Answer :
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:
Area of isosceles trapezoid(A) is given by:
[tex]A = \frac{1}{2} \cdot h \cdot (a+b)[/tex]
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:
[tex]\angle ADE = 45^{\circ}[/tex]
Since, DE=FC = 3 units
We need to find the value of AE:
Use tangent ratio in triangle AED:
[tex]\tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}}[/tex]
Then;
[tex]\tan 45^{\circ} = \frac{AE}{DE}[/tex]
Substitute the given values we have;
[tex]1 = \frac{AE}{3}[/tex]
⇒[tex]AE = 3[/tex] units
In the given isosceles ABCD:
AE = height = 3 units
AB = 6 units and CD = 12 units
then using area formula:
[tex]A = \frac{1}{2} \cdot AE \cdot (AB+CD)[/tex]
Substitute the given values we have;
[tex]A = \frac{1}{2} \cdot 3 \cdot (6+12) = \frac{3}{2} \cdot 18 = 3 \cdot 9 = 27[/tex] sqaure units.
Therefore, the area of the isosceles trapezoid is, 27 square units
