# Find the area of a trapezoid given the length of a diagonal and the altitude of the trapezoid.

In a trapezoid with perpendicular diagonals, the length of a diagonal is 5 and the length of the altitude of the trapezoid is 4. Find the area of the trapezoid.

I have tried using similar triangle properties and using pythagorean theorem but have found no solution. I tried using triangle BPC ~ triangle APD but didn't know where to go after that.

UPDATE: I have gotten to the point using the definition of the height of a triangle and the pythagorean theorem. But I don't know how to find the area of the top left triangle.enter image description here

I used the fact that ACF and ABC are congruent by ASA so I got the area of the trapezoid is 38 but I'm not sure if that is right

• Why is $AF+FD=16$ ? You might have made a mistake here. Commented Oct 4, 2021 at 20:52

Hint: Use this figure to find:

1-In a trapzoid if diagonals are perpendicular then one side is equal to one diagonal(AB=BD).

2- In this case AD=6 (use Pythagoras's theorem).

3-Use the congruence of right angled triangles with vertex E to find the measure of side BC.

4-The area will be $$\frac{(BC+AD)}2\times BH\approx 17$$.

Say diagonal $$AC = 5$$. As diagonals are perpendicular to each other, $$\angle AOD = 90^\circ$$

$$\triangle AOD \sim \triangle AFC$$ and $$\triangle AOD \sim \triangle COB$$. If $$OC = y, AO = 5 - y$$,

$$\displaystyle \frac{BC}{y} = \frac{AD}{5-y} = \frac{5}{3}$$

$$\implies \displaystyle BC = \frac{5y}{3}, AD = \frac{25-5y}{3}$$

$$\displaystyle BC + AD = \frac{25}{3}$$

Area of trapezoid is,

$$\displaystyle A = \frac{1}{2} (BC + AD) \cdot CF = \frac{50}{3}$$

Let $$BF=4$$ and $$AC=5$$

Draw $$BE\perp BD$$, so $$BE \parallel AC$$ and $$BE=AC=5$$

Area of trapezoid$$=\frac{(AD+BC)\times BF}{2}=\frac{(EA+AD)\times BF}{2}=\frac{ED\times BF}{2}=\frac{\frac{25}{3}\times 4}{2}=\frac{50}{3}$$, since $$EA=BC$$

Applying Pytha

$$EF^2=BE^2-BF^2$$, $$EF=3$$

Applying Euclidean Theorems

$$BE^2=EF\times ED$$ then $$25=3\times ED=$$, $$ED=\frac{25}{3}$$