Question

Two parallel chords on the same side of the centre of a circle are 10 cm and 24 cm long and are 7 cm apart. The radius of the circle is (1) 13 cm (2) 8 cm (3) 5 cm (4) 7 cm​

Answer 1

Answer:

$\qquad\qquad\underline{\textsf{\textbf{ \color{magenta}{AnSwEr} }}}$

Consider △ CGE.

(CG)^2 = (EG)^2 + (CE)^2

r^2 = (7 + x)^2 + 5^2

r^2 = 49 + 14x + x^2 + 25

r^2 = x^2 + 14x + 74 :Equation 1

Consider △ FBG.

(BG)^2 = (FB)^2 + (FG)^2

r^2 = 12^2 + x^2

r^2 = 144 + x^2 :Equation 2

Equation 1 = Equation 2

x^2 + 14x + 74 = 144 + x^2

14x = 144 - 74

14x = 70

x = 5

Now, we can compute for the value of the radius, r.

r^2 = 144 + 5^2

r^2 = 169

r = 13 cm

Henceforth, the correct options is (4) 7 cm

Answer 2

• Let PQ=10 cm and RS=24cm be the 2 chords. Lets take a perpendicular line OA to both PQand RSwhere B lies on RSand A on PQ. Since OA is perpendicular to the chords and passes through center, the chords are bisected by OA.
• $\sf{⟹RB=SB}$
• $= \sf\frac{24}{2} = 12cm$
• $\sf{⟹PA=QA}$
• $\sf{ = \frac{10}{2} = 5cm}$
• In △OBS
• $\sf \red{ {OB}^{2} = { OS }^{2} - { SB}^{2} }$
• $OB = \sqrt{ {r}^{2} - {12}^{2} }$
• So,In triangle OAQ
• $OA = \sqrt{ {r}^{2} - {5}^{2} }$
• $OA–OB=7$
• $+ \sqrt{ {r}^{2} - {7}^{2} } - \sqrt{ {r}^{2} - {12}^{2} } = 7$
• $\fbox \red{r = 13}$
• Answer : 13 cm.