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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
A)8
B)8√2
C)12
D)12√2
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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.
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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06 Apr 2019, 10:20
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Chethan92
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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