Problem:
The figure below is constructed from 11 line segments, each of which has length 2 . The area of pentagon ABCDE can be written as m+n, where m and n are positive integers. What is m+n ?
Answer Choices:
A. 20
B. 21
C. 22
D. 23
E. 24 Solution:
Let M be the midpoint of CD. Noting that AED and ABC are 120−30−30 triangles because of the equilateral triangles,
AM=AD2−MD2=12−1=11⟹[ACD]=11
Also, [AED]=2⋅2⋅21⋅sin120∘=3 (or split △AED into two 30−60−90 triangles) and so
[ABCDE]=[ACD]+2[AED]=11+23=11+12⟹(D) 23
OR
Draw diagonals AC and AD to split the pentagon into three parts.
We can compute the area for each triangle and sum them up at the end. For triangles ABC and ADE, they each have area 2⋅21⋅443=3. For triangle ACD, we can see that AC=AD=23 and CD=2. Using Pythagorean Theorem, the altitude for this triangle is 11, so the area is 11. Adding each part up, we get 23+11=12+11⟹(D) 23
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