Carcass wrote:
If AB = BC, which of the following is an expression for the area of quadrilateral ABDE ?
A) \(\frac{a^2}{2} - \frac{b^2}{2}\)
B) \(\frac{a^2}{2} + \frac{b^2}{2}\)
C) \(a^2 - b^2\)
D)\(\frac{a^2}{4} - \frac{ab}{2}\)
E) \(\frac{a^2}{4} + \frac{ab}{2}\)
Area of quadrilateral ABDE =
(Area of ∆ABC) -
(Area of ∆EDC)So, what are the areas of each triangle?
Area of ∆ABCSince AB = BC, then side BC also has length a (Aside: ∆ABC is called an isosceles right triangle)
Since ∆ABC is a right triangle, we can make one of the legs the base (with length a) which makes the other leg the height (height = a)
Area of triangle = (base)(height)/2
So, the area of ∆ABC = (a)(a)/2 =
a²/2Area of ∆EDCWe must first recognize that ∆ABC and ∆EDC are SIMILAR TRIANGLES
This means that ∆EDC is also an isosceles right triangle
In this case, the base and the height are both equal to b.
So, the area of ∆EDC = (b)(b)/2 =
b²/2So, area of quadrilateral ABDE =
a²/2 -
b²/2= (a² - b²)/2
Answer:
Cheers,
Brent