Tangents Questions and Answers Class 10 Maths Chapter 8 Kerala Syllabus Solutions

Students often refer to Kerala Syllabus 10th Standard Maths Textbook Solutions Chapter 8 Tangents Questions and Answers Notes Pdf to clear their doubts.

SSLC Maths Chapter 8 Tangents Questions and Answers

Tangents Class 10 Questions and Answers Kerala State Syllabus

SCERT Class 10 Maths Chapter 8 Tangents Solutions

Class 10 Maths Chapter 8 Kerala Syllabus – Lines and Circles

(Textbook Page No. 165)

Question 1.
In each of the two pictures below, a tangent to the circle, the radius through the point of contact, and another line through the centre are drawn to make a triangle:

Draw these in your notebook.
Answer:
Construction of Figure 1:
Draw a circle of radius 2.5 cm. At a point on the circle, draw a perpendicular to the radius. This line will be the tangent.
Consider the centre of the circle and draw an arc of radius 5 cm that cuts the tangent from the centre.
A triangle is formed.

Construction of Figure 2:
Draw a line segment of length 2 cm slanting as shown in the figure. Draw a perpendicular from the upper end of this line.
Considering the lower end as the centre, draw an arc of radius 4 cm to cut the perpendicular drawn earlier.
Mark the point of intersection as O.
With O as the centre and consider the distance from O to the upper end of the 2 cm line segment as the radius, and draw the circle.

Question 2.
Prove that the tangents drawn at the two ends of a diameter of a circle are parallel.
Answer:

From the figure,
The diameter is perpendicular to the tangent.
The angles formed between the tangent and the ends of the diameter are each 90°.
The perpendiculars drawn to a line through its endpoints are parallel to each other.

Question 3.
If the tangents are drawn to a circle at the ends of a pair of perpendicular diameters, what kind of quadrilateral do they form?
Answer:

From the figure, it forms a square.

SCERT Class 10 Maths Chapter 8 Solutions – Tangents and Angles

(Textbook Page No. 168-169)

Question 1.
Draw a circle of radius 2.5 centimetres. Draw a triangle of angles 40°, 60°, 80°, with its sides touching the circle.
Answer:
Draw a circle.
From the centre, draw radii that divide the angles around the centre into parts measuring 2 × 40° = 80° and 2 × 60° = 120°.
Through the endpoints of those radii, draw tangents to the circle.
The tangents will intersect to form the required triangle of angles 40°, 60°, and 80°.

Question 2.
In the picture, the small (blue) triangle is equilateral. The sides of the larger (green) triangle are tangents at the vertices of the smaller triangle to its circumcircle.

(i) Prove that the larger triangle is also equilateral and its sides are double those of the smaller triangle.
(ii) Draw this picture with the sides of the smaller triangle as 3 centimetres.
Answer:
(i) Construction:

Draw a circle with centre O.
Draw the radii OA, OB, and OC.
∠AOB = 2 × 60 = 120°.
AOBP is a cyclic quadrilateral.
Therefore, ∠P = 180 – 120 = 60°.
Similarly, we get ∠Q = ∠R = 60°.
Triangle PQR is an equilateral triangle.
ABCQ and PBCA are parallelograms.
∠B = ∠Q = 60° and ∠C = ∠P = 60°
BC = AQ = AP.
That means PQ = 2 × BC
Similarly, ABRC is a parallelogram.
AC = PB = BR.
That means PR = 2 × AC and QR = 2 × AB.

(ii) Construction:
Draw an equilateral triangle with sides 3 cm.
Draw the perpendicular bisectors of its sides.
The intersection point of the bisectors is the centre; take that point as the centre and draw the circumcircle of the triangle.
Through the corners, draw tangents to the circle.

Question 3.
The picture shows two tangents to a circle and the radii through the points of contact:

(i) Prove that the lengths of the tangents are equal.
(ii) Prove that the line joining the centre of the circle to the point of intersection of the tangents bisects the angle between the radii.

(iii) Prove that this line is the perpendicular bisector of the chord joining the points of contact.
Answer:
(i) ∆PAO and ∆PBO are right-angled triangles.

∠A = ∠B = 90°
PA² + OA² = OP², PB² + OB² = OP²
PA² + OA² = PB² + OB²
OA = OB (radii of the circle)
PA² = PB²
⇒ PA = PB

(ii) PA = PB, OA = OB and OP is the common side.
The angles opposite to the equal sides are equal.
Hence, the angles opposite to OA and OB are equal.
That is ∠APO = ∠BPO

(iii) Consider the triangles PAQ and PBQ:

PA = PB, PQ is the common side, and the included angles are equal.
Hence, the sides opposite equal angles are equal, i.e,. AQ = BQ.
Also, ∠PQA + ∠PQB = 180°.
So, ∠PQA = ∠PQB = 90°.
Therefore, the line OP is the perpendicular bisector of the chord AB.

Question 4.
The picture shows a rhombus with each side tangent to a circle within it:

Draw this in your notebook.
Answer:
Draw a circle with a radius of 4 cm.
Draw diameters so that the angle between them is 180 – 40 = 140°
Through the endpoints of those diameters, draw tangents to the circle.
A rhombus is formed.

Class 10 Maths Kerala Syllabus Chapter 8 Solutions – Chord and Tangent

(Textbook Page No. 176-177)

Question 1.
In the picture, the sides of the larger triangle are the tangents to the circumference of the smaller triangle at its vertices:

Calculate the angles of the larger triangle.
Answer:
∠QAB = ∠QBA = 60°
∠Q = 180 – 120 = 60°
∠PAC = ∠PCA = 80°
∠P = 180 – 160 = 20°
∠R = 180 – (60 + 20) = 100°

Question 2.
In the picture, the sides of the larger triangle are tangents to the circle. Their points of contact are the vertices of the smaller triangle:

Calculate the angles of the larger triangle.
Answer:

In ∆APQ, ∠P = ∠Q = 60°, ∠R = 60°
In ∆CPR, ∠P = ∠R = 70°, ∠Q = 70°
In ∆PQR, ∠P = 180 – (60 + 70) = 50°

Question 3.
In the picture, PQ, RS, TU are tangents to the circumcircle of triangle ABC at its vertices.

Sort out the equal angles in the picture.
Answer:
∠UCB = ∠RBC = ∠CAB
∠TCA = ∠PAC = ∠CBA
∠BAQ = ∠ABS = ∠ACB

Question 4.
In each picture below, a tangent is drawn to the circumcircle of a regular polygon, at a vertex:

In each picture, calculate the angles between the tangent and the sides of the polygon through the point of contact.
Answer:
In Figure 1:
∠5 = 60°.
Hence x = 60°

In Figure 2:
If AC is drawn, then ∠ACD = ∠CAD =45°.
Hence y = 45°

In Figure 3:
The angle formed by one side with the opposite corner is \(\frac {108}{3}\) = 36°.
Hence z = 36°

SSLC Maths Chapter 8 Questions and Answers – Tangent From Outside

(Textbook Page No. 180)

Question 1.
Draw a circle of radius 2.5 centimetres and draw tangents to it from a point 7 centimetres from the centre.
Answer:
Draw a circle of radius 2.5 cm and mark a point P with 7 cm from the centre O.
Join the centre O and the point P with a dotted line.
Draw another circle with OP as its diameter.
This new circle intersects the first circle at two points.
From the point P, draw two straight lines to the point of intersection.
These lines are the tangents.

Question 2.
Draw a circle of radius 3 centimetres and draw tangents to it from a point 7.5 centimetres from the centre.
Answer:
Draw a circle of radius 3 cm and mark a point P with 7.5 cm from the centre O.
Join the centre O and the point P with a dotted line.
Draw another circle with OP as its diameter.
This new circle intersects the first circle at two points.
From the point P, draw two straight lines to the point of intersection.
These lines are the tangents.

(Textbook Page No. 182)

Question 1.
In the figure, a triangle is formed by two perpendicular tangents and a third tangent to the circle:

(i) Prove that the perimeter of the triangle is the sum of the lengths of the perpendicular tangents.
(ii) Prove that the length of each of the perpendicular tangents is equal to the radius of the circle.
(iii) Prove that the perimeter of the triangle is equal to the diameter of the circle.
Answer:

(i) Perimeter = MA + AB + MB
= MA + AP + PB + MB
= MA + AC + BD + MB
= MC + MD

(ii) The tangent MD is perpendicular to the tangent MC. MC is perpendicular to OC, and MD is perpendicular to OD.
This is because a tangent and the radius drawn through the point of contact are always perpendicular to each other.
Therefore, MCOD is a square.
MC = MD = OD = OC.

(iii) We know that the Perimeter = MC + MD
MC + MD is equal to OC + OD (property of a square).
OC + OD is the sum of two radii, which is equal to the diameter.

Question 2.
The picture shows the triangle formed by three tangents to a circle:

Calculate the length of each tangent from each vertex to the point of contact.
Answer:

Let AP = x then PB = 4 – x
BR = 4 – x
CR = 7 – (4 – x) = 3 + x
CQ = 3 + x
AQ = 5 – (3 + x) = 2 – x
AP = AQ
⇒ x = 2 – x
⇒ 2x = 2
⇒ x = 1
∴ AP = 1, BP = 3, BR = 3, CR = 4, CQ = 4, AQ = 1

Question 3.
The figure shows two circles touching at a point and the common tangent at this point:

(i) Prove that this tangent bisects another common tangent to these circles.

(ii) Show that the triangles formed by joining these points of contact are right-angled.

Answer:
(i) PA, PB, and PC are tangents from the point P.
PA = PC, PB = PC.
Therefore, PA = PB.

(ii) In ∆PXC, PA = PC ⇒ ∠A = ∠C = x
In ∆PBC, PB = PC ⇒ ∠B = ∠C = y
Therefore in ∆ABC
⇒ x + x + y + y = 180
⇒ 2x + 2y = 180
⇒ x + y = 90°
⇒ ∠C = 90°

Kerala Syllabus Class 10 Maths Chapter 8 Solutions – Circle Touching A Line

(Textbook Page No. 187-188)

Question 1.
Draw the triangle with sides of 7 centimeters, 8 centimeters, and 9 centimeters, and draw its incircle.
Answer:

Question 2.
Draw the rhombus with sides of length 5 centimetres and one angle of 50°, and draw its incircle.
Answer:

Question 3.
Draw an equilateral triangle and draw a semicircle touching two of its sides as shown in the figure:

Answer:
Draw a triangle.
Mark the midpoint of one side and draw the perpendicular bisector.
From this midpoint, draw a perpendicular to the other side.
Taking this perpendicular distance as the radius and the midpoint as the centre, draw a semicircle.

Question 4.
Calculate the radius of the incircle of the equilateral triangle with a side of 12 centimetres.
Answer:
Area of the triangle = \(\frac{\sqrt{3}}{4} \times 12^2\) = 36√3 cm²
Perimetre = 3 × 12 = 36 cm
s = \(\frac {36}{2}\) = 18
r = \(\frac{A}{s}=\frac{36 \sqrt{3}}{18}\) = 2√3

Question 5.
Calculate the radius of the incircle of a triangle with sides 13, 14, and 15 centimetres.
Answer:

Question 6.
Prove that the radius of the incircle of any equilateral triangle is half the radius of its circumcircle.
Answer:
Let the side of an equilateral triangle be a,
then the height, h = \(\frac{a}{2} \times \sqrt{3}\)
In an equilateral triangle, the circumcenter, centroid, and incentre coincide.
The height is the same as the median and passes through this common point.
Circumradius R = \(\frac{2}{3} \times \frac{a}{2} \sqrt{3}=\frac{a}{\sqrt{3}}\)
Area of the equilateral triangle = \(\frac{\sqrt{3}}{4} a^2\)
Perimeter = 3a
s = \(\frac {3a}{2}\)
Radius of incircle = \(\frac{A}{s}=\frac{\sqrt{3}}{4} a^2 \div \frac{3 a}{2}=\frac{a}{2 \sqrt{3}}\)
Radius of incircle r = \(\frac {R}{2}\)
Therefore, the radius of the incircle of an equilateral triangle is half the radius of its circumcircle.

Question 7.
Prove that if the hypotenuse of a right triangle is h and the radius of its incircle is r, then its area is r(h + r).
Answer:
Let the perpendicular sides of the right-angled triangle be a and b.
Hypotenuse be h

s = \(\frac{a+b+h}{2}\)
A = rs
From the figure, POQB is a square.
RC = CQ = a – r
PA = RA = b – r
a – r + b – r = h
⇒ a + b – 2r = h
⇒ a + b = h + 2r
A = rs
= r × \(\left(\frac{a+b+h}{2}\right)\)
= r × \(\left(\frac{h+2 r+h}{2}\right)\)
= r × \(\left(\frac{2 r+2 h}{2}\right)\)
∴ A = r(r + h)

Question 8.
The lengths of the perpendicular sides of a right triangle are a, b, and its hypotenuse is c. Prove that the radius of the incircle is \(\frac {1}{2}\)(a + b – c).
Answer:
Let the perpendicular sides of the right-angled triangle be a and b.
Hypotenuse is c

From the figure, OPCQ is a square.
The radius of the incircle is equal to the sides of the square itself.
BQ – BR= a – r
AP = AR = b – r
a – r + b – r = c
⇒ a + b – 2r = c
⇒ 2r = a + b – c
⇒ r = \(\frac{a+b-c}{2}\)

Question 9.
In the picture, two vertices of a triangle are joined to its incenter:

(i) Calculate the angle marked in the picture.
(ii) In any triangle, what is the relation between the angle at a vertex and the angle between the lines joining the other two vertices to the incentre?
(iii) If one of these angles is half the other, what is the angle of the triangle?
Answer:
(i) 70° + ∠B + ∠C = 180°
∠B + ∠C = 110°
∠OBC + ∠OCB = \(\frac {110}{2}\) = 55°
∠BOC = 180 – 55 = 125°

(ii) Let the angles in the triangle ABC be A, B, and C.
Then, B + C = 180 – A
The construction is completed by drawing the angle bisectors of B and C, as the circle is an incircle.
\(\frac{B}{2}+\frac{C}{2}=\frac{180-A}{2}=90-\frac{A}{2}\)
⇒ ∠BOC = 180 – (90 – \(\frac {A}{2}\))
⇒ ∠BOC = 90 + \(\frac {A}{2}\)

(iii) If ∠BOC = x, then ∠BAC = \(\frac {x}{2}\)
∠OBC + ∠OCB = 180 – x
⇒ 2 × (∠OBC + ∠OCB) = 360 – 2x
⇒ A = 180 – (360 – 2x) = 2x – 180
⇒ 2x – A = 180
⇒ 2x – \(\frac {x}{2}\) = 180
⇒ \(\frac {3x}{2}\) = 180
⇒ x = 120°
∴ ∠A = 120°
Therefore, the marked angle in the figure is 120°
That is ∠BOC = 120°.

Tangents Class 10 Notes Pdf

Class 10 Maths Chapter 8 Tangents Notes Kerala Syllabus

Introduction
When a line and a circle are drawn on a paper, they can be related in three different ways. The line may not touch the circle at all, it may intersect the circle at two points, or it may touch the circle at exactly one point. If a line touches a circle at only one point, it is called a tangent to the circle. Tangents can be drawn through every point on a circle, and from any given point on the circle, only one tangent can be drawn.

This unit explores the geometric concepts related to tangents, along with some key geometric constructions. Every geometric construction is based on a clear geometric principle. In other words, a construction is a visual representation of a geometric concept.

→ The tangent at a point to a circle is perpendicular to the diameter through that point.

→ The angle at the centre of a circle between two radii, and the angle between the tangents at the ends of the radii, add up to 180°.

→ The angle between the tangents at the ends of a chord in a circle is the central angle of the chord subtracted from.

→ The angles that the tangents to a circle at the ends of a chord make on one side of the chord are both equal to the angle in the alternate segment.

→ From a point outside a circle, two tangents can be drawn to the circle.

→ The two tangents from a point outside a circle have the same length.

→ In a quadrilateral formed by the tangents at four points on a circle, the sum of the lengths of opposite sides is equal.

→ In any triangle, the bisectors of all three angles intersect at a single point.

→ The radius of the incircle of a triangle is equal to its area divided by half the perimeter.

Lines and Circles
Two lines are drawn through a point on the circle in such a way that they intersect the circle.
In the figure, lines AB and AC are drawn through point A.

If AB is the diameter, mark the centre O on AB. By joining OC, we get the triangle AOC. This is an isosceles triangle.
The angles opposite the equal radii are marked as x, and the angle opposite the third side is marked as y.
We can write it as, 2x + y = 180°

Let the point C move along the circle and approach the point A. Then the measure of angle y decreases. That is, angle y gets closer to zero.

When point C reaches point A, the value of y becomes zero, that is, y = 0.
At this point, AC touches the circle; in other words, AC becomes the tangent to the circle.
2x + y = 180° can be written as 2x + 0 = 180°. That is, x = 90°.
Its diagram looks like this:

It can be seen that the diameter AB is perpendicular to the tangent.
The tangent at a point to a circle is perpendicular to the diameter through that point.

Question 1.
Draw a circle with a radius of 3 cm and draw two parallel tangents to it.
Answer:
Draw a circle and mark the diameter AB.
Draw perpendiculars to AB through points A and B.
These lines are parallel tangents.

Question 2.
Draw a circle with a radius of 3 centimetres and construct a square that encloses the circle.
Answer:
Draw a circle and draw the perpendicular diameters AB and CD.
Draw perpendiculars to AB through A and B.
Draw perpendiculars to CD through C and D.
These tangents join together to form a square that encloses the circle.

Question 3.
In the figure, PA is a tangent, OA is the radius, OP is the line joining the centre O and the point P, ∠OPA = 30°, and OP = 16.

(a) What is the length of the tangent?
(b) What is the radius of the circle?
Answer:
(a) 8√3 cm
(b) 8 cm

Question 4.
PA is a tangent to a circle of radius 10 cm and ∠AOP = 60°.

(a) What is the length of the tangent?
(b) What is OP?
Answer:
(a) 10√3 cm
(b) 20 cm

Tangents and Angles
If two tangents to a circle are not parallel, then they will definitely meet at a point.

Tangents drawn through the points A and B, meet at the point C.
In the figure, the radius OA is perpendicular to AC, and the radius OB is perpendicular to BC.
∠OAC + ∠OBC = 180°
∠AOB + ∠QCB = 180°
OACB is a cyclic quadrilateral.
The angle at the centre of a circle between two radii, and the angle between the tangents at the ends of the radii, add up to 180°.
By using this geometric principle, a triangle can be drawn that covers the circle.

Question 1.
PA and PB are tangents to the circle with centre O. ∠AOC is twice ∠APC.

(a) Find ∠APB?
(b) Find ∠AOB?
Answer:
(a) If ∠APB = x then ∠AOB = 2x
x + 2x = 180
⇒ 3x = 180
⇒ x = 60
∴ ∠APB = 60°
(b) ∠AOB = 180 – 60 = 120°

Question 2.
PA and PB are tangents to the circle with centre O. ∠OAB = 20°

(a) Find the measure of ∠AOB?
(b) Find ∠APB?
Answer:
(a) ∠ABO = 20°
So ∠AOB = 180 – 40 = 140°
(b) ∠APB = 180 – 140 = 40°

Question 3.
ABC is an equilateral triangle. O is the circumcenter. PA, PC are tangents to the circle.

(a) Find ∠AOC?
(b) Find ∠APC?
Answer:
(a) ∠ABC = 60°
∠AOC = 120°
(b) ∠APC = 180 – 120 = 60°

Question 4.
Draw a circle with a radius of 3 cm, and draw an equilateral triangle that encloses the circle.
Answer:
Draw a circle of radius 3 cm.
Divide the angle around the centre into three parts with radii of 120° each.
Draw tangents to the circle along the ends of the radii.
The angle between the tangents is 180 – 120 = 60°.
Hence, the triangle formed is an equilateral triangle with each angle 60°.

Chord and Tangent
The angles that the tangents to a circle at the ends of a chord make on one side of the chord are both equal to the angle in the alternate segment.
Let us analyze this concept. For this, refer to the worksheet and the answer given below.

Let O be the centre of the circle, and AB the diameter. PA and PB are the tangents at the endpoints of the diameter. The angles between the chord and the tangent on the side are equal to x.

(a) Why ∠ABP = ∠BAP?
(b) Why ∠BPA = 180° – 2x
(c) Why ∠BOA = 2x?
(d) Why ∠BCA = y?
(e) Write the conclusion.
Answer:
(a) PA and PB are equal tangents.
In ∆PAB, the angles opposite the equal sides are equal.
(b) The sum of the angles in a triangle is 180°.
Therefore, ∠BPA = 180° – 2x.
(c) PAOB is a cyclic quadrilateral.
The sum of opposite angles is 180°.
(d) The angle in the alternate segment is half of the central angle subtended by the chord.
(e) The tangents drawn at the ends of a chord of 4 a circle make equal angles with the chord on one side, and these are equal to the angles in the opposite segment of the circle.

Question 1.
In the figure, AB is the diameter, and the line AP makes an angle of 40° with the diameter.

(a) Draw a circle with a suitable radius and draw the line AP.
(b) Draw a tangent at P without using the centre of the circle.
Answer:
(a) Draw a circle, its diameter, and a chord.
(b) Draw BP. Take P as the vertex; the other side will be the tangent, making an angle of 40°.

Question 2.
PT is the tangent to the circumcircle of ∆ABC at A. If AC = AB and ∠A = 40°.

(a) What are the measures of ∠C and ∠B?
(b) Is the line PT parallel to BC? Explain.
Answer:
(a) ∠B = ∠C = 70°
(b) The tangent at C and the side AC from an angle equal to ∠B.
Since AC = BC, ∠B = ∠A
That is the angle between the tangent, and AC is equal to ∠A.
From the equality of the alternate angles, the tangent is parallel to AB.

Question 3.
In ∆ABC, O is the circumcenter and ∠BOC = 140°

(a) What is ∠BAC?
(b) If PC is the tangent at C, what is ∠BCP?
Answer:
(a) ∠BAC = 70°
(b) 70°

Question 4.
In ∆ABC, the tangent to the circumcircle at A makes an angle of 70° with side AB.

(a) What is the measure of ∠ACB?
(b) What is the measure of ∠AOB?
Answer:
(a) ∠ACB = 10°
(b) ∠AOB = 140°

Question 5.
In the figure, O is the centre of the circle, and x, y, z are in a parallel sequence.

(a) What is x?
(b) What is ∠BAP?
(c) If the radius of the circle is 10 cm, what is the length of AB?
Answer:
(a) x = \(\frac {y}{2}\) and x + z = 180°
Since x, y, and z are in a parallel sequence,
x + z = 2y
⇒ 2y = 180°
⇒ y = 90°
x = \(\frac {90}{2}\) = 45°
z = 180 – 45 = 135°
(b) 45°
(c) 10√2

Tangent From Outside
From a point outside a circle, two tangents can be drawn to the circle. The lengths of those two tangents are equal.

The construction of tangents to a circle from an external point is a practical activity. This construction involves two geometric ideas.
(a) A tangent and the radius drawn through the point of contact are perpendicular to each other.
(b) An angle in a semicircle is 90°.

Steps of Construction:
Draw a circle and mark a point at a particular distance from the centre.
Join the centre and the external point with a straight line.
Draw another circle with this line as its diameter.
This new circle intersects the first circle at two points.
From the external point, draw two straight lines to the point of intersection.
These lines are the tangents, and each is perpendicular to the radius at the point of contact.

Let’s draw a quadrilateral that encloses a circle. Not just one quadrilateral, but many quadrilaterals can be drawn that enclose a circle.

Consider one of these quadrilaterals, and its sides can be marked as a, b, c, d.
The length of the tangents from each vertex is p, q, r, and s.
r + s = a
p + q = c
p + q + r + s = a + c
Similarly, we can write, p + q + r + s = b + d.
That means the sum of the lengths of opposite sides is equal. This can be written as follows.
In a quadrilateral formed by the tangents at four points on a circle, the sum of the lengths of opposite sides is equal.

Question 1.
PA and PB are perpendicular tangents drawn from an external point P to the circle. O is the centre of the circle.

(a) What is a suitable name for PAOB?
(b) What is the length of AB, if PB = 4 centimetres?
Answer:
(a) Square
(b) 4√2

Question 2.
PA and PB are parallel tangents to a circle of radius 4 cm. Another line, PQ, touches the circle at C.

(a) What is the distance between the parallel lines?
(b) What is the length of AB, if PA = 6 centimetres and QB = 4 centimetres?
Answer:
(a) 8 cm
(b) PA = PC = 6
QB = QC = 4
PQ = 10 cm

Question 3.
A circle touches the sides of a triangle. If AP = 1, BQ = 2, and CR = 3.

(a) Find the perimeter of the triangle.
(b) Calculate the area of the triangle?
Answer:
(a) AR = 1, CD = 3, BP = 2
Perimeter of the triangle ABC = 12 cm
(b) A right-angled triangle with perpendicular sides 3 cm and 4 cm
Area = \(\frac {1}{2}\) × 3 × 4 = 6 cm2

Question 4.
In the figure, PA = 12 centimeters, QA = 3 centimeters, and RB = 4 centimeters.

(a) Find the length of PB?
(b) Find the perimeter of the triangle PQR?
Answer:
(a) PA = 12
(b) QA = QS = 3
RB = RS = 4
PQ = 12 – 13 = 9
PR = 12 – 4 = 8
Perimeter of ∆PQR = 9 + 8 + 7 = 24

Question 5.
In the figure, PA and QB are parallel lines. Another line, PQ, touches the circle at R.

(a) Draw OA, OR, and OB.
(b) Name the equal triangles.
(c) Find ∠POQ?
Answer:
(a) Draw the lines.
(b) PA = PR, OA = OR, OP is common.
Therefore, ∆PAO and ∆PRO are equal triangles.
Similarly, ∆QRO and ∆QBO are also equal triangles.
(c) If ∠POQ = ∠POR = x then
∠QOR = ∠QOB = y
2x + 2y = 180
⇒ x + y = 90
∴ ∠POQ = 90°

Circle Touching A Line
A circle can be drawn that touches two lines meeting at a point. The centre of this circle lies on the bisector of the angle between the two lines. Therefore, to draw a circle that touches the two lines, first draw the bisector of the angle formed by the lines. From a point on the bisector, draw perpendiculars to the lines. Then, taking that point on the bisector as the centre and the perpendicular distance to the lines as the radius, draw the circle. This idea can be used for constructing the incircle of a triangle. The circle that touches all three sides of a triangle is called the incircle. Each side of the triangle can be considered as a tangent to the incircle.

To construct it, draw the triangle with the given measurements, then draw the bisectors of any two angles. The point where the bisectors meet is the centre of the circle. Taking the perpendicular distance from this point to any side as the radius, then draw the circle. The circle thus drawn will touch all three sides of the triangle. The radius of the incircle of a triangle is equal to its area divided by half the perimeter. This can be explained as follows.

Let the sides of the triangle ABC be a, b, and c, and let the radius of the incircle be r. Let S be the half of the triangle. By joining the centre of the circle to the vertices of the triangle, the triangle can be divided into three smaller triangles, ∆OAB, ∆OAC, and ∆OBC, as shown in the figure.

The sum of the areas of these three triangles gives the total area of the triangle ABC.
A = \(\frac{1}{2} a r+\frac{1}{2} b r+\frac{1}{2} c r\)
= \(\frac {1}{2}\)r(a + b + c)
= r × \(\frac{a+b+c}{2}\)
= rs
Which can also be written as, r = \(\frac {A}{s}\)

The area of the triangle with sides a, b, and c is,
A = \(\sqrt{s(s-a)(s-b)(s-c)}\)
Therefore, the radius can be written as,
r = \(\frac{\sqrt{s(s-a)(s-b)(s-c)}}{s}\) = \(\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}\)
This relationship is true for all polygons that can have an incircle.

The radius of the incircle is obtained by dividing the area of the polygon by half of its perimeter. There are quadrilaterals like squares. These have an incircle and a circumcircle. In the case of a square, the centre of both types of circles is the same. But in other quadrilaterals where the incircle and circumcircle can be drawn, the centres of the two circles are different points.

How can such a quadrilateral be drawn?
Draw a circle and then draw two perpendicular chords within it. From the endpoints of these chords, draw tangents to the circle. The quadrilateral formed by the intersection of these tangents is cyclic. Now draw its circumcircle. Thus, a quadrilateral will be obtained that has both an incircle and a circumcircle.

Question 1.
The length of the hypotenuse of a right-angled triangle is 18 cm, and the radius of its incircle is 3 cm.
(a) What is the perimeter of the triangle?
(b) Calculate the area of the triangle.
Answer:
(a) Let a and b be the two perpendicular sides.
r = \(\frac{a+b-c}{2}\)
⇒ 3 = \(\frac{a+b-18}{2}\)
⇒ a + b – 18 = 6
⇒ a + b = 24
So perimeter is 24 + 18 = 42

(b) s = \(\frac {42}{2}\) = 21
A = rs
= 3 × 21
= 63 cm²

Question 2.
(a) A circle is drawn that touches the sides of a quadrilateral. Prove that in such quadrilaterals, the sum of the opposite sides is equal.

(b) Prove that AB + CD = AD + BC.
Answer:
(a) AR = AS
BR = BQ, DP = DS, CP = CQ
(b) Taking the sum of the opposite sides,
AR + BR + DP + CP = AS + BQ + DS + CQ
AB + CD = AD + BC

Question 3.
Draw an angle with suitable measures and draw a circle that touches its sides.

(a) Where is the centre of the circle located?
(b) Write the geometric principle used in the construction.
Answer:
Draw a cone and draw its angle bisector. From a point on the bisector, take the perpendicular distance to the sides of the cone as the radius and draw the circle.
(a) On the angle bisector of the cone.
(b) A tangent to a circle at any point is perpendicular to the radius drawn to that point.

Question 4.
In ∆ABC, AB = 7 cm, ∠A = 40°, ∠B = 60°
(a) Draw the triangle using the given measurements.
(b) Draw the circle that touches the sides of the triangle.
(c) Measure and write down the radius.
Answer:
(a) A triangle is drawn using the given measurements.
(b) Draw the angle bisectors of any two angles.
(c) Using the point where the bisectors meet as the centre, draw the circle by taking the perpendicular distance from this point to any side as the radius.