Plane Geometry
- Two parallel chords of lengths 40 cm and 48 cm are drawn in a circle of radius 25 cm. What will be the distance between the two chords ?
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On the basis of question we draw a figure of a circle with centre O in two different cases ,
Case I,
When chords lie on both sides of centre.
AB = 40 cm.
CD = 48 cm.
CE = DE = 24 cm.
AF = BF = 20 cm.
OA = OC = 25 cm.
In ∆ AOF,
OF = √OA² - AF²
OF = √25² - 20²
OF = √(25 + 20)(25 - 20)
OF = √45 × 5 = √5 × 3 × 3 × 5 = 15 cm.
In ∆ COE,
OE = √OC² - CE²
OE = √25² - 24²
OE = √(25 + 24)(25 - 24)
OE = √49 = 7 cm.
∴ Required distance = EF = OE + OF = (7 + 15) cm = 22 cm.
Case II
When the chords lie on the same side of centre
AF = 20 cm.
CE = 24 cm.
OC = OA = 25 cm.
In ∆ OAF
OF = √OA² - AF²Correct Option: D
On the basis of question we draw a figure of a circle with centre O in two different cases ,
Case I,
When chords lie on both sides of centre.
AB = 40 cm.
CD = 48 cm.
CE = DE = 24 cm.
AF = BF = 20 cm.
OA = OC = 25 cm.
In ∆ AOF,
OF = √OA² - AF²
OF = √25² - 20²
OF = √(25 + 20)(25 - 20)
OF = √45 × 5 = √5 × 3 × 3 × 5 = 15 cm.
In ∆ COE,
OE = √OC² - CE²
OE = √25² - 24²
OE = √(25 + 24)(25 - 24)
OE = √49 = 7 cm.
∴ Required distance = EF = OE + OF = (7 + 15) cm = 22 cm.
Case II
When the chords lie on the same side of centre
AF = 20 cm.
CE = 24 cm.
OC = OA = 25 cm.
In ∆ OAF
OF = √OA² - AF²
OF = √25² - 20²
OF = √625 - 400
OF = √225 = 15 cm.
In ∆ OCE,
OE = √OC² - CE² = √25² - 24²
OE = √(25 + 24)(25 - 24)
OE = √49 = 7 cm.
∴ Required distance = EF = OF – OE = 15 – 7 = 8 cm.
- The perpendicular from the centre of a circle to a chord is 16 cm. If the diameter of the circle is 40 cm, what is the length of the chord ?
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As per the given in question , we draw a figure of a circle with centre O,
Give , AC = CB
OC = 16 cm.
OA = 20 cm.
∴ AC = √OA² - CB²
AC = √20² - 16²
AC = √400 - 256Correct Option: C
As per the given in question , we draw a figure of a circle with centre O,
Give , AC = CB
OC = 16 cm.
OA = 20 cm.
∴ AC = √OA² - CB²
AC = √20² - 16²
AC = √400 - 256
AC = √144 = 12 cm.
∴ AB = 2 AC = 24 cm.
- In a circle, a chord, 52 cm long, makes a right angle at the centre. Then the length of the radius of the circle will be
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On the basis of question we draw a figure of a circle with centre O ,
OA = OB = radius
∴ ∠OAB = ∠OBA = 45°
OC ⊥ AB∴ AC = CB = 5√2 cm. 2 ∴ cos OAC = AC OA ⇒ OA cos 45° = 5√2 2
Correct Option: B
On the basis of question we draw a figure of a circle with centre O ,
OA = OB = radius
∴ ∠OAB = ∠OBA = 45°
OC ⊥ AB∴ AC = CB = 5√2 cm. 2 ∴ cos OAC = AC OA ⇒ OA cos 45° = 5√2 2 ⇒ OA = 5√2 √2 2 ⇒ OA = 5√2 × √2 = 5 cm. 2
- An arc of 30° in one circle is double an arc in a second circle, the radius of which is three times the radius of the first. Then the angles subtended by the arc of the second circle at its centre is
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According to question , we draw a figure of a circle with radius r ,
θ = l ; l1 = 2 ; l2 = 3 r l2 l1 ∴ θ1 = l1r2 = 2 × 3 = 6 θ2 l2r1 ⇒ 30° = 6 θ2 ⇒ θ2 = 30° = 5° 6
Correct Option: C
According to question , we draw a figure of a circle with radius r ,
θ = l ; l1 = 2 ; l2 = 3 r l2 l1 ∴ θ1 = l1r2 = 2 × 3 = 6 θ2 l2r1 ⇒ 30° = 6 θ2 ⇒ θ2 = 30° = 5° 6 
θ (radian) = l . Here, θ1 is a ratio. 
r θ2
- AB is the diameter of a circle with centre O and P is a point on its circumference. If ∠POA = 120°, then the value of ∠PBO is :
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As per the given in question , we draw a figure of a circle with centre O,
∠AOB = 180°
∠ POA = 120°
∴ ∠ POB + ∠ POA = 180°
∴ ∠ POB = 180° – 120° = 60°Correct Option: B
As per the given in question , we draw a figure of a circle with centre O,
∠AOB = 180°
∠ POA = 120°
∴ ∠ POB + ∠ POA = 180°
∴ ∠ POB = 180° – 120° = 60°
In ∆ OPB,
OP = OB = radius
∴ ∠OPB = ∠PBO = 60°
