Plane Geometry
- In the figure below, AB is a chord of a circle with centre O. A tangent AT is drawn at point A so that ∠BAT=50°. Then ∠ADB=?
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As per the given in question ,
∴ Here , ∠BAT = 50°
OA = OB = radii
∠OAT = 90°
∴ ∠OAB = ∠OBA = 90° – 50° = 40°
∴ ∠AOB = 100°Correct Option: B
As per the given in question ,
∴ Here , ∠BAT = 50°
OA = OB = radii
∠OAT = 90°
∴ ∠OAB = ∠OBA = 90° – 50° = 40°
∴ ∠AOB = 100°∴ ∠AEB = 100° = 50° 2
ADBE is a cyclic quadrilateral.
∴ ∠ADB = 180° – 50° = 130°
- Two concentric circles are drawn with radii 12 cm and 13 cm. What will be the length of any chord of the larger circle that is tangent to the smaller circle?
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According to question , we draw a figure of two concentric circles with centre O
AB is tangent at point C.
AC = BC
OA = 13 cm.
OC = 12 cm.
∠ACO = 90°
From ∆ OAC,
AC = √OA² - OC²Correct Option: C
According to question , we draw a figure of two concentric circles with centre O
AB is tangent at point C.
AC = BC
OA = 13 cm.
OC = 12 cm.
∠ACO = 90°
From ∆ OAC,
AC = √OA² - OC²
AC = √13² - 12²
AC = √(13 + 12)(13 -12) = √25 = 5 cm.
∴ AB = 2AC = 2 × 5 = 10 cm.
- O is the centre of a circle and AB is the tangent to it touching at B. If OB = 3 cm. and OA = 5 cm, then the measure of AB in cm is
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On the basis of question we draw a figure of a circle with centre O
Given that , OB = 3 cm. and OA = 5 cm
∠OBA = 90°
In ∆ OAB, OA² = OB² + AB²
⇒ 5² = 3² + AB²Correct Option: D
On the basis of question we draw a figure of a circle with centre O
Given that , OB = 3 cm. and OA = 5 cm
∠OBA = 90°
In ∆ OAB, OA² = OB² + AB²
⇒ 5² = 3² + AB²
⇒ AB² = 5² – 3² = 25 – 9 = 16
⇒ AB = √16 = 4 cm.
- 2 equal tangents PA and PB are drawn from an external point P on a circle with centre O. What is the length of each tangent, if P is 12 cm from the centre and the angle between the tangents is 120°?
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As per the given in question , we draw a figure of a circle with centre O
∠APB = 120°
OA = OB = radii
∴ ∠APO = ∠OPB = 60°
∴ From ∆ OAP,cos 60° = AP OP
Correct Option: B
As per the given in question , we draw a figure of a circle with centre O
∠APB = 120°
OA = OB = radii
∴ ∠APO = ∠OPB = 60°
∴ From ∆ OAP,cos 60° = AP OP ⇒ 1 = AP ⇒ AP = 6cm. 2 12
- A and B are the centres of two circles with radii 11 cm and 6 cm respectively. A common tangent touches these circles at P and Q respectively. If AB = 13 cm., then the length of PQ is
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As per the given in question , we draw a figure of two circles with radii 11 cm and 6 cm respectively
Here , AB = 13 cm.
PQ = √(AB)² - (r1 - r2)²Correct Option: D
As per the given in question , we draw a figure of two circles with radii 11 cm and 6 cm respectively
Here , AB = 13 cm.
PQ = √(AB)² - (r1 - r2)²
PQ = √(13)² - (11 - 6)²
PQ = √169 -25 = √144 = 12 cm.
