Question

O is the centre of the circle (Fig. 9.30). Drop perpendicular from B on CA. Where does it meet CA? ​

Answer 1

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=90o since angle in a semicircle=90o. ∴ΔCAB&ΔCDB are right triangles with BC&DC as hypotenuses.

∴  By Pythagoras theorem, we have  OB=BD2+OD2=52+42cm=41cm.

But BC=2OB(diameter=2radius).

BC=2×41cm.∴AC=BC2−AB2=(41)2−102cm=8cm.  

So CD=AD2+AC2=

Answer 2

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