In the standard (x, y) coordinate plane, an isosceles trapezoid ABCD has vertices at A(0, … — ACT

In the standard $(x, y)$ coordinate plane, an isosceles trapezoid $ABCD$ has vertices at $A(0, 0)$ , $B(8, 0)$ , $C(6, 4)$ , and $D(2, 4)$ . The diagonals $AC$ and $BD$ intersect at point $P$ . What is the area of triangle $APB$ ?

A
8
$\frac{32}{3}$ Answer
C
12
D
$\frac{64}{3}$
E
16

Answer

\frac{32}{3}

The correct answer is obtained by first finding the equations of the lines representing the diagonals

AC

and

BD

, which intersect at

P(4, \frac{8}{3})

. Since

AB

lies on the x-axis from

x = 0

x = 8

, the length of the base of triangle

APB

8

. The height of the triangle is the y-coordinate of

P

, which is

\frac{8}{3}

. Using the formula for the area of a triangle, we get

\frac{1}{2} \times 8 \times \frac{8}{3} = \frac{32}{3}

Step-by-Step Solution

Find the equations of the lines containing the diagonals

AC

and

BD

Line

AC

passes through

(0,0)

and

(6,4)

, so its equation is

y = \frac{2}{3}x

. Line

BD

passes through

(8,0)

and

(2,4)

, so its slope is

\frac{4-0}{2-8} = -\frac{2}{3}

and its equation is

y = -\frac{2}{3}(x-8) = -\frac{2}{3}x + \frac{16}{3}

Determining the equations of the lines allows us to find their intersection point, which is the vertex

P

of triangle

APB

Determine the coordinates of the intersection point

P

by solving the system of equations.

Equating the two expressions for

y

gives

\frac{2}{3}x = -\frac{2}{3}x + \frac{16}{3}

, which simplifies to

\frac{4}{3}x = \frac{16}{3}

, so

x = 4

. Substituting

x = 4

back into the equation for line

AC

gives

y = \frac{2}{3}(4) = \frac{8}{3}

. The intersection point

P

(4, \frac{8}{3})

The

y

-coordinate of point

P

represents the height of triangle

APB

relative to the base

AB

along the x-axis.

Calculate the area of triangle

APB

using the area formula.

The base of triangle

APB

is the segment

AB

, which has length

8 - 0 = 8

. The height is the

y

-coordinate of

P

, which is

\frac{8}{3}

. The area is

\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \times \frac{8}{3} = \frac{32}{3}

Applying the triangle area formula with the correct base and height yields the final answer.

Key Concept

Properties of Quadrilaterals (specifically, trapezoids and their diagonals on the coordinate plane)

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

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