12
44
26
D
AC and BD are diameters of the circle which intersect each other at the
of the circle. If diametre of the circle is 91 cm. Find the area of the
region.
(A) 2266 cm (B) 2366 cm (C) 3266 cm 102466 cm
(E) None of these
Look at the following figure:
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Answer:
ANSWER
Given- O is the centre of a circle whose diameter is BC. AB is a chord and OD⊥ AB. BD=5cm and OD=4cm. CD has been joined.
To find out- CD=?
Solution- OD⊥AB.
∴ D is the mid point of AB since the perpendicular, dropped from the center of a circle to its any chord bisects the latter. So AB=2BD=2×5cm=10cm. And BD=AD=5cm. Now ∠BAC=90
o
since angle in a semicircle=90
o
. ∴ΔCAB&ΔCDB are right triangles with BC&DC as hypotenuses.
∴ By Pythagoras theorem, we have OB=
BD
2
+OD
2
=
5
2
+4
2
cm=
41
cm.
But BC=2OB(diameter=2radius).
BC=2×
41
cm.
∴AC=
BC
2
−AB
2
=
(
41
)
2
−10
2
cm=8cm.
So CD=
AD
2
+AC
2
=
5
2
+8
2
cm=
89
cm.
Ans- Option C.
Step-by-step explanation:
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