Circles and Lines Questions and Answers Class 10 Maths Chapter 10 Kerala Syllabus Solutions

Students often refer to Kerala Syllabus 10th Standard Maths Textbook Solutions Chapter 10 Circles and Lines Questions and Answers Notes Pdf to clear their doubts.

SSLC Maths Chapter 10 Circles and Lines Questions and Answers

Circles and Lines Class 10 Questions and Answers Kerala State Syllabus

SCERT Class 10 Maths Chapter 10 Circles and Lines Solutions

Class 10 Maths Chapter 10 Kerala Syllabus – Areas

(Textbook Page No. 216)

Question 1.
Draw a rectangle with a length of 5 cm and a height of 4 cm.
(i) Draw a rectangle with a length of 6 cm.
(ii) Draw a square with a length of 5 cm.
Answer:
(i) Consider a rectangle with length 5 cm and breadth 4 cm.
Let us also assume that the other side is 6 cm long.
From the original rectangle, we can draw a length of 4 cm from the bottom side to the left and a length of 6 cm from the left bottom side to the bottom side.
Now draw a circle through the left, right, and bottom points.
From the rectangle, draw a line where the left side intersects the circle.
Now, we can draw the specified rectangle by marking the length obtained here.
The rectangle we need is ABCD.

(ii) Consider a rectangle with length 5 cm and breadth 4 cm.
First, extend the length of the original rectangle by the measurement of height.
The new length is 5 + 4 = 9 cm.
Now draw a semicircle as the diameter of the lower line, extend the right side of the rectangle to the right, and meet the semicircle.
This line is the side of the square.
The square we need is ABCD.

Question 2.
Draw a square with an area of 15 square centimeters.
Answer:
15 = 5 × 3
Draw a line of length 3 + 5 = 8 cm.
Mark 5 cm and 3 cm on it.
Draw a circle with a diameter of 8 cm.
Draw a perpendicular from the points 5 cm and 3 cm.
Mark the point where this perpendicular intersects the circle as AB.
ABCD is the required square.

Question 3.
Draw a square with an area of 5 square centimeters in three equal parts (Hint: Recall Pythagoras’ Theorem).
Answer:
Method 1
5 = 1 × 5
1 + 5 = 6 cm.
Draw a line of length 5 cm.
Mark 5 cm and 1 cm.
Draw a semicircle with a diameter of length 6 cm.
Draw a perpendicular from the point of intersection of 5 cm and 1 cm.
Mark the point where this perpendicular intersects the semicircle as AB.
ABCD is the required square.

Method 2
5 = 2.5 × 2
Continue the above process from here.
Method 3
Draw a right-angled triangle with 2 centimeters and 1 centimeter as perpendicular sides.
√3 will be the diagonal.

Draw a square by taking this diagonal as the length of the sides of the square, and that is the required square.

SCERT Class 10 Maths Chapter 1 Solutions – Line and Point

(Textbook Page No. 222-224)

Question 1.
In the picture, a line from the centre of a circle cuts a chord into two parts:

What is the radius of the circle?
Answer:

PX = 3 cm, PY = 1 cm, d = 2 cm
PX × PY = r² – d²
⇒ 3 × 1 = r² – 2²
⇒ 3 = r² – 4
⇒ r² = 3 + 4 = 7
⇒ r = √7
⇒ r = 2.645

Question 2.
In the picture, a line from the centre of a circle meets a chord:

Find 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
OA = OC = 3 cm
PD = 3 – 2 = 1 cm
CP = 3 + 2 = 5 cm
AP × PB = 5 × 1
AP × PB = 5 cm
Also, AP + PB = 4.5 cm
Two numbers that give 5 as their product and the sum as 4.5 are 2.5 and 2.
Which means if we take, a + b = 4.5 and ab = 5 then,
x² – (a + b)x + ab = 0
x² – 4.5x + 5 = 0
If so, D = 4.5² – 4 × 1 × 5
= 20.25 – 20
= 0.25
√D = √0.25 = 0.5
x = \(\frac{-b \pm \sqrt{D}}{2 a}\)
⇒ x = \(\frac{4.5 \pm 0.5}{2}\)
⇒ x = 2.5 or x = 2
⇒ AP = 2 cm and PB = 2.5 cm

Question 3.
In the picture, AB is a diameter of the circle, and it is extended to a point P. The tangent from P touches the circle at Q:

What is the radius of the circle?
Answer:

Considering right right-angled triangle ∆OPQ
QP = 4 cm, OQ = r
AP = 8 cm
OP = 8 – r cm
According to Pythagoras’ theorem,
r² + 4² = (8-r)²
⇒ r² + 16 = 64 – 16r + r²
⇒ 16r = 64 – 16
⇒ 16r = 48
⇒ r = 3 cm
The radius of the circle = 3 cm.

Question 4.
In the first of the two pictures below, the line joining two points of a circle is extended outward to a point, and then a tangent is drawn from this point to the circle.

In the second picture, the same line is extended a bit more to a point, and a tangent is drawn from this point:

What is the length of this tangent?
Answer:

In picture 1:
PX × PY = PT²
PX = 4 cm, PT = 6 cm
⇒ 4 × PY = 6²
⇒ PY = 9 cm
XY = PY – PX
= 9 – 4
= 5 cm
Then, XY = 5 cm
In picture 2:
RX × RY = RT²
RX = 5 cm, XY = 5 cm (in picture 1)
RY = RX + XY
= 5 + 5
= 10 cm
⇒ 5 × 10 = RT²
⇒ 50 = RT²
⇒ RT = √50 = 7.07
Then the length of the tangent = RT = √50 = 7.07

Question 5.
In the picture, the line joining the points of intersection of two circles is drawn, and from a point on it, one tangent is drawn to each circle:

Prove that the lengths of these tangents are equal.
Answer:

The point of intersection of two circles can be called X and Y.
In this picture, the tangent of the first circle can be marked as PQ, and the tangent of the second circle can be marked as PT.
(If a line is drawn from a point outside a circle and intersects two points on the circle, the product of the distances between the point outside the circle to the points of intersection is the square of the length of the tangent to that point.)
Which means PX × PY = PT²
If so,
Picture 1: PX × PY = PT²
Picture 2: PX × PY = PQ²
⇒ PT² = PX × PY = PQ²
⇒ PT² = PQ²
⇒ PT = PQ
Which means both the tangents have the same length.

Circles and Lines Class 10 Notes Pdf

Class 10 Maths Chapter 10 Circles and Lines Notes Kerala Syllabus

Introduction
A circle is a basic geometric shape. From the pizza on our plates to the night sky to the moon, we see circles everywhere. When you study circles in high school geometry, you learn that a circle is defined as the set of all the points on a plane that are a fixed distance from a fixed point called the center. This fixed distance is called the radius. You may already be familiar with basic terms related to circles, such as diameter, perimeter, and area. This chapter will focus more on the relationships and theorems that apply to circles than on introducing new definitions.

→ Two chords that are not the diameter intersecting inside a circle, the two parts are not equal.

→ When two chords of a circle intersect within the circle, the product of the parts of one chord is equal to the product of the parts of the other.

→ Since the diameter is perpendicular to the chord from the centre of the circle, it will equally divide the chord.

→ The product of the parts into which a perpendicular chord cuts a diameter of a circle is the square of half the chord.

→ When two chords intersect within a circle, the rectangles with sides as the parts of each chord have equal areas.

→ For any chord XY of a circle of radius r, passing through a point P within the circle at a distance d from the centre, PX × PY = r² – d²

→ For all lines through a point outside a circle intersecting the circle at two points, the product of the distances from the point to the points of intersection with the circle is the same number.

→ If a line from a point P outside a circle of radius r at a distance d from the center cuts the circle at X and Y, then PX × PY = d² – r²

→ If X and Y are the points of intersection of a circle of radius r with a line through a point P inside or outside the circle, at a distance d from the centre, then PX × PY = |r² – d²|.

Chords
Any two diameters of a circle intersect at the centre of the circle.
All four parts made by the intersection are equal to the radius of the circle.

When two chords that are not diameters intersect each other, those parts are not equal.

If the lengths of the parts of a chord are a, b, and the lengths of the parts of another chord are c, d; if we denote the chords AB, CD, and the point where they intersect P, we get the relationship between them. That is, AP × PB = CP × PD.

When two chords of a circle intersect within the circle, the product of the parts of one chord is equal to the product of the parts of the other.

Question 1.
Find the length of CP from the picture given below?
Answer:
AP = 3 cm
PB = 5 cm
PD = 5 cm
AP × PB = CP × PD
⇒ 3 × 5 = CP × 5
⇒ 15 = 5PC
⇒ PC = 3 cm
If one of the chords intersecting each other is the diameter of the circle, and the other is a chord perpendicular to it:
Since the diameter is the perpendicular from the center of the circle to a chord, it bisects the chord.
If the lengths of the parts of the diameter are a and b, and if the length of the two segments intersecting the chord is c, then according to the general principle, ab = c2.

The product of the parts into which a perpendicular chord cuts a diameter of a circle, is the square of half the chord.

Question 1.
Look at the picture:

A perpendicular from a point on the diameter meets the semicircle. What is the length of this perpendicular?
Answer:
If we draw the complete circle and extend the perpendicular downwards, it becomes a chord, and the perpendicular is half the chord.

That is, ab = c²
3 × 2 = 6
Length of the perpendicular = √6 cm

Question 2.
How do we draw a line of length √5 centimetres?
Can you draw a line of length √6 centimetres in any other way?
Answer:
To draw a line that is √5 centimeters long, the square root of the perpendicular is 1 × 5 = 5
The length of the perpendicular is √5 centimeters
To draw a line that is √6 centimeters long, the square root of the perpendicular is 1 × 6 = 6
The length of the perpendicular is √6 centimeters
The square root of any number can be represented by drawing a line segment.

Areas
When two chords intersect within a circle, the rectangles with sides as the parts of each chord have equal areas.
To draw a rectangle with the same area but a different length than the perimeter of a rectangle:
Consider a rectangle with length a and width b.

Consider another line c.

The base of the rectangle can be drawn by extending the length b from the bottom to the left, and extending the length c from the top to the bottom:

Now, draw a circle through the left, right, and bottom points, and extend the left side of the rectangle to the top till it meets the circle.

Now the length we thus get can be marked horizontally to draw the rectangle we want.

To draw such rectangles, we do not need to know the lengths of the sides of the original rectangle; only how much a side has to be lengthened or shortened.

To draw a square with the same area as the rectangle:
Consider a rectangle with length a and height b.

Extend the base length with the height of the first rectangle; the new length is a + b.

Now draw a semicircle with the bottom line as the diameter, extend the right side of the rectangle to the bottom, and meet it with the semicircle.

This line is the side of the square.

This method can be used to draw a square of specified area.

To draw a line of length \(\sqrt{a b}\):
At first, draw a line of length ‘a’.
Draw another line of length ‘b’ from one end of the line ‘a’.
Draw a perpendicular from the point where the two lines intersect to the semicircle.
The length of that perpendicular will be \(\sqrt{a b}\).

Line and Point
It is possible to draw some chords that pass through a single point inside the circle.

The point divides each chord into two parts. The product of the lengths of each pair of these parts gives the same number.

If we draw a diameter through this point, it is also a chord.
The product of the lengths of the parts into which the points divide the diameter is equal to this number. So, if we denote the radius of the circle as r and the distance of the point from the centre of the circle as d, we can write the lengths of the parts of the diameter as r + d and r – d.

Then the product of the these parts, (r + d)(r – d) = r² – d²
If so, the products of the lengths of the parts of every such chord are PX × PY = r² – d²

For any chord XY of a circle of radius r, passing through a point P within the circle at a distance d from the centre, PX × PY = r² – d².
r² – d² is equal to the square of half the chord through P, perpendicular to the diameter through P.

If we take a point outside the circle and draw lines intersecting the circle:

For all lines through a point outside a circle intersecting the circle at two points, the product of the distances from the point to the points of intersection with the circle is the same number.
PA × PB = PC × PD

When a line that starts from a point outside the circle intersects the circle in two points that pass through the centre of the circle.
Let the radius of the circle be r, and the distance from the point P to the centre be d.

Here, the distances from the point to the points of intersection with the circle can be seen as d – r and d+r; and so their product is (d+r)(d – r) = d² – r²
If we take any line that is drawn through this point, the product of the distances from the point to the points of intersection with the circle would be d² – r².

PX × PY = d² – r²

If a line from a point P outside a circle of radius r at a distance d from the center cuts the circle at X and Y, then PX × PY = d² – r²

Draw the tangent from P to the circle and mark the point of contact as T. Joining T and P with the center of the circle, we get a triangle:
Which means, d² – r² = PT²

If so, whatever line we draw from a point outside a circle, intersecting the circle at two points, the product of the distances from the point to the points of intersection is the square of the length of the tangent from the point.
Which means, PX × PY = PT²

Whether the point is inside or outside the circle, what we do here is to subtract the larger of r² and d² from the smaller; thus, in both cases, we take the absolute value of r² – d².

If X and Y are the points of intersection of a circle of radius r with a line through a point P inside or outside the circle, at a distance d from the centre, then PX × PY = |r² – d²|.