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Bunuel
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siddharthfrancis
Please explain the solution. Thanks. !!

Hi siddharthfrancis,
Detailed explanation has been provided in the above post. You may raise specific queries(if any).
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siddharthfrancis
Please explain the solution. Thanks. !!

Hi siddharthfrancis,
Detailed explanation has been provided in the above post. You may raise specific queries(if any).

trianlge 30-60-90 is fine, how did we get BC=AC/2?

what rule?
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siddharthfrancis
Please explain the solution. Thanks. !!

Hi siddharthfrancis,
Detailed explanation has been provided in the above post. You may raise specific queries(if any).

trianlge 30-60-90 is fine, how did we get BC=AC/2?

what rule?

Hi siddharthfrancis abhishekdadarwal2009,
In the triangle ABC , Angle A=30, Angle C=60, Angle B=90
So the corresponding sides are in the ratio:
BC:AB:AC=x:√3x:2x
Or, BC=x and AC=2x
Or, AC is 2 times BC
Or, BC=AC/2

You may go thru the below link for a detailed insight on SPECIAL TRIANGLES.
https://gmatclub.com/forum/math-triangles-87197.html

Hope your query is nullified.
Thank you.
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Hi siddharthfrancis abhishekdadarwal2009,
In the triangle ABC , Angle A=30, Angle C=60, Angle B=90
So the corresponding sides are in the ratio:
BC:AB:AC=x:√3x:2x
Or, BC=x and AC=2x
Or, AC is 2 times BC
Or, BC=AC/2

You may go thru the below link for a detailed insight on SPECIAL TRIANGLES.
https://gmatclub.com/forum/math-triangles-87197.html

Hope your query is nullified.
Thank you.[/quote]




Now i understand that we use the 30-60-90 rule on the bigger triangle to find one comman side, and then wth that side we use 30-60-90 on the smaller trianlge to find the lenghts of the sides of the triangle and thus finding the area of the smaller triangle.

thanks.
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abhishekdadarwal2009
Hi siddharthfrancis abhishekdadarwal2009,
In the triangle ABC , Angle A=30, Angle C=60, Angle B=90
So the corresponding sides are in the ratio:
BC:AB:AC=x:√3x:2x
Or, BC=x and AC=2x
Or, AC is 2 times BC
Or, BC=AC/2

You may go thru the below link for a detailed insight on SPECIAL TRIANGLES.
https://gmatclub.com/forum/math-triangles-87197.html

Hope your query is nullified.
Thank you.




Now i understand that we use the 30-60-90 rule on the bigger triangle to find one comman side, and then wth that side we use 30-60-90 on the smaller trianlge to find the lenghts of the sides of the triangle and thus finding the area of the smaller triangle.

thanks.[/quote]

Your reasoning bears zero fallacies.
You are welcome.
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Hi Bunuel

Could you please explain where I made a mistake in my solution? I got B as my answer. My strategy was to find the area of ABC and BXY. Then subtracting these two to find the area of BOC.

Area of ABC= 1/2*(4√3)*4 = 8√3
Area of BXY = 1/2*(5√3)*5= 12.5√3

Area of the shaded region: 12.5√3 - 8√3= 4.5√3

Please help!
Thank You!
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csaluja
Hi Bunuel

Could you please explain where I made a mistake in my solution? I got B as my answer. My strategy was to find the area of ABC and BXY. Then subtracting these two to find the area of BOC.
Area of ABC= 1/2*(4√3)*4 = 8√3
Area of BXY = 1/2*(5√3)*5= 12.5√3

Area of the shaded region: 12.5√3 - 8√3= 4.5√3

Please help!
Thank You!

Hi csaluja ,
I have highlighted the mistake in reasoning in red color.
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csaluja
Hi Bunuel

Could you please explain where I made a mistake in my solution? I got B as my answer. My strategy was to find the area of ABC and BXY. Then subtracting these two to find the area of BOC.
Area of ABC= 1/2*(4√3)*4 = 8√3
Area of BXY = 1/2*(5√3)*5= 12.5√3

Area of the shaded region: 12.5√3 - 8√3= 4.5√3

Please help!
Thank You!

Hi csaluja ,
I have highlighted the mistake in reasoning in red color.

Hi PKN,

Could you please explain why the highlighted part is incorrect? Would greatly appreciate it!
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csaluja
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csaluja
Hi Bunuel

Could you please explain where I made a mistake in my solution? I got B as my answer. My strategy was to find the area of ABC and BXY. Then subtracting these two to find the area of BOC.
Area of ABC= 1/2*(4√3)*4 = 8√3
Area of BXY = 1/2*(5√3)*5= 12.5√3

Area of the shaded region: 12.5√3 - 8√3= 4.5√3

Please help!
Thank You!

Hi csaluja ,
I have highlighted the mistake in reasoning in red color.

Hi PKN,

Could you please explain why the highlighted part is incorrect? Would greatly appreciate it!

Hi csaluja,

According to your reasoning ,

Area of BOC= Area of ABC-Area of BXY

Let's break the area of ABC and BXY in terms of area of BOC,

Area of ABC= Area of AOB+Area of BOC--------(1)
Area of BXY=Area of XOCY+Area of BOC--------(2)

Now, subtracting (2) from(1) ,we have
Area of ABC-Area of BXY=Area of AOB-Area of XOCY, which is not equal to area of BOC.

Hope it's clear.
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