let ABCD be a quadilateral inscribd in a circle with diameter AC ,E be the foot of perpendicular from D onto AB as shown in the figure if AD=DC and the area of ABCD is 24 cm² .find the square length of DE(in cm)
Answers
Let CD=x then AB=2CD=2x.
Let r be the radius of the circle inscribed in the quadrilateral ABCD.
ar(□ABCD)=18 and AB∥CD.
⇒
2
1
(x+2x).2r=18
⇒ 3xr=18
⇒ xr=6 ----- ( 1 )
OP=OM=PD=OQ=AM=r
⇒ PC=x−r and MB=2x−r
Let ∠PCO=∠OCQ=θ
In right-angled △OPC,
tanθ=
CP
OP
=
x−r
r
----- ( 2 )
CD∥AB
∴ ∠PCB=∠QOM=2θ
⇒ ∠CBA=180
o
−2θ
⇒ ∠OBM=90
o
−θ
From △OMB,
tan(90
o
−θ)=
MB
OM
=
2x−r
r
Right-angled △OBM,
⇒ tanθ=
r
2x−r
------ ( 3 )
From ( 2 ) and ( 3 ),
⇒
x−r
r
=
r
2x−r
⇒ 2x
2
−3xr=0
⇒ x(2x−3r)=0
⇒ x=
2
3r
---- ( 4 )
From ( 1 ) and ( 4 ), we get
⇒ xr=6
⇒
2
3rr
=6
⇒ r
2
=4
⇒ r=2
ANSWER :
Let CD= x then AB = 2CD = 2x
Let r be the radius of the circle inscribed in the quadrilateral ABCD.
ar ( square ABCD ) = 18 and AB||CD
=> 1/2 (x+2x)2r = 18
=>3xr = 18
=> xr = 6. ------------------(1)
OP = OM = PD = OQ = AM = r
=> PC = x-r and MB = 2x-r
Let angle PCQ = angle OCQ = O
In right - angled ΔOPC,
tanΦ = OP/CP = r/x-r. -------------(2)
CD||AB
.°. angle PCB = angle QOM = 20
=> angle CBA = 180°- 20
=> angle OBM = 90° - 0
From ΔOMB
tan (90° - Φ) = OM/MB = r/2x-r
Right - angled ΔOBM,
tan Φ = 2x-r/r. --------------------(3)
From (2) and (3)
=> r/x-r 2x-r/r
=> 2x² - 3xr = 0
=> x(2x-3r) = 0
=> x=3r/2. ---------------------(4)
From (1) and (4)
=> xr = 6
=> 3xr/2 = 6
=> r² = 4
=> r= 2