Question

What is the area of trapezoid ABCD ?

Answer 1

AD = 5
BC = 15
AB = 5

Area = 1/2(AD + BC)*AB
Area = 1/2(5 + 15)(5)
Area = 1/2(20)(5)
Area = 50

Answer 2

ANSWER

Area of the trap-ezoid is $50$ square units

EXPLANATION

The given trap-ezoid has vertices $A(-2,2)$, $B(2,5)$, $C(11,-7)$ and $D(1,-2)$.

Area of a trap-ezoid is given by

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

We use the distance formula to determine length of all the necessary sides and plug them in to the formula.

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

The vertical height of the trap-ezoid is

$|AB|=\sqrt{(2--2)^2+(5-2)^2}$

$|AB|=\sqrt{(2+2)^2+(5-2)^2}$

$|AB|=\sqrt{(4)^2+(3)^2}$

$|AB|=\sqrt{16+9}$

$|AB|=\sqrt{25}$

$|AB|=5$ units

The length of the parallel sides are;

$|AD|=\sqrt{(1--2)^2+(-2-2)^2}$

$|AD|=\sqrt{(3)^2+(-4)^2}$

$|AD|=\sqrt{9+16}$

$|AD|=\sqrt{25}$

$|AD|=5$

and

$|BC|=\sqrt{(11-2)^2+(-7-5)^2}$

$|BC|=\sqrt{(9)^2+(-12)^2}$

$|BC|=\sqrt{81+144}$

$|BC|=\sqrt{225}$

$|BC|=15$ units

we now substitute all these values to obtain,

$Area =\frac{1}{2}(15+5)\times 5$

$Area =\frac{1}{2}(20)\times 5$

$Area =(10)\times 5$

$Area =50$ square units.