Class 6

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 10 Practical Geometry Ex 10.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 10 Practical Geometry Ex 10.3

1. Draw a line r and mark a point P on it. Construct a line perpendicular to r at point P.

Question (i)
Using a ruler and compasses.
Solution:
Using ruler and compasses

Steps of Construction.

1. Draw a line r and mark a point P on it.

2. Draw an arc from P to the line r of any suitable radius which intersects line r at A and B.

3. Draw arcs of any radius which is more than half of arc made in step (2) from A and B which intersect at Q.

4. Join PQ.

Thus PQ is perpendicular to AB or line l or PQ ⊥ A.
Here P is called foot of perpendicular.

Question (ii)
Using a ruler and a set square.
Solution:
Using a ruler and a set square

Steps of Construction

1. Draw a line r and a point P on it.
2. Place one of the edges of a ruler along the line l and hold if firmly.

3. Place the set square in such a way that one of its edges contaning the right angle coincides with the ruler.
4. Holding the ruler, slide the set square along the line l till the vertical side reaches the point P.

5. Firmly hold the set square in this position. Draw PQ along its vertical edge. Now PQ is the required perpendicular to l ie. PQ ⊥ r.

2. Draw a line p and mark a point z above it. Construct a line perpendicular to p, from the point z.

Question (i)
Using a ruler and compasses.
Solution:
1. Draw a line p and mark a point z not lying on it.
2. From point z draw an arc which intersects line p at two points P and Q.


3. Using any radius and taking P and Q as centre, draw two arcs that intersect at point say B. On the other side (a shown in figure).
4. Join AB to obtain altitude to the line p.

Thus xz is altitude to line p.
i.e. xz ⊥ p.

Question (ii)
Using a ruler and set square
Solution:
Steps of constructions:
1. Draw a line p and mark a point z which is not lying on it.
2. Place one of the edge of a ruler along the line p and hold it firmly.

3. Place the set square in such a way that one of its edges containing the right angle coincides with the ruler.
4. Holding the ruler firmly, slide the set square along the line p till its vertical side reaches the point z.

5. Firmly hold the set square in this position, Draw xz along its vertical edge. Now xz is the required altitude to p i.e. xz ⊥ p.

3. Draw a line AB and mark two points P and Q on either side of line AB, Construct two lines perpendicular to AB, from P and Q using a ruler and compasses.
Solution:
1. Draw a line AB and Mark two points P and Q on either side of AB.

2. From point P draw an arc which intersect line AB at two points C and D.
3. Using any radius and taking C and D as centre draw two arcs that intersects at point say E on the other side as shown in figures.

4. Join PE to obtain perpendicular to AB.
5. From point Q draw an arc which intersects AB at two points X and Y.

6. Using any radius and taking X and Y as centre draw two arcs that intersects at point say R on the other side of line AB as shown in figures.
7. Join QR to obtain perpendicular to AB.
Thus, PE ⊥ AB and QR ⊥ AB

4. Draw a line segment of 7 cm and draw perpendicular bisector of this line segment.
Solution:
Steps of Construction:
1. Draw a line segment AB = 7 cm.
2. With A as centre and radius more than half of AB, draw an arc on both sides of AB.

3. With B as centre and the same radius as in step 2, draw an arc intersecting the first arc at C and D.
4. Join CD intersecting AB at O. Then CD is the perpendicular bisector of AB.

5. Draw a line segment PQ = 6.8 cm and draw its perpendicular bisector XY which bisect PQ at M. Find the length of PM and QM. Is PM = QM ?
Solution:
Steps of Construction:

1. Draw a line segment PQ = 6.8 cm
2. With P as centre and radius more than half of PQ draw arcs on both sides of PQ.

3. Now with Q as centre and the same radius as in step 2 draw arcs intersecting the previous drawn arcs at A and B respectively.
4. Join AB intersecting PQ at M. Then M bisects the line segment.
5. Measure the length of PM and QM
PM = 3.4 cm and QM = 3.4 cm
∴ PM = QM.

6. Draw perpendicular bisector of line segment AB = 5.4 cm. Mark point X anywhere on perpendicular bisector Join X with A and B. Is AX = BX ?
Solution:
Steps of construction.
1. Draw a line segment AB = 5.4 cm.
2. With A as centre and radius more than half of AB, draw an arc in both sides of AB.

3. With B as centre and the same radius as in step 2, draw an arc intersecting the first arc at C and D.
4. Join CD intersecting AB at O.
Then CD is the perpendicular bisector of AB.
Mark any point X on the perpendicular bisector CD. Drawn. Then join AX and BX.
On examination, we find that AX = BX.

7. Draw perpendicular bisectors of line segment of the following lengths.

Question (i)
8.2 cm
Solution:
Steps of Construction.
1. Draw a line regment AB = 8.2 cm

2. With A as centre and radius more than half of AB, draw arcs on both sides of AB.
3. With B as centre and the same radius as in step 2, draw an arcs intersecting the previous arc at C and D.
4. Join CD intersecting AB at O. Then CD is the perpendicular bisector of AB.

Question (ii)
7.8 cm
Solution:
Steps of Construction.

1. Draw a line segment AB = 7.8 cm

2. With A as centre and radius more than half of AB draw arcs on both sides of AB.
3. With B as centre and the same radius as in step 2, draw arcs intersecting the previous arcs at C and D.
4. Join CD intersecting AB at O. Then CD is the perpendicular bisector of AB.

Question (iii)
6.5 cm.
Solution:
Steps of Construction.
1. Draw a line segment AB = 6.5 cm

2. With A as centre and radius more than half of AB draw arcs on both sides of AB.
3. With B as centre and the same radius as in step 2 draw arcs intersecting the previous arcs at C and D.
4. Join CD intersecting AB at O. Then CD is the perpendicular bisector of AB.

8. Draw a line segment of length 8 cm and divide it into four equal parts Using compasses. Measure each part.
Solution:
Steps of construction.

1. Draw a line segment AB of length 8 cm
2. With A as centre and radius more than half of AB, draw arcs on both sides of AB.
3. With B as centre and the same radius as in step 2, draw arcs intersecting the previous arcs at P and Q.
4. Join PQ intersecting AB at C then PQ is the perpendicular bisector of AB intersecting AB at C.
5. Similarly draw the perpendicular bisector of AC intersecting AC at D.
6. Draw the perpendicular bisector of CB intersecting CB at E.
By actual measurement, it can be verified that
AD = DC = CE = EB

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 6th Class Maths Book Solutions Chapter 7 Algebra MCQ Questions with Answers.

PSEB 6th Class Maths Chapter 7 Algebra MCQ Questions

Multiple Choice Questions

Question 1.
Each side of square is represented by ‘s’ then perimeter of square is:
(a) 4 + s
(b) s – 4
(c) 4s
(d) s.
Answer:
(c) 4s

Question 2.
Write commutative property of multiplication using variables x and y:
(a) xy = yx
(b) x + y = y + x
(c) x + y
(d) xy.
Answer:
(a) xy = yx

Question 3.
How many terms in expression 7l – 3l?
(a) 1
(b) 3
(c) 2
(d) 4.
Answer:
(c) 2

Question 4.
5 is subtracted from m = ……………….. .
(a) 5 – m
(b) m + 5
(c) 5 + m
(d) m – 5.
Answer:
(d) m – 5.

Question 5.
Multiply p by 3 then 2 is added = ……………. .
(a) 2p + 3
(b) 3p – 2
(c) 3p + 2
(d) 2p – 3.
Answer:
(c) 3p + 2

Question 6.
If Armaan’s present age is x years then what will be his age after 4 years?
(a) x – 4
(b) x + 4
(c) 4x
(d) 4 – x.
Answer:
(b) x + 4

Question 7.
Write as algebraic equation: 7 more than 4 times ofy gives 23 :
(a) 4 + 7y = 23
(b) 7 + y = 23
(c) 4y – 7 = 23
(d) 4y + 7 = 23.
Answer:
(d) 4y + 7 = 23.

Question 8.
Find x if x – 3 = 2 :
(a) 3
(b) 6
(c) 5
(d) 2.
Answer:
(c) 5

Question 9.
Solve:
4l – 3 = 5
(a) 3
(b) 4
(c) 1
(d) 2.
Answer:
(d) 2.

Question 10.
If \(\frac {a}{4}\) = 5 then a = ……………. .
(a) 5
(b) 20
(c) 4
(d) 18.
Answer:
(b) 20

Question 21.
What is algebraic expression for subtracting 7 from – m?
(a) m – 1
(b) m + 7
(c) 7 – m
(d) – m – 7.
Answer:
(d) – m – 7.

Question 22.
What is algebraic expression for subtracting 7 from p?
(a) p – 7
(b) p + 7
(c) 7 – p
(d) 7 × p.
Answer:
(a) p – 7

Question 23.
What is algebraic expression for multiplying p by 16?
(a) 16 p
(b) p + 6
(c) p – 16
(d) \(\frac {p}{16}\)
Answer:
(a) 16 p

Question (iv)
What is algebraic expression for first multiplying x by 3 and then adding 2 to the product?
(a) x + 6
(b) 3x + 2
(c) 3x – 2
(d) 6x
Answer:
(b) 3x + 2

Question (v)
What is algebraic expression for first multiplying y by 2 and then subtracting 5 from the product?
(a) 2y + 5
(b) y + 10
(c) 2y – 5
(d) 10y.
Answer:
(c) 2y – 5

Fill in the blanks:

Question (i)
The algebraic expression for first multiplying y by 10 and then adding 7 to the product is ………… .
Answer:
10y + 7

Question (ii)
The algebraic expression for first multiplying n by 2 and then subtracting l from the product ………… .
Answer:
2n – l

Question (iii)
7 × 20 – 82 is expression of only ………………. .
Answer:
Numbers

Question (iv)
Each side of a square is l, then perimeter of square is ……………. .
Answer:
4l

Question (v)
5 is added to x = …………….. .
Answer:
x + 5

Write True/False:

Question (i)
If \(\frac {a}{5}\) = 4, then a = 20. (True/False)
Answer:
True

Question (ii)
If x – 3 = 2, then x = l. (True/False)
Answer:
False

Question (iii)
If 4l – 3 = 5, then l = 2. (True/False)
Answer:
True

Question (iv)
Number of terms in expression 3p + 2 is two. (True/False)
Answer:
True

Question (v)
The letters which can take any numerical value are called variables. (True/False)
Answer:
True