solve this? There's obviously something basic I'm just missing here. According to my book, the answer is supposed to be $\frac{1}{4}a^2(14+\sqrt{31})$.

  • $\begingroup$ @RossMillikan Increase the height of the trapezoid. Eventually (certainly before the height is $4a$) the longer diagonal will be long enough. $\endgroup$ – David K Jan 11 '16 at 20:22


Let $b$ the longer basis $AD$. Than, given that the angle $\angle BAH$ is $45°$, the height of the trapezoid is $h= BH=AH=AD-BC=b-a$, and from the diagonal $AC=4a$ ( as hipothenuse of the triangle $ACD$ ) we have:

$$ AC^2=AD^2+CD^2 \qquad \iff \qquad b^2+(b-a)^2=16a^2 $$

solve this equation for $b>0$ and you have done.

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The equation becomes: $$ b^2-2ab-15a^2=0 $$ with solutions : $$ b=\frac{a\pm\sqrt{a^2+30a^2}}{2} $$ and, for $b>0$: $$ b=\frac{a}{2}(1+\sqrt{31}) $$ so the area is: $$ A=\frac{1}{2}(b+a)h=\frac{1}{2}(b+a)(b-a)=\frac{1}{2}\left(\frac{a}{2}(1+\sqrt{31})+a \right)\left(\frac{a}{2}(1+\sqrt{31})-a \right)= $$ $$ =\frac{1}{2}\left(\frac{a^2}{4}(1+\sqrt{31})^2-a^2 \right)=\frac{1}{4}a^2(14+\sqrt{31}) $$

  • $\begingroup$ Thanks you. I was able to get the right answer by using calculator but for some reason I'm unable to solve b by myself. It seems I'm not able to get b only to the other side of the equation. Off topic, but could you give some tips how to solve it? $\endgroup$ – Antti Kivi Jan 11 '16 at 21:18
  • $\begingroup$ $b^2 + b^2 - 2ab + a^2 = 16a^2 \implies 2b^2 - 2ab - 15a^2 = 0$. Use the quadratic equation. I don't think the answer is pretty. But I maybe made an calculation error. $\endgroup$ – fleablood Jan 11 '16 at 21:29
  • $\begingroup$ I've added to my answer. I hope it's useful :) $\endgroup$ – Emilio Novati Jan 11 '16 at 21:30
  • $\begingroup$ Thanks, it really is! I think my problem with this was that I just tried to start thinking it too complexly. $\endgroup$ – Antti Kivi Jan 11 '16 at 22:28

OOOPS. 4a is the diagonal of the trapezoid. Not the long base.

Okay: take 2:

Drop a perpendicular from the top side to the bottom side so that you cut the trapezoid into a rectangle and a right triangle. The rectangle will be $a$ long and $h$ high. The right triangle will be $b$ long and $h$ high and the long side of the trapezoid will be $a + b$.

The right triangle has a 45 degree angle so the other angle is also 45 degrees and the triangle is isosceles. So $b = h$ and the bottom side of the trapezoid is $a + h$.

So what is $h$?

The long diagonal is $4a$. Notice it is the hypotenuse of the right triangle formed by the $a + h$ base and the $h$ side of the trapezoid.

By pythagorean theorem $(a + h)^2 + h^2 = (4a)^2$ so $a^2 + 2ah + 2h^2 = 16a^2$. So $h^2 + ah - 15a^2/2 = 0$. So $h = \frac{a \pm \sqrt{a^2 + 30a^2}}{2} = \frac{a \pm a\sqrt{31}}{2} = \frac{a(\sqrt{31} + 1)}{2}$.

(I hope. With my luck I probably made a really careless arithmetic error.)

So area of trapezoid is $1/2(s_1 + s_2)*h = 1/2(a + a + h)h = ah + h^2/2 = a^2(\frac{\sqrt{31} + 1}{2}) + a^2(31 + 1 + 2\sqrt{31})/8 =a^2(\sqrt{31} + 1/2 + 4 + \sqrt{31}/4) = a^2\frac{5\sqrt{31} + 18}4 $

==== wrong (but much easier) answer follows =====

Imagine you extended the side with the right triangle upward. Imagine you extend the side with the 45 degree angle upward and sideways. Imagine those two extended lines intersect.

The resulting shape is a right triangle with a 45 degree angle. The new angle you just created must also be 45 degrees. So it is an isosceles triangle. It's base is 4a. So it's height is also 4a. So it's area is $1/2*4a*4a = 8a^2$.

Now look at the shorter side, the one that is only a long. It is also a base of a right isosceles triangle. So the area of this little triangle is $1/2*a*a = a^2/2$.

Now the trapezoid is what you get when you "cut off" the small triangle from the big triangle. So the area is $8a^2 - a^2/2 = 15a^2/2$.


Other way. Way your teacher probably wants.

Area of trapeziod = $1/2(s_1 + s_2)*h = 1/2(4a + a)h = 5ah/2$. But what is $h$? Well drop a perpendicular for the top side to the bottom side so that you cut the trapezoid into a rectangle and a triangle. The triangle is a right triangle with a 45 degree angle and a side of $3a$ long.

So the other angle is 45 degree. The triangle is isosceles. So the other side is $3a$. So the height is $3a$.

So the area of the trapezoid is $5ah/2 = 5a*3a/2 = 15a^2/2$.

  • $\begingroup$ Wrong. $4a$ is the longer diagonal not the longer basis !. $\endgroup$ – Emilio Novati Jan 11 '16 at 20:21
  • $\begingroup$ Oh. You are right. $\endgroup$ – fleablood Jan 11 '16 at 20:54

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