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605-655 (Medium)|   Geometry|      
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KSBGC
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In the figure given below, ABC is a right angled , isoceles triangle . Updated on: 28 Sep 2022, 04:34
Originally posted by KSBGC on 11 Oct 2019, 04:11.
Last edited by chetan2u on 28 Sep 2022, 04:34, edited 1 time in total.
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45% (medium)

Question Stats:

67% (02:18) correct 33% (02:41) wrong based on 88 sessions

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In the figure given below, ABC is a right angled , isoceles triangle . If BC||DE, BC =x and DE =y, what is the area of the plot of DBCE?
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A

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In the figure given below, ABC is a right angled , isoceles triangle . 11 Oct 2019, 22:57
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KSBGC
In the figure given below, ABC is a right angled , isoceles triangle . If BC =x and DE =y, what is the area of the plot of DBCE?
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\(\triangle ADE\)and \(\triangle ABC\) are right angled , isoceles triangles...
So Area of \(\triangle ADE=\frac{1}{2}*AD*DE=\frac{y*y}{2}=\frac{y^2}{2}\) and Area of \(\triangle ABC=\frac{1}{2}*AB*BC=\frac{1}{2}*x*x=\frac{x^2}{2}\)

The area of the plot of DBCE=Area of \(\triangle ADE\) - Area of \(\triangle ABC\) = \(\frac{x^2-y^2}{2}\)
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In the figure given below, ABC is a right angled , isoceles triangle . 05 Jun 2020, 10:16
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chetan2u
KSBGC
In the figure given below, ABC is a right angled , isoceles triangle . If BC =x and DE =y, what is the area of the plot of DBCE?

\(\triangle ADE\)and \(\triangle ABC\) are right angled , isoceles triangles...
So Area of \(\triangle ADE=\frac{1}{2}*AD*DE=\frac{y*y}{2}=\frac{y^2}{2}\) and Area of \(\triangle ABC=\frac{1}{2}*AB*BC=\frac{1}{2}*x*x=\frac{x^2}{2}\)

The area of the plot of DBCE=Area of \(\triangle ADE\) - Area of \(\triangle ABC\) = \(\frac{x^2-y^2}{2}\)
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Can you tell me how did we know ADE is a isosceles triangle? Thanks!
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In the figure given below, ABC is a right angled , isoceles triangle . 13 Jun 2020, 03:33
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KSBGC
In the figure given below, ABC is a right angled , isoceles triangle . If BC =x and DE =y, what is the area of the plot of DBCE?

\(\triangle ADE\)and \(\triangle ABC\) are right angled , isoceles triangles...
So Area of \(\triangle ADE=\frac{1}{2}*AD*DE=\frac{y*y}{2}=\frac{y^2}{2}\) and Area of \(\triangle ABC=\frac{1}{2}*AB*BC=\frac{1}{2}*x*x=\frac{x^2}{2}\)

The area of the plot of DBCE=Area of \(\triangle ADE\) - Area of \(\triangle ABC\) = \(\frac{x^2-y^2}{2}\)

Can you tell me how did we know ADE is a isosceles triangle? Thanks!
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aalam383

question does not mentions the part " DE parallel to BC"

above solution assumed the same....

else other fine!!!!! :)
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In the figure given below, ABC is a right angled , isoceles triangle . 13 Jun 2020, 06:17
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if DE parallel to BC then option A else might be other option :)
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In the figure given below, ABC is a right angled , isoceles triangle . 12 Jul 2021, 13:37
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It's a problem of similar triangle.Think the properties of similar triangle & find the solution on your own.

here BC=X & DE=Y
Since ABC is an isosceles triangle, side AC is represented by 90 degree, side AB & BC both are represented by 45 degree.

In triangle ADE,side AE is represented by 90 degree and both AD & DE are by 45 degree.

So we can write, BC=AB=X
DE=AD=Y
So the area of triangle ADE is
1/2(AD*DE)=1/2(Y*Y)=(Y^2)/2
and area of ABC =1/2(AB*BC)
=1/2(X*X)=(X^2)/2
So now area of DBCE=[(X^2)/2]-[(Y^2)/2]
=(X^2-Y^2)/2
Answer:A

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In the figure given below, ABC is a right angled , isoceles triangle . 28 Sep 2022, 04:07
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the question doesn't mention that DE is parallel to BC, so how do we infer that the two triangles are similar? I think without that information the answer should be E.
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In the figure given below, ABC is a right angled , isoceles triangle . 28 Sep 2022, 04:32
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KarishmaB

the question doesn't mention that DE is parallel to BC, so how do we infer that the two triangles are similar? I think without that information the answer should be E.
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You are correct, it is not specifically mentioned. It should have been given accordingly.

But answer cannot be E, as ‘none of these’ means there is some answer possible, which is not given in option.

However, if the lines are not parallel, then the answer cannot be determined.
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