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22 Jul 2014, 03:50
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File comment: Triangle

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Both triangles ABC and ADE are equilateral. The shaded area enclosed in BCED is equal to the area of ΔADE. If AE=5, what is the length of BD?

A. 5(√2 + 1)
B. 2√5
C. 3
D. 5(2 - √2)
E. 5 / (√2 + 1)
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22 Jul 2014, 04:24
1

Both triangles ABC and ADE are equilateral. The shaded area enclosed in BCED is equal to the area of ΔADE. If AE=5, what is the length of BD?

A. 5(√2 + 1)
B. 2√5
C. 3
D. 5(2 - √2)
E. 5 / (√2 + 1)

The area of equilateral triangle is $$side^2*\frac{\sqrt{3}}{4}$$.

Since the area of the shaded region and ΔADE are equal, then the area of the big triangle is twice the area of ΔADE.

Say the side of the big triangle is x. Then $$x^2*\frac{\sqrt{3}}{4}=2*(5^2*\frac{\sqrt{3}}{4})$$ --> $$x=5\sqrt{2}$$.

$$BD = x - 5 = 5\sqrt{2} - 5 = 5(\sqrt{2}-1)$$ --> multiply by $$\frac{\sqrt{2}+1}{\sqrt{2}+1}$$ --> $$BD = \frac{5}{\sqrt{2}+1}$$.

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22 Jul 2014, 04:25
1
Bunuel wrote:

Both triangles ABC and ADE are equilateral. The shaded area enclosed in BCED is equal to the area of ΔADE. If AE=5, what is the length of BD?

A. 5(√2 + 1)
B. 2√5
C. 3
D. 5(2 - √2)
E. 5 / (√2 + 1)

The area of equilateral triangle is $$side^2*\frac{\sqrt{3}}{4}$$.

Since the area of the shaded region and ΔADE are equal, then the area of the big triangle is twice the area of ΔADE.

Say the side of the big triangle is x. Then $$x^2*\frac{\sqrt{3}}{4}=2*(5^2*\frac{\sqrt{3}}{4})$$ --> $$x=5\sqrt{2}$$.

$$BD = x - 5 = 5\sqrt{2} - 5 = 5(\sqrt{2}-1)$$ --> multiply by $$\frac{\sqrt{2}+1}{\sqrt{2}+1}$$ --> $$BD = \frac{5}{\sqrt{2}+1}$$.

Check other Shaded Region Problems in our Special Questions Directory.
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06 Jun 2015, 03:08
Attachment:

T8892.png [ 5.59 KiB | Viewed 4267 times ]

Both triangles ABC and ADE are equilateral. The shaded area enclosed in BCED is equal to the area of ΔADE. If AE=5, what is the length of BD?

A. $$5*(\sqrt{2}-1)$$
B. $$2*\sqrt{5}$$
C. 3
D. $$5*(2-\sqrt{2})$$
E. $$\frac{5}{\sqrt{2}+1}$$

Do you mind solving withouth doing the maths?
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06 Jun 2015, 03:48
1
reto wrote:
Attachment:
T8892.png

Both triangles ABC and ADE are equilateral. The shaded area enclosed in BCED is equal to the area of ΔADE. If AE=5, what is the length of BD?

A. $$5*(\sqrt{2}-1)$$
B. $$2*\sqrt{5}$$
C. 3
D. $$5*(2-\sqrt{2})$$
E. $$\frac{5}{\sqrt{2}+1}$$

Do you mind solving withouth doing the maths?

Ratio of Area of two similar triangles = Square of ratio of their correcponding sides

i.e $$Area of ADE / Area of ABC$$ = $$(AD/AB)^2$$

$$x / 2x$$ = $$[5/(5+BD)]^2$$

1/\sqrt{2} = 5/(5+BD)

i.e. (5+BD) = $$5\sqrt{2}$$

i.e. BD = $$5(\sqrt{2}-1)$$

Multiplying $$(\sqrt{2}+1)$$ in Numerator and denominator gives us

BD = 5 / $$(\sqrt{2}+1)$$

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06 Jun 2015, 04:14
Concept: Ratio of areas of similar triangles is equal to the ratio of squares of the corresponding similar sides.

(x/2x) =(5/(5+y))^2

-> y= 5[sqrt(2)-1]

On Factorizing,

BD = 5/[sqrt(2)+1]

Option E
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06 Jun 2015, 06:50
The shortest way for those who dont know the formula [
Ratio of Area of two similar triangles = Square of ratio of their correcponding sides]

But I doubt this question....as Option A and E are same.

Two step procedure..

Area of triangle= Area of quad

Area of Traiangle ADE= \sqrt{3} *25/4.

Area of Quad= 25* \sqrt{3} /4

Implies Area of bigger equilateral triangle = 50 \sqrt{3}/4

Apply Area of Equilateral again= you will get, AC=AB= 5 \sqrt{2}.

so DB= 5\sqrt{2}-5 => You will get Option A, which is nothing but Option E.

Please let me know if I am wrong.

reto wrote:
Attachment:
T8892.png

Both triangles ABC and ADE are equilateral. The shaded area enclosed in BCED is equal to the area of ΔADE. If AE=5, what is the length of BD?

A. $$5*(\sqrt{2}-1)$$
B. $$2*\sqrt{5}$$
C. 3
D. $$5*(2-\sqrt{2})$$
E. $$\frac{5}{\sqrt{2}+1}$$

Do you mind solving withouth doing the maths?
CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2977
Location: India
GMAT: INSIGHT
Schools: Darden '21
WE: Education (Education)

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06 Jun 2015, 07:16
shriramvelamuri wrote:
The shortest way for those who dont know the formula [
Ratio of Area of two similar triangles = Square of ratio of their correcponding sides]

But I doubt this question....as Option A and E are same.

Two step procedure..

Area of triangle= Area of quad

Area of Traiangle ADE= \sqrt{3} *25/4.

Area of Quad= 25* \sqrt{3} /4

Implies Area of bigger equilateral triangle = 50 \sqrt{3}/4

Apply Area of Equilateral again= you will get, AC=AB= 5 \sqrt{2}.

so DB= 5\sqrt{2}-5 => You will get Option A, which is nothing but Option E.

Please let me know if I am wrong.

reto wrote:
Attachment:
T8892.png

Both triangles ABC and ADE are equilateral. The shaded area enclosed in BCED is equal to the area of ΔADE. If AE=5, what is the length of BD?

A. $$5*(\sqrt{2}-1)$$
B. $$2*\sqrt{5}$$
C. 3
D. $$5*(2-\sqrt{2})$$
E. $$\frac{5}{\sqrt{2}+1}$$

Do you mind solving withouth doing the maths?

You are absolutely CORRECT.

Option A and E are same so both should be correct and your working is absolutely Right.
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06 Jun 2015, 07:49
GMATinsight wrote:
shriramvelamuri wrote:
The shortest way for those who dont know the formula [
Ratio of Area of two similar triangles = Square of ratio of their correcponding sides]

But I doubt this question....as Option A and E are same.

Two step procedure..

Area of triangle= Area of quad

Area of Traiangle ADE= \sqrt{3} *25/4.

Area of Quad= 25* \sqrt{3} /4

Implies Area of bigger equilateral triangle = 50 \sqrt{3}/4

Apply Area of Equilateral again= you will get, AC=AB= 5 \sqrt{2}.

so DB= 5\sqrt{2}-5 => You will get Option A, which is nothing but Option E.

Please let me know if I am wrong.

reto wrote:
Attachment:
T8892.png

Both triangles ABC and ADE are equilateral. The shaded area enclosed in BCED is equal to the area of ΔADE. If AE=5, what is the length of BD?

A. $$5*(\sqrt{2}-1)$$
B. $$2*\sqrt{5}$$
C. 3
D. $$5*(2-\sqrt{2})$$
E. $$\frac{5}{\sqrt{2}+1}$$

Do you mind solving withouth doing the maths?

You are absolutely CORRECT.

Option A and E are same so both should be correct and your working is absolutely Right.

Please tell me more why A and E should be the same Options... because they're not the same value.
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06 Jun 2015, 07:57
GMATinsight wrote:
shriramvelamuri wrote:
The shortest way for those who dont know the formula [
Ratio of Area of two similar triangles = Square of ratio of their correcponding sides]

A. $$5*(\sqrt{2}-1)$$
B. $$2*\sqrt{5}$$
C. 3
D. $$5*(2-\sqrt{2})$$
E. $$\frac{5}{\sqrt{2}+1}$$

You are absolutely CORRECT.

Option A and E are same so both should be correct and your working is absolutely Right.

Please tell me more why A and E should be the same Options... because they're not the same value.[/quote]

Option A : A. $$5*(\sqrt{2}-1)$$

Multiple $$\sqrt{2}+1$$ in both numerator and denominator of Option A, we get,

i.e. Option A becomes $$5*(\sqrt{2}-1)*(\sqrt{2}+1)/(\sqrt{2}+1)$$

Please note the property $$(a-b)*(a+b) = a^2 - b^2$$

i.e. Option A becomes $$5*(\sqrt{2})^2-1^2)/(\sqrt{2}+1)$$

i.e. Option A becomes $$\frac{5*(2-1)}{(\sqrt{2}+1)}$$

i.e. Option A becomes $$\frac{5}{\sqrt{2}+1}$$ which is same as Option E

I hope it clears your doubt!!!
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06 Jun 2015, 08:06
GMATinsight wrote:
GMATinsight wrote:
shriramvelamuri wrote:
The shortest way for those who dont know the formula [
Ratio of Area of two similar triangles = Square of ratio of their correcponding sides]

A. $$5*(\sqrt{2}-1)$$
B. $$2*\sqrt{5}$$
C. 3
D. $$5*(2-\sqrt{2})$$
E. $$\frac{5}{\sqrt{2}+1}$$

You are absolutely CORRECT.

Option A and E are same so both should be correct and your working is absolutely Right.

Please tell me more why A and E should be the same Options... because they're not the same value.

Option A : A. $$5*(\sqrt{2}-1)$$

Multiple $$\sqrt{2}+1$$ in both numerator and denominator of Option A, we get,

i.e. Option A becomes $$5*(\sqrt{2}-1)*(\sqrt{2}+1)/(\sqrt{2}+1)$$

Please note the property $$(a-b)*(a+b) = a^2 - b^2$$

i.e. Option A becomes $$5*(\sqrt{2})^2-1^2)/(\sqrt{2}+1)$$

i.e. Option A becomes $$\frac{5*(2-1)}{(\sqrt{2}+1)}$$

i.e. Option A becomes $$\frac{5}{\sqrt{2}+1}$$ which is same as Option E

I hope it clears your doubt!!![/quote]

Wow, obviously I hope there's no typo in the answer choices. But I should have copied it 1:1 from Economist. Maybe my mistake. Try to check that.
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13 Apr 2016, 18:30
goodyear2013 wrote:
Attachment:
Triangle.png

Both triangles ABC and ADE are equilateral. The shaded area enclosed in BCED is equal to the area of ΔADE. If AE=5, what is the length of BD?

A. 5(√2 + 1)
B. 2√5
C. 3
D. 5(2 - √2)
E. 5 / (√2 + 1)

i really don't get why we need to multiply everything by sqrt(2)+1...
but..that's the only one that makes sense..

area of equilateral triangle is s^2* sqrt(3) / 4
we know that AE=5, thus, area must br 25*sqrt(3)/4
area big triangle is:
25*sqrt(3)/2 or 12.5*sqrt(3).
now..this is equal to side big triangle ^2 * sqrt(3) / 4
cross multiply, and cancel sqrt(3) on both sides:
50=s^2
s=5*sqrt(2)
BD is then 5*sqrt(2) -5 or 5(sqrt2 -1)
this is a perfect answer..why do we need to complicate ourselves???

multiply by sqrt(2)+1 / sqrt(2)+1 and get 5*(2-1)/sqrt(2)+1

E
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28 Apr 2016, 21:08
goodyear2013 wrote:
Attachment:
Triangle.png

Both triangles ABC and ADE are equilateral. The shaded area enclosed in BCED is equal to the area of ΔADE. If AE=5, what is the length of BD?

A. 5(√2 + 1)
B. 2√5
C. 3
D. 5(2 - √2)
E. 5 / (√2 + 1)

Hi debbiem,
SS1983 refer our discussion on similar triangles
an Easier and quick method is to realize that equilateral triangle gives us an important info - both are similar triangles..
ratio of areas of two triangles = ratio of $$side^2$$ of two triangles..

$$\frac{area of ΔADE}{area of ΔABC}=\frac{area of ΔADE}{area of ΔADE + area of BCED}=\frac{X}{2X}=\frac{1}{2} = \frac{side of ΔADE ^2}{side of ΔABC ^2}$$..
$$\sqrt{\frac{1}{2}} = \frac{side of ΔADE}{side of ΔABC}=\frac{AE}{AE+EC}= \frac{5}{5+EC}$$..

or$$\frac{5}{5+EC} = \sqrt{\frac{1}{2}}$$..

$$5+EC = 5\sqrt{2}$$..

$$EC = 5(\sqrt{2}-1) = 5(\sqrt{2}-1)*{\sqrt{2}+1/\sqrt{2}+1}$$= $$\frac{5}{\sqrt{2}+1}$$..

and EC is nothing but BD
E
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25 Aug 2017, 10:42
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