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.
[asy]
unitsize(1 cm);
pair A, B, C, D;
A = (1,2);
B = (3,2);
C = (4,0);
D = (0,0);
draw(A--B--C--D--cycle);
draw(B--D);
label("$A$", A, NW);
label("$B$", B, NE);
label("$C$", C, SE);
label("$D$", D, SW);
label("$7$", (A + D)/2, W);
label("$7$", (B + C)/2, E);
label("$9$", (B + D)/2, SE);
label("$8$", (C + D)/2, S);
[/asy]
I know I can use Heron's Formula, but I still need side AB.
"You even copied the asymptote? How pathetic. And Steve, don't you realize that the answer is a number? You have to find the height. That's wrong anyway." -very funny (aug. 1 2016)
To find the area of trapezoid $ABCD$, you can use the formula: \[ \text{Area} = \frac{1}{2}(AB + CD) \cdot h \] where $h$ is the height.
To find the height, you can use the Pythagorean Theorem and some additional information given in the problem. Since $AD = BC = 7$ and $BD = 9$, we have a right triangle with legs of length 7 and 9. The height of the trapezoid is the length of the perpendicular segment between the bases $AB$ and $CD$.
Let's use the letters $E$ and $F$ to represent the feet of the perpendiculars from $C$ and $D$ to $AB$, respectively. Then, $CE = h$ and $DF = h$. We need to find the length of $EF$.
Since $CE$ and $DF$ are both perpendicular to $AB$, we have $CE \parallel DF$. This means that $\triangle CDE$ and $\triangle DFE$ are similar triangles, with a ratio of similarity equal to $\frac{CE}{DF}$.
Since $CD = 8$ and $BD = 9$, we can use the proportion $\frac{CE}{DF} = \frac{CD}{BD}$. Substituting the given values, we have $\frac{CE}{DF} = \frac{8}{9}$.
Now, let's express $EF$ in terms of $CE$ and $DF$. Since $E$, $F$, and $B$ are collinear, we have $EF = EB - FB$.
Note that $\triangle BCF$ is similar to $\triangle BDA$ because they share an angle ($\angle B$) and have proportional side lengths. Using this similarity, we can set up the equation $\frac{CE}{DF} = \frac{BC}{FD}$. Substituting the given values, we have $\frac{CE}{h} = \frac{7}{DF}$. Rearranging, we get $CE = \frac{7h}{DF}$.
From $\triangle CDE$, using Pythagorean theorem, we have $DE = \sqrt{CD^2 - CE^2} = \sqrt{8^2 - \left(\frac{7h}{DF}\right)^2}$.
Similarly, from $\triangle DBF$, using Pythagorean theorem again, we have $DF = \sqrt{BD^2 - EF^2} = \sqrt{9^2 - \left(\frac{7h}{DF}\right)^2}$.
Substituting these expressions for $CE$ and $DF$ into the proportion, we have $\frac{\frac{7h}{\sqrt{8^2 - \left(\frac{7h}{DF}\right)^2}}}{\frac{7h}{DF}} = \frac{8}{9}$. Simplifying, we get $\frac{DF^2}{\sqrt{8^2 - \left(\frac{7h}{DF}\right)^2}} = \frac{8}{9}$. Cross-multiplying and rearranging, we have $DF^3 = \frac{8}{\sqrt{8^2 - \left(\frac{7h}{DF}\right)^2}}$.
Squaring both sides of this equation, we get $DF^6 = \frac{64}{64 - \frac{49h^2}{DF^2}}$. Rearranging and multiplying both sides by $64 - \frac{49h^2}{DF^2}$, we have $DF^8 - 64DF^2 + 49h^2 = 0$.
This is a quartic equation in terms of $DF^2$, so it can be a bit complicated to solve directly. However, we can take advantage of a clever observation to simplify things. Notice that $\frac{7h}{DF}$ and $\frac{9}{DF}$ are both expressions that can be written in terms of the sides $AB$, $CD$, and $BD$.
Using this information, we can rewrite the equation $DF^3 = \frac{8}{\sqrt{8^2 - \left(\frac{7h}{DF}\right)^2}}$ as $DF^3 = \frac{8}{\sqrt{64 - \left(\frac{9}{DF}\right)^2}}$.
Now, let's introduce a new variable $x = \frac{9}{DF}$. We can rewrite the equation in terms of $x$ as $DF = \frac{9}{x}$. Substituting this into the equation, we have $\left(\frac{9}{x}\right)^3 = \frac{8}{\sqrt{64 - x^2}}$. Simplifying, we get $\frac{729}{x^3} = \frac{8}{\sqrt{64 - x^2}}$.
Cross-multiplying and squaring both sides, we get $729(64 - x^2) = 8^2x^6$. Expanding and rearranging, we have $8x^6 + 729x^2 - 8^2 \cdot 64 = 0$.
This is a cubic equation in terms of $x^2$, so we can solve it using various methods, such as factoring, synthetic division, or using a cubic formula. Once we find the value of $x^2$, we can substitute it back into $DF = \frac{9}{x}$ to find the length of $DF$.
Finally, we can substitute the length of $DF$ into the equation $EF = EB - FB$ to find the length of $EF$.
Once we have the length of $EF$, we can substitute it into the formula $\text{Area} = \frac{1}{2}(AB + CD) \cdot h$ to find the area of the trapezoid.
I hope this explanation helps! Let me know if you have any further questions.
huh? An easy isosceles trapezoid? Mark off P and Q on AB so you have a rectangle PQCD.
That means that AP=QB=(9-8)/2 = 1/2
Now the height h of the trapezoid is
h^2 = 7^2 - (1/2)^2
And the area is h(8+9)/2