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Practice each question, then open the answer or worked solution.
If the radii of two concentric circles are 4 cm and 5 cm, then the length of each chord of one circle, which is tangent to the other circle, is
(A) 3 cm (B) 6 cm
(C) 9 cm (D) 1 cm
According to the question,
OA = 4cm, OB = 5cm
And, OA ⊥ BC
Therefore, OB2 = OA2 + AB2
⇒ 52 = 42 + AB2
⇒ AB = √(25 – 16) = 3cm
⇒ BC = 2AB = 2 × 3cm = 6cm
In Fig. 9.3, if AOB = 125°, then COD is equal to
(A) 62.5° (B) 45°
(C) 35° (D) 55°
ABCD is a quadrilateral circumscribing the circle
We know that, the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle.
So, we have
∠AOB + ∠COD = 180°
125° + ∠COD = 180°
∠COD = 55°
In Fig. 9.4, AB is a chord of the circle and AOC is its diameter such that ACB = 50°. If AT is tangent to the circle at point A, then BAT is equal to
(A) 65° (B) 60°
(C) 50° (D) 40°
According to the question,
A circle with centre O, diameter AC and ∠ACB = 50°
AT is a tangent to the circle at point A
Since, angle in a semicircle is a right angle
∠CBA = 90°
By angle sum property of a triangle,
∠ACB + ∠CAB + ∠CBA = 180°
50° + ∠CAB + 90° = 180°
∠CAB = 40° … (1)
Since tangent to at any point on the circle is perpendicular to the radius through point of contact,
We get,
OA ⏊ AT
∠OAT = 90°
∠OAT + ∠BAT = 90°
∠CAT + ∠BAT = 90°
40° + ∠BAT = 90° [from equation (1)]
∠BAT = 50°
From a point P which is at a distance of 13 cm from the centre O of a circle of radius 5 cm, the pair of tangents PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR is
(A) 60 cm2 (B) 65 cm2
(C) 30 cm2 (D) 32.5 cm2
Construction: Draw a circle of radius 5 cm with centre O.
Let P be a point at a distance of 13 cm from O.
Draw a pair of tangents, PQ and PR.
OQ = OR = radius = 5cm …equation (1)
And OP = 13 cm
We know that, tangent to at any point on the circle is perpendicular to the radius through point of contact,
Hence, we get,
OQ ⏊ PQ and OR ⏊PR
△POQ and △POR are right-angled triangles.
Using Pythagoras Theorem in △PQO,
(Base)2+ (Perpendicular)2= (Hypotenuse)2
(PQ)2 + (OQ)2= (OP)2
(PQ)2 + (5)2 = (13)2
(PQ)2 + 25 = 169
(PQ)2 = 144
PQ = 12 cm
Tangents through an external point to a circle are equal.
So,
PQ = PR = 12 cm … (2)
Therefore, Area of quadrilateral PQRS, A = area of △POQ + area of △POR
Area of right angled triangle = ½ x base x perpendicular
A = (½ x OQ x PQ) + (½ x OR x PR)
A = (½ x 5 x 12) + (½ x 5 x 12)
A = 30 + 30 = 60 cm2
At one end A of a diameter AB of a circle of radius 5 cm, tangent XAY is drawn to the circle. The length of the chord CD parallel to XY and at a distance 8 cm from A is
(A) 4 cm (B) 5 cm
(C) 6 cm (D) 8 cm
According to the question,
Radius of circle, AO=OC = 5cm
AM=8CM
AM=OM+AO
OM =AM-AO
Substituting these values in the equation,
OM= (8-5) =3CM
OM is perpendicular to the chord CD.
In ∆OCM <OMC=90°
By Pythagoras theorem,
OC²=OM²+MC²
Therefore,
CD= 2 ×CM = 8 cm