Bunuel wrote:
Tough and Tricky questions: Geometry.
As shown in the figure above, two sides of triangle BCD are each 9 feet long. Triangle BCD shares side BD with square ABDE, and angle CBD measures 45°. What is the total area of figure ABCDE in square feet? (Note: Figure not drawn to scale.)
A. 121.5
B. 40.5 + 81√2
C. 202.5
D. 221
E. 243
Kudos for a correct solution.Attachment:
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OFFICIAL SOLUTION:(C) Since triangle BCD is an isosceles triangle, and we know that angle CBD measures 45°, we also know that angle CDB measures 45° (since these are the two angles opposite the sides of equal length).
Since all the angles in a triangle must add up to 180°, we know that angle BCD is a right angle. Thus, the triangle is an isosceles right triangle, and its sides are proportional to each other in the ratio of 1 : 1 : √2.
Since each of the legs of the triangle measures 9 feet, the hypotenuse measures 9√2 feet. The length of the hypotenuse of the triangle is also the length of each side of the square. The base and height of the triangle are the legs, both measuring 9 feet. We now have enough information to calculate the area of the figure.
To determine the area of the square portion, we square the length of one side of the square:
Area of square = (9√2)²
= 9 × √2 × 9 × √2
= 9 × 9 × √2 × √2
= 81 × 2 = 162.
We can determine the area of the triangular portion since the base and height are 9 (the hypotenuse is 9√2) using the formula for the area of triangles:
Area of triangle = (1/2)(b × h)
= (1/2)(9 × 9)
= (1/2)(81) = 40.5.
The area of the entire figure is therefore:
40.5 + 162 = 202.5 square feet.
The correct answer is choice (C).