NCERT Exemplar Solutions: Geometry | Mathematics (Maths) Class 6 PDF Download

Ans: (a) 
From the question it is given that, number of lines passing through five points.
We know that, we need tow points to form a line.
Therefore, number of lines passing through five points = 5 × 2
= 10

Ans: (d)
From the formula, diagonal = n(n – 3)/2
Where n = number of sides in a polygon.
There are 7 sides in a heptagon.
So, d = 7(7 – 3)/2
= 7(4)/2
= 28/2
= 14

Ans: (b)
From the figure, number of line segments are, AB, AC, AD, AE, BC, BD, BE, CD, CE, DE.
Therefore, there are 10 line segments in the given figure.

Ans: (b)
We know that, measure of complete angle of clock is 360º.

From the figure, we can say that angle at 2 = (9/12) × 360º= 270º
Then, angle at 1= 360º – 270º
= 90º
Therefore, measures of the two angles between hour and minute hands of a clock at 9 O’ clock are 270º and 90º.

Ans: (b)
We know that, bicycle wheel is in circular shape.
So, total angle of wheel is 360º.
From the question it is given that bicycle wheel has 48 spokes.
Then, the angle between a pair of two consecutive spokes is = 360º/48
= 7.5
= 7(1/2)

Ans: (b)

Ans: (a)

As per the condition given the question, point A is shifted to point B along the ray PX such that PB = 2PA.
So, by shifting the point ‘A’ there is no change at the point ‘P’.
Therefore, the measure of ∠BPY is 45º.

Ans: (d)
From the above figure,Angle are, ∠POS, ∠QOS, ∠ROS, ∠POQ, ∠QOR, ∠ROS.
Therefore, number of angle formed are 6.

Ans: (c)

Obtuse angle is defined as an angle formed is greater than 90^o and less than 180^o.
From the above figure,
Angle are, ∠ROP = ∠ROQ + ∠QOP
= 65º + 30º
= 95º
∠SOP = ∠ROQ + ∠QOP + ∠SOR
= 65º + 30º + 45º
= 140º
∠SOQ = ∠SOR + ∠ROQ
= 45º + 65º
= 110º
∠TOQ = ∠TOS + ∠SOR + ∠ROQ
= 20º+ 45º+ 65º
= 130º
Therefore, number of obtuse angle formed are 4.

Ans: (c)

By observing the above figure, there are 13 triangles.
ΔPVR, ΔPSQ, ΔSQU, ΔSQT, ΔTQU, ΔQUR, ΔVSU, ΔVST, ΔVTU, ΔPQV, ΔRQV, ΔVSQ, ΔVUQ

Ans: (d)
Because, sum of all angles of right angles is must be equal to 180º.

Ans: (d)
Because, sum of all angles of right angles is must be equal to 180º.
Right angle triangle does not contain obtuse angle.
Obtuse angle is defined as an angle formed is greater than 90º and less than 180º.

Ans: (d)
The two least consecutive prime numbers are, 2 and 3
Sum of two numbers = 2 + 3 = 5
By using formula = n(n – 3)/2
= 5(5 – 3)/2
= (5 × 2)/2
= 10/2
= 5

Ans: (c)
From the figure, ΔABC, ΔABD and ΔADC are right triangle.

Ans: (c)
We have, ∠BAC = 90° and AD ⊥ BC
∵ ∠BDA = ∠CDA = ∠BAC = 90°
∴ There are 3 right triangles formed in the given figure.

Ans: (b)
From the question it is given that, PQ ⊥ RQ, PQ = 5 cm and QR = 5 cm.
We know that, isosceles triangle has two equal side.
Therefore, the given ΔPQR is an isosceles right triangle.

An angle greater than 180° and less than a complete angle is called reflex angle.

The number of diagonals in a hexagon is 9.
We know that, hexagon has 6 sides.
By using the formula, number of diagonals = n(n – 3)/2
= 6(6 – 3)/2
= 6(3)/2
= 18/2
= 9

A pair of opposite sides of a trapezium are parallel.

In Fig. points lying in the interior of the triangle PQR are O and S, that in the exterior are N and T and that on the triangle itself are M, P, Q and R.

(a) AD = AB + BD
Where, BD = BC + CD
(b) AD = AC + CD
(c) mid point of AE is c
(d) mid point of CE is D
(e) AE = 4 × AB. [given]

From the figure,
(a) ∠AOD is a/an right angle.
Is anyone angle in the triangle is equal to 90º then the triangle is called right triangle.
∠AOD = ∠AOB + ∠BOC + ∠COD
= 30º + 20º + 40º

The number of triangles in Fig. is 5. Their names are ΔAOC, ΔCOD, ΔAOB, ΔACB, ΔACD.

Number of angles less than 180° in Fig. is 12 and their names are ∠OAB, ∠OBA, ∠OAC, ∠OCA, ∠OCD, ∠ODC, ∠AOB, ∠AOC, ∠COD, ∠DOB, ∠BAC, ∠ACD.

The number of straight angles in Fig. is four.
The angles are, ∠AOD, ∠BOC, ∠COB, ∠DOA.

The number of right angles in a straight angle is tow and that in a complete angle is four.
We know that, angle formed by straight angle = 180º
and angle formed by right angle = 90º
So, number of right angles = 180º/90º = 2
We know that, complete angle = 360º
So, number of right angles = 360º/90º = 4

The number of common points in the two angles marked in Fig. is two.
From the given figure, the common points are P and Q.

The number of common points in the two angles marked in Fig. is one.
From the given figure, the common point is A.

The number of common points in the two angles marked in Fig. three.
From the given figure, the common points are P, Q and R.

The number of common points in the two angles marked in Fig. is four.
From the given figure, the common points are D, E, F and G.

The common part between the two angles BAC and DAB in Fig. is ray AB.

True.

False.
The angle is not affected, if the arms of an angle on the paper are increased or decreased.

False.
The angle is not affected, if the arms of an angle on the paper are increased or decreased.

True.
From the question, line PQ || line m.
Then, parts of those lines are also parallel.
Therefore, line segment PQ || m.

False.
Parallel lines never meet each other.

True.
From the figure, both ∠ABC and ∠CBA contains common angle B.

False.
Because, two line segments are intersecting at only one point.

False.
only one line can pass through two given points.

False.
Infinite number of line can pass through a given point.

False.
Two angles can have either one or two five points in common.

The line segments are AB, BC, CD, DE, AC, AD, AE, BD, BE and CE

The line segments are
AB, BC, CD, DE and EA

X is a mid-point of AC,
Y is a mid-point of BC and
Z is a mid-point of AB.

The vertices are : A, B, C, D and E.
The line segments are : AB, BC, CD, DE, EA, AC and AD.

The names of fifteen angles (less than 180°) involved in figure are :
∠AEC, ∠ADB, ∠EAD, ∠EFD, ∠EFB, ∠DFC, ∠FBC, ∠FCB, ∠BFC, ∠ABC, ∠ACB, ∠DCF, ∠FDC, ∠EBF and ∠BEF.

(a) ∠1 = ∠CBD
(b) ∠2 = ∠DBE
(c) ∠3 = ∠EBA
(d) ∠1 + ∠2 = ∠CBD + ∠DBE = ∠CBE
(e) ∠2 + ∠3 = ∠DBE + ∠EBA = ∠DBA
(f) ∠l + ∠2 + ∠3 = ∠CBD + ∠DBE + ∠EBA
= ∠CBA
(g) ∠CBA – ∠1 = ∠CBA – ∠CBD = ∠DBA

(i) Name of the points ➝ A, B and C. Name of the line segments ➝ AB, BC and CA.
(ii) Name of the points ➝ A, B, C and D. Name of the line segments ➝ AB, BC, CD and DA.
(iii) Name of the points ➝ A, B, C, D and E. Name of the line segments ➝ AB, BC, CD, DE and EA.
(iv) Name of the points ➝ A, B, C, D, E and F.
Name of the line segments ➝ AB, CD and EF.

(i) The given figure shows there is no mid-point.
(ii) The given figure shows that O is the mid-point of ZB and the name of the two equal line segments are AC and OB.
(iii) The given figure shows that D is the mid-point of BC and the name of the two equal line segments are BD and DC.

(a) No, it is not possible for the same line segment to have two different lengths.
(b) No, it is not possible for the same angle to have two different measures.

Yes, ∵ ∠ABD = ∠ABC + ∠CBD
⇒ ∠ABD is the sum of ∠ABC and ∠CBD

Yes, ∵ the length of line segment AC is the sum of the lengths of line segment AB and BC.

Angles are measured in degrees. The symbol for degree is a little circle.
The full circle is 360° (360 degree). A half circle or a straight angle is 180°. A quarter circle or a right angle is 90°.
Place the mid-point of the protractor on the Vertex of the angle. Line up one side of the angle with the Zero line of the protractor (where you see the number 0).
Read the degrees here the other side crosses the number scale.
1. Measure the angles.
2. Measure the angles. Label each angle as acute or obtuse.
3. Tasha measured an acute angle, and got 146°. The teacher pointed out that she had read the wrong set of numbers on the protractor.
4. Measure the following angles using your own protractor. If you need to, make the sides of the angles longer with a ruler.
5. Draw four dots and connect them so that you get a quadrilateral.
Measure all the angles of your quadrilateral. Then add the angle measure.

Yes, the given figure shows that the points B and C lie in the interior of ∠2 also.

All the three statements (a), (b) and (c) are incorrect.
∵ The common initial point of two rays forms an angle.

(a) Figure (ii) shows the perpendicular bisector
(b) Figure (ii) and (iii) shows the bisector.
(c) Figure (iii) shows only the bisector.
(d) Figure (i) shows only the perpendicular.

Both the figures (i) and (ii) have 3 line segments.
No, Fig. (i) is not a triangle since the three line segments does not form a closed figure.

Yes

(a) Acute angles : ∠ADE, ∠AEB, ∠ABE and ∠ECD.
(b) Obtuse angles : ∠BCD and ∠BAD.

(a) Yes, AC + CB = AB
(b) No, AB – AC = CB
(c) No, AB – BC = CA

(a) AE + EC = AC
(b) AC – EC = AE
(c) BD – BE = ED
(d) BD – DE = BE

(a) ∵ BA ⊥ BD,
∴ The right angle is ∠ABD
(b) ∵ RT ⊥ ST, ∴ The right angle is ∠RTS
(c) ∵ AC ⊥ BD,
∴ The right angles are ∠ACD and ∠ACB
(d) ∵ RS ⊥ RW,
∴ The right angle is ∠SRW
(e) ∵ AC ⊥ BD,
∴ The right angles are ∠AED, ∠AEB, ∠BEC and ∠CED
(f) ∵ AE ⊥ CE,
∴ The right angle is ∠AEC
(g) ∵ AC ⊥ CD,
∴ The right angle is ∠ACD
(h) ∵ OP ⊥ AB,
∴ The right angles are ∠AKO, ∠AKP, ∠BKO and ∠BKP.

(a) ∵ DB is the bisector of ∠ADC.
∴ ∠ADB = ∠CDB
(b) ∵ BD bisects ∠ABC.
∴ ∠ABD = ∠CBD
(c) ∵ DC is the bisector of ∠ADB, CA ⊥ DA and CB ⊥ DB
∠ADC = ∠BDC, ∠CAD = ∠CBD = 90°

We have given, ∠BAC = ∠CAD = ∠DAE
∴ There are two trisectors namely, AC and AD.

Two points A and B are marked.

Only one line segment, AB is there.

Three points A, B and C are marked.

Three line segments, namely AB, BC and AC are there.

Four points A, B, C and D are marked.

Six line segments, namely AB, AC, AD, BC, BD and CD.

Five points are marked, namely A, B, D, E and C.

Ten line segments, namely AB, AD, AE, AC, BD, BE, BC, DE, DC and EC.

(a) Name of chords : PC and BA.
(b) Name of radii : PO, OC, OB and OA.
(c) PC is a chord which is not the diameter of the circle.
(d) Shaded sectors OAC and OPB are as:

(e) Shade the smaller segment of the circle formed by CP.

(a) Yes, ∵ the sum of two acute angles can be the acute angle.
E.g., 30° and 40° are two acute angles and their sum = 30° + 40° = 70°, which is also an acute angle.
(b) Yes, ∵ the sum of two acute angles can be a right angle.
E.g., 30° and 60° are two acute angles and their sum = 30° + 60° = 90°, which is a right angle.
(c) Yes, ∵ the sum of two acute angles can be an obtuse angle.
E.g., 45° and 60° are two acute angles and their sum = 45° + 60° = 105°, which is an obtuse angle.
(d) No, ∵ the sum of two acute angles is always less than 180°.
(e) No, ∵ the sum of two acute angles is always less than 180°.

(a) Yes, ∵ the sum of two obtuse angles is always greater than 180°.E.g., 135° and 100° are two obtuse angles and their sum = 135° + 
100° = 235°, which is greater than 180°.
(b) No, ∵ the sum of two obtuse angles is greater than 180° but less than 360°. In the above example, we can see that the sum of 135° and 100° i.e., 235° is greater than 180° but less than 360°.

(a) Vertices : A, B, C, D, E and F.
(b) Edges : AB, BC, AC, DF, FC, BD, EF, ED and AE.
(c) Faces : EACF, EDBA, ABC, DEF and DBCF.

In a sphere, edges – 0, faces – 0 and vertices – 0.

The diagonals of a pentagon ABCDE are AC, AD, BE, BD and EC