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# In the figure above, the area of the shaded region is

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In the figure above, the area of the shaded region is [#permalink]
gmatt1476 wrote:

In the figure above, the area of the shaded region is

A. $$8\sqrt{2}$$

B. $$4\sqrt{3}$$

C. $$4\sqrt{2}$$

D. $$8(\sqrt{3} - 1)$$

E. $$8(\sqrt{2} - 1)$$

PS57302.01

Attachment:
The attachment 2019-09-21_1806.png is no longer available

Join BC as shown, BC=$$4\sqrt{2}$$, as it is diagonal of square of side 4
Area of shaded region=Area of equilateral triangle ABC - Area of isosceles right angled triangle with equal sides as4 -BC = $$(4\sqrt{2})^2*\sqrt{3}/4-\frac{1}{2}*4*4=8\sqrt{3}-8$$=$$8(\sqrt{3} - 1)$$

D
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2019-09-21_1806.png [ 29.74 KiB | Viewed 17437 times ]

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Re: In the figure above, the area of the shaded region is [#permalink]
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gmatt1476 wrote:

In the figure above, the area of the shaded region is

A. $$8\sqrt{2}$$

B. $$4\sqrt{3}$$

C. $$4\sqrt{2}$$

D. $$8(\sqrt{3} - 1)$$

E. $$8(\sqrt{2} - 1)$$

PS57302.01

Attachment:
The attachment 2019-09-21_1806.png is no longer available

Beautiful question!

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Re: In the figure above, the area of the shaded region is [#permalink]
Solution:

Please refer to the image here:

Attachment:

triangle.png [ 51.65 KiB | Viewed 9626 times ]

Area of shaded region = Area of equilateral triangle $$ABC$$ - Area of right isoceles triangle $$BCD$$

$$⇒ \frac{\sqrt3 }{4} \times (4\sqrt2)^2 - \frac{1}{2}\times 4\times 4$$

$$⇒ 8\sqrt3 - 8$$

$$⇒ 8(\sqrt3 -1)$$

Hence the right answer is Option D.
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Re: In the figure above, the area of the shaded region is [#permalink]
Could someone please explain to me how to calculate this step by step? $$⇒ 8\sqrt3 - 8$$

$$⇒ 8(\sqrt3 -1)$$

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gmatt1476

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In the figure above, the area of the shaded region is [#permalink]
Create an equilateral triangle by drawing a diagonal through the square. The diagonal will be 4$$\sqrt{2}$$ because the sides of the square are 4.
The area of an equilateral triangle is $$3^{2}$$$$\sqrt{3}$$/4
so that will be 8$$\sqrt{3}$$.
8$$\sqrt{3}$$ - 8