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# In the figure above, if BD = 6, what is the area of ∆ ABE?

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In the figure above, if BD = 6, what is the area of ∆ ABE? [#permalink]

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08 Aug 2017, 10:37
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In the figure above, if BD = 6, what is the area of ∆ ABE?

(A) 8
(B) 10
(C) 12
(D) 14
(E) 16

[Reveal] Spoiler:
Attachment:

2017-08-08_2136.png [ 15.27 KiB | Viewed 721 times ]
[Reveal] Spoiler: OA

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Re: In the figure above, if BD = 6, what is the area of ∆ ABE? [#permalink]

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08 Aug 2017, 14:05
Bunuel wrote:

In the figure above, if BD = 6, what is the area of ∆ ABE?

(A) 8
(B) 10
(C) 12
(D) 14
(E) 16

[Reveal] Spoiler:
Attachment:
The attachment 2017-08-08_2136.png is no longer available

Attachment:

trapezoidabcd_2136.png [ 35.05 KiB | Viewed 498 times ]

To find area of ∆ABE, use (area of trapezoid) - (area of blue ∆ ADE + area of yellow ∆ BCD)

1. Area of trapezoid

Polygon ABCD is a trapezoid: two right angles mean that side BC is parallel to side AD

BD = height of trapezoid = 6

ABCD area = $$\frac{(8 + 4)}{2}$$ * 6 = 36

2. Area of two colored triangles - start with similar triangles to find height of blue ∆ ADE

Use properties of similar triangles to find height of blue triangle ADE

∆ ADE (blue) and ∆ BCE (light yellow) are similar:

Both have right angles. Their vertical angles are equal. AA = AAA. And the other two angles marked in red are alternate interior angles of parallel lines cut by transversal segment AC.

Corresponding sides of similar triangles will be in the same ratio.

Base AD of blue triangle is 8. Corresponding base BC of light yellow triangle is 4. Ratio is 2:1

Use the ratio to find height, DE, of blue triangle ADE.

Height of light yellow triangle BCE is BE

$$\frac{DE}{BE} = \frac{2x}{1x}$$

BD = 6

BD = sum of the heights of both similar triangles, i.e. BD = DE + BE

6 = 2x + x
x = 2

So BE is 2, and DE (height of blue triangle) is 4. See figure.

3. Find area of the two colored triangles (blue ∆ ADE and two-toned yellow ∆ BCD)

Area of blue ∆ ADE = $$\frac{b*h}{2}$$ = $$\frac{4*8}{2}$$= 16

Area of two-toned yellow triangle BCD = $$\frac{4*6}{2}$$ = 12

Total area of both colored triangles is 16 + 12 = 28

Area of trapezoid = 36

36 - 28 = 8

Area of ∆ ABE = 8

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Re: In the figure above, if BD = 6, what is the area of ∆ ABE? [#permalink]

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08 Aug 2017, 21:08
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Bunuel wrote:

In the figure above, if BD = 6, what is the area of ∆ ABE?

(A) 8
(B) 10
(C) 12
(D) 14
(E) 16

[Reveal] Spoiler:
Attachment:
2017-08-08_2136.png

Area of ∆ BCD = 12
Area of ∆ ABD = 24

From figure ∆ BCE & ∆ AED
BC:AD = 4:8 => BE:ED = 2:4

Area of ∆ AED =16 => Area of ∆ ABE = Area of ∆ ABD - Area of ∆ AED = 24 - 16 =8

A
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In the figure above, if BD = 6, what is the area of ∆ ABE? [#permalink]

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08 Aug 2017, 21:29
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Area of ∆ABD = $$\frac{1}{2}*AD*BD = \frac{1}{2}*6*8 = 24$$

From figure,
Angle B = Angle D
Angle E is common in both the triangles(opposite angles)
Therefore by AA similarity, ∆BCE is similar to ∆AED.

We know that the relative sides are proportional in similar triangle.
Hence,
$$\frac{BC}{AD} = \frac{BE}{ED}$$

$$\frac{4}{8} = \frac{BE}{ED}$$

$$ED = 2BE$$ but we also know that $$BE + ED = 6$$

Therefore, $$BE = 2$$ and $$ED = 4$$

Area of triangle ∆AED : $$\frac{1}{2}*AE*ED = \frac{1}{2}*8*4 = 16$$

Since we need the area of ∆ABE, it must be area of ∆ABD - area of ∆AED
Thus, Area of ∆ABE = 24 - 16 = 8(Option A)
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In the figure above, if BD = 6, what is the area of ∆ ABE?   [#permalink] 08 Aug 2017, 21:29
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