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Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 8 Basic Geometrical Concepts Ex 8.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 8 Basic Geometrical Concepts Ex 8.6

1. In the given figure, write the name of:

Question (i)
Centre
Solution:
Centre: O

Question (ii)
Radii
Solution:
Radii: OX, OY, OP

Question (iii)
Diameter
Solution:
Diameter: XY

Question (iv)
Chord.
Solution:
Chord: QR.

2. In the given figure, write the name

Question (i)
minor arc
Solution:
Minor arc : PAQ

Question (ii)
major arc
Solution:
Major arc : PBQ

Question (iii)
minor sector
Solution:
Minor sector: OPAQ

Question (iv)
major sector.
Solution:
Major sector : OPBQ.

3. In the given figure, write the name

Question (i)
Minor segment
Solution:
Minor segment: ACBA

Question (ii)
Major segment.
Solution:
Major segment: ADBA

4. In the given figure, name the points:

Question (i)
In its interior
Solution:
Points in its interior : O, A, D, F

Question (ii)
On its boundary (circumference)
Solution:
Points on its boundary (circumference) C

Question (iii)
In its exterior.
Solution:
Points in its exterior : B, E

5. Find the diameter of the circle whose radius is:

Question (i)
5 cm
Solution:
Given Radius of the circle = 5 cm
∴ Diameter of the Circle = 2 × radius = 2 × 5 cm = 10 cm

Question (ii)
4 cm
Solution:
Given radius of the circle = 4 cm
∴ Diameter of the circle = 2 × radius = 2 × 4m = 8m

Question (iii)
10 cm.
Solution:
Given radius of the circle = 10 cm
∴ Diameter of circle = 2 × Radius = 2 × 10 cm = 20 cm

6. If the diameter of the circle is 12 cm. Find the radius:
Solution:
Given diameter of a circle = 12 cm
∴ Radius of circle = Diameter + 2
= 12 cm + 2
= 6 cm

7. Fill in the blanks:

Question (i)
The distance around a circle is called ……………… .
Solution:
Circumference

Question (ii)
The diameter of a circle is ……………… times its radius.
Solution:
two

Question (iii)
The longest chord of circle is …………….. .
Solution:
diameter

Question (iv)
All the radii of a circle are of ……………… length.
Solution:
equal

Question (v)
The diameter of a circle passes through ……………. .
Solution:
centre

Question (vi)
A circle divides all the points in a plane into ……………… parts.
Solution:
three.

8. State true or false:

Question (i)
The diameter of a circle is equal to its radius.
Solution:
False

Question (ii)
The diameter is a chord of circle.
Solution:
True

Question (iii)
A radius is a chord of the circle.
Solution:
False

Question (iv)
Every circle has a centre.
Solution:
True

Question (v)
The region enclosed by a chord and arc is called a segment
Solution:
True

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 8 Basic Geometrical Concepts Ex 8.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 8 Basic Geometrical Concepts Ex 8.5

1. Out of the following, Identify the quadrilateral:

Solution:

2. Name the given quadrilaterals:

Solution:
(i) ABCD
(ii) PQRS
(iii) XYZW.

3. Write the name of all vertices, angles, sides, diagonals of the following quadrilaterals:

Solution:
(i) Vertices = O, N, M, L;
Angles = \(\angle \mathrm{O}, \angle \mathrm{N}, \angle \mathrm{M}, \angle \mathrm{L}\)
Sides = ON, NM, ML, LO;
Diagonals = OM, NL

(ii) Vertices = H, G, F, E;
Angles = \(\angle \mathrm{H}, \angle \mathrm{G}, \angle \mathrm{F}, \angle \mathrm{E}\)
Sides = HG, GF, FE, EH;
Diagonals = EG, FH.

4. For the given quadrilateral ABCD, name:

Question (i)
(i) Side opposite to AB
(ii) Angles adjacent to B
(iii) Diagonal joining B and D
(iv) Angle opposite to A
(v) Sides adjacent to CD.

Solution:
(i) CD
(ii) \(\angle \mathrm{A} \text { and } \angle \mathrm{C}\)
(iii) BD
(iv) \(\angle \mathrm{C}\)
(v) AD and BC.

5. In the given quadrilateral JUMP, name the points.

Question (i)
In its interior
Solution:
Points in the interior of quad. JUMP are :
L, N, C

Question (ii)
In its exterior
Solution:
Points in its exterior of quad. JUMP are :
B, O, X

Question (iii)
On its boundary.
Solution:
Points on the boundary of quad. JUMP are :
P, M, U, Y, J, A.

6. Fill in the blanks:

Question (i)
A quadrilateral has …………….. vertices.
Solution:
4

Question (ii)
A quadrilateral has …………… sides.
Solution:
4

Question (iii)
A quadrilateral has …………… angles.
Solution:
4

Question (iv)
A quadrilateral has ………….. diagonals.
Solution:
2

Question (v)
A diagonal divides the quadrilateral into……………… triangles.
Solution:
2

Question (vi)
A line segment joining the opposite vertices of a quadrilateral is called its ………. .
Solution:
Diagonal

Question (vii)
The interior and the boundary of a quadrilateral together constitute the ……………. region.
Solution:
Quadrilateral.

7. State True or False:

Question (i)
A diagonal divides quadrilateral into four triangles.
Solution:
False

Question (ii)
The angle that have a common vertex are called adjacent angles.
Solution:
True

Question (iii)
The sides that have a common vertex are called adjacent sides.
Solution:
True

Question (iv)
A quadrilateral has four diagonals.
Solution:
False

Question (v)
The quadrilateral region consists of the exterior and the boundary of the quadrilateral.
Solution:
False

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 6 Triangles Ex 6.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5

Question 1.
Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.
(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm.
Solution:
(i) Let ∆ABC, with AB = 7 cm BC = 24 cm, AC = 25 cm
AB² + BC² = (7)² + (24)²
= 49 + 576 = 625
AC² = (25)² = 625
Now AB² + BC² = AC²
∴ ∆ABC is right angled triangle. Hyp. AC = 25cm.

(ii) Let ∆PQR with PQ = 3 cm, QR = 8 cm PR = 6 cm
PQ² + PR² = (3)² + (6)²
= 9 + 36 = 45
QR² = (8)² = 64.
Here PQ² + PR² ≠ QR²
∴ ∆PQR is not right angled triangle.

(iii) Let ∆MNP, with MN =50 cm, NP = 80 cm, MP = 100 cm
MN ²+ NP² = (50)² + (80)²
= 2500 + 6400 = 8900
MP² = (100)² = 10000
Here MP² ≠ MN² + NP².
∴ ∆MNP is not right angled triangle.

(iv) Let ∆ABC, AB = 13 cm, BC = 12 cm, AC = 5 cm
BC² + AC² = (12)² + (5)²
= 144 + 25 = 169
AB² = (13)² = 169
∴ AB² = BC² + AC²
∆ABC is right angled triangle.
Hyp. AB = 13 cm.

Question 2.
PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM² = QM . MR.
Solution:
Given: ∆PQR is right angled at P and M is a point on QR such that PM ⊥ QR.
To prove : PM² = QM × MR

Proof: ∠P = 90° (Given)
∴ ∠1 + ∠2 = 90°
∠M = 900 (Given)
In ∆PMQ,
∠1 + ∠3 + ∠5 = 180°
=> ∠1 + ∠3 = 90° [Angle Sum Property] ………….(2) [∠5 = 90°]
From (1) and (2),
∠1 + ∠2 = ∠1 + ∠3
∠2 = ∠3
In ∆QPM and ∆RPM,
∠3 = ∠2 (Proved)
∠5 = ∠6 (Each 90°)
∴ ∆QMP ~ ∆PMR [AA similarity]
\(\frac{{ar} .(\Delta \mathrm{QMP})}{{ar} .(\Delta \mathrm{PMR})}=\frac{\mathrm{PM}^{2}}{\mathrm{MR}^{2}}\)

[If two triangles are similar, ratio o their areas is equal to square of corresponding sides]
\(\frac{\frac{1}{2} \mathrm{QM} \times \mathrm{PM}}{\frac{1}{2} \mathrm{RM} \times \mathrm{PM}}=\frac{\mathrm{PM}^{2}}{\mathrm{MR}^{2}}\)

[area of ∆ = \(\frac{1}{2}\) Base × Altitude]

\(\frac{\mathrm{QM}}{\mathrm{RM}}=\frac{\mathrm{PM}^{2}}{\mathrm{RM}^{2}}\)

PM² = QM × RM Hence proved.

Question 3.
In fig., ABD is a triangle right angled at A and AC ⊥ BD. Show that

(i) AB² = BC.BD
(ii) AC² = BC.DC
(üi) AD² = BD.CD.
Solution:
Given. A right angled ∆ABD in which right angled at A and AC ⊥ BD.
To Prove:
(i) AB² = BC.BD
(ii) AC² = BC.DC .
(iii) AD² = BD.ÇD .
Proof. In ∆DAB and ∆DCA,
∠D = ∠D (common)
∠A = ∠C (each 90°)
∴ ∆DAB ~ ∆DCA [AA similarity]
In ∆DAB and ∆ACB,
∠B = ∠B (common)
∠A = ∠C . (each 90°)
∴ ∆DAB ~ ∆ACB, .
From (1) and (2),
∆DAB ~ ∆ACB ~ ∆DCA.
(i) ∆ACB ~ ∆DAB (proved)
∴ \(\frac{{ar} .(\Delta \mathrm{ACB})}{{ar} .(\Delta \mathrm{DAB})}=\frac{\mathrm{AB}^{2}}{\mathrm{DB}^{2}}\)

[If two triangles are similar corresponding sides are proportional]

\(\frac{\frac{1}{2} \mathrm{BC} \times \mathrm{AC}}{\frac{1}{2} \mathrm{DB} \times \mathrm{AC}}=\frac{\mathrm{AB}^{2}}{\mathrm{DB}^{2}}\)
[Area of triangle = \(\frac{1}{2}\) Base × Altitude]
BC = \(\frac{\mathrm{AB}^{2}}{\mathrm{BD}}\)
AB² = BC × BD.

(iii) ∆ACB ~ ∆DCA (proved)
\(\frac{{ar} .(\Delta \mathrm{DAB})}{{ar} .(\Delta \mathrm{DCA})}=\frac{\mathrm{DA}^{2}}{\mathrm{DB}^{2}}\)
[If two triangles are similar corresponding sidec are proportional]

\(\frac{\frac{1}{2} \mathrm{CD} \times \mathrm{AC}}{\frac{1}{2} \mathrm{BD} \times \mathrm{AC}}=\frac{\mathrm{AD}^{2}}{\mathrm{BD}^{2}}\)

CD = \(\frac{\mathrm{AD}^{2}}{\mathrm{BD}}\)
⇒ AD² = BD × CD.

Question 4.
ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC².
Solution:
Given: ABC is an isosceles triangle right angled at C.
To prove : AB² = 2AC².
Proof: In ∆ACB, ∠C = 90° & AC = BC (given)

AB² = AC² + BC²
[By using Pythagoras Theorem]
=AC² + AC² [BC = AC]
AB² = 2AC²
Hence proved.

Question 5.
ABC is an isosceles triangle with AC = BC. If AB² = 2AC², prove that ABC is right triangle.
Solution:
Given: ∆ABC is an isosceles triangle AC = BC
To prove: ∆ABC is a right triangle.

Proof: AB² = 2AC² (given)
AB² = AC² + AC²
AB² = AC² + BC² [AC = BC]
∴ By Converse of Pythagoras Theorem,
∆ABC is right angled triangle.

Question 6.
ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Solution:
∆ABC is equilateral triangle with each side 2a
AD ⊥ BC
AB = AC = BC = 2a
∆ADB ≅ ∆ADC [By RHS Cong.]
∴ BD = DC = a [c.p.c.t]

In right angled ∆ADB
AB² = AD² + BD²
(2a)² = AD² + (a)²
4a² – a² = AD².
AD² = 3a²
AD = √3a.

Question 7.
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals. [Pb. 2019]
Solution:
Given: Rhombus, ABCD diagonal AC and BD intersect each other at O.
To prove:
AB² + BC² + CD² + AD² = AC² + BD²
Proof:The diagonals of a rhombus bisect each other at right angles.

∴ AO = CO, BO = DO
∴ ∠s at O are rt. ∠s
In ∆AOB, ∠AOB = 90°
∴ AB² = AO² + BO² [By Pythagoras Theorem] …………..(1)
Similarly, BC² = CO² + BO² ……………..(2)
CD² = CO² + DO² ……………(3)
and DA² = DO² + AO² ……………….(4)
Adding. (1), (2), (3) and (4), we get
AB² + BC² + CD² + DA² = 2AO² + 2CO² + 2BO² + 2DO²
= 4AO² + 4BO²
[∵ AO = CO and BO = DO]
= (2AO)² + (2BO)² = AC² + BD².

Question 8.
In fig., O is a point In the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that
(i) OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE²

(ii)AF² + BD² + CE² = AE² + CD² + BF².

Solution:
Given: A ∆ABC in which OD ⊥ BC, 0E ⊥ AC and OF ⊥ AB.

To prove:
(i) AF² + BD² + CE² = OA² + OB² + OC² – OD² – OE² – OF²
(ii) AF² + BD² + CE² = AE² + CD² + BF².
Construction: Join OB, OC and OA.
Proof: (i) In rt. ∠d ∆AFO, we have
OA² = OF² + AF² [By Pythagoras Theorem]
or AF² = OA² – OF² …………..(1)

In rt. ∠d ∆BDO, we have:
OB² = BD²+ OD² [By Pythagoras Theorem]
⇒ BD² = OB² – OD² …………..(2)

In rt. ∠d ∆CEO, we have:
OC² = CE² + OE² [By Pythagoras Theorem]

⇒ CE² = OC² – OE² ……………(3)

∴ AF² + BD² + CE² = OA² – OF² + OB² – OD² + OC² – 0E²
[On adding (1), (2) and (3)]
= OA² + OB² + OC² – OD² – OE² – OF²
which proves part (1).
Again, AF² + BD² + CE² = (OA² – OE²) + (OC² – OD²) + (OB² – OF²)
= AE² + CD² + BF²
[∵AE² = AO² – OE²
CD² = OC² – OD²
BF² = OB² – OF²].

Question 9.
A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.
Solution:
Height of window from ground (AB) = 8m.

Length of ladder (AC) = 10 m
Distance between foot of ladder and foot of wall (BC) = ?
In ∆ABC,
AB² + BC² = AC² [By Pythagoras Theorem]
(8)² + (BC)² = (10)²
64 + BC² = 100
BC² = 100 – 64
BC = √36
BC = 6 cm.
∴ Distance between fóot of ladder and foot of wall = 6 cm.

Question 10.
A guy wire attached to a vertical pole of height 18 m Is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
Solution:
Let AB is height of pole (AB) = 18 m
AC is length of wire = 24 m

C is position of stake AB at ground level.
In right angle triangle ABC,
AB² + BC² = AC² [By Pythagoras Theorem]
(18)² + (BC)² = (24)²
324 + (BC)² = 576
BC² = 576 – 324
BC = \(\sqrt{252}=\sqrt{36 \times 7}\)
BC = 6√7 m.

Question 11.
An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two pLanes after 1\(\frac{1}{2}\) hours?
Solution:
Speed of first aeroplane = 1000km/hr.

Distance covered by first aeroplane due north in 1\(\frac{1}{2}\) hours =1000 × \(\frac{3}{2}\)
OA = 1500 km
Speed of second aeroplane = 1200 km/hr.
Distance covered by second aeroplane in 1\(\frac{1}{2}\) hours = 1200 × \(\frac{3}{2}\)
OB = 1800 km.
In right angle ∆AOB
AB² = AO² + OB² [By Phyrhagoras Theorem]
AB² = (1500)² + (1800)²
AB = \(\sqrt{2250000+3240000}\)
= \(\sqrt{5490000}\)
= \(\sqrt{61 \times 90000}\)
AB = 300√61 km.
Hence, Distance between two aeroplanes = 300√61 km.

Question 12.
Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the
distance between their tops.
Solution:
Height of pole AB = 11 m
Height of pole (CD) = 6 m

Distance between foot of pole = 12 m
from C draw CE ⊥ AB. such that
BE = DC = 6 m
AE = AB – BE = (11 – 6) m = 5 m.
and CE = DB = 12 m.
In rt. ∠d ∆AEC,
AC² = AE² + FC²
[By Phythagoras Theorem)
AC = \(\sqrt{(5)^{2}+(12)^{2}}\)
= \(\sqrt{25+144}\)
= \(\sqrt{169}\) = 13.
Hence, Distance between their top = 13m.

Question 13.
D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C.
Prove that AE² + BD² = AB² + DE².
Solution:
Given: In right angled ∆ABC, ∠C = 90° ;
D and E are points on sides CA & CB respectively.
To prove: AE² + BD² = AB² + DE²
Proof: In rt. ∠d ∆BCA,
AB² = BC² + CA² …………..(1) [By Pythagoras Theorem]

In rt. ∠d ∆ECD,
DE² = EC² + DC² ……………….(2) [By Pythagoras Theorem]
In right angled triangle ∆ACE,
AE² = AC² + CE² ……………….(3)
In right angled triangle ∆BCD
BD² = BC² + CD² ……………….(4)
Adding (3) and (4),
AE² + BD² = AC² + CE² + BC² + CD²
= [AC² + CB²] + [CE² + DC²]
= AB² + DE²
[From (1) and (2)]
Hence 2 + BD² = AB² + DE².
Which is the required result.

Question 14.
The perpendicular from A on side BC of a ∆ABC intersects BC at D such that DB = 3 CD. Prove that 2AB² = 2AC² + BC².

Solution:
Given: ∆ABC, AD ⊥ BC
BD = 3CD.
To prove: 2AB² = 2AC² + BC².

Proof: In rt. ∠d triangles ADB and ADC, we have
AB² = AD² + BD²;
AC² = AD² + DC² [By Pythagoras Theorem]
∴ AB² – AC² = BD² – DC²
= 9 CD² – CD²; [∵ BD = 3CD]
= 8CD² = 8 (\(\frac{\mathrm{BC}}{4}\))²
[∵ BC = DB + CD = 3 CD + CD = 4 CD]
∴ CD = \(\frac{1}{4}\) BC
∴ AB² – AC² = \(\frac{\mathrm{BC}^{2}}{2}\)
⇒ 2(AB² – AC²) = BC²
⇒ 2AB² – 2AC² = BC²
∴ 2AB² = 2AC² + BC².
Which is the required result.

Question 15.
In an equilateral triangle ABC, D is a point on side BC such that BD = \(\frac{1}{3}\) BC. Prove that 9 AD² = 7 AB².
Solution:
Given: Equilateral triangle ABC, D is a point on side BC such that BD = \(\frac{1}{3}\) BC.
To prove: 9AD² = 7 AB².
Construction: AB ⊥ BC.
Proof: ∆AMB ≅ ∆AMC [By R.HS. Rule since AM = AM and AB = AC]

∴ BM = MC = \(\frac{1}{2}\) BC [c.p.c.t.]
Again BD = \(\frac{1}{3}\) BC and DC = \(\frac{1}{3}\) BC (∵ BC is trisected at D)
Now in ∆ADC, ∠C is acute
∴ AD² = 2AC² + DC² – 2 DC × MC
= AC² + \(\left[\frac{2}{3} \mathrm{BC}\right]^{2}\) – 2 \(\left[\frac{2}{3} \mathrm{BC}\right] \frac{1}{2} \mathrm{BC}\)

[∵ DC = \(\frac{2}{3}\) BC and MC = \(\frac{1}{2}\) BC]
= AB² + \(\frac{4}{9}\) AB² – \(\frac{2}{3}\) AB²
[∵ AC = BC = AB]
= (1 + \(\frac{4}{9}\) – \(\frac{2}{3}\)) AB²

= \(\left(\frac{9+4-6}{9}\right) \mathrm{AB}^{2}=\frac{7}{9} \mathrm{AB}^{2}\)

∴ AD² = \(\frac{7}{9}\) AB²
⇒ 9 AD² = 7 AB².

Question 16.
In an equilateral triangle, prove that three times the square of one side Ls equal to four times the square of one of its
altitudes.
Solution:
Given:
ABC is equilateral ∆ in which AB = BC = AC

To prove: 3 AB² = 4 AD²
Proof: In right angled ∆ABD,
AB² = AD² + BD² (Py. theorem)
AB² = A BD² (Py. theorem)

AD² = \(\frac{3}{4}\) AB²
⇒ 4 AD² = 3 AB²
Hence, the result.

Question 17.
Tick the correct answer and justify: In ∆ABC, AB = 6 cm, AC = 12 cm and BC = 6√3 cm. [The angles of B are respectively
(A) 120°
(B) 64°
(C) 90°
(D) 45°
Solution.
AC = 12 cm
AB = 6√3 cm
BC = 6 cm
AC² = (12)² = 144 cm
AB² + BC² = (6√3)² + (6)²
= 108 + 36
AB√3 + BC√3 = 144
∴ AB√3 + BC√3 = AC√3
Hence by converse of pythagoras theorem ∆ABC is right angred triangle right angle at B
∴ ∠B = 90°
∴ correct option is (C).

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 8 Basic Geometrical Concepts Ex 8.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 8 Basic Geometrical Concepts Ex 8.4

1. Write all the names of the following triangles in all orders:

Solution:
(i) ∆ABC, ∆ACB, ∆BAC, ∆BCA, ∆CAB, ∆CBA.
(ii) ∆XYZ, ∆XZY, ∆YZX, ∆YXZ, ∆ZXY, ∆ZYX
(iii) ∆LMN, ∆LNM, ∆MNL, ∆MLN, ∆NML, ∆NLM.

2. Write the name of vertices, sides and angles of the following triangles:

Solution:

	(i)	(ii)	(iii)
Vertices	P, R, Q	D, E, F	T, P, S
Sides	PR, QR, PQ	DE, EF, DF	TP, PS, TS
Angles	\(\angle \mathrm{P}, \angle \mathrm{R}, \angle \mathrm{Q}\)	\(\angle \mathrm{D}, \angle \mathrm{E}, \angle \mathrm{F}\)	\(\angle \mathrm{T}, \angle \mathrm{P}, \angle \mathrm{S}\)

3. In the given figure, name the points that lie:

Question (i)
On the boundary of ∆GEM
Solution:
Points on the boundary of AGEM are: G, A, E, C, M

Question (ii)
In the interior of ∆GEM
Solution:
Points in the interior of AGEM are : P, X, D

Question (iii)
In the exterior of ∆GEM.
Solution:
Points in the exterior of AGEM are : Y, B.

4. In the given figure, write the name of:

Question (i)
All different triangles
Solution:
All different triangles are :
∆AOD, ∆DOC, ∆BOC, ∆AOB, ∆ABD, ∆BCD, ∆ACD, ∆ABC

Question (ii)
Triangles having O as the vertex
Solution:
Triangles having O as the vertex are :
∆AOB, ∆BOC, ∆COD, ∆AOD

Question (iii)
Triangles having A as the vertex.
Solution:
Triangles having A as the vertex are:
∆AOB, ∆AOD, ∆ABD, ∆ABC, ∆ACD.

5. Fill in the blanks of the following:

Question (i)
A triangle has …………… vertices.
Solution:
3

Question (ii)
A triangle has …………… angles.
Solution:
3

Question (iii)
A triangle has ……………. sides.
Solution:
3

Question (iv)
A triangle divide the plane into ……………. parts.
Solution:
3

Question (v)
A triangle has …………… parts.
Solution:
6

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 8 Basic Geometrical Concepts Ex 8.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 8 Basic Geometrical Concepts Ex 8.3

1. Name the given angles in all ways:

Solution:
(i) \(\angle \mathrm{DEF}, \angle \mathrm{FED}, \angle \mathrm{E}, \angle a\)
(ii) \(\angle \mathrm{XOY}, \angle \mathrm{YOX}, \angle \mathrm{O}, \angle 1\)
(iii) \(\angle N O M, \angle M O N, \angle O, \angle x\)

2. Name the vertex and the arms of given angles:

Solution:

	(i)	(ii)	(iii)
Vertex	B	Q	o
Arm	\(\overrightarrow{\mathrm{BC}}, \overrightarrow{\mathrm{BA}}\)	\(\overrightarrow{\mathrm{QP}}, \overrightarrow{\mathrm{QR}}\)	\(\overrightarrow{\mathrm{OS}}, \overrightarrow{\mathrm{OP}}\)

3. Name all the angles of the given figure:

Solution:
(i) \(\angle \mathrm{X}, \angle \mathrm{Y}, \angle \mathrm{Z}\)
(ii) \(\angle \mathrm{P}, \angle \mathrm{Q}, \angle \mathrm{R}, \angle \mathrm{S}\)
(iii) \(\angle \mathrm{AOB}, \angle \mathrm{BOC}, \angle \mathrm{AOC}\)

4. In the given figure, name the points that lie:

Question (i)
In the interior of \(\angle \mathrm{DOE}\)
Solution:
Points in the interior of \(\angle \mathrm{DOE}\) are :
A, X, M

Question (ii)
In the exterior of \(\angle \mathrm{DOE}\)
Solution:
Points in the exterior of \(\angle \mathrm{DOE}\) are :
H, L

Question (iii)
On the \(\angle \mathrm{DOE}\)
Solution:
Points on the \(\angle \mathrm{DOE}\) are :
D, B, O, E.

5. In the given figure, write another name for the following angles :

Question (i)
\(\angle \mathrm{1}\)
Solution:
\(\angle S \text { or } \angle PSR \text { or } \angle RSP\)

Question (ii)
\(\angle \mathrm{2}\)
Solution:
\(\angle \mathrm{RPQ} \text { or } \angle \mathrm{QPR}\)

Question (iii)
\(\angle \mathrm{3}\)
Solution:
\(\angle \mathrm{SRP} \text { or } \angle \mathrm{PRS}\)

Question (iv)
\(\angle \mathrm{a}\)
Solution:
\(\angle \mathrm{Q} \text { or } \angle \mathrm{RQP} \text { or } \angle \mathrm{PQR}\)

Question (v)
\(\angle \mathrm{b}\)
Solution:
\(\angle \mathrm{PRQ} \text { or } \angle \mathrm{QRP}\)

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 8 Basic Geometrical Concepts Ex 8.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 8 Basic Geometrical Concepts Ex 8.2

1.

Question (i)
(a) Which of the following are simple curves?
(b) Classify the following as open or closed curve.

Solution:
(a) Simple curves :
(i), (iii), (iv), (vi), (vii), (iii)

(b) Open curves :
(iii) , (vi), (viii)
Closed curves :
(i), (ii), (iv), (v), (vii)

2. Identify the polygons:

Question (i)

Solution:
(ii), (iii), (v) are polygons.

3. Draw any polygon and shade its interior.
Solution:

4. Name the points which are:

Question (i)
In the interior of the closed figure.
Solution:
Points in the interior of closed figure are :
A B, Q

Question (ii)
In the exterior of the closed figure,
Solution:
Points in the exterior of closed figure are :
R, N

Question (iii)
On the boundary of the closed figure.
Solution:
Points in the boundary of closed figure are :
P, M

5. In the given figure, name :

Question (i)
The vertices
Solution:
Vertices are :
D, E, A, B, C

Question (ii)
The sides
Solution:
Sides are :
AB, BC, CD, DE, EA

Question (iii)
The diagonals
Solution:
Diagonals are :
AC, AD, BE, BD, CE

Question (iv)
Adjacent sides of AB
Solution:
Adjacent sides of AB are :
AE and BC

Question (v)
Adjacent vertices of E.
Solution:
Adjacent vertices of E are :
A and D.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 8 Basic Geometrical Concepts Ex 8.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 8 Basic Geometrical Concepts Ex 8.1

1. Give the examples of:

Question (i)
A point
Solution:
A point. Point A •
Examples:
(i) A small dot marked by a sharp pencil on a sheet of paper.
(ii) A tiny prick made by a fine needle or pin on a paper.
(iii) Bindi.
(iv) A star in the sky.

Question (ii)
A line segment
Solution:
A line segment
Examples:
(i) An edge of a box.
(ii) A tube light.
(iii) The edge of a postcard.
(iii) Parallel lines.

Question (iii)
Parallel lines
Solution:
Examples:
(i) The opposite edges of ruler (scale).
(ii) The crossbars of window.
(iii) The opposite edges of blackboard.
(iv) Rail lines.
(iv) Interescting lines.

Question (iv)
Intersecting lines
Solution:
Examples:
(i) Two adjacent edges of your notebook.
(ii) The letter X of the English alphabet.
(iii) Crossing roads.

Question (v)
Concurrent lines.
Solution:
Concurrent lines.
(i) Three angle bisectors of a triangle.
(ii) Three medians of a triangle.
(iii) Three perpendiculars of a triangle.
(iv) The intersection of the three walls of a room.

2. Name the lines segments in given lines.

Solution:
AB, AC, AD, BC, BD, CD are line segments.

3. How many lines can pass through a point?
Solution:
Infinite lines can pass through a point.

4. How many points lie on line?
Solution:
Infinite points lie on a line.

5. How many lines pass through two points?
Solution:
One and only one line passes through two points.

6. Use the figure to name:

Question (i)
Five Points
Solution:
Five Points are :
O, A, B, C, D, or E

Question (ii)
A line
Solution:
BE is the line.

Question (iii)
Four rays
Solution:
Four rays are :
\(\overrightarrow{\mathrm{OA}}, \overrightarrow{\mathrm{OB}}, \overrightarrow{\mathrm{OC}}, \overrightarrow{\mathrm{OD}} \text { or } \overrightarrow{\mathrm{OE}}\)

Question (iv)
Five line segments.
Solution:
Five line segments are :
OA, OB, OC, OD, OE, DE.

7. Name the given ray in all possible ways.

Solution:
The possible rays are:
\(\overrightarrow{\mathrm{PQ}}, \overrightarrow{\mathrm{PR}}, \overrightarrow{\mathrm{QR}}\)

8. Use the figure to name :

Question (i)
Pair of parallel lines.
Solution:
Pair of parallel lines are :
l and m.

Question (ii)
All pairs of intersecting lines.
Solution:
All pairs of intersecting lines are:
p and n, n and l, n and m, p and l, p and m.

Question (iii)
Lines whose point of intersection is S.
Solution:
Lines whose point of intersection is S :
m and n.

Question (iv)
Collinear points.
Solution:
Collinear points are :
P, Q, S and P, R, T.

9. Use the figure to name:

Question (i)
All pairs of parallel lines.
Solution:
All pairs of parallel lines are : n and p, q and p, n and q.

Question (ii)
All pairs of intersecting lines.
Solution:
All pairs of intersecting lines are :
m and l, m and n, m and p, m and q, l and n, l and p, l and q.

Question (iii)
Lines whose point of intersection is D.
Solution:
Lines whose point of intersection is D are :
p and l.

Question (iv)
Point of intersection of lines m and p.
Solution:
Point of intersection of lines m and p is E.

Question (v)
All sets of collinear points.
Solution:
All sets of collinear points are :
G, E, C, A and F, D, C, B.

10. Use the figure to name:

Question (i)
Line containing point P.
Solution:
Lines containing point are : l, n.

Question (ii)
Lines whose point of intersection is B.
Solution:
Lines whose point of intersection is B are : l and m.

Question (iii)
Point of intersection of lines m and l.
Solution:
Point of intersection of lines m and l is : B.

Question (iv)
All pairs of intersecting lines.
Solution:
All pairs of intersecting lines are : m and l, n and l.

11. State which of the following statements are True (T) or False (F):

Question (i)
Two lines in a plane, always intersect at a point
Solution:
False

Question (ii)
If four lines intersect at a point, those are called concurrent lines.
Solution:
True

Question (iii)
Point has a size because we can see it as a thick dot on the paper.
Solution:
False

Question (iv)
Through a given point, only one line can be drawn.
Solution:
False

Question (v)
Rectangle is a part of the plane.
Solution:
True.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 12 Heron’s Formula Ex 12.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.2

Question 1.
A park, in the shape of a quadrilateral ABCD has ∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?
Answer:

In ∆ BCD, ∠C = 90°
∴ BD² = BC² + CD²
= (12)² + (5)²
= 144 + 25
= 169
= (13)²
∴ BD = 13 m

In ∆ BCD, a = 5 m, b = 12 m and c = 13 m
∴ Semiperimeter s = \(\frac{a+b+c}{2}\)
= \(\frac{5+12+13}{2}\) = \(\frac{30}{2}\) = 15 m
Then, s – a = 15 – 5 = 10m,
s – b = 15 – 12 = 3m and
s – c = 15 – 13 = 2 m.

Area of ∆ BCD = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{15 \times 10 \times 3 \times 2}\) m²
= \(\sqrt{900}\) m²
= 30 m²

Note: ∆ BCD is a right triangle.
∴ Area of ∆ BCD = \(\frac{1}{2}\) × BC × CD
= \(\frac{1}{2}\) × 12 × 5 = 30 m²

Now, in ∆ ABD, a = 9 m, b = 13 m arid c = 8 m
∴ Semiperimeter s = \(\frac{a+b+c}{2}\)
= \(\frac{9+13+8}{2}\) = \(\frac{30}{2}\) = 15 m
Then,
s – a = 15 – 9 = 6m,
s – b = 15 – 13 = 2m and
s – c = 15 – 8 = 7 m.

Area of ∆ ABD = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{15 \times 6 \times 2 \times 7}\) m²
= \(\sqrt{5 \times 3 \times 3 \times 2 \times 2 \times 7}\) m²
= 6 √35 m²
= 35.5 m² (approx.)
Then, the area of park in the shape of quadrilateral ABCD
= Area of ∆ BCD + Area of ∆ ABD
= (30 + 35.5) m² (approx.)
= 65.5 m² (approx.)
Thus, the area of the park is 65.5 m² (approx.)

Question 2.
Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.
Answer:

In ∆ ABC, a = 3 cm; b = 4 cm and c = 5 cm
∴ Semiperimeter s = \(\frac{a+b+c}{2}\)
= \(\frac{3+4+5}{2}\)
= \(\frac{12}{2}\) = 6 cm
Then,
s – a = 6 – 3 = 3 cm,
s – b = 6 – 4 = 2 cm and,
s – c = 6 – 5 = 1 cm.
Area of ∆ ABC = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{6 \times 3 \times 2 \times 1}\) cm²
= 6 cm²
Note: Proving that ∆ ABC is a right triangle, Area of ∆ ABC = \(\frac{1}{2}\) × 3 × 4 = 6 cm² can be obtained easily.
In ∆ ACD, a = 4 cm; b = 5 cm and c = 5 cm

Area of quadrilateral ABCD
= Area of ∆ ABC + Area of ∆ ACD
= (6 + 9.2) cm² (approx.)
= 15.2 cm² (approx.)

Question 3.
Radha made a picture of an aeroplane with coloured paper as shown in the given figure, s Find the total area of the paper used. ;

Answer:
The sides of the triangle in part 1 measure 5 cm, 5 cm and 1 cm.
∴ a = 5 cm, b = 5 cm and c = 1 cm
∴ Semiperimeter s = \(\frac{a+b+c}{2}\)
= \(\frac{5+5+1}{2}\) = \(\frac{11}{2}\) cm
Area of part 1
= Area of triangle

The length and breadth of rectangle in part II are 6.5 cm and 1 cm respectively.
Area of part II = Area of rectangle
= length × breadth
= (6.5 × 1) cm²
= 6.5 cm²
For the trapezium in part III, the parallel sides measure 1 cm and 2 cm, while both the non-parallel sides measure 1 cm each.

Drawing DM ⊥ AB and CN ⊥ AB. we get
AM = BM = \(\frac{2-1}{2}\) = \(\frac{1}{2}\) cm.
In ∆ DMA, ∠M = 90°
Area of trapezium ABCD
= \(\frac{1}{2}\) × Sum of parallel sides X Distance between parallel sides
= \(\frac{1}{2}\) × (AB + CD) × DM
= \(\frac{1}{2}\) × (2 + 1) × \(\frac{\sqrt{3}}{2}\)cm²
= \(\frac{1}{2}\) × 3 × \(\frac{\sqrt{3}}{2}\) cm²
= 1.3 cm² (approx.)
For the right triangle in part IV the sides forming the right angle measure 6 cm and 1.5 cm.
Area of right triangle in part IV.
= \(\frac{1}{2}\) × Product of sides forming the right angle
= \(\frac{1}{2}\) × 6 × 1.5 cm²
= 4.5 cm²
The right triangle in part V is congruent to the right triangle in part IV.
∴ Area of right triangle in part V = 4.5 cm²
Now, total area of the paper used
= Areas of figures in part I to part V
= (2.5 + 6.5 + 1.3 + 4.5 + 4.5) cm²
= 19.3 cm²

Question 4.
A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.
Answer:
In the given triangle, a = 26 cm, b = 28 cm and c = 30 cm
∴ Semiperimeter s = \(\frac{a+b+c}{2}\)
= \(\frac{26+28+30}{2}\) = \(\frac{84}{2}\) = 42 cm
Area of triangle

= 7 × 6 × 4 × 2 cm²
= 336 cm²
The area of the triangle and the area of the parallelogram are equal.
∴ Area of the parallelogram = 336 cm²
∴ Base × Corresponding altitude = 336 cm²
∴ 28 cm × Corresponding altitude = 336 cm²
∴ Corresponding altitude = \(\frac{336}{28}\) cm
∴ Corresponding altitude = 12 cm
Thus, the height of the parallelogram is 12 cm.

Question 5.
A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?
Answer:

Rhombus ABCD in the given figure represents the field.
A diagonal of a rhombus divides it into two congruent triangles.
∴ Area of rhombus ABCD = 2 × Area of ∆ ABC
In ∆ ABC, a = 30 m; b = 30 m; and c = 48 m.
∴ Semiperimeter s = \(\frac{a+b+c}{2}\)
= \(\frac{30+30+48}{2}\) = \(\frac{108}{2}\) = 54 cm
Area of ∆ ABC

= 3 × 6 × 24 m²
= 432 m²
Now, area of the field
= area of rhombus ABCD
= 2 × area of ∆ ABC
= 2 × 432 m²
= 864 m²
Now, area of grass field available for 18 cows to graze = 864 m²
∴ Area of grass field available for 1 cow to graze = \(\frac{864}{18}\) m² = 48 m²
Thus, each cow gets 48 m² of grass field to graze.

Question 6.
An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see the given figure), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?

Answer:
Out of 10 triangular pieces, 5 are dark coloured and 5 are light coloured.
For each triangle, a = 20 cm, b = 50 cm and c = 50 cm

Hence, the total area of 5 dark coloured cloth pieces = 5 × 200 √6 cm² = 1000 √6 cm²
Similarly, the total area of 5 light coloured cloth pieces = 5 × 200 √6 cm² = 1000 √6 cm²

Question 7.
A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in the given figure. How much paper of each shade has been used in it?

Answer:
Let us name the square part as ABCD and the triangular part as CMN.
Suppose the length of square ABCD is xcm.
∴ In ∆ ABD, AB = AD = x cm and ∠A = 90°
The length of hypotenuse BD is given to be 32 cm.
AB² + AD² = BD² (Pythagoras’ theorem)
∴ (x)² + (x)² = (32)²
∴ 2x² = 1024
∴ x² = 512
∴ x = √512
∴ x = \(\sqrt{256 \times 2}\)
∴ x = 16√2
Thus, the length of each side of square ABCD is 16 √2 cm.
Area of part I = Area of ∆ ABD
= \(\frac{1}{2}\) × AB × AD (∠A is a right angle.)
= \(\frac{1}{2}\) × 16 √2 × 16 √2 cm²
= 256 cm²
Area of part II = Area of A BCD
= \(\frac{1}{2}\) × BD × CD (∠A is a right angle.)
= \(\frac{1}{2}\) × 16 √2 × 16 √2 cm²
= 256 cm²
Note: Here, area of square ABCD can easily be found as below:
Area of square ABCD = \(\frac{(\text { Hypotenuse })^{2}}{2}\)
= \(\frac{(32)^{2}}{2}\)
= \(\frac{1024}{2}\)
= 512 cm²
To find the area of part III, we find the area of ∆ CMN.
In ∆ CMN, a = 6 cm, b = 8 cm and c = 6 cm.
∴ Semiperimeter s = \(\frac{a+b+c}{2}\)
= \(\frac{6+8+6}{2}\) = \(\frac{20}{2}\) = 10 cm

Area of part III
= Area of ∆ CMN

= 8 × 2.24 cm² (approx.)
= 17.92 cm² (approx.)

Question 8.
A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see the given figure). Find the cost of polishing the tiles at the rate of 50 p per cm².

Answer:
For each of 16 triangular tiles,
a = 9 cm; b = 28 cm and c = 35 cm

= 88.2 cm² (approx.)
∴ Area of 16 tiles = 16 × 88.2 cm²
= 1411.2 cm²
50 paise = ₹ 0.50
Cost of polishing 1 cm² region = ₹ 0.50
∴ Cost of polishing 1411.2 cm² region
= ₹ (1411.2 × 0.50)
= ₹ 705.60

Question 9.
A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.
Answer:

In the given figure, trapezium ABCD represents the field in which AB || CD,
AB = 25 m, BC = 14 m, CD = 10 m and DA = 13 m.
Through C, draw a line parallel to DA to intersect AB at E.
In quadrilateral AECD, AE || CD and DA || CE
∴ AECD is a parallelogram.
∴ CE = DA = 13 m and AE = CD = 10 m
Now, BE = AB – AE = 25 – 10 = 15 m
In ∆ CEB, a = 13 m; b = 15 m and c = 14 m

In ∆ CEB, draw CM ⊥ BE.
Area of ∆ CEB = \(\frac{1}{2}\) × BE × CM
∴ 84 m² = \(\frac{1}{2}\) × 15 m × CM
∴ CM = \(\frac{84 \times 2}{15}\) m
∴ CM = 11.2 m
Area of parallelogram AECD
= Base × Corresponding altitude
= AE × CM
= 10 × 11.2 m²
= 112 m²
Hence, area of the field
= Area of trapezium ABCD
= Area of ∆ CEB + Area of parallelogram AECD
= 84 m² + 112 m²
= 196 m²
Note: After finding CM = 11.2m, the area of . the field can also be found as below:
Area of the field
= Area of trapezium ABCD
= \(\frac{1}{2}\) × sum of parallel sides × distance between parallel sides
= \(\frac{1}{2}\) × (AB + CD) × CM
= \(\frac{1}{2}\) × (25 + 10) × 11.2 m²
= \(\frac{1}{2}\) × 35 × 11.2 m²
= 196 m²

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 12 Heron’s Formula MCQ Questions with Answers.

PSEB 9th Class Maths Chapter 12 Heron’s Formula MCQ Questions

Multiple Choice Questions and Answer

Answer each question by selecting the proper alternative from those given below each question to make the statement true:

Question 1.
The sides of a triangle measure 8cm, 12cm and 6 cm. Then, the semiperimeter of the triangle is ……………………… cm.
A. 26
B. 52
C. 13
D. 6.5
Answer:
C. 13

Question 2.
Each side of an equilateral triangle measures 8 cm. Then, the semiperimeter of the triangle is ……………………….. cm.
A. 4
B. 24
C. 12
D.36
Answer:
C. 12

Question 3.
In a right angled triangle, the length of the hypotenuse is 15 cm and one of the sides forming right angle is 9 cm. Then, the semiperimeter of the triangle is ……………………….. cm.
A. 36
B. 18
C. 12
D. 15
Answer:
B. 18

Question 4.
The ratio of the measures of the sides of a triangle is 3:4:5. If the semiperimeter of the < triangle is 36 cm, the measure of the longest side of the triangle is ……………………. cm.
A. 12
B. 15
C. 20
D. 30
Answer:
D. 30

Question 5.
The area of a triangle is 48 cm2 and one of its sides measures 12 cm. Then, the length of the altitude corresponding to this side is …………………. cm.
A. 4
B. 8
C. 16
D. 6
Answer:
B. 8

Question 6.
The sides of a triangle measure 12 cm, 17 cm and 25 cm. Then, the area of the triangle is ……………………….. cm².
A. 54
B. 90
C. 180
D. 135
Answer:
B. 90

Question 7.
Two sides of a triangle measure 9 cm and 10 cm. If the perimeter of the triangle is 36cm, then its area is …………………. cm².
A. 17
B. 36
C. 72
D. 18
Answer:
B. 36

Question 8.
The area of an equilateral triangle with each side measuring 10 cm is ………………….. cm².
A. \(\frac{5 \sqrt{3}}{2}\)
B. 25√3
C. 5√3
D. 3√5
Answer:
B. 25√3

Question 9.
∆ ABC is an isosceles triangle in which BC = 8 cm and AB = AC = 5 cm. Then, area of ∆ ABC = ……………………….. cm².
A. 6
B. 12
C. 18
D. 24
Answer:
B. 12

Question 10.
ABCD is a parallelogram. If ar (ABC) = 18 cm², then ar(ABCD) = …………………. cm².
A. 18
B. 9
C. 36
D. 27
Answer:
C. 36

Question 11.
ABCD is a parallelogram. If ar (ABC) = 18 cm², then ar (ABCD) = …………………. cm².
A. 3.6
B. 7.2
C. 7.5
D. 6
Answer:
B. 7.2

Question 12.
In quadrilateral ABCD, AC = 10 cm. BM and DN are altitudes on AC from B and D respectively. If BM = 12cm and DN = 4 cm, then ar (ABCD) = …………………. cm².
A. 160
B. 80
C. 320
D. 480
Answer:
B. 80

Question 13.
The perimeter of rhombus ABCD is 40 cm and BD =16 cm. Then, ar (ABCD) = ……………………. cm².
A. 96
B. 48
C. 24
D. 72
Answer:
A. 96

Question 14.
The area of a rhombus is 72 cm² and one of its diagonals measures 16 cm. Then, the length of the other diagonal is ………………… cm.
A. 12
B. 9
C. 18
D. 15
Answer:
B. 9

Question 15.
PQRS is a square. If PQ = 10 cm, then PR = ……………………….. cm.
A. 10
B. 20
C. 10√2
D. 2√10
Answer:
C. 10√2

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 9 Understanding Elementary Shapes Ex 9.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 9 Understanding Elementary Shapes Ex 9.1

1. Measure the line segments using a ruler and a divider:

Solution:
(i) PQ = 4.4 cm
(ii) CD = 3.6 cm
(iii) XY = 2.5 cm
(iv) AB = 5.8 cm
(v) LM = 5 cm.

2. Compare the line segments in the figure and fill in the blanks:

Question (i)
AB _ AB
Solution:
AB = AB

Question (ii)
CD _ AC
Solution:
CD < AC Question (iii) AC _ AD Solution: AC > AD

Question (iv)
BC _ AC
Solution:
BC < AC Question (v) BD _ CD. Solution: BD > CD.

3. Draw any line segment AB. Take any point C between A and B. Measure the lengths of AB, BC and AC. Is AB = AC + CB?
Solution:
If A, B, C are any three points on a line such that AC + CB = AB, then we are sure that C lies between A and B.

On measuring the lengths of AB, BC and AC, we get
AB = 6 cm, AC = 4 cm, CB = 2 cm
Now, AC + CB = 4 cm + 2 cm = 6 cm
Hence, AB = AC + CB.

4. Draw a line segment AB = 5 cm and AC = 9 cm in such a way that points A, B, C are collinear. What is the length of BC?
Solution:

AB = 5 cm and AC = 9 cm
Since, A, B and C are collinear
∴ AB + BC = AC
⇒ 5 cm + BC = 9 cm
⇒ BC = 9 cm – 5 cm
= 4 cm.
Hence, Length of BC = 4 cm