Plane Geometry
- A 8 cm long perpendicular is drawn from the centre of a circle to a 12 cm long chord. The diameter of the circle is :
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According to question , we draw a figure of a circle with centre O ,
Given that , AB = Length of chord = 12 cm.
OC ⊥ AB
∴ AC = CB = 6 cm.
OC = 8 cm.
In ∆ OAC,
OA = √OC² + CA²
OA = √8² + 6²Correct Option: D
According to question , we draw a figure of a circle with centre O ,
Given that , AB = Length of chord = 12 cm.
OC ⊥ AB
∴ AC = CB = 6 cm.
OC = 8 cm.
In ∆ OAC,
OA = √OC² + CA²
OA = √8² + 6²
OA= √64 + 36 = √100 = 10 cm.
∴ Diametre of circle = 2 × OA = 2 × 10 = 20 cm.
- In a circle, two arcs of unequal length subtend angles in the ratio 5 : 3. If the smaller angle is 45° then the measure of other angle in degrees is :
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We find the required answer with the help of given formula ,
θ = l where θ is in radian measure. r ∴ θ1 = l1 : θ2 = l2 r1 r2 ⇒ θ1 = l1 r2 θ2 r1 l2
Correct Option: A
We find the required answer with the help of given formula ,
θ = l where θ is in radian measure. r ∴ θ1 = l1 : θ2 = l2 r1 r2 ⇒ θ1 = l1 r2 θ2 r1 l2 ⇒ θ1 = 5 (∵ r1 = r2) 45° 3
(∵ Here ratio is given. Hence, θ is not taken in radian.)⇒ θ1 = 5 × 45° = 75° 3
- In the figure ∆ABC is inscribed in a circle with centre O. If∠ABC = 30° then ∠ACB is equal to
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According to given figure in question ,
The angle of a semi–circle is right angle.
∴ ∠BAC = 90°Correct Option: B
According to given figure in question ,
The angle of a semi–circle is right angle.
∴ ∠BAC = 90°
∴ ∠ACB = 90° – ∠ABC
∠ACB = 90° – 30° = 60°
- Two circles of radii 17 cm and 8 cm are concentric. The length of a chord of greater circle which touches the smaller circle is
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As per the given in question , we draw a figure of two concentric circles with centre O,
According to the Question,
OC = radius of smaller circle = 8 cm.
OA = radius of larger circle = 17 cm.
AB = chord of larger circle
OC ⊥ AB
∴ AC = CB
From ∆ OAC,
AC = √OA² - OC² = √17² - 8²Correct Option: C
As per the given in question , we draw a figure of two concentric circles with centre O,
According to the Question,
OC = radius of smaller circle = 8 cm.
OA = radius of larger circle = 17 cm.
AB = chord of larger circle
OC ⊥ AB
∴ AC = CB
From ∆ OAC,
AC = √OA² - OC² = √17² - 8²
AC = √(17 + 8)(17 - 8)
AC = √25 × 9 = 5 × 3 = 15 cm.
∴ AB = 2AC = 30 cm.
- The length of a chord which is at a distance of 5 cm from the centre of a circle of radius 13 cm is :
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On the basis of question we draw a figure of a circle with centre O
Here , OC = 5 cm.
OC ⊥ AB
∴ AC = CB
OA = 13 cm.
In ∆ OAC,
AC = √OA² - OC²Correct Option: B
On the basis of question we draw a figure of a circle with centre O
Here , OC = 5 cm.
OC ⊥ AB
∴ AC = CB
OA = 13 cm.
In ∆ OAC,
AC = √OA² - OC²
AC = √13² - 5²
AC = √169 - 25
AC = √144 = 12 cm.
∴ AB = 2 AC = 2 × 12 = 24 cm.
