Question

An isosceles trapezoid has base angles equal to 45 and bases of lengths 6 and 12. Find the area of the trapezoid.

Answer 1

Area of isosceles trapezoid(A) is given by:

`A = (1)/(2) \cdot h \cdot (a+b)`

where

a and b are the unequal side length and

h is the height of the isosceles trapezoid.

Given that:

An isosceles trapezoid has base angles equal to 45 and bases of lengths 6 and 12.

See the diagram as shown below.

In isosceles trapezoid ABCD

AB = 6 units , CD = 12 units

AB = EF = 6 units

In triangle AED:

`\angle ADE = 45^(\circ)`

Since, DE=FC = 3 units

We need to find the value of AE:

Use tangent ratio in triangle AED:

`\tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}}`

Then;

`\tan 45^(\circ) = (AE)/(DE)`

Substitute the given values we have;

`1 = (AE)/(3)`

⇒
`AE = 3` units

In the given isosceles ABCD:

AE = height = 3 units

AB = 6 units and CD = 12 units

then using area formula:

`A = (1)/(2) \cdot AE \cdot (AB+CD)`

Substitute the given values we have;

`A = (1)/(2) \cdot 3 \cdot (6+12) = (3)/(2) \cdot 18 = 3 \cdot 9 = 27` sqaure units.

Therefore, the area of the isosceles trapezoid is, 27 square units

Answer 2

we are given the dimensions of a trapezoid: base lengths of 6 and 12 and a base angle of 45 degrees. In this case, we can identify the height of the trapezoid by: tan 45 = h/ (12-6)/2 ; h is equal to 2 units. The area of the trapezoid is A = (b1+ b2)*(h/2). Hence, A = 27 unit2