Worksheet Solutions: Circles

True and False

Q1: If two arc of a circle are congruent. Then corresponding chord are unequal.
Ans: False (Corresponding chords are equal)

Q2:  Two perpendicular bisector of chord intersect at center of circle.
Ans: True ( each perpendicular bisector of chord passes through the center so center is common point for the two perpendicular bisectors of the chords)

Q3: The line joining the mid-point of a chord to centre perpendicular to chord. 
Ans: True (A line joining the mid-point of a chord to the centre of circle, perpendicular to the chord)

Q4: It is possible to draw two circles from three non-collinear points.
Ans: False (One and only one circle can be passed through three co-linear points in a plane.)

Answer the following Questions

Q1: If O is the center of circle of radius 5 cm OP perpendicular to AB and OQ perpendicular to CD, AB||CD, AB = 6cm and CD = 8 cm. Determine PQ.
Ans:

Q2: AB and CD are the two chords of the circle such that AB = 6 cm, CD = 12 cm and AB||CD, if the distance between AB and CD is 3 cm, find the radius of the circle.
Ans:

Q3: Prove that the line joining to the centre of circle to the mid-point of a chord, is perpendicular to the chord.
Ans:

O is the centre of the circle and AB is the Chord ,OM is the line segment intersecting at M , Mid point of AB

In ΔAMO and ΔBMO

AM = BM (M is the mid point of AB)

OA = OB (radius of circle)

OM = OM (Same side)

Hence

ΔAMO ≌ ΔBMO (By SSS)

∠AMO = ∠BMO ( BY C.P.C.T)

∠AMO +∠ BMO = 180 (linear pair angles)

∠AMO = 90

Q4: Given an arc of circle how you will find its centre and complete the circle.
Ans:

Construction : 

Step 1: Take 3 points P,Q,R on circumference of arc join P to Q and R to Q

Step 2: Draw Perpendicular bisector of line PQ and RQ these intersect at point O

Step 3: join O to P , O to Q and O to R

O is the centre of given arc where OQ and OR and OP are the radius of circle

Q5: Two equal chord AB and CD of circle with center O, when produced meet at a point E, proving that BE=DE and AE=CE
Ans:

Let OL and OM be two perpendiculars from centre O to chord AB and CD respectively so L and M are the mid points of Chord AB and CD

Since AB = CD (Given)

AB/2 = CD/2 or LB = MD

In ΔOLE and ΔOME

OL = OM (Equal Chords having equal distance from the centre)

<OLE = <OME (both 90 degrees)

OE = OE (common side)

So BY R.H.S

ΔOLE ≌ ΔOME

So LE = ME ( by C.P.C.T)

LE = LB+BE

ME = MD+DE

LB+BE = MD+DE

But LB = MD (proved above)

So BE = DE

Q6: Two equal circles intersect in P and Q. A straight line through P meets the circles in A and B. prove that QA=QB
Ans:

Let O and O’ be the centre of the circle where C(o, r) ≌ C(o’, r) circles are equal

So PQ is the common arc in both the circle

arc(PDQ) = arc(PCQ)

<QAP = <QBP (equal chords make equal angles on the circles)

in triangle ABQ <A = <B

so

Side AQ = Side BQ (Sides opposite to equal sides are equal)

True and False

Q1: If two arc of a circle are congruent. Then corresponding chord are unequal.

Q2:  Two perpendicular bisector of chord intersect at center of circle.

Q3: The line joining the mid-point of a chord to centre perpendicular to chord.

Q4: It is possible to draw two circles from three non-collinear points.

Answer the following Questions

 Q1: If O is the center of circle of radius 5 cm OP perpendicular to AB and OQ perpendicular to CD, AB||CD, AB = 6cm and CD = 8 cm. Determine PQ.

Q2: AB and CD are the two chord of the circle such that AB = 6 cm , CD = 12 cm and AB||CD, if the distance between AB and CD is 3 cm, find the radius of the circle.

Q3: Prove that the line joining to the centre of circle to the mid-point of a chord, is perpendicular to the chord.

Q4: Given an arc of circle how you will find its centre and complete the circle.

Q5: Two equal chord AB and CD of circle with center O, when produced meet at a point E, prove that BE=DE and AE=CE

Q6: Two equal circles intersect in P and Q. A straight line through P meets the circles in A and B. Prove that QA=QB.

Multiple Choice Questions

Q1: ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD
Which of the following is true based on given information
(a) AP = CQ
(b) QD = PB
(c) DP = QB
(d) ΔPAD ≅ ΔQCB
Ans: (a, b, c, d)

Sol: All are correct In Triangle ΔPAD and ΔQCB
AD = CB
∠P = ∠Q = 90º
∠CBQ = ∠ADP (Alternate interior angles of AB||CD)
So AAS congruence
Also as they are congruent, we get AP = CQ and DP = QB
Now Let’s see the triangles ΔAPB and ΔCQD
AB = CD
∠P = ∠Q = 90º (Alternate interior angles of AB||CD)
So QD = PB

Q2: The angles of the quadrilateral are in the ratio 2 : 5 : 4 : 1? Which of the following is true?
(a) Largest angle in the quadrilateral is 150º
(b) Smallest angle is 30º
(c) The second largest angle in the quadrilateral is 80º
(d) None of these
Ans: (a, b)

Sol: Angles are 2 x , 5 x , 4 x , x
Now
2x + 5x + 4x + x = 360
Or x = 30
Angles are 30º, 60º, 120º, 150º

Q3: Two adjacent angles in a parallelogram are in the ratio 2 : 4. Find the values?
(a) 80, 100
(b) 40, 140
(c) 60, 120
(d) None of the above
Ans: (c)

Sol: Adjacent angles 2x + 4x = 180
x = 30
60, 120 are adjacent angles

Q4: ABCD is a trapezium with AB = 10cm, AD = 5 cm, BC = 4 cm and DC = 7 cm?
Find the area of the ABCD
(a) 34 cm²
(b) 28cm²
(c) 20 cm²
(d) None of these
Ans: (a)

Sol: BC is the altitude between the two parallel sides AB and DC
So Area of trapezium will be given by
A = 1/2 BC (AB + DC) = 34cm² 

Q5: ABCD is a trapezium where AB||DC. BD is the diagonal and E is the mid point of AD. A line is draw from point E parallel to AB intersecting BC at F. Which of these is true?
(a) BF = FC
(b)  EA = FB
(c) CF = DE
(d) None of these
Ans: (a)

Sol: Let’s call the point of intersection at diagonal as G
Then in triangle DAB
EG||AB and E is the mid point of DA, So by converse of Midpoint Theorem,
G is the mid point of BD
Now in triangle DBC
GF||CD
G is the mid point of DB
So by converse of mid point theorem
F is the mid point of BC

True or False

Q1: The diagonals of a parallelogram bisect each other.
Ans: True. It is by definition

Q2: In a parallelogram, opposite sides and angle are equal.
Ans: True. It is by definition

Q3: A diagonal of a parallelogram divides it into two congruent triangles.
Ans: True. This can be proved easily using SSS congruence

Q4: The bisectors of the angles of parallelogram create a rectangle.
Ans: True

Q5: Sum of all the internal angles is 360^∘.
Ans: True. This can easily proved by drawing one diagonal and summing all the angles based on triangle angle sum.

Q6: Sum of all the exterior angles is 180^∘.
Ans: False

Q7: Square, rectangle and rhombus are all parallelogram.
Ans: True

Q8: Consecutive angles are supplementary.
Ans: True

Answer the following Questions

Q1: Show that the quadrilateral formed by joining the mid- points of adjacent sides of rectangle is a rhombus.
Ans: The figure is shown as below

To Prove: quadrilateral PQRS is a rhombus
Proof:
In Δ ABC
P and Q are mid points of sides AB and BC
By Mid point theorem
PQ = 1/2 AC and PQ || AC -(X)
In Δ ACD
S and R are mid points of sides AD and DC
By Mid point theorem
SR = 1/2 AC and SR || AC –(Y)
From (X) and (Y), we have
PQ = SR
PQ ||SR
Hence PQRS is a parallelogram
Now in Δ BCD
Q and R are mid points of sides BC and DC
By Mid point theorem
QR = 1/2 BD
Nowe AC = BC
Hence
PQ = SR = QR
Now a parallelogram whose adjacent sides are equal is a rhombus.
Hence proved

Q2: P, Q, R and S are respectively the mid-point of sides AB, BC, CD and DA of a quadrilateral ABCD such that AC = BD. Prove that PQRS is a rhombus.

Ans:
Given AC = BD
Proof:
In Δ ABC
P and Q are mid points of sides AB and BC
By Mid point theorem
PQ = 1/2 AC and PQ || AC -(X)
In Δ ACD
S and R are mid points of sides AD and DC
By Mid point theorem
SR = 1/2 AC and SR || AC –(Y)
From (X) and (Y), we have
PQ = SR = 1/2 AC —-(1)
Similarly in Δ BCD
Q and R are mid point of BC and CD
By Mid point theorem
QR = 1/2 BD
Similarly in Δ ADB
S and P are mid point of AD and AB
By Mid point theorem
SP = 1/2 BD
Therefore
SP = QR = 1/2 BD —(2)
AC = BD
So from (1) and (2)
PQ = SR = SP = QR
Hence PQRS is a rhombus

Q3: l, m and n are three parallel lines intersected by transversal’s p and q such that l, m and n cut off equal intercepts AB and BC on p. Show that l, m and n cut off equal intercepts DE and EF on q also.

Ans:
Given:
AB = BC
To Prove:
DE = EF
Proof:
Let us join A to F intersecting m at G
The trapezium ACFD is divided into two triangles namely Δ ACF and & Δ AFD
In Δ ACF, it is given that B is the mid-point of AC (AB = BC) and BG || CF (since m || n).
So, G is the mid-point of AF (by Mid Point Theorem)
Now, in Δ AFD, we can apply the same argument as G is the mid-point of AF,
GE || AD and so by by Mid Point Theorem, E is the mid-point of DF,
i.e., DE = EF.
In other words, l, m and n cut off equal intercepts on q also.

Q4. Find all the angles of a parallelogram if one angle is 80°.

Ans: For a parallelogram ABCD, opposite angles are equal.

So, the angles opposite to the given 80° angle will also be 80°.

It is also known that the sum of angles of any quadrilateral = 360°.

So, if ∠A = ∠C = 80° then,

∠A + ∠B + ∠C + ∠D = 360°

Also, ∠B = ∠D

Thus,

80° + ∠B + 80° + ∠D = 360°

Hence, 2∠B = ∠D = 200°/2

Now, all angles of the quadrilateral are found which are:

∠A = 80°

∠B = 100°

∠C = 80°

∠D = 100°

Q5: In a trapezium ABCD, AB//CD. Calculate ∠C and ∠D if ∠A = 55° and ∠B = 70°

Ans: In a trapezium ABCD, ∠A + ∠D = 180° and ∠B + ∠C = 180°

So, 55° + ∠D = 180°

Or, ∠D = 125°

Similarly,

70° + ∠C = 180°

Or, ∠C = 110°

Multiple Choice Questions

Q1: ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD
Which of the following is true based on given information
(a) AP = CQ
(b) QD = PB
(c) DP = QB
(d) ΔPAD ≅ ΔQCB

Q2: The angles of the quadrilateral are in the ratio 2 : 5 : 4 : 1? Which of the following is true?
(a) Largest angle in the quadrilateral is 150º
(b) Smallest angle is 30º
(c) The second largest angle in the quadrilateral is 80º
(d) None of these

Q3: Two adjacent angles in a parallelogram are in the ratio 2 : 4. Find the values?
(a) 80, 100
(b) 40, 140
(c) 60, 120
(d) None of the above

Q4: ABCD is a trapezium with AB = 10cm, AD = 5 cm, BC = 4 cm and DC = 7 cm?
Find the area of the ABCD
(a) 34 cm²
(b) 28cm²
(c) 20 cm²
(d) None of these

Q5: ABCD is a trapezium where AB||DC. BD is the diagonal and E is the mid point of AD. A line is draw from point E parallel to AB intersecting BC at F. Which of these is true?
(a) BF = FC
(b) EA = FB
(c) CF = DE
(d) None of these

True or False

Which are these is true or false about parallelogram
Q1: The diagonals of a parallelogram bisect each other.

Q2: In a parallelogram, opposite sides and angle are equal.

Q3: A diagonal of a parallelogram divides it into two congruent triangles.

Q4: The bisectors of the angles of parallelogram create a rectangle.

Q5: Sum of all the internal angles is 3600.

Q6: Sum of all the exterior angles is 1800.

Q7: Square, rectangle and rhombus are all parallelogram.

Q8: Consecutive angles are supplementary.

Answer the following Questions

Q1: Show that the quadrilateral formed by joining the mid-points of adjacent sides of rectangle is a rhombus.

Q2: P, Q, R and S are respectively the mid- point of sides AB, BC, CD and DA of a quadrilateral ABCD such that AC = BD. Prove that PQRS is a rhombus.

Q3: l, m and n are three parallel lines intersected by transversal’s p and q such that l, m and n cut off equal intercepts AB and BC on p. Show that l, m and n cut off equal intercepts DE and EF on q also.

Q4. Find all the angles of a parallelogram if one angle is 80°.

Q5: In a trapezium ABCD, AB∥CD. Calculate ∠C and ∠D if ∠A = 55° and ∠B = 70°

Multiple Choice Questions

Q1: If AD = BC and ∠ BAD = ∠ ABC, then ∠ ACB is equal to
(a) ∠ABD
(b) ∠ BAD 
(c) ∠BAC 
(d) ∠BDA 
Ans: (d)
In △ABC and △ABD
AD =BC     (given)
∠ BAD = ∠ ABC    (Given)
AB = AB   (Common side)
∴ △ABC ≅ △ABD  (by SAS congruency)
By CPCT theorem, ∠ACB=∠BDA.

Q2: If O is a midpoint of AB and ∠BQO = ∠APO, then ∠OAP is equal to
(a) ∠QPA
(b) ∠OQB
(c) ∠QBO
(d) ∠BOQ
Ans: (c)
In △AOP and △BOQ,
AO = BO (O is the midpoint of AB)
∠APO = ∠BQO (given)
∠AOP = ∠BOQ (vertically opposite angles)
∴ △AOP ≅ △BOQ (by AAS congruency)
By CPCT, ∠OAP = ∠QBO.

Q3: If △ABC is an isosceles triangle, AB = AC,∠ B = 65º, then find ∠ A.
(a) 60º
(b) 70º
(c) 50º
(d) none of these
Ans: (c)
Since △ABC is an isosceles triangle,
∴ ∠B = ∠C
∴ ∠B = 65°
∴ ∠C = 65°
Using the angle sum property of a triangle,
∠A + ∠B + ∠C = 180°
∴ ∠A + 65° + 65° = 180°
∴ ∠A + 130° = 180°
∴ ∠A = 180° − 130° = 50°

Q4: An angle is 14º more than its complement. Find its measure.
(a) 42
(b) 32
(c) 52
(d) 62
Ans: (c)
Two angles whose sum is 90° are called complementary angles.
Let the first angle be x.
Its complement = 90° − x.
According to the question,
x = 14° + (90° − x)
x = 104° − x
⇒ 2x = 104°
⇒ x = 52°
∴ The angle is 52°.

Q5: If ABCD is a quadrilateral where AD = CB, AB = CD, and ∠ D = ∠ B, then ∠CAB is equal to
(a) ∠ACD
(b) ∠CAD
(c) ∠ACD 
(d) ∠BAD
Ans: (c)
In △ABC and △CDA,
CB = AD (given)
AB = CD (given)
∠B = ∠D (given)
∴ △ABC ≅ △CDA (by SAS congruency)
By CPCT theorem,
∠CAB = ∠ACD.

Q6: If AB ⊥BC and ∠A =∠C, then the correct statement will 
(a) AB ≠ AC
(b) AB = BC 
(c) AB = AD 
(d) AB = AC
Ans: (b)
In △ABC, ∠A = ∠C.
Opposite sides to equal angles are also equal.
Therefore, AB = BC.

Q7: If AB = AC and ∠BAC = 120°, find ∠B.
(a) 50°
(b) 60°
(c) 70°
(d) none of these
Ans: (b)
Given AB = AC,
Therefore, ∠ABC = ∠ACB = x.
In △ABC,
∠A + ∠B + ∠C = 180° (sum of angles of a triangle)
120° + x + x = 180°
2x = 60°
x = 30°
Then, ∠B = ∠C = 30°.

Answer the following questions

Q1: AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB.

Ans: In ΔAOD and ΔBOC,
BC = AD (given)
∠OAD = ∠OBC = 90°
Since BC || AD 
∠ODA = ∠OCB  (alternate angles are equal)
So, ΔAOD ≅ ΔBOC (by the ASA congruence rule) 
OD = OC (CPCT)
AO = OB  (CPCT)
Therefore, CD bisects AB.

Q2: AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB. Show that
(i) ΔDAP ≌ ΔEBP
(ii) AD = BE
Ans. (i) ΔIn DAP and ΔEBP

P is its mid-point of AB
∴ AP = PB
So, ΔDAP ≌ ΔEBP(by the ASA congruence rule)
(ii) AD = BE(CPCT)

Q3: In an isosceles triangle ABC with AB = AC, D and E are points on BC such that BE = CD. Show that AD = AE.
Ans: In ΔABD and ΔACE,
AB = AC (Given) ..(1)
∠B = ∠C (Angles opposite to equal sides) ..(2)
Also, BE = CD..(3)
So, BE – DE = CD – DE
That is, BD = CE (3)
So, ΔABD ≌ ΔACE  by SAS rule
(Using (1), (2), (3) and SAS rule).
This gives AD = AE (CPCT)

Q4: In Figure OA = OB and OD = OC.
Show that
(i) ΔAOD ≅ ΔBOC
(ii) AD || BC
Ans: (i) In ΔAOD and ΔBOC,
OA = OB (given)
OD = OC (given)
∠AOD = ∠BOC (pair of vertically opposite angles)
So, ΔAOD ≅ ΔBOC(by the SAS congruence rule)
(ii) ∠OAD = ∠OBC (CPCT)
and these form a pair of alternate angles for line segments AD and BC.
Therefore, AD || BC.

Q5: In Fig, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

Ans. In ΔABC and ΔADE,
AB = AD (given)
AC = AE (given)
∠BAD = ∠EAC (given)
∠ADC+ ∠BAD = ∠ADC + ∠EAC
∠BAC = ∠DAE
So, ΔABC ≌ ΔADE (by the SAS congruence rule)
BC = DE (CPCT)

Q6: In ΔABC, the bisector AD of ∠A is perpendicular to side BC. Show that AB = AC and ΔABC is isosceles.
Ans: In ΔABD and ΔADC
AD bisects ∠A
⇒ ∠BAD = ∠CAD
∠ADC = ∠ADB 90º
AD = AD (Common Side)
So, ΔABD ≌ ΔADC (by the SAS congruence rule)
AB = AC(CPCT)
∴ ΔABC is isosceles

Q7: ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal. Show that
(i) ΔABE ≌ ΔACF
(ii) AB = AC, i.e., ABC is an isosceles triangle.
Ans: In ΔABE and ΔACF,
BE = CF (given)
∠A = ∠A (common angle)
∠ABE = ∠ACF (both are right angles)
So, ΔABE ≅ ΔACF (by ASA congruence rule).
This gives AB = AC (CPCT).
∴ ABC is an isosceles triangle.

Multiple Choice Questions

Q1: If AD = BC and ∠ BAD = ∠ ABC, then ∠ ACB is equal to
(a) ∠ABD
(b) ∠ BAD
(c) ∠BAC
(d) ∠BDA

Q2: If O is a midpoint of AB and ∠BQO = ∠APO, then ∠OAP is equal to
(a) ∠QPA
(b) ∠OQB
(c) ∠QBO
(d) ∠BOQ

Q3: If △ABC is an isosceles triangle, ∠ B = 65º, AB = AC,∠ B = 65º, then find ∠ A.
(a) 60º
(b) 70º
(c) 50º
(d) none of these

Q4: An angle is 14​º more than its complement. Find its measure.
(a) 42º
(b) 32º
(c) 52º
(d) 62º

Q5: If ABCD is a quadrilateral where AD= CB, AB=CD, and ∠ D= ∠ B, then ∠CAB is equal to
(a) ∠ACD
(b) ∠CAD
(c) ∠ACD
(d) ∠BAD

Q6: If AB ⊥BC and ∠A =∠C, then the correct statement will
(a) AB ≠ AC
(b) AB = BC
(c) AB = AD
(d) AB = AC

Q7: If AB = AC and ∠ ACD = 120^​º, find ∠A.
(a) 50º
(b) 60º
(c) 70º
(d) None of these

Answer the following questions

Q1: AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB.

Q2: AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB. Show that
(i) ΔDAP ≌ ΔEBP
(ii) AD = BE

Q3: In an isosceles triangle ABC with AB = AC, D and E are points on BC such that BE = CD. Show that AD = AE.

Q4: In Figure OA = OB and OD = OC.
Show that
(i) ΔAOD ≅ ΔBOC
(ii) AD || BC

Q5: In Fig, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

Q6: In ΔABC, the bisector AD of ∠A is perpendicular to side BC. Show that AB = AC and ΔABC is isosceles.

Q7: ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal. Show that
(i) ΔABE ≌ ΔACF
(ii) AB = AC, i.e., ABC is an isosceles triangle

Multiple Choice Questions

Q1: Find the value of x(a) 98º

(b) 100º

(c) 108º

(d)96º

Ans: (c)

x + 36 + 36 = 180 ⇒ x = 108º

Q2: A pair of angles is called linear pair if the sum of two adjacent angles is?

(a) 180º

(b) 90º

(c) 270º

(d) 360º

Ans: (a)

Q3: Find the value of x, y and z(a) x = 110º , y = 70º, z = 80º(b) x = 70º , y = 110º, z = 60º(c) x = 70º, y = 100º, z = 70º(d) x = 70º, y = 110º, z = 70º

Ans: (d)

Vertically opposite angle theorem and linear pair axiom can be used to find the answer

Q4: An exterior angle of the triangle is 110º. And its two opposite interior angles are in the ratio 5:6. What are the values of those angles?

(a) 50º, 60º

(b) 25º, 30º

(c) 35º, 42º

(d) 40º, 48º

Ans: (a)

Q5: Lines l || k and m || n. Find the value of angle z(a) 45º

(b) 60º

(c) 70º

(d) 50º

Ans: (b)

Q6:Find the value of x(a) 67º(b) 71º(c) 57º(d) None of these

Ans: (a)

Q7: The side BC, AB and AC of the triangle ABC are produced in order forming exterior angles ∠ACD = x, ∠BAE = y and ∠CBF = z then the value of 2x + 2y + 2z is

(a) 180°

(b) 360°

(c) 540°

(d) 720°

Ans: (d)

Let a, b,c be the angles of triangle, then a + b + c = 180Now x = a + b, y = b + c, z = a + cTherefore 2x + 2y + 2z = 4(a + b + c) = 720

Q8: The angles of the triangles are in the ratio 1 : 2 : 3. Then the triangle is

(a) scalene

(b) obtuse angled

(c) acute angled

(d) right angles

Ans: (d)

Let x, 2x, 3x be the angles of triangle, then x + 2x + 3x = 180Therefore x = 30So angles are 30, 60 and 90

True or False

Q1: Pairs of vertically opposite angles is always equal

Ans: True

Q2: The sum of the angles of a triangle is 180º

Ans: True

Q3: If the sum of two adjacent angles is 45º, then two adjacent angles are acute angles

Ans: True

Q4: If a line is perpendicular is one of two parallel lines, then it is also perpendicular to the other

Ans: True

Q5: Two lines are intersected by the transversal, and then the corresponding angles are equal

Ans: False

Q6: Can we have a triangle where all the interior angles are more than 60º

Ans: False

Q7: Sum of two complimentary angles is equal to 90º

Ans: True

Q8: Sum of all the exterior angles of any polygon is always 360º

Ans: True

Fill in the blanks 

Q1: Sum of two supplementary angles is ______.

Ans: 180º

Q2: Two lines parallel to the same line is ____ each other.

Ans: parallel

Q3: An acute angle is always less than _____.

Ans: 90º

Q4: Angles forming a linear pair are ______.

Ans: supplementary

Q5: If one angle of triangle is equal to the sum of other two angles, then the triangle is ______.

Ans: right angle triangle

Q6: If two straight lines intersect, the adjacent angles are ______.

Ans: supplementary

Table Type Question

Ans: Complementary = 90 − xSupplementary = 180 − x

Short Answer Type Questions

Q: Write the type of angles

(i) 

(ii) 

(iii) 

(iv) 

(v) 

(vi) 

(vii) 

Ans:

(i) Acute angle

(ii) Right angle

(iii) Obtuse angle

(iv) Straight angle

(v) Reflex angle

(vi) Vertically opposite angle

(vii) Alternate interior angles

Multiple Choice Questions

Q1: Find the value of x(a) 98º

(b) 100º

(c) 108º

(d) 96º

Q2: A pair of angles is called linear pair if the sum of two adjacent angles is?

(a) 180º

(b) 90º

(c) 270º

d) 360º

Q3: Find the value of x, y and z(a) x = 110º , y = 70º, z = 80º

(b) x = 70º , y = 110º, z = 60º(c) x = 70º, y = 100º, z = 70º(d) x = 70º, y = 110º, z = 70ºQ4: An exterior angle of the triangle is 110º. And its two opposite interior angles are in the ratio 5:6. What are the values of those angles?(a) 50º, 60º(b) 25º, 30º(c) 35º, 42º(d) 40º, 48ºQ5: Lines l || k and m || n. Find the value of angle z(a) 45º

(b) 60º

(c) 70º

(d) 50ºQ6:Find the value of x

(a) 67º

(b) 71º

(c) 57º

(d) None of these

Q7: The side BC, AB and AC of the triangle ABC are produced in order forming exterior angles ∠ACD = x, ∠BAE = y and ∠CBF = z then the value of 2x + 2y + 2z is

(a) 180°

(b) 360°

(c) 540°

(d) 720°

Q8: The angles of the triangles are in the ratio 1 : 2 : 3. Then the triangle is

(a) scalene

(b) obtuse angled

(c) acute angled

(d) right angles

True or False

Q1: Pairs of vertically opposite angles is always equal

Q2: The sum of the angles of a triangle is 180º

Q3: If the sum of two adjacent angles is 45º, then two adjacent angles are acute angles

Q4: If a line is perpendicular is one of two parallel lines, then it is also perpendicular to the other

Q5: Two lines are intersected by the transversal, and then the corresponding angles are equal

Q6: Can we have a triangle where all the interior angles are more than 60º

Q7: Sum of two complimentary angles is equal to 90º

Q8: Sum of all the exterior angles of any polygon is always 360º

Fill in the blanks 

Q1: Sum of two supplementary angles is ______.

Q2: Two lines parallel to the same line is ____ each other.

Q3: An acute angle is always less than _____.

Q4: Angles forming a linear pair are ______.

Q5: If one angle of triangle is equal to the sum of other two angles, then the triangle is ______.

Q6: If two straight lines intersect, the adjacent angles are ______.

Table Type QuestionShort Answer Type Questions

Q: Write the type of angles

(i) 

(ii) 

(iii) 

(iv) 

(v) 

(vi) 

(vii) 

Q1. If A, B and C are three points on a line, and ‘B’ lies between ‘A’ and ‘C’ (as shown in the figure), then prove that: AB + BC = AC

Sol.

In the figure given above, AC coincides with AB+ BC

From Euclid’s Axiom 4: Things which coincide with one another are equal to each other.

So, we write AB+BC=AC

Q2. Prove that an equilateral triangle can be constructed on any given line segment. 

Sol. Equilateral triangle is a angle with all sides are equal

1. Draw a line segment AB of the any length.

2. Take compass put the pointy end at point A & pencile at point B.

3.Draw an arc.

Here we draw an arc of radii AB

4. Now put the pointy end at B and pencil at A.

5. Draw another arc.

Here we draw an arc of radii BA .

6. Mark the intersecting point as C.

7. Join point A to point C by a straight line .

8. Join point B to point C by a straight line

ABC is the triangle.

Q3. Prove that two distinct lines cannot have more than one point in common.

Ans 3. 

Given: Two distinct line l and m

To Prove: Lines l and m have at most one point in common.

Proof: Two distinct lines l and m intersect at a point P.

Let us suppose they will interact at another point, say Q (different from P).

It means two lines l and m passing through two distinct point P and Q.

But it is contrary to the axiom 5.1 which states that “Given two distinct points, there exists one and only one line pass through them”

So our supposition is wrong

Hence, two distinct lines cannot have more than one point in common

Q4. Is the following statement true? “Attempts to prove Euclid’s fifth postulate using  the other postulate and axioms led to the discovery of several other geometries.”

True

Q5. Fill in the blanks to complete the following axioms :
(i) Things, which are equal to the same things, are ………………………….
(ii) If equals are added to equals, the ………………………….
(iii) If equals are subtracted from equals, ………………………….
(iv) Things which coicide with one another are ………………………….
(v) The whole is greater than the ………………………….
(vi) Things which are double of the same things are ………………………….
(vii) Things which are halves of the same things, are ………………………….

Sol (i) equal to one another
(ii) wholes are equal
(iii) the remainders are equal
(iv) equal to one another
(v) part
(vi) equal to one another
(vii) equal to one another

Q6. In the figure, line PQ falls on AB and CD such that (∠1 + ∠2) < 180°. So, lines AB and CD, if produced will intersect on the left of PQ. This is an example of which Postulate of Euclid?

Sol. Fifth postulate

Q7. In the given figure, if AC = BD, then prove that AB = CD

Hint: AC = BC   [given]     …(1)
∴ AC = AB + CD [∵ B lies between A and C]    …(2)
BD = BC + CD [∵ C lies between B and D]      …(3)
From (1), (2) and (3) we have: AB + BC = BC + CD or AB = CD

Sol 

From the figure, it can be observed that

AC = AB + BC

BD = BC + CD

It is given that AC = BD

AB + BC = BC + CD (1)

According to Euclid’s axiom, when equals are subtracted from equals, the remainders are also equal.

Subtracting BC from equation (1), we obtain

AB + BC − BC = BC + CD − BC

AB = CD

Q8. Write ‘true’ or ‘false’ for the following statement:
(i) Three lines are concurrent if they have a common point.
(ii) A line separates a plane into three parts, namely the two half-planes and the line itself. (iii) Two distinct lines in a plane cannot have more than one point in common.
(iv) A ray has two endpoints.
(v) A line has indefinite length.

Sol. (i),  (ii), (iii) and  (v) are true.   

Q9. How many lines can pass through:
(i) one point
(ii) two distinct points?

Sol (i) infinite  
(ii) one only

Q10. AB and CD are two distinct lines. In how many points can they at the most intersect? 

Sol. one point only

Q11. Prove that any line segment has one and one mid-point.

Sol. Let AB be a line segment

and let D and E be its two midpoints

now, since D is the midpoints of AB

so, AD=DB

AB=AD+DB=2AD-(1)

Also E is a point of AB

So, AE=EB

AB=AE+EB=2AE-(2)

From eq 1 &2

2AD=2AE

D and E coincide to each other

AB has one and only one mid points

Hence every line segments has one and only one midpoint.

Q12. In the given figure, AB = BC, P is midpoint of AB and Q is midpoint of BC. Show that AP = QC
Hint: Things which are halves of the same thing (or equal things) are equal to one another.

Sol. We have AB = BC…(1) [Given]

Now, A, M. B are the three points on a line, and M lies between A and B, then

AM + MB = AB …(2)

Similarly, BN * NC = BC….(3)

So. we get AM + MB = BN * NC

From (1), (2). (3) and Euclid’s axiom 1

Since M Is the mid-point of AB and N is the mid-point of BC. therefore

2AM = 2NC

Using axiom 6, things which are double of the same thing are equal to one another.

Hence, AM = NC.

Q1. Write each of the following is an equation in two variables:
(i) x = –3
(ii)y = 2
(iii) 2x = 3
(iv) 2y = 5

Sol. 

(i)Given equation, x = -3

The above equation can be written in two variables as,

x + 0.y + 3 = 0

(ii) Given equation, y =2

The above equation can be written in two variables as,

0.x + y – 2 = 0

(iii) Given equation, 2x =3

The above equation can be written in two variables as,

2x + 0.y – 3 = 0

(iv) Given equation, 2y =5

The above equation can be written in two variables as,

2y-5= 0

(0)x + 2y- 5= 0

Q2. Write each of the following equations in the form ax + by + c = 0 and also write the values of a, b and c in each case:
(i) 2x + 3y = 3.47
(ii) x – 9 = √3 y
(iii) 4 = 5x – 8y
(iv) y = 2x

Sol. 

(i) 2x + 3y – 3.47 = 0; a = 2, b = 3 and c = –3.47
(ii) x – √3y – 9 = 0; a = 1, b = – √3 and c = –9
(iii) –5x + 8y + 4 = 0; a = –5, b = 8 and c = 4
(iv) –2x + y + 0 = 0; a = –2, b = 1 and c = 0

Q3. (a) Is (3, 2) a solution of 2x + 3y = 12?
(b) Is (1, 4) a solution of 2x + 3y = 12?
(c) Is  a solution of 2x + 3y = 12?
(d) Is  a solution of 2x + 3y = 12?

Sol.

(a) Yes, 

2(3)+ 3(2)= 6+6 =12

(b) No, 

2(1)+ 3(4)= 2+12 =14

(c) Yes, 

2(-5)+ 3(22/3)= -10+ 22 =12

(d) Yes, 

2(2)+ 3(8/3)= 4+8 =12

Q4. Find four different solutions of the equation x + 2y = 6.

Sol. To find the solutions, substitute different values of y and calculate the corresponding values of x.Hence, four different solutions are:
(6,0), (4,1), (2,2), (0,3)

Q5. Find two solutions for each of the following equations:
(i) 4x + 3y = 12
(ii) 2x + 5y = 0
(iii) 3y + 4 = 0

Sol. 1) 4x+3y=12

for y=4

4x+12=12 x=0

for y=0

4x+0=12 x=3

(0,4) & (3,0) are 2 solution

2) 2x+5y=0

for y=−2

2x−10=0 x=5

for y=−4

2x−20=0 x=10

(5,−2) and (10,−4) are 2 solutions

3) 3y+4=0

y=−4/3 is only solution

Q6. Find the value of k such that x = 2 and y = 1 is a solution of the linear equation 2x – ky + 7 = 8

Sol. We can find the value of k by substituting the values of x and y in the given equation.

By substituting the values of x = 2 and y = 1 in the given equation

2x – ky + 7 = 8

⇒ 2(2) – k(1) + 7 = 8

⇒ 4- k+ 7=8

⇒ -k=8-11

k=3

Therefore, the value of k is 3.

Q7. Draw the graph of  y+x = 4.

Sol. Let x be 0 = (0,4)

Let y be 0 = (4,0)

Q8. Force applied on a body is directly proportional to the acceleration produced in the body.
Write an equation to express this situation and plot the graph of the equation.

Sol. Given that, the force (F) is directly proportional to the acceleration (a).

i.e., F∝a

⇒F=ma [where, ,m=arbitrary constant and take value 6 kg of mass ]

∴                           F=6a

(i) If a=5m/s2, then from Eq. (i), we get

F=6×5=30N

(ii) If a=6m/s2, then from Eq. (i), we get

F=6×6=36N

Here, we find two points A (5, 30) and B (6, 36). So draw the graph by plotting the points and joining the line AB. 

Q9. For each of the graph given in the following figure select the equation whose graph it is from the choices given below:
(i) x + y = 0

(ii) x – y = 0

(iii) 2x = y
(iv) y = 2x + 1 

(i) x + y = 0

(ii) x – y = 0
(iii) y = 2x + 4
(iv) y = x – 4 

(i) x + y = 0

(ii) x – y = 0
(iii) y = 2x + 1
(iv) y = 2x – 4 

(i) x + y = 0
(ii) x – y = 0
(iii) 2x + y = –4
(iv) 2x + y = 4

Sol.
(a) x – y = 0
(b) y = 2x + 4
(c) y = 2x – 4
(d) 2x + y = –4

Q10. Which of the following is not a linear equation in two variables?
(i) px + qy + c = 0
(ii) ax² + bx + c = 0
(iii) 3x + 2y = 5

Sol. 

(ii) ax² + bx + c = 0 

(ii) is not a linear equation because it consists x² in it. Linear equation will not contain any exponent to variables

Q11. One of the solutions of the linear equation 4x – 3y + 6 = 0 is
(i) (3, 2)
(ii) (–3, 2)
(iii) (–3, –2)

Sol.  Option (iii) –3, –2 

Q12. lx + my + c = 0 is a linear equation in x and y. For which of the following, the ordered pair (p, q) satisfies it:
(i) lp + mq + c = 0
(ii) y = 0
(iii) x + y = 0
(iv) x = y

Sol. 

 $lp+mq+c=0$lp+mq+c=0

To check if $(p,q)$(p,q) satisfies the equation: $l\cdot p+m\cdot q+c=0$l⋅p+m⋅q+c=0

This matches the form of the linear equation $lx+my+c=0$lx+my+c=0, so statement (i) is correct.

Q13. What is the equation of the x-axis? 

Sol. The x-axis is the horizontal line where $y=0$y=0.

Equation of the x-axis: y=0.

Q14. What is the equation of the y-axis? 

Sol. The y-axis is the vertical line where $x=0$x=0.

Equation of the y-axis: x=0.

Q15. How many solutions do a linear equation in two variables x and y have?

Sol. 

A linear equation in two variables will have infinite solutions