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

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

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New post 05 Jan 2015, 06:50
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Tough and Tricky questions: Geometry.



Image
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

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Attachment:
2015-01-05_1750.png
2015-01-05_1750.png [ 2.55 KiB | Viewed 3630 times ]

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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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New post 05 Jan 2015, 08:57
The length of the hypothenuse of the triangle is the length of one of the shorter sides multiplied by \(\sqrt{2}\), so \(9\sqrt{2}\). Therefore the area of the square is \(9\sqrt{2}*9\sqrt{2}=81*2=162\).

The area of the triangle must be exactly one quarter of the area of the square. If you don't see that directly, imagine the triangle being flipped down into the square. Therefore, the total area is 162 plus one quarter. Answer 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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New post 05 Jan 2015, 17:09
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Bunuel wrote:

Tough and Tricky questions: Geometry.



Image
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


BD = \sqrt{9^2 + 9^2}
= 9\sqrt{2}

Area of the square = (9\sqrt{2})^2 = 162
Area of the traingle = 1/2 * 9 * 9 = 40.5

Area of the figure = 162+40.5 = 202.5

Ans 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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New post 06 Jan 2015, 00:50
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Answer = C. 202.5

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]

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New post 06 Jan 2015, 22:14
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]

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New post 08 Jan 2015, 00:16
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]

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New post 08 Jan 2015, 14:27
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
=1/2*9*9+(9 *sqrt2)^2=40.5+162=202.5 (Answer is 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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New post 08 Jan 2015, 15:27
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

(9rad2)^2 + (9^2)/2

162 + 40.5

202.5

The answer is 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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New post 09 Jan 2015, 06:36
Bunuel wrote:

Tough and Tricky questions: Geometry.



Image
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] 17 Dec 2018, 23:31
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