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Standard IX

Mathematics

Perpendicular from the Center to a Chord Bisects the Chord

In a circle w...

Question

In a circle with centre O, AB and CD are two diameters perpendicular to each other. The length of chord AC is

2AB

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$\sqrt{2}$

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$\frac{1}{2} A B$

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$\frac{1}{\sqrt{2}} A B$

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Solution

The correct option is D
$\frac{1}{\sqrt{2}} A B$

AB and CD are two diameters of a circle with centre O.

Length of chord $A C = \sqrt{O A^{2} + O C^{2}}$

$= \sqrt{r^{2} + r^{2}} = \sqrt{2 r^{2}} = \sqrt{2} r$

$= \sqrt{2} \times \frac{1}{2} A B = \frac{1}{\sqrt{2}} A B$

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Similar questions

Q. In a circle with centre O, AB and CD are two diameters perpendicular to each other. The length of chord AC is

(a) 2AB

(b)

\sqrt{2}

(c)

\frac{1}{2} A B

(d)

\frac{1}{\sqrt{2}} A B

Q. In the figure given,

O

is the centre of the circle.

A B

and

C D

are two chords of the circle.

O M

is perpendicular to

C D

and

O N

is perpendicular to

A B

A B = 24 c m, O N = 5 c m, O M = 12 c m

. Find the length of chord

C D

Q. In fig.,

A B

and

C D

are two diameters of a circle (with centre

O

) perpendicular to each other and

O D

is the diameter of the smaller circle. If

O A = 14 c m

, find the area of the shaded region

Q. In Fig.,

A B

and

C D

are two diameters of a circle with centre

O

, which are perpendicular to each other.

O B

is the diameter of the smaller circle. If

O A = 7 c m

, find the area of the shaded region.

[U s e π = \frac{22}{7}]

In figure, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region. [4 MARKS]

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In a circle with centre O, AB and CD are two diameters perpendicular to each other. The length of chord AC is

The correct option is D 1√2AB AB and CD are two diameters of a circle with centre O. Length of chord AC=√OA2+OC2 =√r2+r2=√2r2=√2r =√2×12AB=1√2AB

The correct option is D
$\frac{1}{\sqrt{2}} A B$

AB and CD are two diameters of a circle with centre O.

Length of chord $A C = \sqrt{O A^{2} + O C^{2}}$

$= \sqrt{r^{2} + r^{2}} = \sqrt{2 r^{2}} = \sqrt{2} r$

$= \sqrt{2} \times \frac{1}{2} A B = \frac{1}{\sqrt{2}} A B$