Circles and Lines Class 10 Extra Questions Kerala State Syllabus Maths Chapter 10

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SSLC Maths Chapter 10 Circles and Lines Important Questions and Answers

Circles and Lines Class 10 Extra Questions Kerala Syllabus

Circles and Lines Class 10 Kerala Syllabus Extra Questions

Question 1.
AB and CD intersect at P. O is the centre of the circle. CD is the diameter of the circle. OP = 2, PA = 9, PB = 5, then what is its radius?

(a) 7
(b) 8
(c) 4
(d) 3
Answer:
(a) 7
(r + 2)(r – 2) = 9 × 5
⇒ r² – 2² = 45
⇒ r² = 49
⇒ r = 7

Question 2.
AB is the diameter of the circle. P divides the diameter of the circle into a and b as shown in the diagram. PC is perpendicular to AB. The point P divides AB in the ratio a : b. The area of the square with side PC is:

(a) a × b
(b) a + b
(c) a / b
(d) ab
Answer:
(a) a × b
a × b = PC²

Question 3.
In the diagram, AB is the diameter of the semicircle, PQ = √14, and RS = √18 are perpendicular to AB. What is the length of the line AB?

(a) 10
(b) 9
(c) 12
(d) 15
Answer:
(b) 9
\(\sqrt{18}=\sqrt{6 \times 3}\)
\(\sqrt{14}=\sqrt{7 \times 2}\)
AB = 9

Question 4.
Read the two statements given below:
In a circle, the length of the chord AB is 16 cm. The length of the perpendicular drawn from the center of the circle to AB is 6 cm.
Statement (1): The radius of the circle is 10 cm.
Statement (2): The perpendicular drawn from the center to the circle to the chord bisects the chord.
Choose the correct answer from those given below.
(a) Statement (1) is true, Statement 2 is false
(b) Statement (2) is true, Statement 1 is false
(c) Both are true. Statement 1 is the reason for Statement 2
(d) Both are true. Statement 1 is not the reason for Statement 2
Answer:
(c) Both are true. Statement 1 is the reason for Statement 2.

Question 5.
Read the two statements given below:
In a circle, two chords AB and CD meet at a point P inside the circle. AP = 4 cm, PB = 6 cm, CP = 3 cm, PD = ?
Statement (1): PD = 8 cm
Statement (2): If two chords intersect inside a circle, then AP × PB = CP × PD
Choose the correct answer from those given below.
(a) Statement 1 is correct, Statement 2 is incorrect
(b) Statement 2 is correct, Statement 1 is incorrect
(c) Both are correct. Statement A is the reason for Statement 2
(d) Both are correct. Statement 1 is not the reason for Statement 2
Answer:
(c) Both are correct. Statement 1 is the reason for Statement 2.

Question 6.
In the diagram, a line passing through the center of a circle divides a chord into two. What is the radius of the circle?

Answer:
If we extend the line OP that intersects the two ends of the circle, we get another chord CD.

AB and CD are two chords that meet at the point P.
Therefore, PA × PB = PC × PD.
That is, 4 × 6 = (r + 5) × (r – 5)
⇒ 24 = r² – 5²
⇒ 24 = r² – 25
⇒ 24 + 25 = r²
⇒ 49 = r²
⇒ r = ±7
r is the radius of the circle; it will always be a positive number.
⇒ r = 7 cm

Question 7.
In the diagram, a line drawn from the center of a circle intersects a chord of the circle. What are the lengths of the two parts of the chord?

Answer:

If we extend the line OP that intersects the two ends of the circle, we get a chord CD.
AB and CD are two chords that meet at the point P.
So, AP × PB = CP × PD
AP × PB = 4 × 10
AP × PB = 40
Also AP + PB = 13 cm
The product of two numbers is 40, and their sum is 13 is 8 and 5.
Therefore AP = 5 cm, PB = 8 cm

Question 8.
The chords AB and CD intersect at point P. AB = 17 cm, PA = 9 cm, PD = 12 cm.

(a) What is the length of PB?
(b) Calculate the length of PC?
Answer:
(a) AB = PA + PB
⇒ 17 = 9 + PB
⇒ PB = 17 – 9 = 8 cm

(b) PA × PB = PC × PD
⇒ 9 × 8 = PC × 12
⇒ PC = 6 cm

Question 9.

(a) In the diagram, AB is the diameter of the semicircle. PC is perpendicular to AB. If AP = 5 cm, PB = 3 cm, calculate the length of PC.
(b) Draw a square with an area of 15 square centimeters.
Answer:
(a) AP × PB = PC × PC
⇒ 5 × 3 = PC × PC
⇒ 15 = PC²
⇒ PC = √15
(b)

Question 10.
Chords AB and CD intersect at point P. AB = 13 cm, CD = 15 cm, PD = 3 cm.

(a) What is the length of PC?
(b) If PB = x, find PA?
(c) What is the length of PB?
Answer:
CD = PC + PD
⇒ 15 = PC + 3
⇒ PC = 15 – 3 = 12
(a) AB = PA + PB
⇒ 13 = PA + x
⇒ PA = 13 – x

(b) PC × PD = PA × PB
⇒ 12 × 3 = (13 – x) × x
⇒ 36 = 13x – x²
⇒ x² – 13x + 36 = 0
⇒ x = 4 or x = 9
Therefore PB = 4 cm

Question 11.
Draw a rectangle with an area of 24 square centimeters. Draw a square of area equal to the area of this rectangle.
Answer:
Draw a rectangle with length 6 cm and height 4 cm.
Add the height of the rectangle to the length of the base.
The new length is 6 + 4 = 10 cm.
Now draw a semicircle with the bottom line as the diameter, extend the right side of the square, and meet it with the semicircle.
This line is the side of the square.
BSTF is the required square.

Question 12.
In the diagram, AB is the diameter of the semicircle, and PC is the perpendicular to AB. AC = 5√29 cm, PA = 25 cm.

(a) What is the length of PC?
(b) What is the length of PB?
(c) What is the radius of the circle?
Answer:
(a) PC = \(\sqrt{(5 \sqrt{29})^2-25^2}\) = 10 cm

(b) PA × PB = PC²
⇒ 25 × PB = 10²
⇒ PB = 4 cm

(c) AB = 25 + 4 = 29
Radius = 14.5 cm

Question 13.
If the chords AB and CD are extended, they intersect at P outside the circle. If PA = PC, then prove AB = CD.

Answer:
PA × PB = PC × PD
PA and PC on both sides
⇒ PB = PD,
⇒ PA – AB = PC – CD
PA and PC on both sides
⇒ -AB = -CD
⇒ AB = CD

Question 14.
The chords AB and CD intersect at a point inside a circle at P as shown in the figure.

(a) Draw a diagram. Join AC and BD. Establish the similarity of triangles PAC and PBD and the relationship between PA, PB, PC, and PD.
(b) PA = 3 cm, PB = 5 cm. What is the area of the rectangle with sides PC and PD?
(c) Draw a rectangle with sides of length 5 cm and 3 cm. Draw another rectangle with one side 6 cm and an area equal to the area of the first rectangle.
Answer:
(a) Consider ∆PCA and ∆PBD
∠C = ∠B, ∠A = ∠D (Angles at the end of an arc are equal)
\(\frac{P A}{P D}=\frac{P C}{P B}\)
PA × PB = PC × PD

(b) 3 × 5 = PC × PD
PC × PD = 15
The area of the rectangle with sides PC, PD is 15 sq.cm

(c) Consider a rectangle with length 5 cm and width 3 cm.
Consider another length of 6 cm.
In the square, we can draw a line extending 3 cm from the bottom edge to the left side and 6 cm from the left edge to the bottom:
Now draw a circle through the left, right, and bottom corners.
Draw a line through the rectangle where the left side intersects the circle.
Now, let’s draw the required square by marking the length of the rectangle we have obtained.

Here, ABCD is the required rectangle.

Question 15.
See the diagram below. AB is the diameter ofthe  semicircle, and PC is perpendicular to AB. You can see the triangles ACB, PCB.

(a) ∆ACB and ∆PCB are similar triangles, and prove PA × PB = PC².
(b) Construct a diagram such that PC = √12 cm, and a square of area 12 sq.cm
(c) Draw a chord of length √48 cm in a circle in your construction.
Answer:
Draw AC and BC.
Consider ∆ACB and ∆ACP
In triangle ACB, ∠C = 90°
(a) If ∠PAC = x, then ∠ACP = 90 – x, ∠BCP = x, ∠PBC = 90 – x
The two angles of the triangle ABC are equal to the two angles of the triangle ACP.
The triangle ABC and the triangle ACP are similar triangles.
Sides opposite to equal angles are proportional.
\(\frac{P A}{P C}=\frac{P C}{P B}\)
⇒ PA × PB = PC²

(b) \(\sqrt{12}=\sqrt{6 \times 2}\)
Draw a semicircle with diameter AB = 6 + 2 = 8 cm
Mark the point P.
Length of PC = √12

Complete the square with side PC.
The area of the square will be (√12)² = 12

(c) Draw the circle.
Extend the line CP to D.
Length of the chord CD is √48.

CD = 2 × √12 = √48.