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# As shown in the figure above, two sides of triangle BCD are each 9 fee

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Re: As shown in the figure above, two sides of triangle BCD are each 9 fee [#permalink]
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Side of the square $$= \sqrt{2*9^2} = 9\sqrt{2}$$

Area of square = 81*2 = 162

Area of right triangle $$= \frac{1}{2} * 81 = 40.5$$

Total area = 162+40.5 = 202.5
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Re: As shown in the figure above, two sides of triangle BCD are each 9 fee [#permalink]
AngleCBD=45 and triangle CBD is isosceles hense Angle CDB=45 as well. Hence Triangle CBD is right angled triangle with BD as hypotenuse.

BD = 9*sqrt2 (From Pythagores Theoram)

Area of Square ABDE = BD^2 = 162
Area of Triangle CBD = 1/2 * CB * CD = 40.5
Area of figure ABCDE = 162 + 40.5 = 202.5

Option C is correct.
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Re: As shown in the figure above, two sides of triangle BCD are each 9 fee [#permalink]
Triangle is an isosceles right angle triangle.

Side of the square = 9(sqrt2) -(by Pythogorus Theorem)
Area of Square = side * side = 81*2 = 162
Area of triangle = 1/2( 9*9) Sin90 = 40.5

Total Area = 162 + 40.5 = 202.5

Option C
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Re: As shown in the figure above, two sides of triangle BCD are each 9 fee [#permalink]
BC=CD=9ft
CBD =45 deg (given)
therefore CDB=45 deg(Isosceles triangle)
that means BCD=90 deg
so triangle BCD is 90:45:45 traingle
=> sides will be in the rattio 1:1:\sqrt{2}
so side BD=9 *sqrt 2
Now the total area of ABCDE=Area of DCD +Area of ABDE
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Re: As shown in the figure above, two sides of triangle BCD are each 9 fee [#permalink]
We can reason that the triangle is 45-45-90 (isosceles right) with a hypotenuse of 9*rad2. The hypotenuse is also the side of the square.

Square + Triangle

162 + 40.5

202.5

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Re: As shown in the figure above, two sides of triangle BCD are each 9 fee [#permalink]
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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:
2015-01-05_1750.png

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).
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Re: As shown in the figure above, two sides of triangle BCD are each 9 fee [#permalink]
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Re: As shown in the figure above, two sides of triangle BCD are each 9 fee [#permalink]
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