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Intern
Joined: 26 Mar 2019
Posts: 28

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06 Apr 2019, 07:17
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in the figure angle <ABE=<BDC=90°;AB=4cm and angle <BAE=45°.If CD=AE.what is the area of the triangle BCE in square centimetre?
A)8
B)8√2
C)12
D)12√2
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Director
Joined: 18 Jul 2018
Posts: 907
Location: India
Concentration: Finance, Marketing
WE: Engineering (Energy and Utilities)

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06 Apr 2019, 09:54
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Hey! Wasif007, Let me give a try.
Given info AE = CD
AB = 4
Angle B = Angle D = 90
Also, Angle A = 45, then In triangle ABE, Angle E = 45 (isosceles triangle), then side BE = 4, as both AB and BE are same.
Then the hypotenuse in triangle ABE becomes $$4\sqrt{2}$$ (as per 45-45-90 triangle)
Then CD = $$4\sqrt{2}$$, as AE = CD.
Now, Area of triangle BCE = $$\frac{1}{2}*Base*Height$$
Base = BE = 4
Height = CD = $$4\sqrt{2}$$
Area = $$\frac{1}{2}*4*4\sqrt{2}$$
Area = $$8\sqrt{2}$$

Hope it is clear.
Also, PFA for better understanding.
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Intern
Joined: 26 Mar 2019
Posts: 28

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06 Apr 2019, 10:20
Chethan92 wrote:
Hey! Wasif007, Let me give a try.
Given info AE = CD
AB = 4
Angle B = Angle D = 90
Also, Angle A = 45, then In triangle ABE, Angle E = 45 (isosceles triangle), then side BE = 4, as both AB and BE are same.
Then the hypotenuse in triangle ABE becomes $$4\sqrt{2}$$ (as per 45-45-90 triangle)
Then CD = $$4\sqrt{2}$$, as AE = CD.
Now, Area of triangle BCE = $$\frac{1}{2}*Base*Height$$
Base = BE = 4
Height = CD = $$4\sqrt{2}$$
Area = $$\frac{1}{2}*4*4\sqrt{2}$$
Area = $$8\sqrt{2}$$

Hope it is clear.
Also, PFA for better understanding.

Why CD is the height of triangle BCE??? is not CD the height of triangle DCE?

And How you get <CED and <DCE 45°??? Those clue are not given??

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Director
Joined: 18 Jul 2018
Posts: 907
Location: India
Concentration: Finance, Marketing
WE: Engineering (Energy and Utilities)

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06 Apr 2019, 10:33
Given that CD = DE, then the angle must also be the same.
As angle D = 90, Angle E and C should be 45 each.
Height = Perpendicular distance from BASE, which is BE.
CD is the height of both the triangles.

Is it clear?
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Intern
Joined: 26 Mar 2019
Posts: 28

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06 Apr 2019, 11:16
Chethan92 wrote:
Given that CD = DE, then the angle must also be the same.
As angle D = 90, Angle E and C should be 45 each.
Height = Perpendicular distance from BASE, which is BE.
CD is the height of both the triangles.

Is it clear?

I am actually a bit of confused here...<D=90° that i understand. But in question CD=AE is given not CD=DE....so wont those two angles are different??

And as we know area=1/2 x base x height.
So isn't CD the height of triangle of DCB OR CDE? But the triangle BCE is different.

I am having a little bit of trouble understanding here

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Manager
Joined: 29 Nov 2018
Posts: 67
Location: India
GPA: 4
WE: Information Technology (Consulting)

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06 Apr 2019, 23:20
In Triangle ABE , AB = 4 cm (given) and <ABE = 90 , <BAE = 45 (given)
from this , <BEA = 45 and also AE = 4\sqrt{2} ( by applying pythagoras theorm)
and AB = BE = 4 ( sides opposite to equal angles are equal).

Now , Area of triangle BEC = 1/2 * base * height

= 1/2 * BE * CD
= 1/2 *4 * 4\sqrt{2} ( AE = CD given )
= 8\sqrt{2}
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Hope it helps,

Thanks,
Mofe Bhatia
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