Class 9 Math All topics Circles

Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius $5m$ drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is $6m$ each, what is the distance between Reshma and Mandip?

Solution:

Let the three girls Reshma, Salma and Mandip are standing on the circle of radius $5$ cm at points $R,S$ and $M$ respectively.

Here, chord $RS=$ chord $SM$.

$∴$center $O$ lies on bisector of $∠RSM$ i.e. ray $OS$

Now, $RS=SM=6$ m.

Let $OA⊥$ chord $RM$

$RA=AM$ ....perpendicular from center to the chord, bisects the chord.

In $△SAR$, $∠A=90_{o}$

$SR_{2}=SA_{2}+RA_{2}$ ....Pyhtagoras theorem

$⇒36=SA_{2}+RA_{2}$

$⇒RA_{2}=36−SA_{2}$ ... (1)

In the $△OAR$,

$RO_{2}=AO_{2}+RA_{2}$ ....Pythagoras theorem

$⇒25=(OS−AS)_{2}+RA_{2}$

$⇒RA_{2}=25−(OS−AS)_{2}$

$RA_{2}=25−(5−AS)_{2}$ ... (2)

From (1) and (2) , we get

$36−SA_{2}=25−(5−SA)_{2}$

$⇒11−SA_{2}+(5−SA)_{2}=0$

$⇒11−SA_{2}+25−10SA+AS_{2}=0$

$⇒10SA=36$

$⇒SA=3.6$

Putting $SA=3.6$ in (1) we get

$RA_{2}=3.6−(3.6)_{2}=36−12.96$

$⇒36−12.96=23.04$ m

$∴RA=4.8$m

$⇒RM=2RA=2×4.8=9.6$ m

Hence, the distance between Reshma and Mandip $=9.6$ m

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