Plane Geometry
- Chords AB and CD of a circle intersect at E and are perpendicular to each other. Segments AE, EB and ED are of lengths 2 cm, 6 cm and 3 cm respectively. Then the length of the diameter of the circle (in cm) is
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As per the given in question , we draw a figure circle
Given , AE = 2 cm
EB = 6 cm
ED = 3 cm
∴ AE × EB = DE × EC⇒ EC = 2 × 6 = 4 cm. 3
Correct Option: A
As per the given in question , we draw a figure circle
Given , AE = 2 cm
EB = 6 cm
ED = 3 cm
∴ AE × EB = DE × EC⇒ EC = 2 × 6 = 4 cm. 3
∴ Diameter = √7² + 4² = √49 + 16 = √65 cm
- The length of the common chord of two circles of radii 30 cm and 40 cm whose centres are 50 cm apart, is (in cm)
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On the basis of question we draw a figure of a circle with centre O ,
Given , BD = 50 cm , AB = 30 cm and AD = 40 cm
Let BC = y ⇒ CD = 50 – y
AC² = 30² – y² = 40² – (50 – y)²
⇒ 900 – y² = 1600 – 2500 + 100y – y²
⇒ 100y = 1800
⇒ y = 18
∴ AC = √30² - 18² = √( 30 +18 ) ( 30 -18 )Correct Option: D
On the basis of question we draw a figure of a circle with centre O ,
Given , BD = 50 cm , AB = 30 cm and AD = 40 cm
Let BC = y ⇒ CD = 50 – y
AC² = 30² – y² = 40² – (50 – y)²
⇒ 900 – y² = 1600 – 2500 + 100y – y²
⇒ 100y = 1800
⇒ y = 18
∴ AC = √30² - 18² = √( 30 +18 ) ( 30 -18 )
AC = √48 × 12 = 24
∴ AE = 2Ac = 2 × 24 = 48 cm
- A chord AB of a circle C1 of radius (√3 + 1)cm touches a circle C2 which is concentric to C1. If the radius of C2 is (√3 - 1)cm., the length of AB is :
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According to question , we draw a figure of two concentric circles with centre O ,
Given , OC = √3 - 1
and OA = √3 + 1
From ∆ AOC
AC = √OC² + OA²
AC = √(√3 + 1)² - (√3 - 1)²Correct Option: C
According to question , we draw a figure of two concentric circles with centre O ,
Given , OC = √3 - 1
and OA = √3 + 1
From ∆ AOC
AC = √OC² + OA²
AC = √(√3 + 1)² - (√3 - 1)²
AC = √4√3 = 2. 4√3
∴ AB = 2AC = 4. 4√3 cm
- AB and CD are two parallel chords of a circle such that AB = 10 cm and CD = 24 cm. If the chords are on the opposite sides of the centre and distance between them is 17 cm, then the radius of the circle is :
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On the basis of question we draw a figure of a circle with centre O ,
Given , AB = 10 cm, AE = 5 cm
Let OE = y
CD = 24 cm , DF = 12 cm
OF = 17 – y
OA = OD
⇒ 5² + y² = 12² + (17 – y)²
⇒ 25 + y² = 144 + 289 – 34y + y²
⇒ 34y = 408Correct Option: C
On the basis of question we draw a figure of a circle with centre O ,
Given , AB = 10 cm, AE = 5 cm
Let OE = y
CD = 24 cm , DF = 12 cm
OF = 17 – y
OA = OD
⇒ 5² + y² = 12² + (17 – y)²
⇒ 25 + y² = 144 + 289 – 34y + y²
⇒ 34y = 408⇒ y = 408 = 12 34
∴ OA = √5² + 12² = 13 cm
- AB and CD are two parallel chords on the opposite sides of the centre of the circle. If AB = 10 cm, CD = 24 cm and the radius of the circle is 13 cm, the distance between the chords is
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According to question , we draw a figure of a circle with two parallel opposite chords AB and AC ,
OE ⊥ AB and OF ⊥ CD
AE = EB = 5 cm
CF = FD = 12 cm
AO = OC = 13 cm
From ∆ AOE,
OE = √13² - 5² = √169 - 25
OE = √144 = 12 cm
From ∆ COF,Correct Option: A
According to question , we draw a figure of a circle with two parallel opposite chords AB and AC ,
OE ⊥ AB and OF ⊥ CD
AE = EB = 5 cm
CF = FD = 12 cm
AO = OC = 13 cm
From ∆ AOE,
OE = √13² - 5² = √169 - 25
OE = √144 = 12 cm
From ∆ COF,
OF = √13² - 12² = √25 = 5 cm
∴ EF = OE + OF = 17 cm
