CRE 2 - Tangents & Secants | Geometry - Circles
Join our Telegram Channel for CAT/MBA Preparation.
1. CRE 2 - Tangents & Secants | Geometry - Circles
Two chords AB and CD of a circle intersect at a point P, which lies outside the circle. If PA = 6 cm, PB = 12 cm and PC = 3 cm, what is the length of CD in cm?
(a) 21
(b) 24
(c) 6
(d) 12
Answer: Option
A
Join our Telegram Channel for CAT/MBA Preparation.
Explanation :
For a circle, if two chords AB and CD intersect at a single point P outside the circle, then,
PA × PB = PC × PD
∴ 6 × 12 = 3 × PD
∴ PD = 24 cm
∴ CD = PD – PC = 24 – 3 = 21 cm.
Hence, option (a).
2. CRE 2 - Tangents & Secants | Geometry - Circles
In the given figure, PBQ is a straight line while PA, PB, QB, and QC are tangents to the circle with center O. Find the value of angle AOC.
(a) 70°
(b) 80°
(c) 60°
(d) 100°
(e) None of these
Answer: Option
E
Join our Telegram Channel for CAT/MBA Preparation.
Explanation :
In ∆PAB,
PA = PB …(tangents from an external point to a circle are congruent)
∴ m ∠PBA = m ∠PAB; …(angles opposite to congruent sides)
∴ m ∠PBA = 60° …(sum of angles in a triangle is 180°)
Similarly, in ∆QBC
m ∠CBQ = m ∠BCQ = 50°
∵ m ∠PBA + m ∠CBQ + x = 180° …(linear angles)
∴ x = 180° – 60° – 50° = 70°
Now, m ∠AOC = 2 × x
∴ m ∠AOC = 140°
Hence, option (e).
3. CRE 2 - Tangents & Secants | Geometry - Circles
In the figure, PA and PC are tangents to the circle ABC. If ∠P = 48°, then ∠ABC = ?
Answer: 114
Join our Telegram Channel for CAT/MBA Preparation.
Explanation :
Draw two lines joining A and C with center O.
Now in quadrilateral AOCP, sum of all four angles should be 360°.
Also, ∠OAP = ∠OCP = 90° (radius is line joining center and point of tangency is perpendicular to the tangent)
∴ 90° + ∠AOC + 90° + 48° = 360°
⇒ ∠AOC = 132°
Angle subtended by chord AC in the major segment is half of the angle it subtends at the center.
∴ ∠ADC = ½ × ∠AOC = 66°
Now, angles subtended by a chord in different segments are supplementary.
∴ ∠ADC + ∠ABC = 180°
⇒ ∠ABC = 180° - 66° = 114°.
Hence, 114.
4. CRE 2 - Tangents & Secants | Geometry - Circles
In the following figure, lines AP, AQ and BC are tangent to the circle. The length of AP = 22. Find the perimeter of triangle ABC.
Answer: 44
Join our Telegram Channel for CAT/MBA Preparation.
Explanation :
We know 2 tangents drawn from an external point to a circle are equal in length.
⇒ AP = AQ = 22
∴ If we consider point B as the external point, we have two tangents i.e., BP and BD.
⇒ BP = BD
Similarly, if we consider point C as the external point, we have two tangents i.e., CD and CQ.
⇒ CD = CQ
Now, Perimeter of ∆ABC = AB + BD + CD + AC
= AB + BP + CQ + AC
= AP + AQ
= 22 + 22 = 44
Hence, 44.
5. CRE 2 - Tangents & Secants | Geometry - Circles
In the following figure, if PC = 6, CD = 9, PA = 5 and AB = x, find the value of x?
Answer: 13
Join our Telegram Channel for CAT/MBA Preparation.
Explanation :
When two secants are drawn from an external point to a circle
⇒ PC × PD = PA × PB
∴ 6 × 15 = 5 × (5 + AB)
⇒ AB = 13
Hence, 13.
6. CRE 2 - Tangents & Secants | Geometry - Circles
In the following figure, PC = 9, PB = 12, PA = 18, and PF = 8. Then, find the length of DE.
Answer: 2.5
Join our Telegram Channel for CAT/MBA Preparation.
Explanation :
If we consider the smaller circle, two secants are drawn from P.
∴ PC × PB = PF × PE
⇒ 9 × 12 = 8 × PE
⇒ PE = 27/2
Now, if we consider the bigger circle, two secants are drawn from P.
∴ PB × PA = PE × PD
⇒ 12 × 18 = 27/2 × PD
⇒ PD = 16
∴ DE = PD – PE = 16 – 27/2 = 16 – 13.5 = 2.5
Hence, 2.5.
7. CRE 2 - Tangents & Secants | Geometry - Circles
In the figure given below (not drawn to scale), A, B and C are three points on a circle with center O. The chord BA is extended to a point T such that CT becomes a tangent to the circle at point C. If ∠ATC = 30° and ∠ACT = 50°, then ∠BOA is (in degrees)?
Answer: 100
Join our Telegram Channel for CAT/MBA Preparation.
Explanation :
Given, ∠ATC = 30° and ∠ACT = 50°
In ∆ACT, external angle ,
∠CAB = ∠ACT + ∠ATC = 50° + 30° = 80° …(1)
Also, chord CA and tangent CT make an angle of 50°, hence ∠CBA = 50°.
[Angle made by a chord and tangent is same as the angle subtended by the chord in opposite segment.]
Now, in ∆ABC, sum of all angles = 180°
⇒ ∠CBA + ∠BAC + ∠ACB = 180°
⇒ 50° + 30° + ∠ACB = 180°
⇒ ∠ACB = 100°
Now, angle make by a chord in major segment is half the angle it subtends at the center.
∴ ∠AOB = 2 × ∠ACB
⇒ ∠AOB 2 × 100° = 200°
Hence, 100.
8. CRE 2 - Tangents & Secants | Geometry - Circles
In the following figure, if PT = 6 and AB = 5, find PB?
Answer: 9
Join our Telegram Channel for CAT/MBA Preparation.
Explanation :
Let PA = x
When a secant and a tangent are drawn from an external point to a circle
⇒ PA × PB = PT2
∴ x × (x + 5) = 62
⇒ x2 + 5x – 36 = 0
⇒ (x + 9)(x - 4) = 0
∴ x = -9 or 4. (x = -9 is rejected)
∴ PB = x + 5 = 9.
Hence, 9.
9. CRE 2 - Tangents & Secants | Geometry - Circles
Two circles of radius 4 and 6 cm and centers P and Q respectively, touch each other externally. From the center of first circle a tangent PR is drawn to second circle which touches the second circle at R. Find PR (in cm)?
[Type in your answer as the nearest possible integer]
Answer: 8
Join our Telegram Channel for CAT/MBA Preparation.
Explanation :
We can draw the following figure from the given information.
In ∆PQR,∠PRQ = 90° (tangent makes an angle of 90° with the radius.)
∴ ∆PQR is a right triangle with PQ as the hypotenuse
⇒ PQ2 = PR2 + QR2
⇒ 102 = PR2 + 62
⇒ PR2 = 100 - 36 = 64
⇒ PR = 8
Hence, 8.
10. CRE 2 - Tangents & Secants | Geometry - Circles
PQ is the chord of a circle whose center is O. ROS is a line segment originating from a point R on the circle that intersect PQ produced at point S such that QS = OR. If ∠QSR = 30°, then what is the value (in degrees) of POR?
(a) 30
(b) 45
(c) 60
(d) 90
Answer: Option
D
Join our Telegram Channel for CAT/MBA Preparation.
Explanation :
We can draw the following figure from the given information.
Since OR = OP = OQ = OS (radius)
⇒ ∠OPQ = ∠OQP & ∠QSO = ∠QOS
⇒ ∠POQ = 60°
⇒ ∠POR = 180 - (60° + 30°) = 90°
Hence, option (d).
11. CRE 2 - Tangents & Secants | Geometry - Circles
In the figure given below, O is the center of the circle. OQ is perpendicular to RS and ∠SRT = 30°. If RS = 10√2, then what is the value of PR2 ?
(a) 200(1 + √3)
(b) 300(2 + √3)
(c) 200(2 + √3)
(d) 100(3 + 2√3)
Answer: Option
C
Join our Telegram Channel for CAT/MBA Preparation.
Explanation :
RQ = QS = 1/2 × 10√2 = 5√2
∴ ∠ORQ = 90 - 30 = 60°
∆ORQ is a 30°-60°-90° triangle
∴ Hypotenuse will be twice the side opposite to 30°
In ∆PRQ,2 = PQ2 + RQ2 2 = (10√2 + 5√6)2 + (5√2)2 2 = 200 + 150 + 100√12 + 502 = 400 + 200√3 = 200(2 + √3)
Hence, option (c).
Solution
Discuss
Report
