Advertisements
Advertisements
प्रश्न
Prove that, of any two chords of a circle, the greater chord is nearer to the centre.
उत्तर
Given: A circle with centre O and radius r. OM ⊥ AB and ON ⊥ CD Also AB > CD
To prove: OM < ON
Proof: Join OA and OC.
In right ΔAOM,
AO2 = AM2 + OM2
`=> r^2 = (1/2 AB)^2 + OM^2`
`=> r^2 = 1/4 AB^2 + OM^2` ...(i)
Again in right ΔONC,
OC2 = NC2 + ON2
`=> r^2 = (1/2 CD )^2 + ON^2`
`=> r^2 = 1/4 CD^2 + ON^2` ...(ii)
From (i) and (ii)
`1/4 AB^2 + OM^2 = 1/4 CD^2 + ON^2`
But, AB > CD ...(Given)
∴ ON > OM
`=>` OM < ON
Hence, AB is nearer to the centre than CD.
APPEARS IN
संबंधित प्रश्न
AB and CD are two chords of a circle intersecting at P. Prove that AP x PB = CP x PD
In the figure given below, O is the centre of the circle. AB and CD are two chords of the circle. OM is perpendicular to AB and ON is perpendicular to CD.
AB = 24 cm, OM= 5 cm, ON= 12 cm. Find the:
1) radius of the circle
2) length of chord CD.
In the given figure 3 × CP = PD = 9 cm and AP = 4.5 cm. Find BP.
In the given figure, 5 × PA = 3 × AB = 30 cm and PC = 4 cm. Find CD.
In the given figure, tangent PT = 12.5 cm and PA = 10 cm; find AB.
Chords AB and CD of a circle intersect each other at point P such that AP = CP
Show that: AB = CD
In two concentric circles, prove that all chords of the outer circle, which touch the inner circle, are of equal length.
Two tangents are drawn to a circle from an external point P. touching the circle at the points A and B. A third tangent intersects segment PA in C and segment PB in D and touches the circle at Q. if PA = 20 units, find the perimeter of Δ PCD.
If AB and CD are two chords which when produced meet at P and if AP = CP, show that AB = CD.
Two circles of radii 10 cm and 8 cm intersect and the length of the common chord is 12 cm. Find the distance between their centres.
