In the standard (x, y) coordinate plane, trapezoid ABCD has vertices at A(0, 0), B(4, 8), … — ACT

In the standard $(x, y)$ coordinate plane, trapezoid $ABCD$ has vertices at $A(0, 0)$ , $B(4, 8)$ , $C(12, 8)$ , and $D(16, 0)$ . Point $P$ is the midpoint of diagonal $AC$ , and point $Q$ is the midpoint of diagonal $BD$ . What is the distance, in coordinate units, between point $P$ and point $Q$ ?

A
$2$
$4$ Cevap
C
$6$
D
$8$
E
$12$

Cevap

The distance between point

P

and point

Q

4

units.

The correct answer is

4

. Using the midpoint formula, the midpoint of diagonal

AC

P(6, 4)

and the midpoint of diagonal

BD

Q(10, 4)

. Because both points lie on the horizontal line

y = 4

, the distance between them is

|10 - 6| = 4

units.

Adım Adım Çözüm

Calculate the coordinates of midpoint

P

of diagonal

AC

P = \left(\frac{0 + 12}{2}, \frac{0 + 8}{2}\right) = (6, 4)

The midpoint formula is

\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)

Calculate the coordinates of midpoint

Q

of diagonal

BD

Q = \left(\frac{4 + 16}{2}, \frac{8 + 0}{2}\right) = (10, 4)

Applying the midpoint formula to vertices

B(4, 8)

and

D(16, 0)

Calculate the horizontal distance between

P(6, 4)

and

Q(10, 4)

\text{Distance } PQ = \sqrt{(10 - 6)^2 + (4 - 4)^2} = 10 - 6 = 4

Since both midpoints share the same

y

-coordinate (

y = 4

), the distance is simply the absolute difference between their

x

-coordinates.

Anahtar Kavram

Midpoints of Diagonals in a Trapezoid

Alternatif Yöntem

For any trapezoid with parallel bases of lengths

b_1

and

b_2

(where

b_1 > b_2

), the length of the segment connecting the midpoints of the diagonals is given by the formula

\frac{b_1 - b_2}{2}

. Here

b_1 = 16 - 0 = 16

and

b_2 = 12 - 4 = 8

, so the length is

\frac{16 - 8}{2} = 4

Tahmini Süre:1m 15s

Soru

Cevap

Adım Adım Çözüm

Anahtar Kavram

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Alternatif Yöntem

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