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In the figure above, what is the area of the quadrilateral?

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Joined: 02 Sep 2009
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In the figure above, what is the area of the quadrilateral? [#permalink]

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06 Nov 2017, 22:54
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65% (01:18) correct 35% (01:14) wrong based on 37 sessions

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In the figure above, what is the area of the quadrilateral?

(A) 6 + 2√6
(B) 2√30
(C) 11
(D) 13
(E) cannot be determined

Attachment:

2017-11-07_0941_001.png [ 3.92 KiB | Viewed 596 times ]

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In the figure above, what is the area of the quadrilateral? [#permalink]

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07 Nov 2017, 01:17
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Attachment:

2017-11-07_0941_001.png [ 4.47 KiB | Viewed 477 times ]

Since we have been asked to find the length of the quadrilateral,
we can join the line(as done in our figure), to form the hypotenuse
of the right angled triangle.

$$Hyp^2 = 3^2 + 4^2 = 9 + 16 = 25 => Hyp = \sqrt{25} = 5$$

The area of the right angled triangle is $$\frac{1}{2}*3*4 = 6$$

Since the hypotenuse of the right angled triangle is the length of the third side
of the other triangle, the area of the triangle can be calculated using the formula
Area of the triangle = $$\sqrt{s(s-a)(s-b)(s-c)}$$
where a,b,and c are the three sides and s(semi perimeter) = $$\frac{a+b+c}{2}$$

For this question, the three sides of the triangle are a=5,b=5,and c=2
The semi perimeter(s) is $$\frac{2+5+5}{2} = 6$$

Area = $$\sqrt{6(6-2)(6-5)(6-5)} = \sqrt{6(4)} = 2√6$$

Therefore, the area of the quadrilateral(the sum of the areas of the two triangles) is 6 + 2√6(Option A)
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In the figure above, what is the area of the quadrilateral? [#permalink]

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08 Nov 2017, 18:10
Bunuel wrote:

In the figure above, what is the area of the quadrilateral?

(A) 6 + 2√6
(B) 2√30
(C) 11
(D) 13
(E) cannot be determined

Attachment:

nnnnn.png [ 6.3 KiB | Viewed 369 times ]

Divide the original figure into two triangles. One is a right triangle, the other, isosceles. Sum their areas.

On original figure, draw a line to connect points A and B, which creates a right triangle and another triangle.

1) Length of hypotenuse, $$c$$, of right triangle (AB, the two triangles' shared side)

The right triangle is a $$3-4-5$$ right triangle. Hypotenuse = $$5$$. OR
$$3^2 + 4^2 = c^2$$
$$25 = c^2$$
$$c = 5$$

2) Area of the right triangle

$$\frac{b * h}{2} =\frac{3 * 4}{2} =6$$

3) Area of the other triangle ABC - it is isosceles. One side is the hypotenuse of the other triangle, so side lengths are 5 - 5 - 2

Drop an altitude from top vertex to base.
The altitude of an isosceles triangle divides it into two congruent right triangles. Base of length 2 becomes 1 and 1.

Height? Use one right triangle

$$1^2 + x^2 = 5^2$$
$$x^2 = 24$$
$$x = \sqrt{24} = \sqrt{4 * 6}$$
$$x = 2\sqrt{6}$$

Area of the isosceles triangle ABC:

$$\frac{b*h}{2} = \frac{(2 * 2\sqrt{6})}{2}= 2\sqrt{6}$$

$$6 + 2\sqrt{6}$$

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In the figure above, what is the area of the quadrilateral?   [#permalink] 08 Nov 2017, 18:10
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