sheniblog-EM-SSLC Worksheets - Questions and Answers - John P A

Answers
⋆ Draw a circle of radius 3cm
⋆ Draw radii which divides the angle around the centre as 180 − 120 = 60◦, 180 −
40 = 140◦, 160◦
⋆ Draw tangents at the ends of the diametre which makes the triangle.
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2020-21 Academic year Worksheets

Mathematics X
Tangents

Concepts

⋆ Two tangents from an outer point and two radii to the point where
the tangents touch the circle form a cyclic quadrilateral.

1) P is a point at the distance 7cm from the centre of a circle of radius 3 cm
and centre O.
a) Draw two tangents fro P to the circle.
b) Measure the length of radius and write aside.
c) Write the geometric concept of this construction.
Answers
a) ⋆ Draw a circle of radius 3cm, centre O . Mark a point P at the
distance 7cm from the centre.
⋆ Draw a circle with OP as the center.The second circle intersect
the first circle at A and B
⋆ Draw the lines P A and P B. These lines are the tangents to the
circle of radius 3cm from P
b) C√onstruct geometrically and write the lengths .P A = P B =
40cm
c) Angle in the semicircle is 90◦, tangent is perpendicular to the
radius.

2) A circle touches two sides of an equilateral triangle, the centre of the circle
is a point on the third side.
a) If the side of the triangle is 4cm , complete the construction.
b) What is the radius of the semicircle?
c) What is the distance from base1 vertex of the triangle to the point where
the side touches the circle?

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Answers
a) See the rough diagram. Construct geometrically.

Draw equilateral triangle ABC. Mark the mid point of AB as
O. Draw perpendicular from O to AB.It cut AC at P . Draw a
semicircle with O as the centre and OP as the radius.

b) Write the length OP by measuring it.△AOP is a 30◦, 60◦, 90◦

right triangle.AO = 2 cm .Side opposite to√90◦ is 2cm .
Therefore side opposite to 30◦ is 1cm , OP = 3cm

c) AP = 1cm

3) In the figure P A and P B are the tangents to the circle.O is the centre of
the circle.

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a) If ∠ACB : ∠AP B = 2 : 5then find these angles.
b) What is the measure of ∠AOB?
c) What is the measure of ∠ADB?

Answers
a) Let ∠ACB = 2x, ∠AP B = 5x .
Since ∠ACB = 2x, ∠AOB = 4x. OAP B is cyclic
.Therefore ∠AOB + ∠AP B = 180◦
4x + 5x = 180, 9x = 180, x = 20
∠ACB = 2 × 20 = 40◦, ∠AP B = 5 × 20 = 100◦
b) ∠AOB = 4x = 4 × 20 = 80◦
c) ACBD is cyclic . ∠ACB + ∠ADB = 180◦ , ∠ADB =
180 − 40 = 140◦

4) Manju is drawing a circle touches the arms of an angle . The angle taken
for this construction is 40◦.
a) Draw the bisector of the angle.
b) Mark a point on the bisector of the angle.
c) Draw a perpendicular from this point to an arm
d) Draw a circle with the point marked on the bisector as the centre and
the perpendicular distance to the arm as the radius
e) Measure the length of tangent from the vertex to the point where the
circle touces the arm
Answers
See rough diagram. Construct yourself by the steps given below

3

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5) Prove that the length of tangents from an exterior point to a circle are equal.
Answers

⋆ In the figure P A, P B are tangents from P to the circle.
OA, OB radii.

⋆ △OAP, △OBP are the right triangles.
⋆ P A2 = OP 2 − OA2 → OP 2 − OB2 = P B2

P A2 = P B2, P A = P B

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2020-21 Academic year Worksheets

Mathematics X
Tangents

Concepts
Prove that the length of tangents from an exterior point to a circle are equal.

⋆ In the figure P A, P B are tangents from P to the circle. OA, OB radii.
⋆ △OAP, △OBP are the right triangles.
⋆ P A2 = OP 2 − OA2 → OP 2 − OB2 = P B2

P A2 = P B2, P A = P B
1) In the figure AB = AC, the circle touches the sides at P, Q, R.

a) AP = AQwhy?
b) Prove that BR = CR

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Answers
a) Two tangents from an outer point to a circle are equal . ∴ AP = AQ
b) AB = AC, AB − AP = AC − AQ
BP = CQ → BR = CR

2) In the figure ∠B = 90◦,AB = 15cm , BC = 8 cm .

a) Draw a rough figure , mark O as the centre . Suggest a suitable name to P ORB
b) If P B = x then fin the length AP, AQ, CR, CQ
c) What is the radius of the circle.
Answers

a) See the figure

P ORB is a square
b) AP = 15 − x, AQ = 15 − x, CR = 8 − x, CQ = 8 − x

√
c) Hypotenuse of the right triangle is AC = 152 + 82 = 17

15 − x + 8 − x = 17, 23 − 17 = 2x, 2x = 6, x = 3
Radius of the circle is 3cm

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3) In the figure P Q and P R are the tangents from P outside the circle. P Q = 24cm AQ = 10cm
, BR = 8cm then

a) What is the length of P R
b) What is the length of AB?
c) What is the perimetre of △P AB
d) Prove that P Q + P R = △P AB.
Answers

a) P Q = P R( Tangents from the outer point to the circle are equal)
∴ P R = 24cm

b) Mark C , the point where the line AB touches the circle.
AQ = AC = 10
BR = BC = 8
AB = 10 + 8 = 18

c) P A + AB + P B = 14 + 18 + 16 = 48cm
d) P Q + P R = (P A + AQ) + (P B + BR)

(P A + AC) + (P B + BC)
(P A + P B + AC + BC)
P A + P B + AB
Perimetre of the circle.
4) In the figure P A, QB are the parallel tangents.P Q touches the circle at R

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a) Prove that △P AO and △P ROare equal triangles
b) Prove that △QBO and △QROare equal triangles
c) Find ∠P OQ
Answers

a) OA = OR, P A = P R, OP is common. △P AO and △P ROare equal triangles
b) OB = OR, QB = QR, OQ is common. △QBO and △QROare equal triangles
c) ∠AOP = ∠ROP , ∠QOR = ∠QOB

∠AOP + ∠ROP + ∠QOR + ∠QOB = 180◦
2 × ∠ROP + 2 × ∠QOR = 180◦
∠ROP + ∠QOR = 90◦, ∠P OQ = 90◦
5) XP, XQare the tangents to the circle from X outside the circle. The line AB touches the circle
at R

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Prove that XA + AR = XB + BR
Answers
⋆ XP = XQTangents from outer point to the circle are equal.
⋆ XA + AP = XB + BQ
⋆ Since AP = AR and BQ = BR, XA + AR = XB + BR
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2020-21 Academic year Worksheets

Mathematics X
െതാ വരകൾ

Concepts
⋆ In the figure P A, P B are tangents from P to the circle. OA, OB radii.
⋆ △OAP, △OBP are the right triangles.
⋆ P A2 = OP 2 − OA2 → OP 2 − OB2 = P B2

P A2 = P B2, P A = P B

1) In the right triangle △ABC, ∠C = 90◦. a, b, care the sides opposite to A, B and C.A circle
touches sides of the triangle.

a) If the radius of the circle is r write the lengths P B and AP

b) Prove that r = a+b−c
2

c) If the perpendicular sides are 6cm and 8 cm then find the length of the hypotenuse

d) If the perpendicular sides are 6 and 8 cm mthen find the radius of the circle .

Answers
a) See diagram

ORCQ is a square.OR = OQ = r. ∴ CR = CQ = r

BR = a − r, BP = a − r

AQ = AP = b − r 1

b) c = P A + P B = b − r + a − r

c = a + b − 2r, 2r = a + b − c, r = a+b−c
2
√
c) AB = 62 + 82 = 10
d) r =vai+s2bi−tcw=w6+w82−.s10h=en2cismchool.in or whatsapp 7012498606

2) In the figure P M, P N are the tangents to the circle.The distance from P to the centre of the circle
is 13cm, radius of the circle is 5cm . The line AB touches the circle at C

a) Find the length of P M and P N
b) If AM = x then find AC and AP
c) Find x
d) What is the length of AB

Answers
√

a) P M = P N = 132 − 52 = 12cm

b) If AM = x then AC = x, AP = 12 − x

c) Since OC is perpendicular to AB, △ACP is a right triangle.

(12 − x)2 = x2 + (13 − 5)2

122 + x2 − 24x = x2 + 82,24x = 80, x = 80 cm
24

d) AB = 2 × 80 = 20 cm
24 3

3) The sides of ABCD touches the circle at P, Q, R, S

a) Prove that AB + CD = AD + BC
b) If AB = 12cm CD = 8cm , AD = 14cm then find BC.

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Answers (1)
(2)
a) (3)
AP = AS (4)

BP = BQ

DR = DS

CR = CQ

Adding these equations , AP + BP + DR + CR = AS + BQ + DS + CQ
(AP + BP ) + (DR + CR) = (AS + DS) + (BQ + CQ)
AB + CD = AD + BC

b) 12 + 8 = 14 + BC, BC = 20 − 14 = 6cm

4) In the figure ,the line ABtouches a circle.CP is the common tangent .

a) Prove that P A = P B
b) Prove that △ABC is a right triangle.
c) If AC = BC = 10cm then find the length AB
Answers

a) P A = P C, P B = P C(Tangents from outer point to a circle are equal)
PA = PB

b) In △AP C, two sides P A = P C. ∴ the angles opposite to equal sides are equal. ∠A =
∠C = x
In △BP C, two sides P B = P C. ∴ the angles opposite to equal sides are equal. ∠B =
∠C = y
Consider △ABC, ∠A+∠B +∠C = 180◦, x+x+y+y = 180, 2x+2y = 180, x+y =
90◦.△ABC is a right triangle.
√

c) △ABC is a 45circ − 45◦ − 90◦ triangle.AB = 10 2cm

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5) Draw 60◦angle , construct two circles touches the arms of the angle and touch eachother.
Answers
a) Draw the bisector of the angle.
b) Mark a point on the bisector of the angle.
c) Draw a perpendicular from this point to an arm
d) Draw a circle with the point marked on the bisector as the centre and the perpendicular
distance to the arm as the radius
e) Measure the length of tangent from the vertex to the point where the circle touces the arm
f) Mark the point P where the bisector cut the circle. The perpendicular at that point to the
bisector will be the tangent
g) Draw the bisector of the tangent and arm. It will meet bisector the first angle drawn at a
point which will be the centre of the second circle.

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2020-21 Academic year Worksheets

Mathematics X
Tangents

Concepts
⋆ Tangents from an outer point to a circle are equal
⋆ The circle which touches the sides of a quadrilateral inside is called incircle.All triangles will

have an incircle.
⋆ In the quadrialterals having incircle , sum of the opposite sides are equal.

1) In the figure ABCD is a parallelogram .The circle touches the sides at P, Q, R, S

a) Prove that AD + BC = AB + CD
b) Prove that ABCD is a rhombus
Answers

a) BQ = BP, CQ = CR, AS = AP, DS = DR
Adding these equations,
(BQ + CQ) + (AS + DS) = (BP + CR) + (AP + DR)
BC + AD = AB + CD

b) Since ABCD is a parallelogram , AD = BC, AB = CD
BC + AD = AB + CD becomes 2AD = 2AB, AD = AB
That is ,
AB = BC = CD = AD
ABCD is a rhombus.

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2) In the quadrilateral ABCD, ∠D = 90◦
The sides AB, BC, CD, DA touches the circle at P, Q, R, S.
BC = 38 cm , CD = 25 cm , BP = 27cm

a) Prove that ORDS is a square
b) Find the length of CQ
c) What is the side of ORDS ?
d) What is the radius of the circle which touches the sides?
Answers

a) CD is tangent and OR is radius . Therefore CD is perpendicular to OR.
Similarly AD is perpendicular to OS, ∠D = 90◦. In ORDS, ∠O will be 90◦
Also, DR = DS. All sides of ORDS are equal, all angles are 90◦. That is ORDS is a
square.

b) BP = BQ = 27, BC = 38,QC = 38 − 27 = 11cm
c) CQ = CR = 11cm, DR = CD − 11 = 25 − 11 = 14cm

Side of ORDS is 14 cm
d) Since ORDS is a square radius is its side. Radius = 14cm
3) In the figure P Qis the common tangent to the circles . Radius of the big circle is 6cm , radius of
the small circle is 3cm . The diatance between the centres is 15cm.

a) Are the traingles AP C and BQC similar ?
b) What is the length AC and BC?
c) What is the length of P Q?

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Answers

a) Consider △AP C and △BQC
∠P = ∠Q, ∠ACP = ∠BCQ. So △AP C and △BQC are similar

b) AP = AC
BQ BC
If AC = x then BC = 15 − x

6 = x ,6(15 − x) = 3x, 90 − 6x = 3x, 90 = 9x, x = 10cm
3 15−x

AC = 10cm , BC = 5cm
√√
c) P C = 102 − 62 = 8cm, CQ = 52 − 32 = 4.

P Q = 8 + 4 = 12cm

4) In the traingle ABC, ∠B = 90◦, area of the triangle 30cm, sum of the perpendicular sides is
17cm

a) What is the length of AC?
b) What is the radius of the circle ?

Answers

a) a + c = 17, 1 × a × c = 30, ac = 60
2
(a + c)2 = a2 + c2 + 2ac, 172 = a2 + c2 + 120, a2 + c2 = 169

Since a2 + c2 = b2, b2 = 169, b = 13

b) r= a+c−b = 17−13 = 2cm
2 2

5) In △ABC , a, b, c are the sides opposite to A, B and C .
r is the radius of the circle touches the sides, area of the triangle is A,half of its perimetre is s

a) Prove that A = rs
b) Is this relation true in the case of all quarilaterals having incircle?

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Answers
a) see diagarm

In the diagram

Area of △ABC= area △BOC+area of △OAC+area of △OAB

A = 1 a × r + 1 b × r + 1 c × r
2 2 2
a+b+c
A = r( 2 ) = rs

b) see diagram

Try yourself as above

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2020-21 Academic year Worksheets

Mathematics X
Tangents

Concepts
⋆ Tangents from an outer point to a circle are equal

⋆ The circle which touches the sides of a quadrilateral inside is called incircle.All triangles will
have an incircle.

⋆ In the quadrialterals having incircle , sum of the opposite sides are equal.

1) Sides of a triangular metal sheets are 26cm, 24cm and 10cm

a) What kind of triangle is this ?
b) What is the perimetre of this triangle?
c) What is the area of this triangle?
d) Can this metal sheet is used to cover the upper open face of a cyclindrical vessel of radius

5cm?

Answers

a) 242 + 102 = 576 + 100 = 676 = 262.
This is a right triangle

b) Perimeter26 + 24 + 10 = 60cm

c) Area 1 × 24 × 10 = 120 sq.cm
2

d) Radius of incircle r = A
s
26+24+10
s = 2 = 30

r = 120 = 4cm
30

Radius of the upper end of the cylinder is 5cm .It is more than radius of incircle. Not possible

to cover.

2) Side of an equilateral triangle is 10cm

a) What is the altitude of this triangle?
b) Find the perimetre and area of the triangle
c) Find the radius of the incircle of this triangle.

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Answers
a) Look at the picture

△ABD is a 30◦ − 60◦ − 90◦ right triangle . AD = √
5 3cm
√ √
b) Perimetre 30cm , area 1 × 10 × 53 = 25 3sq.cm
2

√√
A 25 3 5 3 3 cm
c) A = rs → r = s = 15 =

3) Draw the incircle of an equilateral triangle using the instructions given below .

a) Draw an equilateral triangle of side 4cm
b) Draw the bisector of two angles
c) Draw a perpendicular to the side of the triangle from the point of intersection of the bisectors.
d) Draw a circle with centre at the intersecting point and the perpendicular distance to the side

as the radius

Answers

Draw yourself as mentioned above

4) In triangle ABC, BC = 15cm , AB = 12cm , ∠B = 30◦

a) What is the perpendicular distance from A to BC?
b) Calculate the area of the triangle?
c) What is the length of AC?
d) Find the radius of the incircle.

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Answers

a) sin 30◦ = AD , 1 = AD , AD = 6 cm
12 2 12

b) Area 1 × 15 × 6 = 45sq.cm
2

c) cos 30◦ = BD , √ = BD , BD = √ = 6 × 1.732 = 10.39െസമീ ർ
12 3 12 63
2
CD = √15 − 10.39 = 4.√61െസ ീമീ ർ √
AC = 62 + 4.612 = 36 + 21.25 = 57.25 = 7.5cm

d) Perimetre of ABC = 12 + 15 + 7.5 = 34.5cm

s = 17.25cm

r = A = 45 = 2.6cm
s 17.25

5) There are quadrilaterals having incircle and circumcircle.Construct such a quadrilateral as follows

a) Draw a circle and two perpendicular chords.
b) Draw tangents at the ends of the chords to the circle.
c) Mark the quadrilaetarl formed by the tangents .
d) This will be a cyclic quadrilaetral. Draw the circle passing through the vertices.
e) Draw this figure using geogebra software .

Answers

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2020-21 Academic year Worksheets

Mathematics X
Tangents

Concepts
Angle between a tangent to a circle and the chord from the point where the tangent touches the
circle is equal to angle in the other segment of the circle.
We can prove it · · ·

⋆ If ∠BAY = x, ∠ACB = y then we have to prove x = y
⋆ AP is a diametre of the circle.Draw line P B, ∠AP B = ∠ACB = y , ∠P BA =

90◦.∠P AB = 90 − y
⋆ Line AP is perpendicular to XY
⋆ x = 90 − ∠P AB = 90 − (90 − y) = y

x=y
1) In the figure AB is a chord , line XY is a tangent at A. If ∠Y AB = 40◦then

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a) Find ∠ACB
b) Find ∠AOB?
c) Find ∠ADB?
Answers

a) ∠ACB = 40◦
b) ∠AOB = 2 × 40 = 80◦
c) ∠ADB = 180 − 40 = 140◦
2) ABCD is a square .The vertices of the square are on the circle. Tangent at A meet CBproduced
at P .

a) What is ∠BAP ?
b) What is ∠ABP ?
c) What is ∠AP B?
d) If AP = 20cm then what is the area of the square ?

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Answers

a) Draw diagonalAC of the square
∠ACB = 45◦, ∠BAP = 45◦

b) Since ABCD is a square ∠ABC = 90◦, ∴ ∠ABP = 90◦

c) ∠AP B = 180 − (90 + 45) = 45◦ √
= 10 2cm
d) Since △ABP is a 45◦ − 45◦ − 90◦ triangle and AP = 20cm, AB = √20
√2
Area = (10 2)2 = 200 sq.cm

3) AP is the tangent of a circle with centre O. The angle between AB and tangent is 140◦

a) What is the measure of ∠ACB
b) What is the central angle of arc ADB?
c) What is the measure of ∠ADB
d) Name an angle in the figure equal to∠ADB

Answers
a) ∠ACB = 140◦
b) The central angle of ADB is 2 × 140 = 280◦
c) ∠ADB = 40◦
d) ∠ADB = ∠QAB

4) In the figure AB is the diametre of the circle, AP is a chord. Draw tangent at P using the steps
given below P

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a) Draw a circle, chord AP and the diametre AB
b) Draw BP
c) Draw an angle eqaul to ∠P AB with P as the vertex and P B an arm
d) Why the other arm become tangent to the circle?
Answers

5) Draw a circle , mark a point P on it
a) Using the centre of the circle draw a tangent at P
b) Without using centre draw the tangent at P
c) Write the geometric principle of these two constructions.

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Answers
a) Tangent using centre O Draw a circle of suitable radius , centre O. mark a point P on the
circle. Draw radius OP and a line through P perpendicular to the radiusOP . This line will
be the tangent.
b) Without using centre. Draw a circle with centre O, Mark a point P on the circle .
Draw a chord AB , join AP and BP .Now you can see an angle P AB .
Draw an angle with vertex at P and P B as an arm. This angle should be equal to ∠P AB
in the figure.The other arm of the angle will be a tangent at P to the circle.
c) ⋆ For the tangent with using centre , principle is radius is perpendicular to the tangent
⋆ For the tangent without using centre, angle between the chord (say AB) and tangent is
equal to angle in the other segment of the circle . It is the angle P AB.

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2020-21 Academic year Worksheets

Mathematics X
Tangents

Concepts
⋆ The angle between the tangent and the chord to the point of tangency is equal to the chord

in the other segment of the circle.
⋆ Tangents from an outer point to a circle are equal in length.
⋆ Two tangents from an outer point to a circle and two radii to the touching point make a

cyclic quadrilateral.
⋆ If A is the area of the circle , r is the radius of the incircle and s is the semiperimetre then

A = rs
⋆ The sum of the opposite sides of a quadrilateral having incircle are equal.
1) In the figure ABCD is a rectangle .A circle touches the triangle formed by two sides and diagonal
at P, Q, R. If AP = 2cm, DQ = 3cm then

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a) What is AD?
b) What is the length of the side AB?
c) What is the length of the diagonal of the rectangle?
d) What is the radius of the circle?

Answers
a) DR = DQ = 3cm , AP = AQ = 2cm
AD = 3 + 2 = 5 cm
b) Let BP = BR = x, AB = x + 2, BD = x + 3
(x + 3)2 = (x + 2)2 + 52, x2 + 6x + 9 = x2 + 4x + 4 + 25
2x = 20, x = 10cm
AB = 10 + 2 = 12cm
c) BD = 13cm
d) Mark the centre of the circle O.
OQAP is a square.OP = OQ = 2cm , Radius of the circle is 2cm

2) ABC is an equilateral triangle.Tangents are drawn at the vertices to the circumcircle.These tangents
form another triangle P QR.

a) Prove that P QR is an equilateral triangle.
b) If the perimetre of ABC is 12cm then what is the perimetre of △P QR.
c) How many times the area of P QR is that of ABC?

Answers
a) △ABC is an equilateral triangle.
∠A = ∠CBR = ∠BCR = 60◦ , ∠R = 60◦
∠B = ∠CAQ = ∠ACQ = 60◦, ∠Q = 60◦,∠P = 60◦. △P QR is an equilateral
triangle.
b) P ACB is a parallelogram.BC = P A
QABC is a parallelogram BC = AQ ∴ P Q = 2 × BC,
Similarly P R = 2 × AC, QR = 2 × AB
Perimetre of △P QR = 2 × 12 = 24cm
c) P ACB, QABC, RBAC are equal parallelograms. Each one can be divided into two equal
triangles.We can see four equal triangles in the picture.
Area of △P QR = 4 times the area of △ABC

√
3) In the figure AP is the diametre of the circle.AB = 6 3cm P B = 6 cm

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a) What is the radius of the circle?
b) What are the angles of △AP B?
c) What is the measure of ∠ACB?
d) What is the measure of ∠BAQ?

Answers

√√
a) AP = (6 3)2 + 62 = 12. Radius of the circle is 6 cm

b) Since AP is the diametre ∠B = 90◦ . √ It is a
Sides of △AP B are in the ratio 1 : 3 : 2.

30◦ − 60◦ − 90◦ triangle.

∠A = 30◦, ∠P = 60◦, ∠B = 90◦

c) ∠ACB = 60◦(angle in the same arc)

d) 60◦

4) ⋆ ⋆ ⋆ Three measurements are needed to construct a triangle.Only two measurements are given
below
In △ABC AB = 6cm , ∠C = 60◦.

a) Draw a triangle using the concepts we discuss in tangents .
b) How many triangles are possible?
c) If AB = 6cm , ∠C = 60◦, the altitude to AB is 4cm, construct the triangle.

Answers

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Answers

a) ⋆ Draw a line AB of length 6 cm.
⋆ At A mark an angle 60◦ with AB and AX as the arms.Extent the arm as XY .

⋆ We have to draw a circle with AB as the chord and XY as a tengent.For this draw
perpendicular bisector of AB and perpendicular to XY through A These lines meet at
O, the centre of the circle.

⋆ Draw the circle with O as the centre and OA as the radius . Mark an point C on the circle,
△ABC will be a triangle satisfying the given condition.

b) We can take any point C in the larger part of the circle. So infinite number of triangles are
possible.

c) On the perpendicular bisector of AB, mark a point 4cm above AB . Draw a line parallel to
AB passing through this point, cut the circle at two points . P, Q. Complete △AP B and
△AQB

5) In the figure QRis the diametre of the circle, P A is the tangent ,∠RP A = 30◦.

a) What is the measure of ∠P QR?
b) What is the measure of ∠P RQ?
c) What is the acute angle formed by P A with P Q?

Answers
a) ∠P QR = 30◦
b) ∠P RQ = 60◦
c) Acute angle is 60◦

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2020-21 Academic year Worksheets

Mathematics X
Tangents

Concepts
⋆ The angle between the tangent and the chord to the point of tangency is equal to the chord

in the other segment of the circle.
⋆ Thangents from an outer point to a circle are equal in length.
⋆ Two tangents from an outer point to a circle and two radii to the touching point make a

cyclic quadrilateral.
⋆ If A is the area of the circle , r is the radius of the incircle and s is the semiperimetre then

A = rs
⋆ The sum of the opposite sides of a quadrilateral having incircle are equal.

1) In △ABC AB = AC, a tangent P Q is drawn through A to its circumcircle.Prove that P Q is
parallel to BC.

Answers
⋆ Since AB = AC opposite angles are equal. ∠B = ∠C
⋆ ∠P AB = ∠C( In a circle the angle between a chord and tangent at its end is equal to the
angle in the other side of the chord on the circle.)
⋆ Since ∠B = ∠C , ∠P AB = ∠B.The equality of altrenate angles shows that BC is
parallel to the tangent at A.
P Q is parallel BC

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2) In △ABC a tangentP Q is drawn through A to the circumcircle of the triangle.If BC is parallel to
P Q then prove that AB = AC

Answers
⋆ ∠P AB = ∠C( In a circle the angle between a chord and tangent at its end is equal to the
angle in the other side of the chord on the circle.)
⋆ ∠P AB = ∠B (Line P Q is parallel to BC, alternate angles are equal )
⋆ ∠P AB = ∠B = ∠C
∠B = ∠C
Sides opposite to equal angles are equal. AB = AC

3) In the figure P Q is the diametre of the circle , M N is the tangent to the circle at P .
If ∠RP N = 50◦

a) What is the measure of ∠P QR?
b) What is the measure of ∠P RQ?
c) What is the measure of ∠QP M ?
Answers

a) ∠P QR = 50◦
( In a circle the angle between a chord and tangent at its end is equal to the angle in the
other side of the chord on the circle.)

b) ∠QP R = 90◦, ∠P RQ = 90 − 50 = 40◦
c) ∠QP M = ∠P RQ = 40◦
4) In the figure BC is the diametre of the circle, P A is a tangent .If ∠AP B = x, ∠P AB = y then

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a) What is the measure of ∠BCA and ∠CAQ
b) Whatn is the measure of ∠ABC?
c) Find x + 2y
Answers

a) ∠BCA = y ( In a circle the angle between a chord and tangent at its end is equal to the
angle in the other side of the chord on the circle.)
Since OC = OA opposite angles are equal .∠OAC = y.
Radius is perpendicular to the tangent ∠CAQ = 90 − y

b) ∠ABC = 90 − y
c) ∠P BA = 180 − (90 − y) = 90 + y

ിേകാണം P BAയിൽ 90 + y + x + y = 180, x + 2y = 90◦

5) ABCD is a cyclic quadrilaterl. The diagonal AC bisects ∠C. Prove that the diagonal BD is
parallel to the tangent at A

Answers
⋆ Line AC is the bisector of ∠C.If ∠ACB = ∠ACD = ythen ∠QAB = y
⋆ ∠ACD = ∠P AD = y
∠ABD = y
⋆ ∠QAB = ∠ABD → Line P Q is parallel to BD.

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2020-21 Academic year Worksheets

Mathematics X
Tangents

Concepts
⋆ The angle between the tangent and the chord to the point of tangency is equal to the chord

in the other segment of the circle.
⋆ Thangents from an outer point to a circle are equal in length.
⋆ Two tangents from an outer point to a circle and two radii to the touching point make a

cyclic quadrilateral.
⋆ If A is the area of the circle , r is the radius of the incircle and s is the semiperimetre then

A = rs
⋆ The sum of the opposite sides of a quadrilateral having incircle are equal.

1) ABCD is a cyclic quadrilaeral . P Q is a tangent atC.BD is the diametre of the circle.
∠DCP = 40◦, ∠ABD = 60◦

a) What is the measure of angle DBC?
b) What is the measure of angle BCQ?
c) What is the measure of angle BDC?
d) What is the measure of ADB?
Answers

a) ∠DBC = 40◦
b) ∠BCD = 90◦, ∠BDC = 90 − 40 = 50◦, ∠BCQ = 50◦
c) ∠BDC = 50◦
d) Since DAB = 90◦ , ∠ADB = 90 − 60 = 30◦

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2) Tangent from an outer point T to the circle is AT . B and C are the points a line from T cut the
circle.In triangle ACB, AD is the bisector of ∠A, ∠A = 70◦,∠CAD = 40◦

a) What is the measure of ∠ADB?
b) What is the measure of ∠BAT ?
c) Find the angles of △DAT
Answers

a) AD is the bisector of ∠A. ∴ ∠CAD = 35◦, ∠ADB = 35 + 40 = 75◦
b) ∠BAT = 40◦
c) In △DAT , ∠A = 35 + 40 = 75◦,( Sum of the two angles of a triangle is equal to the

exterior angle in the other vertex) ∠D = 75◦ ∠T = 180 − 150 = 30◦
3) AB is the diametre of the circle, P A is a tangent . The line P B cut the circle at C, also CQ is

the tangent at C

a) If is drawn AC whatn is the measure of ∠ACB?
b) If ∠ACQ = x then what are the acute angles of △ABC?
c) Is AQ = QC? Why?
d) Prove that the line CQ bisects AP .

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Answers
a) Draw AC , ∠ACB = 90◦
b) If ∠ACQ = xthen in △ABC , ∠B = x, ∠BAC = 90 − x
c) Tangents from outer point Q to the circle are equal. . QC = QA
d) ∠QCP = 90 − x, ∠CP Q = 90 − x.Opposite sides are equal . QP = QC
AQ = QC = P Q → AQ = P Q

4) Two circles intersect at P, C.AB is the common tangent.

Prove that ∠AP C + ∠ACB = 180◦
Answers
⋆ Draw P C in the figure.If ∠BAP = xthen ∠ACP = x
⋆ If ∠ABP = ythen ∠BCP = y
⋆ In △ABP , ∠AP C = 180 − (x + y)
∠ACB = x + y
⋆ ∠AP C + ∠ACB = 180 − (x + y) + (x + y) = 180◦

5) In the figure AB is the diameter of the circle. P is a point on ABproduced.The line from P
touches the circle at C.If ∠CAB = 30◦ and the radius of the circle is 6cm

a) Find the lengths AC and BC
b) Prove that BP = BC.

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Answers

⋆ AB = 12 cm . △AC B is a 30◦ − 60◦ − 90◦ right triangle. BC = 6cm , AC √
= 6 3cm

⋆ ∠BCP = 30◦, ∠CBP = 180 − 60 = 120◦,
In △P BC , ∠P = ∠C = 30◦.

Opposite sides are equal. BP = BC

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2020-21 Academic year Worksheets

Mathematics X
Tangents

Concepts
Let P be a point outside the circle. P T is a tangent to the circle and another line from P cut the
circle at A and B.

PA×PB = PT2

1) Let P be a point outside the circle. P T is a tangent to the circle and another line from P cut the
circle at A and B.

a) What is the relation between ∠P T A, ∠P BT ?
b) Are △P T A, △P BT similar
c) Prove that P A × P B = P T 2

Answers

a) ∠P T A = ∠P BT (Angle between a chord of a circle and tangent at the end in one side is
equal to angle in the other part of the circle)

b) ∠P T A = ∠P BT , ∠P is ommon .അതിനാൽ△P T A ം △P T B are similar triangles

c) Sides opposite to the equal angles are proportional.

PT = P A , PA × PB = PT2
PB P T

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2) P T is a tangent from an outer point P to the circle.
Another line from P intersect the circle at A and B. If the length of the chord P B is 16cm and
AB = 7 cm then

a) What is the length P A?
b) What is the relation between P A, P B, P T ?
c) What is the length of the tangent P T ?
d) What is the length of the other tangent from P to the circle.
Answers

a) P A = 16 − 7 = 9cm
b) P A × P B = P T 2
c) 9 × 16 = P T 2, P T = 3 × 4 = 12 cm
d) 12cm

3) BCis the diametre of the circle.P is a point on BC produced.
TangentP A is drawn from P to the circle. If P A = 6cm and P C = 3cm then

a) What is the length P B?
b) Find the radius of the circle.

Answers

a) P B × P C = P A2
P B × 3 = 62, P B = 12cm

b) BC = 12 − 3 = 9cm

radius = 9 = 4.5cm
2

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4) In △ABC, AB = AC, A circle passing through B intersect AB at P . The circle touches AC
at its mid point D

Prove that 4AP = AB

Answers

⋆ AB × AP = AD2

⋆ AB × AP = ( AC )2
2
AC 2
AB × AP = 4

⋆ AB = AC → AB × AP = AB2
4

⋆ AP = AB
4
AB = 4 × AP

5) In the figure BC is the diametre of the circle and AB is a tangent.

a) Write the relation between AC, AD and AB
b) Prove that AC × CD = BC2

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Answers

a) AC × AD = AB2

b) Consider△ABC and △BDC.

∠ABC = 90◦(Angle between diameter and tangent)

∠BDC = 90◦(angle in the semicircle )

∠ABC = ∠BDC, ∠C is common

△ABC are similar △BDC

BC = AC
CD BC
AC × CD = BC2

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2020-21 Academic year Worksheets

Mathematics X
Tangents

Concepts

If P T is a tangent to a circle from the outer point P and another line from P intersect the cxircle

at A and B then

PA×PB = PT2

This can be proved by Pythagorous theorem also.

⋆ P A × P B = (P M − AM )(P M + BM ) = (P M − AM )(P M + AM )
= (P M 2 − AM 2) = (P M 2 − (OA2 − OM 2)) = P M 2 + OM 2 − OA2
= OP 2 − OA2 = OP 2 − OT 2 = P T 2

1) In the figure AB = BD, also the line AD is a tangent from A.

a) What is the relation between AB, AC and AD
b) Prove that AB × AC = CD2
c) What kind of triangle is △ACD?
d) If ∠BAD = 30◦ and perpendicular distance from D to AB is 12 then what is the length of

tangent AD?
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Answers

a) AB × AC = AD2
b) Consider △ABD and △ACD.

∠ADB = ∠ACD(Angle between chord and the tangent at the end is equal to angle in the
other side of the chord)
Since AB = BD, opposite angles are equal. .∠BAD = ∠ADB
Therefore ∠ADB = ∠ACD → AD = CD AB × AC = AD2 → AB × AC = CD2
c) In △ACD, ∠A = ∠C opposite angles are equal . This is an isosceles triangle.
d) △AP Dis a 30◦ −60◦ −90◦triangle. Side opposite to 30◦ is 12cm .Therefore AD = 24cm
. Length of tangent is24cm

2) In the figure ACis the diametre and BA is a tangent to the circle. The line BC intersect the circle
at P
If the radius of the circle is 2.5cm and the length of tangent is 12cm

a) What is the length BC?
b) What is the length P C?
c) What is the length AP ?

Answers

a) △CAB is a right triangle . ( Diameter is perpendicular to the tangent )
BC2 = AC2 + AB2 = 52 + 122 = 169, BC = 13cm

b) BP × BC = BA2, BP × 13 = 122, BP = 144 = 11.08 cm
13
P C = 13 − 11.08 = 1.92cm

c) AP 2 = AC2 − P C2, AP 2 = 52 − 1.922 = 25 − 3.68 = 21.32cm, AP = 4.6cm

3) P A is a tangent from the outer point to a circle of diametre AB.The line P B intersect the circle
at C.If the radius of the circle is 5cm and AC = 6cm then

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a) What is the length BC?
b) Find P C

Answers
√√

a) △ACB is a right triangle . BC = AB2 − AC2 = 102 − 62 = 8cm

b) △BAP is a right triangle .Take P C = x, P A = y

(8 + x)2 = y2 + 102

△ACP is a right triangle . y2 = 62 + x2

102 + y2 = (8 + x)2 → 102 + 62 + x2 = (8 + x)2, 136 + x2 = 64 + x2 + 16x,

16x = 72, x = 9 = 4.5cm . PC = 4.5cm
2
√
c) y = 62 + 4.52 = 7.5cm . Length of tangent 7.5 cm

4) Ois the centre of a circle of diametre AB. P A is a tangent from P to the circle, line P B intersect
the circle atC.If ∠AOC = 60◦, AC = 6 cm then

a) What is the measure of ∠ABC?
b) What is the diametre of the circle?
c) What is the length BC?
d) What is the length P C?

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Answers

a) ∠ABC = 30◦

b) △OAC is an equilateral triangle .OA = OC = AC = 6cm . Diametre 12cm

c) √△ACB is√a right triangle . AB2 = AC2 + BC2, BC2 = 144 − 36 = 108, BC =
108 = 6 3cm

d) Consider △AP C. ∠A = 90 − 60 = 30◦, ∠C = 90◦, ∠P = 60◦.This is a 30◦ − 60◦ −

90◦right triangle . √√
Side opposite to 60◦ is 6cm . P C = √6 = 2 3cm . P C = 2 3cm note : Length of
√3
tangent is 4 3cm

5) Draw a square of side 4cm , construct a rectangle having area equal to the area of the square and
one side 6cm . Write the principle of this construction.

Answers

⋆ Draw a circle of centre O. Draw the radius OA. Produce OA and mark B such that
AB = 4cm

⋆ Draw a square with side AB. D is an outer point of the circle and DA is a tangent of length
4cm

⋆ Draw an arc with centre D and radius 6cm which cut the circle at P and Q.
⋆ Draw a rectangle with sides DQ and DP . DP × DQ = DA2

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2020-21 Academic year Worksheets

Mathematics X
Tangents

Concepts

If P T is a tangent to a circle from the outer point P and another line from P intersect the cxircle

at A and B then

PA×PB = PT2

This can be proved by Pythagorous theorem also.

PA×PB = PT2

⋆ P A × P B = (P M − AM )(P M + BM ) = (P M − AM )(P M + AM )
= (P M 2 − AM 2) = (P M 2 − (OA2 − OM 2)) = P M 2 + OM 2 − OA2
= OP 2 − OA2 = OP 2 − OT 2 = P T 2

1) ABCDE is a regular pentagon and its circumcircle. The tangents to the circumcircle at A and B
intersect at P .

a) Draw AD, what are the angles of △ADE?
b) What is the measure of ∠ADB?
c) Tangents at A, Bintersect at P . What is the measure of ∠BAP ?
d) What is the measure of ∠AP B? 1

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Answers

a) Angle sum of an n sided polygon = (n − 2) × 180◦

∠AED = (5−2)×180 = 108◦
5
ED = EA → ∠EAD = ∠EDA = 36◦

b) ∠ADB = 108 = 36◦
3

c) ∠BAP = 36◦(Angle between chord AB and tangent AP is equal to angle in the other

side of the chord on the circle)

d) ∠AP B = 180 − (36 + 36) = 108◦

2) In △ABC the sides AB, BC and AC touches a circle at D, E, F .
If AB = 12cm ,BC = 8 cm ,AC = 10cm then find AD, BE and CF .

Answers
⋆ Since AD = x ,AF = x.(Tangents from outer point to the circle are equal)
⋆ BD = 12 − x, BE = 12 − x, CF = 10 − x, CE = 10 − x
⋆ BC = BE + EC, 8 = 12 − x + 10 − x, 8 = 22 − 2x, 2x = 14, x = 7cm
⋆ AD = 7cm , BE = 12 − x = 5cm , CF = 10 − x = 10 − 7 = 3cm

3) M is the mid point of AB.Semicircles are drawn with AM ,BM and AB as the diametre .Another
circle of radius r touches the semicircles.

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a) Prov that r = AB
6

b) If AB = 12then what is the radius of the circle?

c) If AB = 12 what is the perimetre of the triangle formed by joining the centre of the circle
and small semicircles?

Answers

a) AC = CM = M D = DB = DP = DQ = x, OP = OQ = r

△OCD is an isosceles triangle .OC = r + x, OD = r + x

OM is the altitude, OC2 = OM 2 + CM 2,

(r + x)2 = (2x − r)2 + x2

r2 + 2rx + x2 = 4x2 + r2 − 4rx + x2, 3r = 2x, r = 2 x = 2 × AB = AB
3 3 4 6

b) r= AB = 12 =2
6 6

c) If AB = 12 cm x = 12 , r = 2
3
Perimeter of △COD = 4x + 2r = 12 + 4 = 16cm

4) A semicircle is drawn with AB as the diametre in the square ABCD. DE touches the semicircle
at P . If the side of the square is of length 1unit

a) What is the length DP ?
b) If P E = x then find the equation connecting DE, CD and CE
c) Find the length of the line DE.

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Answers

a) side of the square is 1. Since DA = 1 , DP = 1

b) If P E = xthen DE = 1 + x.CE = 1 − x, CD = 1
(1 + x)2 = (1 − x)2 + 12

c) 4x = 1 → x = 1 = 0.25, DE = 1.25
4

5) In the figure AB = AC, BC = 10 cm , altitude from A to BC is 12 cm.The centre of the
semicircle is on BC and the semicircle touches the sides AB and AC.

a) What is the perimetre of △ABC?
b) What is the area of triangle ABC?
c) What is the radius of semicircle?

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