Question

Find the area of the trapezoid.
Answer option: 64ft, 48sqrt3, 32sqrt3, 54sqrt3

Answer 1

Answer: OPTION 2

Step-by-step explanation:

1. Calculate the missing leg of the right triangle shown in the figure, with Pythagorean Theorem:

$x=\sqrt{(8ft)^{2}-(4\sqrt{3}ft)^{2}}=4ft$

3. The area of the triangle is:

$A=\frac{4ft*4\sqrt{3}ft}{2}=8\sqrt{3}ft^2$

4. The area of the rectangle is:

$A=10*4\sqrt{3}=40\sqrt{3}ft^{2}$

5. The total area is:

$A_t=40\sqrt{3}+8\sqrt{3}=48\sqrt{3}ft^{2}$

Answer 2

ANSWER

$Area=32 \sqrt{3} \: \: {ft}^{2}$


EXPLANATION

The area of a tra-pezoid is given by


$Area= \frac{1}{2} (sum \:of \: parallel \: sides) \times height$

Let the base of the triangular portion be x.


From Pythagoras Theorem,

${x}^{2} + {(4 \sqrt{3})}^{2} = {8}^{2}$




${x}^{2} + 48 = 64$




${x}^{2}=64 - 48$



${x}^{2}=16$


$x = \sqrt{16}$

$x = 4 \: ft$


This implies that the shorter parallel side of the tra-pezoid

$= 10 - 4=6 \: ft$


We now substitute into the formula to obtain,


$Area= \frac{1}{2} (10 + 6) \times 4 \sqrt{3} \: \: {ft}^{2}$


$Area= \frac{1}{2} (16) \times 4 \sqrt{3} \: \: {ft}^{2}$


$Area=32 \sqrt{3} \: \: {ft}^{2}$