In the standard (x,y) coordinate plane, a triangle has vertices at A(1, 2), B(9, 2), and C… — ACT

In the standard $(x,y)$ coordinate plane, a triangle has vertices at $A(1, 2)$ , $B(9, 2)$ , and $C(5, 8)$ . A horizontal line defined by the equation $y = k$ divides the area of the triangle into two regions of equal area. What is the value of $k$ ?

A
$5$
$8 - 3\sqrt{2}$ Cevap
C
$2 + 3\sqrt{2}$
D
$8 - 2\sqrt{3}$
E
$4$

Cevap

8 - 3\sqrt{2}

The correct answer is

8 - 3\sqrt{2}

. Because the horizontal line

y=k

is parallel to the base of the triangle, it creates a smaller top triangle similar to the original one. The area of the original triangle is

24

and the area of the smaller triangle is

12

, yielding an area ratio of

\frac{1}{2}

. The height of the original triangle is

6

, and the height of the smaller triangle is

8-k

. Since the ratio of the areas of similar triangles is the square of the ratio of their heights, we write

\frac{1}{2} = \left(\frac{8-k}{6}\right)^2

. Solving this equation yields

k = 8 - 3\sqrt{2}

Adım Adım Çözüm

Calculate the area of the entire triangle

ABC

The base

\overline{AB}

is horizontal along

y = 2

with a length of

9 - 1 = 8

. The height is the vertical distance from

y = 2

to the vertex

C(5, 8)

, which is

8 - 2 = 6

. The area of

\triangle ABC

\frac{1}{2} \times 8 \times 6 = 24

Finding the total area is necessary to determine the target area of the divided regions.

Find the area of the smaller triangle formed above the line

y = k

The line

y = k

divides the triangle into a smaller top triangle and a bottom trapezoid. Since both regions have equal areas, the area of the smaller top triangle is

\frac{24}{2} = 12

The line divides the original area of

24

into two equal parts of

12

each.

Set up the area ratio equation using the properties of similar triangles.

The smaller top triangle is similar to

\triangle ABC

because its base is parallel to

\overline{AB}

. The ratio of their areas is the square of the ratio of their heights:

\frac{12}{24} = \left(\frac{8 - k}{6}\right)^2 \implies \frac{1}{2} = \left(\frac{8 - k}{6}\right)^2

For similar figures, the area ratio is equal to the square of the scale factor.

Solve the ratio equation for

k

Taking the square root of both sides gives

\frac{1}{\sqrt{2}} = \frac{8 - k}{6}

, which simplifies to

\frac{\sqrt{2}}{2} = \frac{8 - k}{6}

. Multiplying both sides by

6

gives

3\sqrt{2} = 8 - k

, which yields

k = 8 - 3\sqrt{2}

This isolates

k

to find the exact vertical coordinate of the dividing line.

Anahtar Kavram

Using properties of similar figures to determine areas and coordinates on the coordinate plane.

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