Question

in the figure O is the center of the circle and OP⊥AB. If AB =8cm and OP=3cm find the diameter of the circle.

Answer 1

Given :-

• OP⊥AB
• AB = 8cm.
• OP = 3cm.

To Find :-

• Diameter of circle ?

Concept used :-

• If a line is drawn from the centre of a circle to the chord , the the line will bisect the chord at right Angle .
• Pythagoras Theoram .

Construction :-

• Join OA .

Solution :-

in Right ∆OAP , we have :-

→ OP = 3cm.

→ AP = (AB/2) = (8/2) = 4cm.

So , By Pythagoras Theoram ,

→ AO = √[(3)² + (4)²]

→ AO = √[9 + 16]

→ AO = √25

→ AO = 5cm. = Radius of circle.

So,

→ Diameter of circle = 2 * Radius = 2 * 5 = 10cm. (Ans).

Hence , Diameter of Given Circle is 10cm.

Answer 2

Answer:

The value of Diameter is 10cm.

Step-by-step explanation:

Given:

AB = 8cm,

OP = 3cm,

OP is perpendicular to AB.

construction:

Join OA and then extend AO to A' such that AA' becomes the longest chord which is diameter.

Join OB.

Here , OA and OB becomes radius of the circle.

To find: AA' distance.

Solution:

OP divides AB into two equal halves, such that,

AP = PB.

( Line drawn from the centre of the circle to the chord of the circle perpendicularly then , the drawn perpendicular line divides the chord into two equal parts.)

Now , by applying this theorem we have,

AP = PB

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=> AP + PB = 8cm

=> AP + AP = 8cm

=> 2 (AP) = 8cm

=> AP = 4cm

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=> AP = PB = 4cm.

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Now, consider traingle, APO

=> AP = 4cm , OP = 3cm

now by using Pythagorean theorem we get,

$AO \: = \sqrt{( {4}^{2}) + {(3)}^{2} }$

$AO \: = \sqrt{16 + 9}$

$AO \: = \sqrt{25}$

$AO \: = 5cm$

=> radius of circle = 5cm

=> Diameter of circle = 2 × radius = 2 × 5cm = 10cm.

therefore, the diameter of circle is 10cm.

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In triangle APO and traingle BPO

we have, AP = PB ( side )

OA = OB ( radius ) ( side )

OP is common side.

Now, by using S.S.S axiom , we can say that

traingle APO is congurent to traingle BPO.

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