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605-655 (Medium)|   Geometry|      
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Bunuel
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If in the figure above AE=AD, FB=6 and the area of the parallelogram 07 Jul 2021, 01:30
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If in the figure above AE=AD, FB=6 and the area of the parallelogram DEBF is 36, what is the area of the rectangle ABCD?

A. 45
B. 54
C. 72
D. 90
E. 108


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If in the figure above AE=AD, FB=6 and the area of the parallelogram 07 Jul 2021, 02:59
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Given ABCD is a rectangle
DEBF is a ||gm
Also given ,
AE = AD and FB =6
FB = DE (opposite sides of the ||gm)
Let AE = x
In triangle AED, (which is right triangle)
2x^2 = 36
x^2 = 18
x = 3√2

Now area of triangle
0.5 * b* h = 0.5 * 3√2 * 3√2 = 9

Area of two triangle = 9 * 2 =18

Area of rectangle = Area of two triangles + Area of ||gm
18 + 36 = 54

IMO B
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If in the figure above AE=AD, FB=6 and the area of the parallelogram 07 Jul 2021, 03:03
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Area of Triangle ADE + BCF = \( 2*\frac{1}{2}*\frac{ 6}{\sqrt{2}} * \frac{ 6}{\sqrt{2}} = 18\\
\)

Area of rectangle = 18 +36 = 54

Option B

Bunuel

If in the figure above AE=AD, FB=6 and the area of the parallelogram DEBF is 36, what is the area of the rectangle ABCD?

A. 45
B. 54
C. 72
D. 90
E. 108


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If in the figure above AE=AD, FB=6 and the area of the parallelogram 07 Jul 2021, 05:02
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Since DEBF is parallelogram and area of 36, BF = EB = ED = DF = 6
Since AE = AD, then AE = AD = 6/sqrt(2) (since ABCD is rectangle)
Therefore, BC = FC = 6/sqrt(2)

Area = 6x6 + (6/sqrt(2) x 6/sqrt(2)) = 36 + 18 = 54 (B)
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If in the figure above AE=AD, FB=6 and the area of the parallelogram 07 Jul 2021, 10:16
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for the given figure
the ∆ AED is 45:45:90 ; x:x:x√2
given FB is 6 so sides of the ∆ AE & AD =3√2
now drop a line perpendicular from E on DC and from F to AB
we get two ∆ whose area = 1/2 * 3√2* 3√2 * 2 ; 18
left with center square of side each 3√2
we now know
width of rectangle ABCD ; 3√2 and length ; ( 3√2+3√2+3√2 ; 9√2)
area of ABCD ; 3√2 *9√2 ; 54
option B


Bunuel

If in the figure above AE=AD, FB=6 and the area of the parallelogram DEBF is 36, what is the area of the rectangle ABCD?

A. 45
B. 54
C. 72
D. 90
E. 108


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If in the figure above AE=AD, FB=6 and the area of the parallelogram 07 Jul 2021, 12:32
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Area of rectangel ABCD= Area of parallelogram DEBF + area of ADE + are of triangle BCF
Are of ||gm DEBF= 36
In traingle ADE, AD=DE=6/rt(2)=3rt(2) using sin45°= AD/DE=AD/6,
SO,AD=3*rt(2)
Area of triangle ADE= 1/2 * 3*rt(2)* 3*rt(2)=9
Area of triangle BCF= 1/2* 3*rt(2)* 3*rt(2)= 9

Hence areaABCD=36+9+9=54
Hence,E.

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