S 1- AM - AP -2- BN - BP -2S 2- OO - AM - BNChoose the correct option-A- Only S 1 is true-B- S1 is true - S2 is false-C- Both statements are true-D- Only S2 is true-

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Theorem 2: Perpendicular from the Center to a Chord Bisects the Chord

S 1: AM = AP ...

Question

S1: $A M = \frac{A P}{2}, B N = \frac{B P}{2}$

S2: $O O^{'} = A M + B N$

Choose the correct option.

Only S1 is true.

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Only S2 is true.

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Both statements are true.

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S1 is true & S2 is false.

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Solution

The correct option is C
Both statements are true.

S1 is true as the perpendicular from the centre of the circle to the chord, bisects the chord into two parts.

Therefore, $A M = \frac{A P}{2}, B N = \frac{B P}{2}$ and

$O O^{'} = M N$

$= P M + P N$

$= A M + B N$ [as $P M = A M$ and $P N = B N$ ]

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Similar questions

S1: AM = $\frac{A P}{2}$ , BN= $\frac{B P}{2}$

S2: OO' = AM + BN

Which of the above statements are true ?

S1: AM = $\frac{A P}{2}$ , BN= $\frac{B P}{2}$

S2: OO' = AM + BN

Which of the above statements are true ?

A has a pair of triangles with corresponding sides proportional, and B has a pair of pentagons with corresponding sides proportional.

$S_{1}$ - A’s triangles must be similar.

$S_{2}$ - B’s pentagons must be similar.

Which of the following statement is correct?

Q. Let A, B, C be three mutually independent events. Consider the two statements

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and

S_{2}

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A figure has an area of $196 c m^{2}$
S1 : The figure can be a square of side 14 cm.
S2 : The figure can be a rectangle if and only if the dimensions are $49 c m \times 4 c m$ .

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S1: AM=AP2, BN=BP2 S2: OO′=AM+BN Choose the correct option.

The correct option is C Both statements are true. S1 is true as the perpendicular from the centre of the circle to the chord, bisects the chord into two parts. Therefore, AM=AP2, BN=BP2 and OO′=MN =PM+PN =AM+BN [as PM=AM and PN=BN]

S1: $A M = \frac{A P}{2}, B N = \frac{B P}{2}$

S2: $O O^{'} = A M + B N$

Choose the correct option.