Isosceles triangle $ABE$ of area 100 square inches is cut by $\overline{CD}$ into an isosceles trapezoid and a smaller isosceles triangle. The area of the trapezoid is 75 square inches. If the altitude of triangle $ABE$ from $A$ is 20 inches, what is the number of inches in the length of $\overline{CD}$?

[asy]
draw((-9,0)--(0,24)--(9,0)--cycle);
draw((-6,8)--(6,8));
label("$A$",(0,24),N);
label("$B$",(-9,0),W);
label("$C$",(-6,8),NW);
label("$D$",(6,8),NE);
label("$E$",(9,0),E);
[/asy]
The area of triangle $ABE$ is $\frac{1}{2}(\text{base})(\text{height})=\frac{1}{2}(BE)(20\text{ in.})$. Setting this equal to $100$ square inches we find $BE=10$ inches.  The area of triangle $ACD$ is $100-75=25$ square inches.  Since triangle $ACD$ is similar to triangle $ABE$ and the ratio of their areas is $\frac{1}{4}$, the ratio of corresponding side lengths is $\sqrt{\frac{1}{4}}=\frac{1}{2}$.  Therefore, $CD=\frac{1}{2}BE=\boxed{5}$ inches.

Alternatively, because triangles $ACD$ and $ABE$ are similar, the height-to-base ratio is the same for each of them.  In triangle $ABE$, this ratio is $\frac{20\text{ in.}}{10\text{ in.}}=2$.  Therefore, the height of $ACD$ is $2\cdot CD$.  Solving $\frac{1}{2}(2\cdot CD)(CD)=25\text{ in.}^2$ we find $CD=5$ inches.
5
Level 4
Geometry
Problem: Isosceles triangle $ABE$ of area 100 square inches is cut by $\overline{CD}$ into an isosceles trapezoid and a smaller isosceles triangle. The area of the trapezoid is 75 square inches. If the altitude of triangle $ABE$ from $A$ is 20 inches, what is the number of inches in the length of $\overline{CD}$?

[asy]
draw((-9,0)--(0,24)--(9,0)--cycle);
draw((-6,8)--(6,8));
label("$A$",(0,24),N);
label("$B$",(-9,0),W);
label("$C$",(-6,8),NW);
label("$D$",(6,8),NE);
label("$E$",(9,0),E);
[/asy]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the length of line segment \(\overline{CD}\), we first need to determine the area of the smaller isosceles triangle \(ACE\). Given that the area of triangle \(ABE\) is 100 square inches and the area of the trapezoid \(BCDE\) is 75 square inches, we can calculate the area of triangle \(ACE\) as follows:

Area of triangle \(ACE\) = Area of triangle \(ABE\) - Area of trapezoid \(BCDE\)  
Area of triangle \(ACE\) = 100 square inches - 75 square inches = 25 square inches.

Next, we know that the area of a triangle can be calculated using the formula:  
Area = \(\frac{1}{2} \times \text{base} \times \text{height}\).

For triangle \(ACE\), let the base \(CE\) be denoted as \(b\) and the height from point \(A\) to line \(CE\) be denoted as \(h\). The area of triangle \(ACE\) can thus be expressed as:  
25 = \(\frac{1}{2} \times b \times h\).

Since the altitude of triangle \(ABE\) from point \(A\) is given as 20 inches, and the height from \(A\) to line \(CD\) (which is the same as the height from \(A\) to line \(CE\)) can be determined by the position of line \(CD\) relative to the base of triangle \(ABE\). The height of trapezoid \(BCDE\) is the difference between the total height of triangle \(ABE\) and the height of triangle \(ACE\).

Let \(h_{CD}\) be the height from \(A\) to line \(CD\). The height of trapezoid \(BCDE\) is \(20 - h_{CD}\). The area of trapezoid \(BCDE\) can also be expressed in terms of its bases \(b_1\) and \(b_2\) (where \(b_1\) is the length of \(BE\) and \(b_2\) is the length of \(CD\)) and its height:  
Area = \(\frac{1}{2} \times (b_1 + b_2) \times (20 - h_{CD})\).

Since \(b_1\) is equal to the base of triangle \(ABE\), we can find \(b_1\) using the area of triangle \(ABE\):  
100 = \(\frac{1}{2} \times b_1 \times 20\)  
=> \(b_1 = 10\).

Now substituting \(b_1\) into the area formula for trapezoid \(BCDE\):  
75 = \(\frac{1}{2} \times (10 + b_2) \times (20 - h_{CD})\).

We also know that the height \(h_{CD}\) can be derived from the area of triangle \(ACE\):  
25 = \(\frac{1}{2} \times b \times h_{CD}\).

Now we have two equations involving \(b_2\) and \(h_{CD}\). We can solve for \(h_{CD}\) in terms of \(b\) and substitute it back into the trapezoid area equation. 

After solving these equations, we find that the length of \(\overline{CD}\) is 5 inches.<|END_OF_LATENT|> The final answer is $5$.