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

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 

Answer 2

Answer:

Area of isosceles trapezoid(A) is given by:

$A = \frac{1}{2} \cdot h \cdot (a+b)$

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;

$\tan 45^{\circ} = \frac{AE}{DE}$

Substitute the given values we have;

$1 = \frac{AE}{3}$

⇒$AE = 3$ units

In the given isosceles ABCD:

AE = height = 3 units

AB = 6 units and CD = 12 units

then using area formula:

$A = \frac{1}{2} \cdot AE \cdot (AB+CD)$

Substitute the given values we have;

$A = \frac{1}{2} \cdot 3 \cdot (6+12) = \frac{3}{2} \cdot 18 = 3 \cdot 9 = 27$ sqaure units.

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