Question

Three girls Ishita, Isha and Nisha are playing a game by standing on a circle of
radius 20 m drawn in a park. Ishita throws a ball to Isha, Isha to Nisha and Nisha to
Ishita. If the distance between Ishita and Isha and between Isha and Nisha is 24 m
each, what is the distance between Ishita and Nisha.
​

Answer 1

Answer:

It is asking Diameter So Twice of Radius is equals to Diameter =20×2=40m

Answer 2

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Given:-

• Ishita, Isha and Nisha are standing on a circle of Radius 20m
• Distance between Isha and Itisha and Isha and Nisha is 24m each

Find:-

• Distance between Ishita and Nisha.

Diagram:-

Let, us assume that Itisha is standing at a point A, Ishita at point B and Nisha at point C.

Construction:- Construct a perpendicular OP to AB and perpendicular OQ to BC

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\qbezier(2.3,0)(2.121,2.121)(0,2.3)\qbezier(-2.3,0)(-2.121,2.121)(0,2.3)\qbezier(-2.3,0)(-2.121,-2.121)(0,-2.3)\qbezier(2.3,0)(2.121,-2.121)(-0,-2.3)\qbezier(2.1,-0.8)(0,0)(0,0)\qbezier(-2.1,-0.8)(0,0)(0,0) \qbezier( -2.1,-0.8)(2.1,-0.8)(2.1,-0.8)\qbezier(0,2.3)(0,-2.3)(0,-2.3)\qbezier(-2.1,-0.8)(0,-2.3)(0,-2.3)\qbezier(2.1,-0.8)(0,-2.3)(0,-2.3)\qbezier(0,0)(-1.3,-1.4)(-1.3,-1.4)\qbezier(0,0)(1.4,-1.3)(1.4,-1.3) \put(0,0){\circle*{0.16}} \put(-2.6,-1){\bf A} \put(0,-2.7){\bf B} \put(2.3,-1){\bf C}\put(-1.5,-1.7){\bf P} \put(1.4,-1.51){\bf Q}\put(0,-1.1){\bf R}\put(0.1,0.1){\bf O}\put(-3,-1.3){\bf Ishita}\put(-0.2,-3){\bf Isha} \put(2.1,-1.3){\bf Nisha} \end{picture}$

Note:- See the diagram from web on desktop mode.

Solution:-

From Diagram:-

Radius of circle = 20m

So, OA = OB = OC = 20m

Distance between Ishita and Isha and Isha and Nisha is 24m each.

Thus, AB = BC = 24m

Considering Chord AB

OP is perpendicular to chord and passes through the centre.

$\longmapsto$ AP = PB .......[Perpendicular from the centre bisect the chord]

Now,

↬ AP + PB = AB

↬ AP + AP = AB.....[∵ AP = PB]

↬ 2AP = AB

↬ AP = $\sf{\frac{AB}{2}}$

↬ AP = $\sf{\frac{24}{2}}$ ......[∵ AB = 24m]

↬ AP = 12m

Now, In ∆OAP

➠ H² = P² + B² ......[Pythogoras Theorem]

➠ OA² = AP² + OP²

where,

• OA = 20m
• AP = 12m

• Substituting these values •

➠ 20² = 12² + OP²

➠ 400 = 144 + OP²

➠ 400 - 144 = OP²

➠ 256 = OP²

➠ √(256) = OP

➠ 16m = OP

➠ OP = 16m

Similarly in ∆OQC

$\longmapsto$ OQ = 16m

Now, OABC forms a quadrilateral.

Where,

ↁ OA = OC ....[Radius]

ↁ AB = BC.....[Radius]

Thus, diagonal of a Quadrilateral are perpendicular here.

So, using

➨ Area of Triangle = $\sf{\frac{1}{2}}$ × b × h

➨ Area of ∆AOB = $\sf{\frac{1}{2}}$ × OP × PB

➨ Area of ∆AOB = $\sf{\frac{1}{2}}$ × AR × OB

Therefore,

➮ $\sf{\frac{1}{2}}$ × AR × OB = $\sf{\frac{1}{2}}$ × OP × PB

➮ $\sf{\frac{1}{2}}$ × AR × 20 = $\sf{\frac{1}{2}}$ × 16 × 24

➮ $\sf{\frac{20}{2}}$ × AR = $\sf{\frac{384}{2}}$

➮ 10AR = 192

➮ AR = $\sf{\frac{192}{10}}$

➮ AR = 19.2m

Now, the distance between Nisha and Nitisha

» AC = AR + RC

» AC = AR + AR ......[ AR = RC]

» AC = 2AR

» AC = 2×19.2......[AR = 19.2m]

» AC = 38.4m

Hence, the Distance between Nisha and Nitisha is 38.4m

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━