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In trapezoid ABCD, \(\overline{AB}\) \(\parallel \overline{CD}\), and AB < CD. The base \(\overline{CD}\) has a length of 8. The legs \(\overline{AD}\) and \(\overline{BC}\) have lengths of 7, and the diagonal \(\overline{BD} \) has a length of 9. Find the area of the trapezoid.

 Apr 29, 2020
 #1
avatar+12208 
+3

Find the area of the trapezoid.

 

laugh

 Apr 29, 2020
 #2
avatar+24983 
+2

In trapezoid ABCD, \(\overline{AB}\parallel \overline{CD}\), and \(AB < CD\).
The base \(\overline{CD}\) has a length of \(8\).
The legs \(\overline{AD}\) and \(\overline{BC}\) have lengths of \(7\), and the diagonal \overline{BD} has a length of \(9\).
Find the area of the trapezoid.

 

 

\(\text{Let $\overline{CE}=\dfrac{8-\color{red}x}{2}$} \\ \text{Let $\overline{DE}=8-\dfrac{8-\color{red}x}{2}$} \\ \text{Let $\overline{BE}=h$} \\ \text{Let $\overline{AB}=\color{red}x$} \\ \text{Let area of the trapezoid $=A $ } \)

 

Pythagorean Theorem:

\(\begin{array}{|rcll|} \hline \mathbf{\left(\dfrac{8-x}{2}\right)^2 + h^2} &=& \mathbf{7^2} \\\\ h^2 &=& 7^2 - \left(\dfrac{8-x}{2}\right)^2 \\\\ \mathbf{h^2} &=& \mathbf{7^2 - \dfrac{\Big(8-x\Big)^2}{4}} \qquad (1) \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{\Big(8-\left(\dfrac{8-x}{2}\right)\Big)^2 + h^2} &=& \mathbf{9^2} \\\\ \left(\dfrac{16-(8-x)}{2}\right)^2 + h^2 &=& 9^2 \\\\ \left(\dfrac{8+x}{2}\right)^2 + h^2 &=& 9^2 \\\\ \dfrac{\Big(8+x\Big)^2}{4} + h^2 &=& 9^2 \\\\ \mathbf{h^2} &=& \mathbf{9^2 - \dfrac{\Big(8+x\Big)^2}{4}} \qquad (2) \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline (1)=(2): & h^2 = 7^2 - \dfrac{\Big(8-x\Big)^2}{4} &=& 9^2 - \dfrac{\Big(8+x\Big)^2}{4} \\\\ & 7^2 - \dfrac{\Big(8-x\Big)^2}{4} &=& 9^2 - \dfrac{\Big(8+x\Big)^2}{4} \\\\ & \dfrac{\Big(8+x\Big)^2}{4} - \dfrac{\Big(8-x\Big)^2}{4} &=& 9^2-7^2 \\\\ & \dfrac{\Big(8+x\Big)^2-\Big(8-x\Big)^2}{4} &=& 32 \quad | \quad \cdot 4 \\\\ & \Big(8+x\Big)^2-\Big(8-x\Big)^2 &=& 32\cdot 4 \\\\ & 64+16x+x^2-(64-16x+x^2) &=& 32\cdot 4 \\ & 64+16x+x^2-64+16x-x^2 &=& 32\cdot 4 \\ & 16x +16x &=& 32\cdot 4 \\ & 32x &=& 32\cdot 4 \quad | \quad :32 \\ & \mathbf{ x } &=& \mathbf{4} \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline (1): & \mathbf{h^2} &=& \mathbf{7^2 - \dfrac{\Big(8-x\Big)^2}{4}} \quad | \quad \mathbf{x=4} \\\\ & h^2 &=& 7^2 - \dfrac{\Big(8-4\Big)^2}{4} \\\\ & h^2 &=& 7^2 - \dfrac{4^2}{4} \\\\ & h^2 &=& 7^2 - 4 \\ & h^2 &=& 45 \\ & h &=& \sqrt{45} \\ & \mathbf{h} &=& \mathbf{3\sqrt{5}} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{A} &=& \mathbf{\left(\dfrac{8+x}{2}\right)\cdot h} \quad | \quad \mathbf{x=4},\ \mathbf{h=3\sqrt{5}} \\\\ A &=& \left(\dfrac{8+4}{2}\right)\cdot 3\sqrt{5} \\\\ A &=& \left(\dfrac{12}{2}\right)\cdot 3\sqrt{5} \\\\ A &=& 6\cdot 3\sqrt{5} \\\\ \mathbf{A} &=& \mathbf{18\sqrt{5}} \\\\ \mathbf{A} &=& \mathbf{40.2492235950} \\ \hline \end{array} \)

 

laugh

 Apr 29, 2020

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