One base of a trapezoid is twice as long as the other. The height is the average of the two bases. The are is 324 cm2. Find the height and lengths of the bases.

Question

Let the lengths of the bases be 2x and 4x

Answer 1

bases

b₁ = 2x, b₂ = 4x

height

h = ½(b₁ + b₂) = ½(2x + 4x) = ½(6x) = 3x

area

To find the area of a trapezoid we multiply the average of the two base lengths ½(b₁ + b₂) by the height h.

A = ½(b₁ + b₂) × h

½(b₁ + b₂) = 3x as calculated above

and we're given that

h = ½(b₁ + b₂) = 3x

Our formula for area simplifies to

A = ½(b₁ + b₂) × ½(b₁ + b₂) = 3x × 3x = 9x²

solving for x

Since we're given that A = 324 cm²

then we can write 9x² = 324 and solve for x.

Diving both sides by 9 gives

x² = 324/9 = 36 so

x = 6

answers

b₁ = 2x = 12 cm

b₂ = 4x = 24 cm

h = ½(b₁ + b₂) = 3x = 18 cm

Answer 2

Hi Audrey,

If the bases are 2x and 4x and the height of the trapezoid is the average of the bases, then the height is (2x+4x)/2 which is 3x.

The formula for the area of a trapezoid is A = h/2 (B + b) where h is the height and B and b are the bases.

Since the area is 324 cm^2, we have

324 = (3x/2) (4x + 2x)

324 = (3x/2) (6x)

324 = 18x^2 / 2

Multiply by 2

648 = 18x^2

Divide by 18

36 = x^2

6 = x

Now, the side length are 2x and 4x, which means they are 12 cm and 24 cm.

The height is 3x, which is 18 cm.

Hope this helps,

Jim