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

(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).

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With this question, how do you know that angle CBD measures 45 degrees? What if the two sides were double the length they are now? Wouldn't that change the angle?

Thanks.