A metallic plate is in the shape of a trapezoid. The parallel sides of the plate have leng… — SAT

A metallic plate is in the shape of a trapezoid. The parallel sides of the plate have lengths of $9$ centimeters and $30$ centimeters. The two non-parallel sides have lengths of $10$ centimeters and $17$ centimeters. What is the area, in square centimeters, of the metallic plate?

Answer: 156 square centimeters

Answer

156

To find the area of the trapezoid, we can find its height by drawing perpendicular lines from the vertices of the shorter base to the longer base. This splits the trapezoid into a rectangle of width

9\text{ cm}

and two right triangles with hypotenuses of

10\text{ cm}

and

17\text{ cm}

. The sum of the bases of these two triangles is

30 - 9 = 21\text{ cm}

. Letting their bases be

x

and

y

(where

x + y = 21

), we apply the Pythagorean theorem:

h^2 + x^2 = 100

and

h^2 + y^2 = 289

. Subtracting the equations yields

y^2 - x^2 = 189

. Since

y^2 - x^2 = (y - x)(y + x)

, we have

21(y - x) = 189

, which simplifies to

y - x = 9

. Solving the system

x + y = 21

and

y - x = 9

yields

x = 6

and

y = 15

. Using the Pythagorean triple

6

8

10

, the height

h

8\text{ cm}

. The area is

\frac{1}{2}(9 + 30)(8) = 156

square centimeters.

Step-by-Step Solution

Drop perpendicular lines (heights) from the endpoints of the shorter base to the longer base.

The trapezoid is decomposed into a central rectangle with a width of

9

centimeters and two right triangles with bases

x

and

y

and hypotenuses of

10

centimeters and

17

centimeters, respectively.

This sets up a system of equations relating the heights and bases of the right triangles.

Determine the relationship between the bases of the two right triangles.

The sum of the bases of the two right triangles is

x + y = 30 - 9 = 21

centimeters.

The sum of the bases of the right triangles plus the width of the rectangle equals the total length of the longer base of the trapezoid.

Set up equations using the Pythagorean theorem for the two right triangles.

h^2 + x^2 = 100

and

h^2 + y^2 = 289

Both right triangles share the same height

h

of the trapezoid.

Solve for the difference between the two triangle bases.

Subtracting the first equation from the second gives

y^2 - x^2 = 189

. Factoring yields

(y - x)(y + x) = 189

. Substituting

y + x = 21

yields

21(y - x) = 189 \implies y - x = 9

Subtracting the equations eliminates the height variable, allowing us to find the difference between the bases.

Solve the system of linear equations for

x

and

y

Adding

x + y = 21

and

y - x = 9

gives

2y = 30 \implies y = 15

. Substituting back gives

x = 6

This determines the exact base segments of both right triangles.

Calculate the height of the trapezoid.

h = \sqrt{10^2 - 6^2} = \sqrt{64} = 8

centimeters.

The height of the trapezoid is required to calculate its area.

Compute the area of the trapezoid.

\text{Area} = \frac{1}{2}(9 + 30)(8) = 156

square centimeters.

This uses the standard formula for the area of a trapezoid.

Key Concept

Decomposing a non-isosceles trapezoid into a rectangle and two right triangles to solve for the height using systems of quadratic equations derived from the Pythagorean theorem.

Estimated Time:1m 30s

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