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I have a problem involving isosceles trapezoids. I have a shape as in the picture with all dimensions known in black or I have the ability to calculate the rest of the black ones as there are enough known.

The problem is I need a way to find out the height ($h_2$) of a trapezoid that is similar given side $b$ and the angles will stay the same, but the only other unit I know is the area ($Y$) of the similar shape.

So the question is can the height ($h_2$) be obtained from the original, given the area ($Y$) of the new trapezoid?

Any help or hints appreciated as this problem is driving me mad.

enter image description here

Community
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Matt_1107
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2 Answers2

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The area of the smaller trapezoid $A_2=\frac{b+a_2}{2}h_2$, The area of the bigger trapezoid is $A_1=\frac{b+2}{2}h$. Thus, $A_1/A_2=\frac{(b+a)h}{(b+a_2)h_2}$. Thus, we can find $a_2h_2$. Also, from similar triangles $h_2/h=(a_2-b)/(a-b)$. Two variables and two equations, we should be able to solve them.

Vasili
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You have $$\frac{b+a_2}{2}h_2=Y,$$ and $$\frac{b+a_2}{2}h_2+\frac{a+a_2}{2}(h-h_2)=\frac{a+b}{2}h.$$ Solving these equations for $h_2$ results in $$h_2 = \frac{-bh+\sqrt{b^2 h^2+2(a-b)Yh}}{a-b}.$$

Verification:

If $Y=0$ then $h_2 = 0$, and $Y=(a+b)h/2 \implies h_2=h$.

Math Lover
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