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
