Question

2) O is the centre of the circle, and length of the chord is 8 cm segOP1 chord AB If I (OP) = 3 cm then find radius of the circle.
​

Answer 1

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☞OA is the radius of the circle

Perpendicular drawn from centre to a chord bisects the chord.

Therefore AM=

2

1

AB

AM=10cm

Thus, triangle OMA is right angled triangled at ∠OMA.

∠OMA=90

∘

.

By Pythagoras theorem,

OA

2

=OM

2

+MA

2

OA=

OM

2

+MA

2

OA=

(2

11

)

2

+10

2

OA=

144

cm

OA=12cm

Radius of a circle is 12cm.

Hope it's helpful↑(◍•ᴗ•◍)❤࿐

Answer 2

Concept Introduction:-

It might resemble a word or a number representation of the quantity's arithmetic value.

Given Information:-

We have been given that O is the centre of the circle, and length of the chord is $8$ cm segOP1 chord AB If I $(OP) = 3$ cm

To Find:-

We have to find that radius of the circle.

Solution:-

According to the problem

Given $o$ is the center of circle, $OP=3 \mathrm{~cm}$ and $\mathrm{AB}=8 \mathrm{~cm}$ Join $\mathrm{O}$ to $\mathrm{A}$ and $\mathrm{B}$

In figure $\mathrm{OP}$ show that the perpendicular from center $\mathrm{O}$ on $\mathrm{OP}$

We know that Perpendicular from the Center of a Circle to a Chord Bisects the Chord.

Then $\mathrm{AB}=\mathrm{OP}$

So $\mathrm{AB}=\mathrm{OP}=\frac{\mathrm{AB}}{2}=\frac{8}{2}=4 \mathrm{~cm}$

$\triangle \mathrm{AOM}$

$(\mathrm{AO})^{2}=(\mathrm{AM})^{2}+(\mathrm{OM})^{2}=(4)^{2}+(3)^{2}=16+9=25$

Final Answer:-

The correct answer is the radius of circle is $5 \mathrm{~cm}$.

#SPJ3