20. F is a point taken on side AD of a square ABCD. CE is drawn perpendicular to CF, meeting AB extended to point E. If the area of CEF = 200cm and area of the square ABCD is 256cm 'then the length in cm) of BE is a) 10 b 11 c) 12 d) 16
Answers
Answer:
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Answer:
Correct option is
C
12
solution: CE⊥CF
area of △CEF=200 cm
2
area of square ABCD=256 cm
2
(AB)
2
=256 cm
2
⇒AB=16 cm=AD=CD=BC
area of △CEF=200 cm
2
2
1
×CE×CF=200 cm
2
⇒CE×CF=400 cm
2
In △AFE
AF
2
+AE
2
=EF
2
(AD−FD)
2
+(AB+BE)
2
=EF
2
AD
2
+FD
2
−2(AD)(FD)+AB
2
+BE
2
+2(AB)(BE)=EF
2
(16)
2
+FD
2
−2(16)(FD)+(16)
2
+BE
2
+2(16)(BE)=EF
2
256+256+FD
2
+BE
2
−32(FD)+32(BE)=CF
2
+EC
2
(∵△ CFE is right angled at C
so (EF
2
=CF
2
+CE
2
) ⇒512+FD
2
+BE
2
−32(FD)+32(BE)=FD
2
+CD
2
+BC
2
+BE
2
(: △CDF is right angled
so,
CF
2
=DF
2
+CD
2
and △CBE is right angled so,CE
2
=CB
2
+BE
2
)
⇒512−32(FD)+32(BE)=(16)
2
+(16)
2
⇒BE=FD.
In △ CDF and △ CBE
CD=CB
∠CDF=∠CBE
DF=BE
so △CDF≅△CBE by R-H-S rule
so CF=CE
(CF)⋅(EC)=400 cm
2
⇒CE=20 cm
In △CBE
CE
2
=CB
2
+BE
2
⇒400 cm
2
=256 cm
2
+BE
2
⇒144 cm
2
=BE
2
⇒12 cm=BE
Answer : option (c).