The bases of trapezoid $ABCD$ are $\overline{AB}$ and $\overline{CD}$.We are given that CD = 8, AD = BC = 7,and BD = 9.Find the area of the trapezoid.

Question

The bases of trapezoid $ABCD$ are $\overline{AB}$ and $\overline{CD}$. We are given that $CD = 8$, $AD = BC = 7$, and $BD = 9$. Find the area of the trapezoid.

Answer 1

• Draw a parallel segment of lenght $7$ to $AB$ from $B$ to $E$
• $DE=DC=4$ and angles $ADC=BEC=BCD$
• Find the height of $BEC$ isosceles triangle using Pythagoras theorem (height is $\sqrt{45}$
• Find the area, which is $6\sqrt{45}$

Answer 2

Using Heron's formula to find the area of $ABC$:

$$A_{\Delta}=\sqrt{(12)(5)(4)(3)}=12\sqrt5$$

We also know that $A_{\Delta}=\frac{bh}{2}$. Using base $b=8$:

$$12\sqrt5=4h$$

$$h=3\sqrt5$$

The segment extending from point $B$ perpendicular to $DC$ has height $3\sqrt5$ and intersects $DC$ at a point we'll call $E$. Thus:

$$DE^2+(3\sqrt5)^2=9^2$$

$$DE=6$$

$$\therefore AB=DE-2=4$$

Area of this trapezoid is the height times average of top segment length and bottom segment length. Thus:

$$A_{trap}=\frac{8+4}{2}(3\sqrt5)$$

$$A_{trap}=18\sqrt5$$

Answer 3

Applying successively formulas of Theron and the sinus for the area of a triangle, then the cosinus, then by projection calculate lenght of side $AB$, then again Theron, we have:

$$\triangle BDC=\sqrt{12\cdot3\cdot4\cdot5}=12\sqrt 5=\frac{8\cdot7\sin\alpha}{2}=28\sin \alpha$$ $$\cos \alpha=\sqrt{1-\frac{45}{49}}=\frac 27$$ $$AB+2\cdot7\cos \alpha=4$$ $$\triangle ABD=\sqrt{10\cdot3\cdot6\cdot1}=6\sqrt 5$$ Thus the area of the trapezoid is equal to $$12\sqrt 5+6\sqrt 5=18\sqrt 5$$