Carcass wrote:
In rectangle ABCD below, AB=5,AF=7, FD=3. Find the following.
(a) Area of ABCD
(b) Area of triangle AEF
(c) Length of BD
(d) Perimeter of ABCD
Attachment:
#GREexcercise In rectangle ABCD below.jpg
(a) 50 (b) 17.5 (c) 5 \(\sqrt{5}\) (d) 30
Math Review
Question: 10
Page: 260
Difficulty: medium
GIVEN: AB = 5, AF = 7, and FD = 3
Add the information to the diagram
(a) Area of ABCD Area of rectangle = (base)(height) = (10)(5) =
50(b) Area of triangle AEF 
Area of triangle = (base)(height)/2
= (7)(5)/2 =
17.5(c) Length of BD
Since ∆ABD is a RIGHT triangle, we can apply the Pythagorean Theorem.
We get: 5² + 10² = (side BD)²
Simplify: 25 + 100 = (side BD)²
Simplify: 125 = (side BD)²
Solve: side BD = √125 = √25 x √5 =
5√5 (d) Perimeter of ABCDAdd lengths of 4 sides
Perimeter = 5 + 10 + 5 + 10 =
30Cheers,
Brent
_________________
Brent Hanneson – Creator of greenlighttestprep.com
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