Question

In the given figure, O is the centre of the circle of radius 10 cm. AB and AC are two chords such that

$AB=AC=4\sqrt{5}cm$.

The length of chord BC is

• A.
  12.4 cm
• B.
  24 cm
• C.
  8 cm
• D.
  16 cm

Answer 1

The correct option is D

16 cm
Join OB, OC, OA.



In $\Delta OBAand\Delta OCA,OB=OC\left(Radii\right)AB=AC\left(Given\right)OA=OA\left(Common\right)\therefore \Delta OBA\cong \Delta OCA\text{(By SSS congruence rule)}\Rightarrow \angle BOM=\angle COM\left(CPCT\right)\dots ..\left(i\right)In\Delta BOMand\Delta COM,OB=OC\left(Radii\right)\angle BOM=\angle COM\left[From\right(i\left)\right]OM=OM\left(Common\right)\therefore \Delta BOM\cong \Delta COM\text{(By SAS congruence rule)}\Rightarrow \angle BMO=\angle CMO\left(CPCT\right)\dots ..\left(ii\right)Now,\angle BMO+\angle CMO={180}^{\circ }\left(Linearpair\right)\Rightarrow 2\angle BMO={180}^{\circ }\left[From\right(ii\left)\right]\Rightarrow \angle BMO={90}^{\circ }LetBM=xandOM=yIn\Delta BMO,\text{by Pythagoras theorem,}(BM{)}^2+(OM{)}^2=(OB{)}^2\Rightarrow x^2+y^2=100(\because OB=10cm)\dots .(iii)In\Delta BMA,\text{by Pythagoras theorem,}(BM{)}^2+(AM{)}^2=(AB{)}^2\Rightarrow (BM{)}^2+(OA–OM{)}^2=(4\sqrt{5}{)}^2\Rightarrow x^2+(10–y{)}^2=80\dots .\left(iv\right)\text{Subtracting (iv) from (iii), we get}{y}^2–(10–y{)}^2=20\Rightarrow y^2–(100–20y+y^2)=20\Rightarrow 20y–100=20\Rightarrow 20y=120\Rightarrow y=6cm\text{Putting value of y in (iii), we get}{x}^2+36=100\Rightarrow x^2=64\Rightarrow x=8cm$

Since the perpendicular from the centre of a circle to a chord bisects the chord.

$\therefore BC=2x=16cm$

Hence, the correct answer is option (d).