Question

$\large{\underline{\underline{\sf{\orange{Heya\:Folks!}}}}}$
$\Large\mathfrak{\boxed{Question:}}$
{ Refer to the image attached! }

Answer of the question - 38.4m
I wanna know how and where do these girls stand?

Class 9th maths RD sharma book's question! (CBSE syllabus)
Chapter name - Circles
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Answer 1

$\large{\boxed{\rm{Topic \: - \: Circle}}}$

$\large{\boxed{\rm{Book \: - \: R.D \: Sharma}}}$

$\large{\boxed{\rm{Class \: 9th}}}$

$\huge{\boxed{\rm{Question}}}$

Three girls Ishita, Isha and Nisha are playing in a ground by standing on a circle of radius 20 metres drawn in a park. Ishita throw a ball to Isha. Isha to Nisha. Nisha to Ishita. If the distance between Ishita and Isha and Nisha 24 metres each. What is the distance between Ishita and Nisha.

$\huge{\boxed{\rm{Answer}}}$

$\large{\boxed{\boxed{\sf{Given \: that}}}}$

• Three girls Ishita, Isha and Nisha are playing in a ground by standing on a circle of radius 20 metres drawn in a park.

• Ishita throw a ball to Isha.

• Isha throw a ball to Nisha.

• Nisha throw a ball to Ishita.

• The distance between Ishita and Isha and Nisha 24 metres each.

$\large{\boxed{\boxed{\sf{To \: find}}}}$

• The distance between Ishita and Nisha.

$\large{\boxed{\boxed{\sf{Solution}}}}$

• The distance between Ishita and Nisha = 38.4m

$\large{\boxed{\boxed{\sf{Let's \: understand \: the \: concept}}}}$

• Firstly this question says that there are three girls named Ishita, Isha and Nisha. They are playing in a ground by standing on a circle of radius 20 metres drawn in a park. (Kindly see the diagram to understand the concept properly and clearly). Afterwards Ishita throw a ball to Isha. Isha to Nisha. Nisha to Ishita. Means the girl Ishita throw the ball to Isha . Afterthat the girl that Isha throw the ball to Nisha. Then, the girl that Nisha send it to Ishita. If the distance between Ishita and Isha and Nisha 24 metres each. What is the distance between Ishita and Nisha.

$\small{\boxed{\boxed{\sf{How \: and \: where \: do \: these \: girls \: stand}}}}$

There are 4 directions as we already know that. According to the diagram Nisha stand in East. Ishita stand in West. Isha stand in the South direction. Means noone stand in the north side.

How the balls come like this, how the girls stand like this ? As u see in the diagram that there is point named as O. And O is the centre point. Let A as Ishita. Let C as Nisha. Let's B as Isha. And the point P is in centre of A and B. Q is in centre of B and C. R is the point in front of B.

Means the ball firstly throw from point A or direction West from Ishita. And it send to point B or direction South to Isha. Afterwards, 2ndly the ball thrown from point B or direction South from Isha. And it send to Nisha or point C or direction East. Afterwards, it send from Nisha or point C or direction East to Ishita or point A or direction West.

Hence, this is all the conclusion of your question.

$\footnotesize{\boxed{\boxed{\sf{Done \: but \: let's \: see \: the \: procedure \: also}}}}$

$\large{\boxed{\boxed{\sf{Full \: solution}}}}$

$\normalsize{\boxed{\boxed{\sf{Using \: concept}}}}$

• Phythagoras theorm.

$\normalsize{\boxed{\boxed{\sf{Using \: formula}}}}$

• Phythagoras theorm =

Hypotenuse² = Perpendicular² + Base²

$\large{\boxed{\boxed{\sf{Full \: solution}}}}$

In ∆OAP and ∆OPB

____________________________

Firstly using phythagoras theorm we have to put the values

OA² = AP² + OP²

20² = 12² + OP² (24/2 = 12)

400 = 144 + OP²

OP² = 400-144

OP² = 256

OP = √256

OP = 16 cm

____________________________

Now, as we see in the diagram that the points form a kite like shape.

Therefore, OA = OC and AB = BC

As we see that the diagonal of the kite are perpendicular to each other.

Area of ∆AOB = 1/2 × OP × OB

Area of ∆AOB = 1/2 × AR × 20 = 1/2 × OP × OP

Area of ∆AOB = 1/2 × AR × 20 = 1/2 × 16 × 24

• {Cancelling 1/2 by 1/2}

AR = 20 = 16 × 24

AR = (16×24)/20

AR = 19.2

____________________________

Distance between Ishita nd Nisha =2

Hence, 2 × 19.2

38.4 m.

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.

↦ AP = PB .......[Perpendicular from the centre bisect the chord]

Now,

↬ AP + PB = AB

↬ AP + AP = AB.....[∵ AP = PB]

↬ 2AP = AB

↬ AP = AB/2

↬ AP = 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

↦ OQ = 16m

Now, OABC forms a quadrilateral.

Where,

ↁ OA = OC ....[Radius]

ↁ AB = BC.....[Radius]

Thu, diagonal of a Quadrilateral are perpendicular here.

So, using

➨ Area of Triangle = 1/2 × b × h

➨ Area of ∆AOB = 1/2 × OP × PB

➨ Area of ∆AOB = 1/2 × AR × OB

Therefore,

➮ 1/2 × AR × OB = 1/2 × OP × PB

➮ 1/2 × AR × 20 = 1/2 × 16 × 24

➮ 20/2 × AR = 384/2

➮ 10AR = 192

➮ AR = 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