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# If triangle{AHE} is a right angle triangle and AH || BG || CF

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Math Expert
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If triangle{AHE} is a right angle triangle and AH || BG || CF  [#permalink]

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Updated on: 24 Jul 2018, 23:22
1
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Difficulty:

75% (hard)

Question Stats:

55% (03:07) correct 45% (02:36) wrong based on 106 sessions

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If $$\triangle{AHE}$$ is a right angle triangle and AH || BG || CF and AB=BC=CE=√3, what is the ratio of area of coloured portion to the area of uncoloured portion]?

(A) $$\frac{1}{5}$$

(B) $$\frac{1}{3}$$

(C) $$\frac{1}{2}$$

(D) $$\frac{2}{3}$$

(E) 1

New question

Attachment:

TRIANGLE.png [ 3.23 KiB | Viewed 1917 times ]

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Originally posted by chetan2u on 24 Jul 2018, 23:00.
Last edited by chetan2u on 24 Jul 2018, 23:22, edited 2 times in total.
Edited the question.
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Re: If triangle{AHE} is a right angle triangle and AH || BG || CF  [#permalink]

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25 Jul 2018, 00:18
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chetan2u wrote:

If $$\triangle{AHE}$$ is a right angle triangle and AH || BG || CF and AB=BC=CE=√3, what is the ratio of area of coloured portion to the area of uncoloured portion]?

(A) $$\frac{1}{5}$$

(B) $$\frac{1}{3}$$

(C) $$\frac{1}{2}$$

(D) $$\frac{2}{3}$$

(E) 1

New question

Attachment:
TRIANGLE.png

OA: C
$$\triangle{HAE}$$,$$\triangle{GBE}$$ and $$\triangle{FCE}$$ are similar as $$AH || BG || CF$$ and $$\angle E$$ is common in three $$\triangle$$.
AB=BC=CE=√3 , BE = CE+BC = 2√3 , AE = AB+BC+CE = 3√3.
$$\frac{{Area of \triangle {FCE}}}{{Area of \triangle {HAE}}}=\frac{{CE^2}}{{AE^2}}= \frac{{(\sqrt[]{3})^2}}{{(3\sqrt[]{3})^2}}=\frac{1}{9}$$
$$\frac{{Area of \triangle {GBE}}}{{Area of \triangle {HAE}}}=\frac{{BE^2}}{{AE^2}}= \frac{{(2\sqrt[]{3})^2}}{{(3\sqrt[]{3})^2}}=\frac{4}{9}$$
Area of colored portion $$GBCF$$ = Area of $$\triangle {GBE}$$ -Area of $$\triangle {FCE}$$ = $$(\frac{4}{9}-\frac{1}{9})$$Area of $$\triangle {HAE}$$=$$\frac{1}{3}$$Area of $$\triangle {HAE}$$
Area of Uncolored Portion = Area of $$\triangle {HAE}$$ - Area of colored Portion = $$\frac{2}{3}$$Area of $$\triangle {HAE}$$
$$\frac{{Area of Colored portion}}{{Area of Uncolored Portion}}=\frac{\frac{1}{3}Area of \triangle {HAE}}{\frac{2}{3}Area of \triangle {HAE}} =\frac{1}{2}$$
##### General Discussion
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Re: If triangle{AHE} is a right angle triangle and AH || BG || CF  [#permalink]

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24 Jul 2018, 23:04
kudos to best solutions..
reserved for OE
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Re: If triangle{AHE} is a right angle triangle and AH || BG || CF  [#permalink]

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24 Jul 2018, 23:12
1
chetan2u wrote:

If $$\triangle{AHE}$$ is a right angle triangle and AH || BG || CF and AH=BG=CF=√3, what is the ratio of area of coloured portion to the area of uncoloured portion]?

(A) $$\frac{1}{5}$$

(B) $$\frac{1}{3}$$

(C) $$\frac{1}{2}$$

(D) $$\frac{2}{3}$$

(E) 1

New question

Attachment:
TRIANGLE.png

The problem statement and the diagram are little inconsistent. Do you mean to say AB = BC = CE = $$\sqrt{3}$$ ?
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Posts: 8309
Re: If triangle{AHE} is a right angle triangle and AH || BG || CF  [#permalink]

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24 Jul 2018, 23:23
workout wrote:
chetan2u wrote:

If $$\triangle{AHE}$$ is a right angle triangle and AH || BG || CF and AH=BG=CF=√3, what is the ratio of area of coloured portion to the area of uncoloured portion]?

(A) $$\frac{1}{5}$$

(B) $$\frac{1}{3}$$

(C) $$\frac{1}{2}$$

(D) $$\frac{2}{3}$$

(E) 1

New question

Attachment:
TRIANGLE.png

The problem statement and the diagram are little inconsistent. Do you mean to say AB = BC = CE = $$\sqrt{3}$$ ?

sorry for the typo
edited
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If triangle{AHE} is a right angle triangle and AH || BG || CF  [#permalink]

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24 Jul 2018, 23:46
Area of trapezium BGCF = $$(BG + CF) * \frac{\sqrt{{3}}}{2}$$

Area of trapezium AHBG = $$(AH + BG) * \frac{\sqrt{{3}}}{2}$$

Area of triangle CFE = $$CF * \frac{\sqrt{{3}}}{2}$$

Area of trapezium AHBG + Area of triangle CFE = $$(AH + BG) * \frac{\sqrt{{3}}}{2}$$ + $$CF * \frac{\sqrt{{3}}}{2}$$

Ratio of area of colored portion to non-colored portion = $$\frac{(BG + CF) * \frac{\sqrt{{3}}}{2}}{{(AH + BG) * \frac{\sqrt{{3}}}{2} + CF * \frac{\sqrt{{3}}}{2}$$

Ratio of area of colored portion to non-colored portion = $$\frac{BG + CF}{BG + CF + AH}$$

Triangles AHE and CFE are similar since $$\angle HAE = \angle FCE = 90$$ and $$\angle AEH = \angle CEF$$

=> The sides of these triangles are in proportion => $$\frac{AH}{CF} = \frac{AE}{CE} = 3$$

=> AH = 3 CF

Triangles BGE and CFE are similar since $$\angle GBE = \angle FCE = 90$$ and $$\angle BEG = \angle CEF$$

=> The sides of these triangles are in proportion => $$\frac{BG}{CF} = \frac{BE}{CE} = 2$$

=> BG = 2 CF

Substituting AH and BG values in the ratio, we get

Ratio of area of colored portion to non-colored portion = $$\frac{3CF}{6CF}$$ = $$\frac{1}{2}$$

Hence option C
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If triangle{AHE} is a right angle triangle and AH || BG || CF  [#permalink]

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25 Jul 2018, 00:23

If $$\triangle{AHE}$$ is a right angle triangle and AH || BG || CF and AB=BC=CE=√3, what is the ratio of area of coloured portion to the area of uncoloured portion]?

(A) $$\frac{1}{5}$$

(B) $$\frac{1}{3}$$

(C) $$\frac{1}{2}$$

(D) $$\frac{2}{3}$$

(E) 1

Since it is right angles triangle and sides AH || BG || CF,the two triangle are similar(△AHE ,△CFE ).
Ratio of Area of similar triangles=Area formula for 2 similar triangles =>
ΔAHE/ΔCFE=AE²/CE²=1/3

Some points to remember about trapezoid properties:
• Median - The median of a trapezoid is a line joining the midpoints of the two legs.
• The median line is always parallel to the bases.
• The median line is halfway between the bases.
• The median divides the trapezoid into two smaller trapezoids each with half the altitude of the original.
Since the median of a trapezoid is a line joining the midpoints of the two legs and B is the midpoint of the leg AC ,so BG is median,Also AH || BG || CF.
And since median divides the trapezoid into two smaller trapezoids each with half the altitude of the original.
So ratio of trapezoids=equal
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If triangle{AHE} is a right angle triangle and AH || BG || CF  [#permalink]

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25 Jul 2018, 09:14
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Can be solved with ratio property. Area of triangle "ABC"/Area of triangle "DEF" = (Side of ABC)^2/(Side of DEF)^2
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Joined: 23 Jul 2018
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Re: If triangle{AHE} is a right angle triangle and AH || BG || CF  [#permalink]

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29 Jul 2018, 13:00
1
For similar triangles: if corresponding sides increase by a scale factor of k, the area increases by a scale factor of k^2.

ΔHAE, ΔGBE, and ΔFCE are all similar triangles because AH || BG || CF.

Say area of ΔFCE is x.

Scale factor of BE/CE is 2, so area of these corresponding triangles must increase by a factor of 4. If area(ΔFCE)=x, than area(ΔGBE)=4x. Therefore, the area of quadrilateral GBCF is 3x.

Scale factor of AE/CE is 3, so area must increase by a factor of 9. If area(ΔFCE)=x, than area(ΔHAE)=9x. Therefore, the area of quadrilateral HACF is 8x. If quadrilateral GBCF is 3x, then the area of quadrilateral HABG is 5x.

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Re: If triangle{AHE} is a right angle triangle and AH || BG || CF  [#permalink]

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02 Aug 2019, 11:11
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Re: If triangle{AHE} is a right angle triangle and AH || BG || CF   [#permalink] 02 Aug 2019, 11:11
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