Bhagya

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 15 Visualising Solid Shapes Ex 15.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.2

1. Use isometric dot paper and make an isometric sketch of the following figures :
Question (i).

Solution:

Question (ii).

Solution:

Question (iii).

Solution:

Question (iv).

Solution:

2. Draw (i) an oblique sketch (ii) Isometric sketch for :
(a) A cube with a edge of 4 cm long
(b) A cuboid of length 6 cm, breadth 4 cm and height 3 cm
Solution:
(a) (i) Oblique sketch of cube with a edge 4 cm long.

(ii) Iso metric sketch for a cube with a edge of 4 cm long.

(b) (i) an oblique sketch for a cuboid of length 6 cm, breadth 4 cm and height 3 cm.

(ii) Isometric sketch for a cuboid of length 6 cm, breadth 4 cm and height 3 cm.

3. Two cubes each with edge 3 cm are placed side by side to form a cuboid, sketch oblique and isometric sketch of this cuboid.
Solution:

4. Draw an Isometric sketch of triangular pyramid with base as equilateral triangle of 6 cm and height 4 cm.
Solution:

5. Draw an Isometric sketch of square pyramid.
Solution:

6. Make an oblique sketch for each of the given Isometric shapes.

Question (i).

Solution:

Question (ii).

Solution:

7. Using an isometric dot paper draw the solid shape formed by the given net.

Solution:

8. Multiple choice questions:

Question (i).
An oblique sheet is made up of:
(a) Rectangles
(b) Squares
(c) Right angled triangles
(d) Equilateral triangles.
Answer:
(b) Squares

Question (ii).
An isometric sheet is made up of dots forming :
(a) Squares
(b) Rectangles
(c) Equilateral triangles
(d) Right angled triangle
Answer:
(c) Equilateral triangles

Question (iii).
An oblique sketch has :
(a) Proportional lengths
(b) Parallel lengths
(c) Non-Proportional lengths
(d) Perpendicular lengths.
Answer:
(c) Non-Proportional lengths

Question (iv).
An Isometric sketch has :
(a) Non proportional lengths
(b) Parallel lengths
(c) Perpendicular lengths
(d) Proportional lengths.
Answer:
(d) Proportional lengths.

Question (v).
Isometric sketches shows objects of:
(a) Two dimensions
(b) Shadows
(c) Three dimensions
(d) One dimension.
Answer:
(c) Three dimensions

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 10 Practical Geometry Ex 10.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 10 Practical Geometry Ex 10.1

1. What is the use of instrument ruler?
Solution:
We use instrument ruler to draw line segment and to measure their lengths.

2. What is the use of protractor?
Solution:
We use a protractor to draw and measure angle.

3. What is the use of compasses?
Solution:
We use compasses to mark equal lengths, draw arcs and circles.

4. Construct the following angles using set squares.

Question (i)
(i) 30°
(ii) 45°
(iii) 60°
(iv) 75°
(v) 90°.
Solution:
(i) Steps of Construction:

1. Draw a ray OA.

2. To construct an angle of 30° we use 30° set square. Place the set square in such a way that one of its edges containing the 30° angle coincides with the ray OA as shown in the figure.

3. Draw a ray OB starting from the vertex O along the 30° edge of the set square as shown in figure.
4. Remove the set square.
Thus, the required \(\angle \mathrm{AOB}\) = 30°.

(ii) Steps of Construction:

1. Draw a ray OA.

2. To construct an angle of 45° we use 45° set square.
Place a 45° set square along ray OA as shown in the figure.

3. Draw a ray OB starting from the vertex O along 45° the, edge of set square.
4. Remove the 45° set square. Thus, the required \(\angle \mathrm{AOB}\) = 45°

(iii) Steps of Construction.

1. Draw a ray OA.

2. To construct an angle of 60°. We use 30° set square.

Place a 30° set square with 60° edge along ray OA as shown in the figure.
3. Draw a ray OB starting from the vertex O along 60° edge of the set square.
4. Remove the 30° set square. Thus the required \(\angle \mathrm{AOB}\) = 60°.

(iv) Steps of Construction

1. Draw a ray of OA.

2. To draw angle of 75° we use both set squares in combination as 45° + 30° = 75°.
Place 45° set square with 45° edge along OA

3. Place 30° set square adjacent to 45° set square as show. Draw a ray starting from the vertex O along the edge of 30° set square.
4. Remove both the set squares. Thus required \(\angle \mathrm{AOB}\) = 75°.

(v) Steps of Construction

1. Draw a ray OA.

2. To construct angle of 90° place any set square with 90° corner at O along OA.

3. Draw a ray OB starting from the vertex O along 90° edge of the set square.
4. Remove the set square. Thus, the required \(\angle \mathrm{AOB}\) = 90°.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 9 Understanding Elementary Shapes MCQ Questions with Answers.

PSEB 6th Class Maths Chapter 9 Understanding Elementary Shapes MCQ Questions

Multiple Choice Questions

Question 1.
In the figure, which of the following is true?
(a) PR = PQ
(b) PR > QR
(c) PS > PR
(d) PR < PQ.
Answer:
(b) PR > QR

Question 2.
Which angle is represented in the given figure?
(a) Reflex
(b) Acute
(c) Obtuse
(d) Right angle.
Answer:
(a) Reflex

Question 3.
Which angle is represented in the given figure?
(a) Acute
(b) Right angle
(c) Obtuse
(d) Reflex
Answer:
(b) Right angle

Question 4.
Which of the following is the example of perpendicular lines?
(a) Railway lines
(b) Line segment forming letter ‘X’
(c) Adjacent edges of a table
(d) Line segment forming line ‘M’.
Answer:
(c) Adjacent edges of a table

Question 5.
Which of the following forms triangles?
(a) 60°, 72°, 48°
(b) 73°, 54°, 59°
(c) 60°, 51°, 70°
(d) 100°, 42°, 39°.
Answer:
(a) 60°, 72°, 48°

Question 6.
Which of the following are sides of a triangle?
(a) 1, 2, 3
(b) 2, 2,1
(c) 3, 4, 2
(d) 5, 6, 12.
Answer:
(c) 3, 4, 2

Question 7.
A parallelogram having adjacent sides equal is called a …………… .
(a) Trapezium
(b) Rhombus
(c) Rectangle
(d) Square.
Answer:

Question 8.
Which of the following is not true for rectangle?
(a) Diagonals are equal
(b) Diagonals bisect each other
(c) Each angle is 90°
(d) All sides are equal.
Answer:
(d) All sides are equal

Question 9.
Which of the following is not true?
(a) Every rhombus is a parallelogram
(b) Each square is rhombus.
(c) Each rectangle is a square.
(d) Each square is parallelogram.
Answer:
(c) Each rectangle is a square

Question 10.
A cuboid has ………….. edges.
(a) 10
(b) 6
(c) 12
(d) 8.
Answer:
(c) 12

Question 11.
The following angle is an:

(a) acute angle
(b) obtuse angle
(c) right angle
(d) straight angle.
Answer:
(a) acute angle

Question (ii)
What is the angle name for half a revolution?
(a) acute angle
(b) obtuse angle
(c) straight angle
(d) right angle.
Answer:
(c) straight angle

Question (iii)
What is the angle name for one-fourth revolution?
(a) right angle
(b) straight angle
(c) complete angle
(d) acute angle.
Answer:
(a) right angle

Question (iv)
An angle whose measure is less than that of a right angle is …………….. .
(a) complete angle
(b) acute angle
(c) obtuse angle
(d) straight angle.
Answer:
(b) acute angle

Question (v)
An angle whose measure is greater than that of a right angle:
(a) acute angle
(b) complete angle
(c) obtuse angle
(d) straight angle.
Answer:
(c) obtuse angle

Fill in the blanks:

Question (i)
An angle whose measure is equal to 90°, is called ……………….. .
Answer:
right angle

Question (ii)
An angle whose measure is the sum of the measure of two right angle is ………………. .
Answer:
straight angle

Question (iii)
Your instrument box looks like a ………………. .
Answer:
cuboid

Question (iv)
A road-roller looks like a ……………… .
Answer:
cylinder

Question (v)
A cuboid has ……………. faces.
Answer:
six

Write True/False:

Question (i)
Every rhombus is a parallelogram. (True/False)
Answer:
True

Question (ii)
Every square is a rhombus. (True/False)
Answer:
True

Question (iii)
All sides of a rectangle are equal. (True/False)
Answer:
False

Question (iv)
Each rectangle is a square. (True/False)
Answer:
False

Question (v)
Each angle of a rectangle is of 90°. (True/False)
Answer:
True.

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.