47. ABCD is a rectangle of dimensions 12 cm and 5 cm.
AEC is a rectangle drawn in such a way that the
diagonal AC of the first rectangle is one of its sides
and side opposite to it is touching the first rectangle
at D as shown in figure. What is the ratio of the area
of rectangle ABCD to AEFC?
В B.
12 cm
5 cm
C С
ED
F
A.
3:1
B.
2:3
C.
1:1
D.
5:4
Answers
Answered by
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Step-by-step explanation:
From the above figure, we can say that AC=10cm by Pythagoras theorem.
So, length of rectangle EACF=10cm.
Let DF=x cm so, then DE=(10−x) cm.
In △CDF,
CD
2
=CF
2
+FD
2
⟹64=CF
2
+x
2
⟹CF
2
=64−x
2
Similarly, in △ADE,
AD
2
=AE
2
+ED
2
⟹36=AE
2
+(10−x)
2
⟹AE
2
=36−(10−x)
2
⟹AE
2
=36−(100+x
2
−20x)
⟹AE
2
=−64−x
2
+20x
Equating, squares of AE and CF,
−64−x
2
+20x=64−x
2
⟹20x=128orx=6.4
DF=6.4 cm so, DE=3.6 cm.
That gives us CF=4.8cm=AE.
The area of ABCD =8×6=48 sq cm.
The area of EACF=4.8×10=48 sq cm.
The ratio of the areas =1.
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