Author Archives: Rohit

Ex 14.1 Class 6 Maths Question 1.
Draw a circle of radius 3.2 cm.
Solution:
Steps of Construction:

1. 1. Mark a point 0 on the paper.
   2. Open the compasses for the required radius 3.2 cm.
      

1. Keep the steel end of the compasses fixed at the point O and rotate the pencil end at 3.2 cm distance from O till it returns to the starting point P.
   Then, the figure so obtained is the required circle.

Ex 14.1 Class 6 Maths Question 2.
With the same centre O, draw two circles of radii 4 cm and 2.5 cm.
Solution:
Steps of Construction:

1. Mark a point 0 on the paper.
2. Open the compasses for the radii 4 cm/2.5 cm (tumwise).
   
3. Keep the steel end of the compasses fixed at the point 0 and rotate the pencil end at 4 cm/2.5 cm (turnwise)distance from O till it returns to their starting points respectively.
   The circles obtained are the required circles.
   Note: Circles with the same centre are called concentric circles.

Ex 14.1 Class 6 Maths Question 3.
Draw a circle and any two ofits diameters. If you join the ends of these diameters, what is the figure obtained? What figure is obtained if the diameters are perpendicular to each other? How do you check your answer?
Solution:
Steps of Construction:

1. Draw a circle with O as centre of any radius.
2. Draw any two diameters AOB and COD.
   
3.  Join AC, CB, BD and DA.
   Clearly, the figure ACBD is a rectangle.
   When the diameter AOB and COD are perpendicular to each other:
   
   Then, figure obtained on joining AC, CB, BD and DA is a square ACBD.

Ex 14.1 Class 6 Maths Question 4.
Draw any circle and mark points A, B and C such that:
(a) A is on the circle.
(b) B is in the interior of the circle.
(c) C is in the exterior of the circle.
Solution:
Steps of Construction:

1. Draw a circle with O as centre of any radius.
2. Mark point
(a) A on the circle,
(b) B in the interior of the circle, and
(c) C in the exterior of the circle.

Ex 14.1 Class 6 Maths Question 5.
Let A, B be the centres of two circles of equal radii; draw them so that each one of them passes through the centre of the other. Let them intersect at C and D.
Examine whether AB and CD are at right angles.
Solution:
Steps of Construction:

1. Mark points A and Bona paper such that AB = radii (i.e., 3 cm, say).
   
2. With A as centre draw a circle of any radius (say, 3 cm).
3. With B as centre draw a circle of radius 3 cm as shown.
4. Let these circles intersect at C and D.
   Clearly, on measuring we find ∠AMC = 90°, so AB ⊥ CD.

Ex 14.2 Class 6 Maths Question 1.
Draw a line segment of length 7.3 cm, using a ruler.
Solution:
Steps of Construction:

1. 1. Mark a point A on the plane of the paper and place the ruler so that zero mark of the ruler is at A.
      

1. Mark with pencil a point B against the mark on the ruler which indicates 7.3 cm.
2. Join points A and B by moving the tip of the pencil against the straight edge of the ruler.
   The line segment AB so obtained is the required line segment.

Ex 14.2 Class 6 Maths Question 2.
Construct a line segment of length 5.6 cm using ruler and compasses.
Solution:
Steps of Construction:

1. Mark a point A on the plane of the paper and draw a line, say l, passing through it.
   
2. Place the steel end of the compasses at zero mark on the ruler and open out it such that the pencil end on the mark indicates 5.6 cm.
3. Transfer the compasses as it is to the line l so that the steel end is on A.
4. With the pencil end make a small stroke on l so as to cut it at B.
5. The segment AB so obtained is the required line segment.

Ex 14.2 Class 6 Maths Question 3.
Construct AB¯¯¯¯¯¯¯¯ of length 7.8 cm. From this, cut off AC¯¯¯¯¯¯¯¯ of length 4.7 cm. Measure BC¯¯¯¯¯¯¯¯.
Solution:
Steps of Construction:

1. Draw a line segment AB of length 7.8 cm.
   
2. Using compasses find a point C on the line segment AB so that segment AC = 4.7 cm.
3. On measuring BC, we find that BC = 3.1 cm.

Ex 14.2 Class 6 Maths Question 4.
Given AB¯¯¯¯¯¯¯¯ of length 3.9 cm, construct PQ¯¯¯¯¯¯¯¯ such that the length of PQ¯¯¯¯¯¯¯¯ is twice that of AB¯¯¯¯¯¯¯¯. Verify by measurement.

Solution:
Steps of Construction:

1. Draw a line l and mark a point P on it.
   
2. Using compasses find a point X so that PX(= AB) = 3.9 cm on the line l.
3. Using compasses find a point Q so that XQ = 3.9 cm on the line l.
   Thus, PQ = PX + XQ = 3.9 cm + 3.9 cm
   = 2(3.9 cm) = 2AB.

Ex 14.2 Class 6 Maths Question 5.
Given AB¯¯¯¯¯¯¯¯ of length 7.3 cm and CD¯¯¯¯¯¯¯¯ of length 3.4 cm, construct a line segment XY¯¯¯¯¯¯¯¯ such that the length of XY¯¯¯¯¯¯¯¯ is equal to the difference between the lengths of AB¯¯¯¯¯¯¯¯ and CD¯¯¯¯¯¯¯¯. Verify by measurement
Solution:
Steps of Construction:

1. Draw line segments AB = 7.3 cm and CD = 3.4 cm.
2. Draw a line l and mark a point X on it.
3. Using compasses find a point P on the line l so that segment XP = segment AB (i.e., 7.3 cm).
4. Using compasses find a point Y so that the segment PY = segment CD (i.e., 3.4 cm). The segment XY so obtained is the required segment, because XY = OP -PY – AB -CD.

Ex 14.3 Class 6 Maths Question 1.
Draw any line segment PQ¯¯¯¯¯¯¯¯. Without measuring PQ¯¯¯¯¯¯¯¯, construct a copy of PQ¯¯¯¯¯¯¯¯.
Solution:
Steps of Construction:
Let PQ be the given line segment and l be a given line.

1. Mark a point 0 on the line l.
2. Open out the compasses and adjust so that the steel end is on A and the pencil end is A on B.
3. Transfer the compasses to the line l without disturbing their opening so that the steel end is on 0.
4. With the pencil end make a small stroke on the line l to cut it at the point P.
   Then, the fine segment OP so obtained = the given line segment AB.

Ex 14.3 Class 6 Maths Question 2.
Given some line segment AB¯¯¯¯¯¯¯¯, whose length you do not know, construct PQ¯¯¯¯¯¯¯¯ such that the length of PQ¯¯¯¯¯¯¯¯ is twice that of AB¯¯¯¯¯¯¯¯.
Solution:
Steps of Construction:
Let AB be the given line segment.

1. Draw any line l and make a point 0 on it.
2. Open out the compasses in such way that the steel end is on A and the pencil end is on B.
3. Transfer the compasses without disturbing their opening to the line l so that the steel end is on O.
4. With the pencil end make a small stroke on -the line l to cut it at the point P.
5. Repeat the steps 3 and 4 with same opening having P as the initial point and Q as the terminal point.
   Then, the segment OQ = OP + PQ = AB + AB= 2 AB.

Ex 14.4 Class 6 Maths Question 1.
Draw any line segment AB¯¯¯¯¯¯¯¯,. Mark any point M on it. Through M, draw a perpendicular to AB¯¯¯¯¯¯¯¯,. (use ruler and compasses)
Solution:
Steps of Construction:
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry 14

Draw a line segment AB, and mark any point M on it.
With centre M and any radius, cut off MX and MY of equal lengths on both sides of M.
With centre X and any radius > MX, draw an arc.
With centre Y and the same radius draw another arc, cutting the previously drawn arc at P
Join MP
Then, the segment PM so obtained is the required perpendicular.
Ex 14.4 Class 6 Maths Question 2.
Draw any line segment PQ¯¯¯¯¯¯¯¯,. Take any point R not on it. Through R, draw a perpendicular to PQ¯¯¯¯¯¯¯¯,. (use ruler and set-square)
Solution:
Steps of Construction:

Let PQ be the line and R is any point not lying on PQ.
Place the set-square so that the base AB of the set-square lies exactly on the line PQ.
Hold the set-square fixed and place a ruler so that its edge position lies along the side AC of the set-square.
Holding the ruler fixed, slide the set-square along the ruler till the point R coincides with the point B of the set-square.
Keeping the set-square fixed in this position, draw a line RT along the edge BC of the set-square through R.
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry 15
Thus, RT is the required perpendicular line to the line PQ passing through R.
Ex 14.4 Class 6 Maths Question 3.
Draw a line l and a point X on it. Through X, draw a line segment XY¯¯¯¯¯¯¯¯ perpendicular to l. Now draw a perpendicular to XY at Y. (use ruler and compasses)
Solution:
Steps of Construction:
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry 16

Draw a line l and mark any point X on it. ,
With centre X and any radius, cut off XA = XB on both sides of X.
With centre A and any radius > XA, draw an arc.
With centre B and the same radius draw another arc, cutting the previously drawn arc at Y.
Join XY. Then XY is perpendicular to line 1.
By proceeding as above draw a perpendicular YZ to XY.

Ex 14.5 Class 6 Maths Question 1.
Draw AB¯¯¯¯¯¯¯¯ of length 7.3 cm and find its axis of symmetry.
Solution:
Steps of Construction:

1. Draw a line segment AB = 7.3 cm.
2. With centre A and radius > 12 AB, draw arcs one on each side of AB.
3. With centre B and the same radius as before, draw arcs cutting the previously drawn arcs at C and D respectively.
4. Join CD intersecting AB at M. Then M bisects the line segment AB as shown.
   
   The line segment so obtained is the required axis of symmetry.

Ex 14.5 Class 6 Maths Question 2.
Draw a line segment of length 9.5 cm and construct its perpendicular bisector.
Solution:
Steps of Construction:

1. Draw a line segment AB = 9.5 cm.
2. With centre A and radius > 12 AB, draw arcs one on each side of AB. . A
3. With centre B and the same radius as before, draw arcs cutting the previously drawn arcs at C and D respectively.
4. Join CD. Then the line segment CD is the required perpendicular bisector of AB.

Ex 14.5 Class 6 Maths Question 3.
Draw the perpendicular bisector of XY¯¯¯¯¯¯¯¯ whose length is 10.3 cm.
(a) Take any point P on the bisector drawn. Examine whether PX = PY.
(b) If M is the mid-point of XY¯¯¯¯¯¯¯¯ , what can you say about the lengths MX and XY?
Solution:
Steps of Construction:

1. Draw a line segment XY =10.3 cm.
   
2. With centre X and radius > 12 XY, draw arcs one on each side of XY.
3. With centre Y and the same radius as before, draw arcs cutting the previously drawn arcs at A and B respectively.
4. oin .AB intersecting XY at M. Then, AB is the perpendicular bisector of XY.
   (a) Mark any point P on AB, the perpendicular bisector. On measuring, we find that PX = PY.
   (b) Since M is the mid-point of the segment XY. Therefore,
   MX = 12XY = 12 x 10.3 cm
   = 5.15 cm

Ex 14.5 Class 6 Maths Question 4.
Draw a line segment of length 12.8 cm. Using compasses, divide it into four equal parts. Verify by actual measurement.
Solution:
Steps of Construction:

1. Draw a line segment AB = 12.8 cm.
2. With centre A and radius > 12 AB, draw arcs one on each side of AB.
3. With centre B and the same radius as before, draw arcs cutting the previously drawn arcs at C and D respectively.
4. Join CD intersecting AB at M.
5. Further find the mid-points M₁ and M₂ of AM and MB respectively proceeding in the same way qs-before.
   ∴ AM₁ = M₁M = MM₂ = M₂B, On measuring, we find that each part = 3.2 cm.

Ex 14.5 Class 6 Maths Question 5.
With PQ¯¯¯¯¯¯¯¯ of length 6.1 cm as diameter draw a circle.
Solution:
Steps of Construction:

1. Draw a line segment PQ = 6.1 cm.
2. Bisect the segment PQ by drawing its perpendicular bisector. Let M be its mid-point.
3. M as centre and radius = MP draw a circle.
   The circle so obtained is the required circle.

Ex 14.5 Class 6 Maths Question 6.
Draw a circle with centre C and radius 3.4 cm. Draw any chord AB¯¯¯¯¯¯¯¯. Construct the perpendicular bisector of AB¯¯¯¯¯¯¯¯ and examine if it passes through C.
Solution:
Steps of Construction:

1. Mark a point C on the plane of a paper.
2.  With C as centre and radius 3.4 cm, draw a circle.
3. Let AB be any chord to this circle.
4. Draw PQ, the perpendicular bisector of chord AB.
   Clearly, this perpendicular bisector passes through C, the centre of the circle.

Ex 14.5 Class 6 Maths Question 7.
Repeat Question 6, if AB happens to he a diameter.
Solution:
If AB happens to be a diameter then C will be the mid-point of the diameter AB.

Ex 14.5 Class 6 Maths Question 8.
Draw a circle of radius 4 cm. Draw any two of its chords. Construct the perpendicular bisectors of these chords. Where do they meet?
Solution:
Steps of Construction:

1. Mark a point O on the plane of paper.
2. With O as centre, draw a circle of radius 4 cm.
3. Let AB and CD be any two chords of this circle.
4. Draw PQ and RS the perpendicular bisectors of chords AB and CD respectively.
   Clearly, these perpendicular bisectors pass through 0, the centre of the circle.

Ex 14.5 Class 6 Maths Question 9.
Draw any angle with vertex O. Take a point A on one of its arms and B on another such that OA = OB. Draw the perpendicular bisectors of OA¯¯¯¯¯¯¯¯ and OB¯¯¯¯¯¯¯¯ . Let them meet at P. Is PA = PB?
Solution:
Steps of Construction:

1. Draw any angle XOY.
2. Take a point A on OX and a point B on OY such that OA = OB.
3. Draw CD and EF, the perpendicular bisectors of OA and OB respectively. Let them meet at P.
   On measuring, we find that PA = PB.

Ex 14.6 Class 6 Maths Question 1.
Draw ZPOQ of measure 75° and find its line of symmetry.
Solution:
Steps of Construction:

1. Draw a ray OP.
2. Draw ∠POR = 60° and ∠POS = 90°.
3. Draw OQ, the bisector of ∠ROS.
   Then,
   

Ex 14.6 Class 6 Maths Question 2.
Draw an angle of measure 147° and construct its bisector.
Solution:
Steps of Construction:

1. Draw a ray OA.
2. Place the protractor on OA such that its centre falls on the initial point O and 0 -180 line lies along OA.
   
3. Mark a point B on the paper against the mark of 147° on the protractor.
4. Remove the protractor and draw OB. Then, the ∠AOB so obtained is the required angle such that ZAOB = 147°.

To construct its bisector:

1. With centre O and a convenient radius draw an arc cutting sides OA and OB at P and Q respectively.
2. With centre P and radius > 12 PQ, draw an arc.
3. With centre Q and the same radius, as in the previous step, draw another arc intersecting the arc drawn in the previous step at R.
4. Join OR and produce it to form ray OC.
   Then, the ∠AOC so obtained is the bisector of ∠AOB.

Ex 14.6 Class 6 Maths Question 3.
Draw a right angle and construct its bisector.
Solution:
Steps of Construction:

To draw an angle of 90°:

1. Draw a ray OA.
2. With O as centre and any convenient radius, draw an arc cutting at P.
3. With P as centre and same radius, draw an arc cutting the arc of step 2 at Q.
4. With Q as centre and the same radius as in steps 2 and 3, draw an arc cutting the arc drawn in step 3 at R.
5. With Q as centre and the same radius, draw an arc.
6. With R as centre and the same radius, draw an arc cutting the arc drawn in step 5 at B.

To draw its bisector:

1. With P as centre and radius > 12 PT, draw an arc in the interior of ∠AOC.
2. With T as centre and the same radius, an in step 1, draw another arc intersecting the arc in step 1 at D.
3. Join OD and produce it to any point E.
   Then, ∠AOE so obtained is the bisector of ∠AOC.

Ex 14.6 Class 6 Maths Question 4.
Draw an angle of measure 153° and divide it into four equal parts.
Solution:
Steps of Construction:

1. With the help of the protractor, draw ∠AOB = 153°.
2. With centre O and any convenient radius, draw an arc cutting OA and OB at P and Q respectively.
3. With centre P and radius > 12 PQ, draw an arc in the interior of ∠AOB.
4. With centre Q and the same radius, as in step 3, draw another arc intersecting the arc in step 3 at B₁.
5. Join OB₁ and produce it to any point C.
   Then, ∠AOC = 12 x ∠AOB i.e., bisector of ∠AOB.
6. Draw OD, the bisector of ∠AOC. Then ∠AOD = ∠DOC.
7. Draw OE, the bisector of ∠COB. Then, ∠COE = ∠COB.
   Combining these results, we have
   ∠AOD = ∠DOC = ∠COE = ∠EOB.
   Thus, ∠AOB is divided into four equal parts by the rays OD,OC and OE.

Ex 14.6 Class 6 Maths Question 5.
Construct with ruler and compasses, angles of following measures:
(a) 60°
(b) 30°
(c) 90°
(d) 45°
(f) 135°
Solution:
(a) Steps of Construction:

1. Draw a ray OA.
2. With centre O and any radius, draw an arc PQ with the help of compasses, cutting the ray at P.
3. With centre P and the same radius draw an arc cutting the arc PQ at R.
4. Join OR and produce it to obtain ray OB.
   Then, ∠AOB so obtained is of 60°.

(b) Steps of Construction:

1. Draw a ray OA.
2. With centre O any radius draw an arc PT with the help of compasses, cutting ray OA at P.
3. With centre P and the same radius draw an arc cutting the arc PT at Q.
4. Join OQ and produce it to obtain ray OB.
   Then, ∠AOB = 60°.
5. With centre P and radius > 12 PQ, draw an arc in the interior of ∠AOB.
6. With centre Q and the same radius, as in step 5, draw another arc intersecting the arc in step 5 at R.
7. Join OR and produce it on any point C.
   Then, ∠AOC = 30°

(c) Step of Construction:

1. Draw a ray OA.
2. With O as centre and any convenient radius, draw an arc cutting at P.
3. With P as centre and same radius, draw an arc cutting the arc of step 2 at Q.
4. With Q as centre and the same radius as in steps 2 and 3, draw an arc cutting the arc drawn in step 3 at R.
5. With Q as centre and the same radius, draw an arc.
6. With R as centre and the same radius, draw an arc cutting the arc drawn in step 5 at B.
7. Join O to B and produce it to any point C.
   Then, ∠AOC =90°

(d) Steps of Construction:

1. Draw a ray OA.
2. With O as centre and any convenient radius, draw an arc cutting OA at P.
3. With P as centre and the same radius draw an arc cutting the first arc at Q.
4. With Q as centre and the same radius, draw an cutting the arc drawn in step 2 at R.
5. Join AR and produce it to any point C.
   Then, ∠AOC so obtained is of 120°.

(e) Steps of Construction:

1. Draw ∠AOC = 90° by following the steps given in part (iii) above,
2. Draw OE, the bisector of ∠AOC.
   
   Then, ∠AOD so obtained is the required angle of 45°.

(f) Steps of Construction:

1. Draw ∠AOC = 120°
   ∠AOB =150°.
2. Draw OD, bisector of ∠COB.
   Then,
   

Or

Steps of Construction:

1. Draw a line AB and mark a point O on it.
2. With centre O and any convenient radius draw a semi-circle cutting OA and OB at P and Q respectively.
3. With Q as centre and same radius, draw an arc cutting the semi-circle as R.
4. With R as centre and same radius, draw an arc cutting the semi-circle of step 2 at S.
5. With R as centre and same radius draw an arc.
6. With S as centre and same radius draw an arc cutting the arc drawn in step 5 at T. Join OT and produce it to D such that ∠BOD – ∠AOD = 90°.
7. Draw OE, the bisector of ∠AOD.
   

Ex 14.6 Class 6 Maths Question 6.
Draw an angle of measure 45° and bisect it.
Solution:
Steps of Construction:

1. Draw ∠AOB = 90° by the steps given in question 5 (c).
   
2. Draw OC, the bisector of ∠AOB. Then, ∠AOC = 45°.
3. Draw OD, the bisector of ∠AOC. Then, ∠AOD = ∠DOC.

Ex 14.6 Class 6 Maths Question 7.
Draw an angle of measure 135° and bisect it.
Solution:
Steps of Construction:

1. Draw ∠EOB = 135° by the steps given in question 5 (f).
   
2.  Draw OF, the bisector of ∠EOB.
   Then, ∠BOF = ∠FOE.

Ex 14.6 Class 6 Maths Question 8.
Draw an angle of 70°. Make a copy of it using only a straight edge and compasses.
Solution:
Steps of Construction:

1. Draw an angle 70° with protractor, i. e. ∠POQ = 70°
2. Draw a ray AB¯¯¯¯¯¯¯¯
3. Place the compasses at O and draw an arc to cut the ray of ∠POQ at I and M.
4. Use the same compasses, setting to draw an arc with A as centre, cutting AB at X.
5. Set your compasses setting to the length LM with the same radius.
   
6.  Place the compasses pointer at X and draw the arc to cut the arc drawn earlier at Y.
7. Join A7.
   Thus, ∠YAX =70°

Ex 14.6 Class 6 Maths Question 9.
Draw an angle of 40° copy its supplementary angle.
Solution:
Steps of construction:

1. Draw an angle of 40° with the help of protractor, naming ∠ AOB.
2. Draw a line PQ.
   
3. Take any point M on PQ.
4. Place the compasses at O and draw an arc to cut the rays of ∠ AOB at L and N.
5. Use the same compasses setting to draw an arc O as centre, cutting MQ at X.
6.  Place the compasses at X and draw the arc to cut the arc drawn earlier Y.
7. Join MY.
   Thus, ∠QMY = 40° and ∠PMY is supplementary of it.

Ex 13.1 Class 6 Maths Question 1.
List any four symmetrical objects from your home or school.
Solution:
List of four symmetrical objects from home or school are
(i) An electric tube
(ii) A glass
(iii) An electric bulb
(iv) A fan

Ex 13.1 Class 6 Maths Question 2.
For the given figure, which one is the mirror line, l₁ or l₂

Solution:
From the given figure, clearly l₂ is the mirror line.

Ex 13.1 Class 6 Maths Question 3.
Identify the shapes given below. Check whether they are symmetric or not. Draw the line of symmetry as well.

Solution:
Figures (a), (b), (d), (e) and (f) are symmetrical and their line of symmetry is shown as a dotted line.

Ex 13.1 Class 6 Maths Question 4.
Copy the following on a squared paper. A square paper is what you would have used in your arithmetic notebook in earlier classes. Then complete them such that the dotted line is the line of symmetry.

Solution:
Keeping the dotted line as the line of symmetry and completing the figure, we have

Ex 13.1 Class 6 Maths Question 5.
In the figure, l is the line of symmetry. Complete the diagram to make it symmetric.

Solution:
Complete diagram is as under:

Ex 13.1 Class 6 Maths Question 6.
In the figure, l is the line of symmetry. Draw the image of the triangle and complete the diagram so that it becomes symmetric.

Solution:
The image of the triangle is shown such that the complete diagram becomes symmetric.

Ex 13.2 Class 6 Maths Question 1.
Find the number of lines of symmetry for each of the following shapes.

Solution:
Each one of the given figures are symmetrical about the dotted line(s) drawn. The number lines of symmetry are indicated against each figure:

Ex 13.2 Class 6 Maths Question 2.
Copy the triangle in each of the following figures on squared paper. In each case, draw the line(s) of symmetry, if any and identify the type of triangle. (Some of you may like to trace the figures and try paper-folding first!)

Solution:
(a), (b) and (d) are isosceles triangles, (c) is a right angled isosceles triangle. Their line(s) of symmetry marked with dotted line(s) in each case as under:

Ex 13.2 Class 6 Maths Question 3.
Complete the following table :

Solution:
Complete table with rough figures duly filled is as under:

Ex 13.2 Class 6 Maths Question 4.
Can you draw a triangle which has
(a) exactly one. line of symmetry?
(b) exactly two lines of symmetry?
(c) exactly three lines of symmetry?
(d) no lines of symmetry? Sketch a rough figure in each case.
Solution:
(a) Yes, it is an isosceles triangle. Its rough sketch is as shown.

(b) No.
(c) Yes, it is an equilateral triangle. Its rough sketch is as shown.

(d) Yes, it is a scalene triangle. Its rough sketch is as shown.

Ex 13.2 Class 6 Maths Question 5.
On a squared paper, sketch the following:
(a) A triangle with a horizontal line of symmetry but no vertical line of symmetry.
(b) A quadrilateral with both horizontal and vertical lines of symmetry.
(c) A quadrilateral with a horizontal line of symmetry but no vertical line of symmetry.
(d) A hexagon with exactly two lines of symmetry.
(e) A hexagon with six lines of symmetry.
Solution:
Sketches of the required figure with their line(s) of symmetry are shown as under:

Ex 13.2 Class 6 Maths Question 6.
Trace each figure and draw the lines of symmetry, if any:

Solution:
The line(s) of symmetry of the given figures are shown as dotted lines as under:

Ex 13.2 Class 6 Maths Question 7.
Consider the letters of English alphabets, A to Z.
List among them the letters which have

(a) vertical lines of symmetry (like A)
(b) horizontal lines of symmetry (like B)
(c) no lines of symmetry (like Q)
Solution:
The English alphabets A to Z having
(a) vertical lines of symmetry (like A) are
A, H, I, M, O, T, U, V, W, X and Y.
(b) horizontal lines of symmetry (like B) are
B, C, D, E, H, I, K, O and X.
(c) no lines of symmetry (like Q) are G, J, L, P, Q, R, S and Z.

Ex 13.2 Class 6 Maths Question 8.
Given here are figures of a few folded sheets and designs drawn about the fold. In each case, draw a rough diagram of the complete figure that would be seen when the design is cut off.

Solution:
The rough diagram of the complete figure that would be seen when the design is cut off is as under:

Ex 13.3 Class 6 Maths Question 1.
Find the number of lines of symmetry in each of the following shapes: .
How will you check your answers?

Solution:
By drawing the line(s) of symmetry, we find that the number of line(s) possessed by them are

Ex 13.3 Class 6 Maths Question 2.
Copy the following drawing on squared paper. Complete each one of them such that the resulting figure has two dotted lines as two lines of symmetry:

How did you go about completing the picture?
Solution:
Completing the figures using the given line(s) of symmetry. The completed figures are as under:

Ex 13.3 Class 6 Maths Question 3.
In each figure alongside, a letter of the alphabet is shown along with a vertical line. Take the mirror image of the letter in the given line. Find which letters look the same after reflection (i.e. which letters look the same in the image) and which do not. Can you guess why?

Try for O E M N P H L T S V X
Solution:
Taking the mirror image of the letters A and B in the given line. These will look as shown.

Clearly, A after reflection looks same but B does not.
It is due to the reason that the shape is preserved but sense is not.
Out of the given letters:
O, M, H, T, V, and X look as before after reflection, whereas E, P, N, L, and S does not.

Ex 12.1 Class 6 Maths Question 1.
There are 20 girls and 15 boys in a class,
(a) What is the ratio of number of girls to the number of boys?
(b) What is the ratio of number of girls to the total number of students in the class?
Solution:
We have,
Number of girls = 20
Number of boys =15
Number of students in the class = (20 + 15) = 35.
(a) Ratio of number of girls to the number of boys
=20 : 15 = 4 : 3
[Dividing the first and second term by their H.C.F. = 5]

(b) Ratio of number of girls to the total number of students
= 20 : 35
= 4 : 7
[Dividing the first and second term by their H.C.F = 5]

Ex 12.1 Class 6 Maths Question 2.
Out of 30 students in a class, 6 like football, 12 like cricket and remaining like tennis. Find the ratio of:
(a) Number of students liking football to number of students liking tennis.
(b) Number of students liking cricket to total number of students.
Solution:
We have,
Total number of students = 30
Number of students who like football = 6
Number of students who like cricket = 12
Number of students who like tennis = 30 – (6 +12)
= 30 – 18 = 12
(a) Ratio of number of students liking football to the number of students liking tennis
= 6 : 12 = 1 : 2
[Dividing the first and second term by their H.C.F. = 6]

(b) Ratio of number of students liking cricket to total number of students
= 12 : 30
= 2 : 5
[Dividing the first and second term by their H.C.F. = 6]

Ex 12.1 Class 6 Maths Question 3.
See the figure and find the ratio of:

(a) Number of triangles to the number of circles inside the rectangle.
(b) Number of square to all the figures inside the rectangle.
(c) Number of circles to all the figures inside the rectangle.
Solution:
From the given figure, we have
Number of triangle =3
Number of circles =2
Number of square =2
Total number of figures = 3+ 2 + 2 = 7
(a) Ratio of the number of triangles to the number of circles = 3 : 2
(b) Ratio of the number of squares to all the figures =2 : 7
(c) Ratio of the number of .circles it) all the figures = 2 : 7

Ex 12.1 Class 6 Maths Question 4.
Distance travelled by Hamid and Akhtar in an hour are 9 km and 12 km. Find the ratio of speed of Hamid to the speed of Akhtar.
Solution:
We know that,

∴ Ratio of the speed of Hamid to the speed of Akhtar
= 9 : 12
= 3 : 4
[Dividing the first term and the second term by their H.C.F. = 3

Ex 12.1 Class 6 Maths Question 5.
Fill in the following blanks:

Solution:

These are equivalent fractions.

Ex 12.1 Class 6 Maths Question 6.
Find the ratio of the following:
(a) 81 to 108
(b) 98 to 63
(c) 33 km to 121 km
(d) 30 minutes to 45 minutes.
Solution:
(a) 81 to 108 = 81 : 108 = 3 : 4
[Dividing the first and second terms by their H.C.F. = 27]
(b) 98 to 63 = 98 : 63
= 14 : 9
[Dividing the first and the second terms by their H.C.F-= 7]
(c) 33 km to 121 km = 33 : 121
= 3 : 11
[Dividing the first and the second terms by their H.C.F. =11]
(d) 30 minutes to 45 minutes = 30 : 45
= 2:3
[Dividing the first and the second terms by their H.C.F. = 15]

Ex 12.1 Class 6 Maths Question 7.
Find the ratio of the following:
(a) 30 minutes to 1.5 hours
(b) 40 cm to 1.5 m
(c) 55 paise to ₹1
(d) 500 ml to 2 litres
Solution:
(a) 30 minutes to 1.5 hours = 30 minutes: 1.5 x 60 minutes
[∵ 1 hour = 60 minutes]
= 30 min : 90 min = 30 :90 = 1:3
[Dividing the first and the second terms by their H.C.F = 30]

(b) 40 cm to 1.5 m = 40 cm : 1.5 x 100 cm [∵ 1 m = 100 cm]
= 40 cm : 150 cm
= 40 :150 = 4:15
[Dividing the first and the second terms by their H.C.F. = 10]

(c) 55 paise to ₹ 1 = 55 paise : 100 paise [∵ ₹ 1 =100 paise]
= 55 : 100
= 11 : 20
[Dividing the first and the second terms by their H.C.F = 5]

(d) 500 ml to 2 litres = 500 ml: 2 x 1000 ml [∵ 1 litre = 1000 ml]
= 500 : 2000
= 1 : 4
[Dividing the first term and the second terms by their H.C.F = 500]

Ex 12.1 Class 6 Maths Question 8.
In a year, Seema earns ? 1,50,000 and saves ? 50,000. Find the ratio of:
(a) Money that Seema earns to the money she saves.
(b) Money that she saves to the money she spends.
Solution:
Seema’s income = ₹150000
Seema’s savings = ₹ 50000
Seema’s expenditure = ₹ 150000 – ₹ 50000
= ₹ 100000
(a) Ratio of Seema’s earnings to her savings
= ₹ 150000 : ₹ 50000
= 150000 : 50000
= 3 : 1
[Dividing the first term and the second term by their H.C.F = 50000]

(b) Ratio of Seema’s savings to her expenditure
= ₹ 50000 : ₹ 100000
= 50000 : 100000 = 1 : 2
[Dividing the first term and the second term by their H.C.F. = 50000]

Ex 12.1 Class 6 Maths Question 9.
There are 102 teachers in a school of 3300 students. Find the ratio of the number of teachers to the number of students.
Solution:
We have,
Number of teachers =102
Number of students = 3300
∴ Ratio of the number of teachers to the number of students
= 102 : 3300
= 17 : 550
[Dividing the first term and the second term by their H.C.F = 6]

Ex 12.1 Class 6 Maths Question 10.
In a college out of 4320 students, 2300 are girls. Find the ratio of:
(a) Number of girls to the total number of students.
(b) Number of boys to the number of girls.
(c) Number of boys to the total number of students.
Solution:
We have,
Total number of students = 4320
Number of girls = 2300
∴ Number of boys = (4320 – 2300) = 2020
(a) Ratio of number of girls to the total number of students
= 2300 : 4320
= 115 : 216
[Dividing the first term and the second term by their H.C.F = 20]

(b) Ratio of number of boys to number of girls
= 2020 : 2300
= 101 : 115
[Dividing the first term and the second term by their H.C.F = 20]

(c) Ratio of number of boys to the total number of students
= 2020 : 4320
= 101 : 216
[Dividing the first term and the second term by their H.C.F. = 20]

Ex 12.1 Class 6 Maths Question 11.
Out of 1800 students in a school, 750 opted basketball, 800 opted cricket and remaining opted table tennis. If a student can opt only one game, find the ratio of:
(a) Number of students who opted basketball to the number of students who opted table tennis.
(b) Number of students who opted cricket to the number of students opting basketball.
(c) Number of students who opted basketball to the total number of students.
Solution:
We have,
Total number of students = 1800
Number of students who opted basketball = 750
Number of students who opted cricket = 800
∴ Number of students who opted table tennis
= 1800 – (750 + 800)
= 1800 – 1550 = 250
(a) Ratio of number of students opting basketball to number of students opting table tennis
= 750 : 250
= 3 : 1
[Dividing the first term and the second term by their H.C.F. = 250]

(b) Ratio of number of students opting cricket to number of students opting basketball
= 800 : 750
= 16 : 15
[Dividing the first term and the second term by their H.C.F. = 50]

(c) Ratio of number of students opting basketball to the total number of students
= 750 : 1800
= 5 : 12
[Dividing the first term and the second term by their H.C.F = 150]

Ex 12.1 Class 6 Maths Question 12.
Cost of a dozen pens is ? 180 and cost of 8 ball pens is ? 56. Find the ratio of the cost of a pen to the cost of a ball pen.
Solution:
In order to compare the cost of pen with that of ball pen, we must first find the cost of the same quantity of each of them.
Let us find the cost of one item of each.
Cost of 1 dozen i.e., 12 pens
= ₹ 180
∴ Cost of 1 pen = ₹ 180 -12 = ₹ 15
Cost of 8 ball pens = ₹ 56
∴ Cost of 1 ball pen = ₹ 56 – 8 = ₹ 7
∴ Ratio of the cost of a pen to the cost of a ball pen
= Cost of 1 pen : Cost of 1 ball pen
= ₹ 15 : ₹ 7 = 15 : 7.

Ex 12.1 Class 6 Maths Question 13.
Consider the statement: Ratio of breadth and length of a hall is 2:5.
Complete the following table that shows some possible breadths and lengths of the hall.

Solution:

Ex 12.1 Class 6 Maths Question 14.
Divide 20 pens between Sheela and Sangeeta in the ratio of 3:2.
Solution:
Sum of the terms of the ratio = (3 + 2) = 5
Total number of pens = 20

Ex 12.1 Class 6 Maths Question 15.
Mother wants to divide ? 36 between her daughters Shreya and Bhoomika in the ratio of their ages. If age of Shreya is 15 years and age of Bhoomika is 12 years, find how much Shreya and Bhoomika will get.
Solution:
Shreya’s age : Bhoomika’s age =15 : 12
= 5 : 4
[Dividing the 1st and the 2nd term by their H.C.F = 3]
Since mother wants to divide ₹ 36 between Shreya and Bhoomika in the ratio of their ages. So, the money will be divided in the ratio 5 : 4.
Sum of the terms of the ratio = (5 + 4) = 9

Ex 12.1 Class 6 Maths Question 16.
Present age of father is 42 years and that of his son is 14 years. Find the ratio of:
(a) Present age of father to the present age of son.
(b) Age of the father to the age of son, when son was 12 years old.
(c) Age of father after 10 years to the age of son after 10 years. Age of father to the age of son when father was 30 years old.
Solution:
We have,
Present age of father = 42 years
Present age pf son = 14 years.
(a) Ratio of present age of father to the present age of son
= 42 years : 14 years
= 42 : 14
= 3 : 1
[Dividing the first term and the second terms by their H.C.F. = 14]

(b) When son’s age is 12 years (i.e., 2 years ago). Then, father’s age
= (42 – 2) = 40 years.
Required ratio = 40 years : 12 years = 40 :12
= 10 : 3
[Dividing the first term and the second terms by their H.C.F = 4]

(c) After 10 years:
Father’s age = (42+10) years = 52 years
Son’s age = (14 +10) years = 24 years
∴ Required ratio = 52 years : 24 years
= 52 : 24
= 13 : 6
[Dividing the first term and the second terms by their H.C.F = 4]

(d) When father’s age is 30 years (i.e., 12 years age), then son’s age
= (14 -12) years = 2 years.
∴ Required ratio = 30 years : 2 years
= 30 : 2
= 15 : 1
[Dividing the first term and the second terms by their H.C.F = 2]

Ex 12.2 Class 6 Maths Question 1.
Determine if the following are in proportion:
(a) 15, 45, 40, 120
(b) 33, 121, 9, 96
(c) 24, 28, 36, 48
(d) 32, 48, 70, 210
(e) 4, 6, 8, 12
(f) 33, 44, 75, 100
Solution:
We have,

Ex 12.2 Class 6 Maths Question 2.
Write True (T) or False (F) against each of the following statements:
(a) 16: 24:: 20: 30
(b) 21 : 6 :: 35 : 10
(c) 12 : 18 :: 28 : 12
(d) 8 : 9 :: 24 : 27
(e) 5.2 : 3.9 :: 3 : 4
(f) 0.9 : 0.36 :: 10 : 4
Solution:

Ex 12.2 Class 6 Maths Question 3.
Are the foUowing statements true?
(a) 40 persons : 200 persons =₹ 15 : ₹ 75
(b) 7.5 litres : 15 litres = 5 kg : 10 kg
(c) 99 kg: 45 kg = ₹ 44 : ₹ 20
(d) 32 m : 64 m = 6 sec : 12 sec
(e) 45 km : 60 km = 12 hours : 15 hours,
Solution:

Ex 12.2 Class 6 Maths Question 4.
Determine if the following ratios form a proportion. Also, write the middle terms and extreme terms where the ratios form a proportion.
(a) 25 cm : 1 m and ₹ 40: ₹ 160
(b) 39 litres : 65 litres and 6 bottles : 10 bottles
(c) 2 kg : 80 kg and 25 g : 625 g
(d) 200 ml: 2.5 litre and ₹ 4 : ₹ 50.
Solution:

Question 1.
If the cost of 7 m of cloth is X 294, find the cost of 5 m of cloth.
Solution:
Cost of 7 m of cloth = ₹ 294
∴ Cost of 1 m of cloth = ₹ (294 7)
= ₹ 42
∴ Cost of 5 m of cloth = ₹ (42 x 5)
= ₹ 210.

Question 2.
Ekta earns X1500 in 10 days. How much will she earn in 30 days?
Solution:
Ekta’s earning of 10 days = ₹ 1500
Ekta’s earning of 1 day = ₹ (1500 ÷ 10)
= ₹ 150
∴ Ekta’s earning of 30 days = ₹ (150 x 30)
= ₹ 4500

Question 3.
If it has rained 276 mm in the last 3 days, how many cm of rain will fall in one full week (7 days) ? Assume that the rain continues to fall at the same rate.
Solution:
Rainfall in 3 days = 276 mm
∴ Rainfall in 1 day = (276 + 3) mm = 92 mm
Rainfall in 1 week (i.e., 7 days) = (92 x 7) mm
= 644 mm.
In cm, rainfall in one full week = 644 x 110 cm [1 m = 110 cm]
= 64.4 cm

Question 4.
Cost of 5 kg of wheat is ? 30.50.
(a) What will be the cost of 8 kg of wheat?
(b) What quantity of wheat can be purchased in ₹ 61?
Solution:
(a) Cost of 5 kg of wheat = ₹ 30.50
∴ Cost of 1 kg of wheat = ₹ (30.50 ÷ 5) = ₹ 6.10
∴ Cost of 8 kg of wheat = ₹ (6,10 x8) = ₹ 48.80
(b) For ₹ 30.50, wheat purchased= 5 kg
For ₹ 1, wheat purchased = (5 ÷30.50) kg

Question 5.
The temperature dropped 15 degree celsius in the last 30 days. If the rate of temperature drop remains the same, how many degrees will the temperature drop in the next ten days?
Solution:
In 30 days, drop in temperature =15 degrees
∴ In 1 day, drop in temperature = 1530 degree = 12 degree
∴ In the next 10 days, drop in temperature = (12×10) degree = 5 degrees.

Question 6.
Shaina pays ? 7500 as rent for 3 months. How much does she has to pay for a whole year, if the rent per month remains same?
Solution:
Rent payment for 3 months = ₹ 7500
∴ Rent payment for 1 month = ₹ (7500 + 3) = ₹ 2500
∴ Rent payment for 1 year i.e., 12 months
= ₹ (2500 x 12) = ₹ 30000.

Question 7.
Cost of 4 dozens of bananas is ₹ 60. How many bananas can be purchased for ₹ 12.50?
Solution:
For ₹ 60, number of bananas purchased = 4 dozens i.e., 48
∴ For ₹ 1, number of bananas purchased = 4860=45
∴ For ₹ 12.50, number of bananas purchased = (45×12.50)
= 4 x 2.50 = 10.

Question 8.
The weight of 72 books is 9 kg. What is the weight of 40 such books?
Solution:
Weight of 72 books = 9 kg
∴ Weight of 1 book = 972 kg = 18kg
∴ Weight of 40 books = (18×40) = 5 kg.

Question 9.
A truck requires 108 litres of diesel for covering a distance of 594 km. How much diesel will be required by the truck to cover a distance of 1650 km?
Solution:
For 594 km, diesel required = 108 litres
∴ For 1 km, diesel required = (108594) litres = (211) litres
∴ For 1650 km, diesel required = (211×1650) litres
= (2 x 150) litres = 300 litres.

Question 10.
Raju purchases 10 pens for ^₹ 150 and Manish buys 7 pens for ₹ 84. Can you say who got the pens cheaper?
Solution:
In case of Raju:
Cost of 10 pens = ₹ 150
∴ Cost of 1 pen = ₹ (15010) = ₹ 15
In case of Manish :
Cost of 7 pens = ₹ 84
∴ Cost of 1 pen = ₹ (847) = ₹ 12
Since 12 < 15 ∴ Manish got the pen cheaper. Question 11. Anish made 42 runs in 6 overs and Anup made 63 runs in 7 overs. Who made more runs per over? Solution: In case of Anish: Runs made in 6 overs = 42 Runs made in 1 over = (426) = 7 In case of Anup: Runs made in 7 overs = 63 Runs made in 1 over = (637) = 9 Since 9 >7
∴ Anup made more runs per over.

Ex 11.1 Class 6 Maths Question 1.
Find the rule, which gives the number of matchsticks required to make the following matchstick patterns. Use a variable to write the rule.
(a) A pattern of letter T as

(b) A pattern of letter Z as

(c) A pattern of letter U as

(d) A pattern of letter V as

(e) A pattern of letter E as

(f) A pattern of letter S as

(g) A pattern of letter A as

Solution:
(a) For T:

Clearly, to make one T, we use 2 matchsticks as shown in the figure. Rule is given as
“Number of matchsticks required = 2n”, where n can take any value 1, 2, 3, …
(b) For Z :

Clearly, to make one Z, we use 3 matchsticks as shown in the figure. Rule is given as
“Number of matchsticks required = 3 n”, where n can take any value 1, 2, 3, …
(c) For U :

Clearly, to make one U, we use 3 matchsticks as shown in the figure. Rule is given as
“Number of matchsticks required = 3n”, where rt can take any value 1, 2, 3, …
(d) For V :

Clearly, to make one V, we use 2 matchsticks as shown in the figure. Rule is given as
“Number of matchsticks required = 2n”, where n can take any value 1, 2,3,…
(e) For E :

Clearly, to make one E, we use 5 matchsticks as shown in the figure. Rule is given as
“Number of matchsticks required = 5n”, where n can take any value 1, 2, 3, …
(f) For S :

Clearly, to make one S, we use 5 matchsticks as shown in the figure. Rule is given as
“Number of matchsticks required = 5n”, where n can take any value 1, 2, 3, …
(g) For A :

Clearly, to make one A, we use 6 matchsticks as shown in the figure. Rule is given as
“Number of matchsticks required = 6 n”, where n can take any value 1, 2, 3, …

Ex 11.1 Class 6 Maths Question 2.
We already know the rule for the pattern of letters L, C and F. Some of the letters from Q. 1 (given above) give us the same rule as that given by L. Which are these? Why does this happen?
Solution:
(a) and (d) parts of Q. 1. give the same rule as for L. It happens as the same number of matchsticks are being used in these cases.

Ex 11.1 Class 6 Maths Question 3.
Cadets are marching in a parade. There are 5 cadets in a row. What is the rule, which gives the number of cadets, given the number of rows? (Use n for the number of rows.)
Solution:
Since, there are 5 cadets in a row and number of rows are n
∴ Rule is given as
Number of cadets in the parade = 5n

Ex 11.1 Class 6 Maths Question 4.
If there are 50 mangoes in a box, how will you write the total number of mangoes in terms of the number of boxes? (Use b for the number of boxes.)
Solution:
Since, there are 50 mangoes in a box and b is the number of boxes.
∴ Total number of mangoes = 50b

Ex 11.1 Class 6 Maths Question 5.
The teacher distributes 5 pencils per student. Can you tell how many pencils are needed, given the number of students? (Use s for the number of students.)
Solution:
Since, there are s number of students and each student having 5 pencils.
∴ Total number of pencils needed = 5n

Ex 11.1 Class 6 Maths Question 6.
A bird flies 1 kilometre in one minute. Can you express the distance covered by the bird in terms of its flying time in minutes? (Use t for flying time in minutes.)
Solution:
Since, a bird flies at 1 kilometre in one minute. If it flies for t minutes. Total distance covered by bird in t minutes = 1 x t km = t km

Ex 11.1 Class 6 Maths Question 7.
Radha is drawing a dot Rangoli (a beautiful pattern of lines joining dots with chalk powder. She has 9 dots in a row. How many dots will her Rangoli have for r rows? How many dots are there if there are 8 rows? If there are 10 rows?

Solution:
Since, a row contains 9 dots.
Therefore, number of dots in r rows = 9 r.
Number of dots in 8 rows = 9 x 8 = 72
and, number of dots in 10 rows = 9 x 10 = 90

Ex 11.1 Class 6 Maths Question 8.
Leela is Radha’s younger sister. Leela is 4 years younger than Radha. Can you write Leela’s age in terms of Radha’s age? Take Radha’s age to be x years.
Solution:
Let Radha’a age be x years. Since Radha’s younger sister Leela is 4 years younger than Radha.
∴ Leela’s age = (x – 4) year

Ex 11.1 Class 6 Maths Question 9.
Mother has made laddus. She gives some laddus to gilests and family members; still 5 laddus remain. If the number of laddus mother gave away is l, how many laddus did she make?
Solution:
Number of laddus given to guests and family members = l
Number of laddus left over = 5
Number of laddus made by mother = 1 + 5

Ex 11.1 Class 6 Maths Question 10.
Oranges are to be transferred from larger boxes into smaller boxes. When a large box is emptied, the oranges from it fill two smaller boxes and still 10 oranges remain outside. If the number of oranges in a small box are taken to be x, what is the number of oranges in the larger box?
Solution:
Number of oranges in the larger box
= 2 x Number of oranges iR small box + oranges left over
= 2x +10
where x is. the number of oranges in the small box.

Ex 11.1 Class 6 Maths Question 11.
(a) Look at the following matchstick pattern of squares (see figure). The squares are not separate. Two neighbouring squares have a common matchstick. Observe the pattern and find the rule that gives the number of matchsticks in terms of the number of squares.

(b) Figure gives a matchstick pattern of triangles. As in Exercise 11(a) above, find the general rule that gives the number of matchsticks in terms of the number of triangles.

Solution:
(a) Clearly, from the figure:

Thus, rule is as under:
Number of matchsticks = 3x +1, where x is the number of squares,
(b) Clearly, from the figure :

Thus, rule is as under: ,
Number of matchsticks = 2x +1, where x is the number of triangles.

Ex 11.2 Class 6 Maths Question 1.
The side of an equilateral triangle is shown by l. Express the perimeter of the equilateral triangle using l.
Solution:
If P represents the perimeter of an equilateral triangle of side l, then P = l + l + l= 3 x l

Ex 11.2 Class 6 Maths Question 2.
The side of a regular hexagon (see figure) is denoted by l. Express the perimeter of the hexagon using l.

Solution:
If P represents the perimeter of a regular hexagon of side l, then P = l + l + l + l + l + l = 6 x l.

Ex 11.2 Class 6 Maths Question 3.
A cube is a three-dimensional figure as shown in figure. It has six faces and all of them are identical squares. The length of an edge of the cube is given by l. Find the formula for the total length of the edges of a cube.

Solution:
Since all the 12 edges of a cube are of same length, l. Then, their total length = 12 x l.

Ex 11.2 Class 6 Maths Question 4.
The diameter of a circle is a line, which joins two points on the circle and also passes through the centre of the circle. (In the adjoining figure AB is a diameter of the circle;
C is its centre). Express the diameter of the circle (d) in terms of its radius (r).

Solution:
Diameter AB = AC + CB
If d represents the diameter of the circle and r is the radius of circle, then
d = CP + CP [∵ CP = r]
d = r + r = 2r

Ex 11.2 Class 6 Maths Question 5.
To find sum of three numbers 14, 27 and 13, we can have two ways.
(a) We may first add 14 and 27 to get 41 and then add 13 to it to get the total sum 54 or
(b) We may add 27 and 13 to get 40 and then add 14 to get the sum 54.
Thus, (14 + 27) + 13 = 14 + (27 + 13).
This can be done for any three numbers. This property is known as the associativity of addition of numbers. Express this property which we have already studied in the chapter on Whole Numbers, in a general way, by using variables a, b and c.
Solution:
The given properly is stated as
For 3 variables a, b and c, we have
(a + b) + c = a + (b + c)

Ex 11.3 Class 6 Maths Question 1.
Make up as many expressions with numbers (no variables) as you can from three numbers 5, 7 and 8. Every number should be used not more than once. Use only addition, subtraction and multiplication.
Solution:
Some possible expressions using three numbers 5, 7 and 8 are
(i) 5 + (8 – 7)
(ii) 5 -(8 – 7)
(iii) 5 +(8 + 7)
(iv) (5 + 8)+ 7
(v) 5 x 8 + 7
(vi) 5 x 7 + 8
(vii) 5 x 8 – 7
(viii) 5 x 7 – 8
(ix) 5 x (8 – 7)
(x) 5 x (8 + 7)
(xi) 8 x (7 – 5)
(xii) 8 x (7 + 5) etc.

Ex 11.3 Class 6 Maths Question 2.
Which out of the following are expressions with numbers only?
(a) y + 3
(b) (7 x 20) – 8z
(c) 5(21 – 7) + 7 x 2
(d) 5
(e) 3x
(f) 5 – 5n
(g) (7 x 20) (5 x 10) – 45 + p
Solution:
(c) and (d) are expressions with numbers only.

Ex 11.3 Class 6 Maths Question 3.
Identify the operations (addition, subtraction, division, multiplication) in forming the following expressions and tell how the expressions have been formed:
(a) z +1, z – 1 y +17, y -17
(b) 17y, y17, 5z
(c) 2y +17, 2y -17
(d) 7m, -7m + 3, -7m – 3
Solution:

Ex 11.3 Class 6 Maths Question 4.
Give expressions for the following cases:
(a) 7 added to p
(b) 7 subtracted from p
(c) p multiplied by 7
(d) p divided by 7
(e) 7 subtracted from -m
(f) -p multiplied by 5
(g) -p divided by 5
(h) p multiplied by -5
Solution:

Ex 11.3 Class 6 Maths Question 5.
Give expressions in the following cases:
(a) 11 added to 2m
(b) 11 subtracted from 2m
(c) 5 times y to which 3 is added
(d) 5 times y from which 3 is subtracted
(e) y is multiplied by -8
(f) y is multiplied by -8 and then 5 is added to the result
(g) y is multiplied by 5 and the result is subtracted from 16
(h) y is multiplied by -5 and the result is added to 16
Solution:

Ex 11.3 Class 6 Maths Question 6.
(a) Form expressions using t and 4. Use not more than one number operation. Every expression must have t in it.
(b) Form expressions using y, 2 and 7. Every expression must have y in it. Use only two number operations. These should be different.
Solution:
(a) Possible expressions using t and 4 are
t + 4, t – 4, 4t and t4.
(b) Possible expressions using y, 2 and 7 (having only two different number operations) are
2y + 7, 2y – 7, 7y + 2, 7y – 2, y2 + 7, y2 – 7, y7+2 and y7 – 2.

Ex 11.4 Class 6 Maths Question 1.
Answer the following:
(a) Take Sarita’s present age to be y years

• What will be her age 5 years from now?
• What was her age 3 years back?
• Sarita’s grandfather is 6 times her age. What is the age of her grandfather?
• Grandmother is 2 years younger than grandfather. What is grandmother’s age?
• Sarita’s father’s age is 5 years more than 3 times Sarita’s age. What is her father’s age?

(b) The length of a rectangular hall is 4 metres less than 3 times the breadth of the hall. What is the length, if the breadth is b metres?
(c) A rectangular box has height h cm. Its length is 5 times the height and breadth is 10 cm less than the length. Express the length and the breadth of the box in terms of the height.
(d) Meena, Beena and Leena are climbing the steps io the hill top. Meena is at step s, Beena is 8 steps ahead and Leena 7 steps behind. Where are Beena and Meena? The total number of steps to the hill top is 10 less than 4 times what Meena has reached. Express the total number of steps using s.
(e) A bus travels at v km per hour. It is going from L>aspur to Beespur. After the bus has travelled 5 hours, Beespuris still 20 km away. What is the distance from Daspur to Beespur? Express it using v.
Solution:
(a) Let Santa’s present age be y years.

• Her age 5 years from now = (y + 5) years.
• Her age 3 years back = (y – 3) years.
• Since Sarita’s grandfather’s age is 6 times that of Sarita’s age.
  ∴ Her grandfather’s age = 6y years.
• Since grandmother is 2 years younger than grandfather. Therefore,
  Grandmother’s age = (6 y – 2) years.
• Since Sarita’s father’s age is 5 years more than 3 times Sarita’s age.
  Therefore, her father’s age = (3y + 5) years.

(b) Let the breadth of the hall be b metres
Since its length is 4 metres less than 3 times the breadth of the hall.
∴ Length of the hall = (3b – 4) metres
(c) Let the height of the rectangular box = h cm
It is given that its length is 5 times the height and breadth is 10 cm less than the length. Therefore,
Its length = 5h cm
and, breadth = (5h-10) cm
(d) Let Meena be at step s while climbing the steps to the hill top.
Since Beena is 8 steps ahead and Leena 7 steps behind Meena.
∴ Beena is at step (s + 8) and Leena is at step (s – 7)
Also, the total number of steps to the hill top is 10 less than 4 times what Meena has reached. .
∴ Total number of steps to the hill top = 4s-10
(e) Speed of the bus = v km per hour
Distance travelled by the bus in 5 hours = 5v km
According to question,
Distance from Daspur to Beespur
= Distance travelled by bus in 5 hours +20 km
= 5km +20 km = (5v +20) km.

Ex 11.4 Class 6 Maths Question 2.
Change the following statements using expressions into statements in ordinary language.
(For example, Given Salim scores r runs in a cricket match, Nalin scores (r +15) runs. In ordinary language-Nalin scores 15 runs more than Salim.)
(a) A notebook costs ₹ p.A book costs ₹ 3p.
(b) Tony puts q marbles on the table. He has 8 q marbles in his box.
(c) Our class has n students. The school has 20n students.
(d) Jaggu is z years old. His uncle is 4z years old and his aunt is (4z – 3) years old.
(e) In an arrangement of dots there are r rows. Each row contains 5 dots.
Solution:
(a) A book costs 3 times the cost of a notebook.
(b) Tony’s box contains 8 times the marbles on the table.
(c) Total number of students in the school is 20 times that of our class.
(d) Jaggu’s uncle is 4 times older than Jaggu and Jaggu’s aunt is 3 years younger than his uncle.
(e) The number of dots in a row is 5 times the number of rows.

Ex 11.4 Class 6 Maths Question 3.
(a) Given, Munnu’s age %p be x years, can you guess what (x – 2) may show?
Can you guess what (x + 4) may show? What (3x + 7) may show?
(b) Given Sara’s age today to be y years. Think of her age in the future or in the past. What will the following expression indicate?
y + 7, y-3, y + 412, y – 212.
(c) Given n students in the class like football, what may 2n show? What may n2 show?
Solution:
(a) (x – 2) may show the age of his younger sister.
(x + 4) may show the age of his elder brother.
(3x + 7) may show the age of his grandfather.
(b) The expressions (y + 7), (y + 412) may indicate Sara’s age in future.
The expressions (y – 3), (y – 212) may indicate Sara’s past age.
(c) The expression 2n may show the number of students who like hockey. The expression n2 may show the number of students who like basketball.

Ex 11.5 Class 6 Maths Question 1.
State which of the following are equations (with a variable). Give reason for your answer. Identify the variable from the equations with a variable.
(a) 17 = x + 7
(b) (t – 7) > 5
(c) 42 = 2
(d) 7 x 3 – 19 = 8
(e) 5 x 4 – 8 = 2x
(f) x – 2 = 0
(g) 2m < 30
(h) 2n + 1 = 11
(i) 7 = (11 x 5) – (12 x 4)
(j) 7 = 11 x 2 + p
(k) 20 = 5y
(l) 3q2 < 5
(m) z + 12 > 24
(n) 20 – (10 – 5) = 3 x 5
(o) 7 – x = 5
Solution:
(a) An equation, equation with variable (x).
(b) Not an equation (as have no = sign).
(c) An equation, equation with numbers.
(d) An equation, equation with numbers.
(e) An equation, equation with variable (x).
(f) An equation, equation with variable (x).
(g) Not an equation (as have no = sign).
(h) An equation, equation with variable (n).
(i) An equation, equation with numbers.
(j) An equation, equation with variable (p).
(k) An equation, equation with variable (y).
(l) Not an equation (as have no = sign).
(m) Not an equation (as have no = sign),
(n) An equation, equation with numbers.
(o) An equation, equation with variable (x).

Ex 11.5 Class 6 Maths Question 2.
Complete the entries in the third column of the table.
Solution:
The table duly completed is as under :

Ex 11.5 Class 6 Maths Question 3.
Pick out the solution from the values given in the bracket next to each equation. Show that the other values do not satisfy the equation.
(a) 5m = 60 (10, 5, 12, 15)
(b) n + 12 = 20 (12, 8, 20, 0)
(c) p – 5 = 5 (7, 2, 10, 14)
(d) q2 = 7 (4, -2, 8, 0)
(e) r – 4 = 0 (4, -4, 8, 0)
(f) x + 4 = 2 (-2, 0, 2, 4)
Solution:
(a) Given equation is 5m = 60.
For m=10: 5m = 5 x 10 = 50 ≠ 60. So, m=10 does not satisfy the given equation.
For m = 5 5m = 5 x 5 = 25 ≠ 60. So, m = 5 does not satisfy the given equation.
For m=12: 5m = 5 x 12 = 60. So, m=12 does not satisfy the given equation.
Thus, n = 12 is its solution.
For m = 15 5m = 5 x 15 = 75 ≠ 60. So, m = 15 does not satisfy the given equation.

(b) Given equation is n +12 = 20.
For n=12: n +12 = 12 + 12 = 24 ≠ 20. So, n = 12 does not satisfy the the given equation.
For n = 8: n +12 = 8 + 12 = 20. So, n = 8 satisfies the given equation.
Thus, n = 8 is its solution.
For n = 20: n + 12 = 20 + 12 = 32 ≠ 20. So, n = 20 does not satisfy the given equation.
For n = 0: n + 12 = 0 + 12 = 12 ≠ 20. So, n = 0 does not satisfy the given equation.

(c) Given equation is p – 5 = 5.
For p = 0: p – 5 = 0 – 5 = -5 ≠ 5. So, p = 0 does not satisfy the given equation.
For p = 10: p – 5 = 10 – 5 = 5. So, p = 10 satisfies the given equation.
Thus, its solution is p = 10.
For p = 5: p – 5 = 5 – 5 = 0≠ 5. So, p = 5 does not satisfy the given equation.
For p = -5: p – 5 = -5 – 5 = -10 ≠ 5. So, p = -5 does not satisfy the given equation.

(d) Given equation is q2 = 7.
For q = 7: q2 = 72 ≠ 7. So, q = 7 does not satisfy the given equation.
For q = 2: q2 = 22 = 1 ≠ 7. So, q = 2 does not satisfy the given equation.
For q = 10: q2 = 102 = 5 ≠ 7. So, q = 10 does not satisfy the given equation.
For q = 14: q2 = 142 = 7. So, q = 14 satisfies the given equation. Thus, q = 14 is its solution.

(e) Given equation is r – 4 = 0.
For r = 4: r – 4 = 4 – 4 = 0. So, r = 4 satisfies the given equation. Thus, r = 4 is its solution.
For r = -4: r – 4 = -4 – 4 = 8 ≠ 0. So, r = -4 does not satisfy the given equation.
For r = 8: r – 4 = 8 – 4 = 4 ≠ 0. So, r-8 does not satisfy the given equation.
For r = 0: r – 4 = 0 – 4 = -4 ≠ 0. So, r = 0 does hot satisfy thh given equation.

(f) Given equation is x + 4 = 2.
For x = -2: x + 4 = -2 + 4 = 2. So, x = -2 satisfies the given equation.
Thus, x = -2 is its solution.
For x = 0: x + 4 = 0 + 4 = 4 ≠ 2. So,- x =0 does not satisfy the given equation.
For x=2: x + 4= 2 + 4= 6 ≠ 2. So, x = 2 does not satisfy the given equation.
For x = 4: x + 4 = 4 + 4= 8 ≠ 2. So, x = 4 does not satisfy the given equation.

Ex 11.5 Class 6 Maths Question 4.
Question (a)
Complete the table and by inspection of the table find the solution to the equation m +10 = 16.

Solution:
Completing the table, we have

By inspection of the above table, we find that m = 6 satisfies the equation m+10 = 16. So, m = 6 is its solution.

Question (b)
Complete the table and by inspection of the table find the solution to the equation 51 = 35.

Solution:

By inspection of the above table, we find that t = 7 satisfies the equation 5t = 35. So, t = 7 is its solution.

Question (c)
Complete the table and find the solution of the equation z3 = 4 using the table.

Solution:
Completing the table, we have


By inspection of the above table, we find that t=12 satisfies the equation z3 = 4. So, t = 12 is its solution.

Question (d)
Complete the table and find the solution to .the equation m – 7 = 3.

Solution:
Completing the table, we have

By inspection of the above table, we find that m = 10 satisfies the equation m – 7 = 3. So, m = 10 is its solution.

Ex 11.5 Class 6 Maths Question 5.
Solve the following riddles, you may yourself construct such riddles.
Who am I?

(i) Go round a square
Counting every comer Thrice and no more!
Add the count to me
To get exactly thirty four!
(ii) For each day of the week
Make an upcount from me
If you make no mistake
You will get twenty three!
(iii) I am a special number
Take away from me a six!
A whole cricket team
You will still be able to fix!
(iv) Tell me who I am
I shall give a pretty clue!
You will get me hack
If you take me out of twenty two!
Solution:
(i) Let I be denoted by x. There are 4 comers of a square.
On counting each comer thrice, we get 3 x 4 = 12
As per problem:
x + 12 = 34 ⇒ x +12 – 12 = 34 – 12
⇒ x + 0 = 22 ⇒ x = 22
Thus, I am 22.

(ii) According to the question:
x + 7 =23
x + 7 – 7 = 23 – 7
⇒ x + 0 =16 ⇒ x =16

(iii) Let the special number be x. Then, according to the problem, we have
x – 6 = 11 [ In a cricket team, no. of players =11]
⇒ x – 6 + 6 = 11 + 6
⇒ x + 0 = 17 ⇒ x = 17
Thus, the special number is 17.

(iv) Let I be denoted by x.
According to the question:
22 – x = x
⇒ 22 – x + x = x + x
⇒ 22 + 0 =2x
⇒ 2x = 22 ⇒ 2x2=222
⇒ x =11
Thus, I am 11.

Ex 10.1 Class 6 Maths Question 1.
Find the perimeter of each of the following figures:

Solution:
(a) Perimeter = the sum of the lengths of sides
= 5 cm +1 cm +2 cm + 4 cm = 12cm
(b) Perimeter = the sum of the lengths of sides
= 40 cm +35 cm + 23 cm +35 cm = 133 cm
(c) Perimeter = 4x the length of one side
= 4 x 15 cm = 60 cm
(d) Perimeter = 5 x the length of one side
4 = 5 x 4 cm = 20 cm
(e) Perimeter = the sum of the lengths of sides
= 4 cm + 0.5 cm + 2.5 cm + 2.5 cm + 0.5 cm + 4 cm + 1 cm = 15 cm
(f) Perimeter = the sum of the lengths of sides
= 4 cm + 3 cm + 2 cm + 3 cm + 1 cm + 4 cm + 3 cm + 2 cm + 3 cm + 1 cm + 4 cm + 3 cm + 2 cm +3 cm +1 cm + 4 cm + 3 cm + 2 cm + 3 cm + 1 cm = 52 cm

Ex 10.1 Class 6 Maths Question 2.
The lid of a rectangular box of sides 40 cm by 10 cm is sealed all round with tape. What is the length of the tape required?
Solution:
Length of the tape required
= Perimeter of the lid of a rectangular box = 2 x ( length + breadth)
= 2 x (40 cm +10 cm)
= 2 x 50 cm = 100 cm or 1 m

Ex 10.1 Class 6 Maths Question 3.
A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top?
Solution:
Perimeter of the table-top
= 2 x (length + breadth)
= 2 x (2 m 25 cm +1 m 50 cm)
= 2 x (3 m 75 cm)
= 2 x 3.75 m = 7.50 m

Ex 10.1 Class 6 Maths Question 4.
What is the length of the wooden strip required to frame a photograph of length and breadth 32 cm and 21 cm, respectively?
Solution:
The length of the wooden strip required to frame a photograph is the perimeter of the photograph.
Perimeter of the photograph = 2 x (length + breadth)
= 2 x (32 cm +21 cm)
= 2 x 53 cm = 106 cm
∴ The length of the wooden strip required is 106 cm.

Ex 10.1 Class 6 Maths Question 5.
A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What is the length of the wire needed?
Solution:
We have to cover 4 times the perimeter of the land measuring 0.7 km by 0.5 km.
∴ Total length of wire required is 4 times its perimeter.
Perimeter of the land = 2 x (length + breadth)
= 2x (0.7 km+0.5 km)
= 2 x 1.2 km = 2.4 km
∴ Total length of wire required = 4 x 2.4 km = 9.6 km

Ex 10.1 Class 6 Maths Question 6.
Find the perimeter of each of the following shapes:
(a) A triangle of sides 3 cm, 4 cm and 5 cm.
(b) An equilateral triangle of side 9 cm.
(c) An isosceles triangle with equal sides 8 cm each and third side 6 cm.
Solution:
(a) Perimeter = the sum of the sides
= 3 cm + 4 cm + 5cm = 12 cm
(b) Perimeter = 3 x the length of one side
= 3 x 9 cm = 27 cm
(c) Perimeter = the sum of the lengths of the sides
= 8 cm + 8 cm + 6 cm = 22 cm

Ex 10.1 Class 6 Maths Question 7.
Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.
Solution:
Perimeter of the triangle = the sum of the lengths of its sides
= 10 cm + 14 cm + 15 cm = 39 cm

Ex 10.1 Class 6 Maths Question 8.
Find the perimeter of a regular hexagon with each side measuring 8 m.
Solution:
A regular hexagon has 6 sides, so its perimeter
= 6 x length of its one side = 6 x 8 m
= 48 m

Ex 10.1 Class 6 Maths Question 9.
Find the side of the square whose perimeter is 20 m.
Solution:
Perimeter = 20 m
A square has 4 equal sides, so we can divide the perimeter by 4 to get the length of one side.
One side of the square = 20 m + 4 = 5m

Ex 10.1 Class 6 Maths Question 10.
The perimeter of a regular pentagon is 100 cm. How long is its each side?
Solution:
Perimeter = 100 cm
A regular pentagon has 5 equal sides, so we can divide the perimeter by
5 to get the length of one side
∴ length of one side = 100 cm + 5 = 20 cm

Ex 10.1 Class 6 Maths Question 11.
A piece of string is 30 cm long. What will be the length of each side if the string is used to forms
(a) a square?
(b) an equilateral triangle?
(c) a regular hexagon?
Solution:
(a) Perimeter = Length of the string = 30 cm
A square has 4 equal sides, so we can divide the perimeter by 4 to get the length of one side.
One side of the square = 30 cm + 4 = 7.5 cm

(b) Perimeter = Length of the string = 30 cm
An equilateral triangle has 3 equal sides, so we can divide the perimeter by 3 to get the length of one side.
∴ One side of an equilateral triangle = 30 cm ÷ 3 = 10 cm
(c) Perimeter = Length of the string = 30 cm
A regular hexagon has 6 equal sides, so we can divide the perimeter by
6 to get the length of one side.
One side of a regular hexagon = 30 cm + 6 = 5 cm

Ex 10.1 Class 6 Maths Question 12.
Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is its third side?
Solution:
Let ABC be the given triangle such that AB =12 cm, BC = 14 cm and its perimeter = 36 cm.
i.e., AB + BC + CA = 36 cm
or 12 cm + 14 cm + CA = 36 cm
or 26 cm + CA = 36 cm
or CA =36 cm – 26 cm = 10 cm
∴ The third side of triangle is 10 cm.

Ex 10.1 Class 6 Maths Question 13.
Find the cost offencing a square park ofside 250 mat the rate of ? 20 per metre.
Solution:
Side of the square park = 250 m
∴ Perimeter of the square park
= 4 x side
= 4 x 250 m = 1000 m
∴ Cost of fencing = ₹ (1000 x 20)
= ₹ 20000

Ex 10.1 Class 6 Maths Question 14.
Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of? 12 per metre.
Solution:
Length of the rectangular park = 175 m Breadth of the rectangular park 125 m .-. Perimeter of the park = 2 x (length + breadth)
= 2 x (175m + 125m)
= 2 x 300 m = 600 m
∴ Cost of fencing = ₹ (600 x 12) = ₹ 7200

Ex 10.1 Class 6 Maths Question 15.
Sweety runs around a square park of side 75 m. Bulbul runs around a rectangular park with length 60 m and breadth 45 m. Who covers less distance?
Solution:
The distance each girl covers in one round is the same as the perimeter of the respective field. Therefore, the distance that Sweety covers in one round
= 4 x side
= 4 x 75 m = 300 m
Also, the distance that Bulbul covers in one round
= 2 x (length + breadth)
= 2 x (60 m + 45 m)
= 2 x 105 m = 210 m
This shows that Bulbul covers less distance than Sweety.

Ex 10.1 Class 6 Maths Question 16.
What is the perimeter of each of the following figures? What do you infer from the answer?


Solution:
(a) Perimeter = 4 x side
= 4 x 25 cm
= 100 cm
(b) Perimeter = 2 x (length + breadth)
= 2 x (40 cm +10 cm)
= 2 x 50 cm = 100 cm
(c) Perimeter = 2 x (length + breadth)
2 x (30 cm +20 cm)
= 2 x 50 cm = 100 cm
(d) Perimeter = 30 cm +30 cm +40 cm = 100 cm
Thus, we observe that the perimeter of each figure is 100 cm i.e., they have equal perimeters.

Ex 10.1 Class 6 Maths Question 17.
Avneet buys 9 square paving slabs, each with a side of 1/2 m. He lays them in the form of a square.

(a) What is the perimeter of his arrangement [Fig. (i)]?
(b) Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement [Fig. (ii)]?
(c) Which has greater perimeter?
(d) Avneet wonders if there is a way of getting an even greater perimeter. Can you find a way of doing this? (The paving slabs must meet along complete edges i.e. they cannot be broken.)
Solution:
(a) In case of Avneet’s arrangement:

(b) In case of Shari’s arrangement:

(c) Clearly, perimeter in case of Shari is greater.
(d) Yes, there is a way shown in the figure in which we get a greater perimeter

Perimeter = 2 x (9 + 1) units = 2 x 10 units = 20 units

Ex 10.2 Class 6 Maths Question 1.
Find the areas of the following figures by counting square:

Solution:
Placing these figures on the centimetre square, we have
(a) Full squares = 9
∴ Area covered by the figure = 9 x 1 sq. cm = 9 sq. cm

(b) Full squares = 5
∴ Area covered by the figure = 5 x 1 sq. cm = 5 sq. cm

(c) Full squares = 2
Half squares = 4
∴ Area covered by the figure (2×1+4×12) sq. cm
= (2 + 2) sq cm = 4 sq. cm

(d) Full squares = 10
∴ Area covered by the figure = 10 x 1 sq. cm = 10 sq. cm

(e) Full squares = 10
∴ Area covered by the figure = 10 x 1 sq. cm = 10 sq. cm

(f) Full squares = 2
Half squares = 4
∴ Area covered by the figure = (2×1+4×12) sq. cm
= (2 + 2)sq. cm = 4 sq. cm

(g) Full squares = 4
Half squares = 4
∴ Area covered by the figure = (2×1+4×12) sq. cm
= (4 + 2)sq. cm = 6 sq. cm

(h) Full squares = 5
∴ Area covered by the figure = 5 x 1 sq. cm
= 5 sq. cm

(i) Full squares = 9
∴ Area covered by the figure = 9 x 1 sq. cm
= 9 sq. cm

(j) Full squares = 2
Half squares = 4
∴ Area covered by the figure = (2×1+4×12) sq. cm
= (2 + 2)sq. cm = 4 sq. cm

(k) Full squares = 4
Half squares = 2
∴ Area covered by the figure = (4×1+2×12) sq. cm
= (4 + 1)sq. cm = 5 sq. cm

(l) Full squares = 4
More than half squares = 3
Half square = 2
∴ Area covered by the figure = (4×1+3×1+2×12) sq. cm
= (4 + 3 + 1)sq. cm
= 8 sq. cm

(m) Full squares = 7
More than half squares = 7
Half square = 0
∴ Area covered by the figure = (7×1+7×1+0×12) sq. cm
= (7 + 7 + 0)sq. cm
= 14 sq. cm

(n) Full squares = 10
More than half squares = 8
Half square = 0
∴ Area covered by the figure = (10×1+8×1+0×12) sq. cm
= (10 + 8 + 0)sq. cm
= 18 sq. cm

Ex 10.3 Class 6 Maths Question 1.
Find the areas of the rectangles whose sides are:
(a) 3 cm and 4 cm
(b) 12 m and 21 m
(c) 2 km and 3 km
(d) 2 m and 70 cm
Solution:
(a) Length of the rectangle = 4 cm
Breadth of the rectangle = 3 cm
Area = Length x Breadth
= 4 cm x 3 cm
= 12 sq. cm

(b) Length of the rectangle = 21 m
Breadth of the rectangle = 12 m
Area = Length x Breadth
= 21 m xl2 m
= 252 sq. m.

(c) Length of the rectangle = 3 km
Breadth of the rectangle =2 km
Area = Length x Breadth
= 3 km x 2km
= 6 sq. km

(d) Length of the rectangle =2m = 2 x l00= 200 cm
Breadth of the rectangle = 70 cm
Area = Length x Breadth
= 200 cm x 70 cm
= 14000 sq. cm

Ex 10.3 Class 6 Maths Question 2.
Find the areas of the squares whose sides are:
(a) 10 cm
(b) 14 cm
(c) 5 m
Solution:
(a) Side of the square = 10 cm
Area = (side)²
= (10)² sq. cm
= 100 sq. cm

(b) Side of the square = 14 cm
Area = (side)²
= (14)² sq. cm
= 196 sq. cm

(c) Side of the square = 5 m
Area = (side)²
= (5)² sq. m = 25sq. m

Ex 10.3 Class 6 Maths Question 3.
The length and breadth of three rectangles are as given below:
(a) 9 m and 6 m
(b) 17 m and 3 m
(c) 4 m and 14 m
Which one has the largest area and which one has the smallest?
Solution:
In rectangle (a), Area = (9 x 6) sq. m
= 54 sq. m
In rectangle (b), Area =(3 x 17) sq. m = 51 sq. m
In rectangle (c), Area = (4 x 14) sq. m = 56 sq. m
Clearly, 56 > 54 > 51 i.e., 56 is the largest number and 51 is the smallest number.
∴ The rectangle (c) has the largest area and the rectangle (a) has the smallest area.

Ex 10.3 Class 6 Maths Question 4.
The area of a rectangular garden 50 m long is 300 sq m. Find the width of the garden.
Solution:
We use the formula A = l x b, where l is the length and b is the breadth of the rectangle in metre and its area in m2.
Here, A = 300 m², and l = 50 m
Therefore, 300 =50 x b [ ∵ A = l x b]
or b=30050=6
Hence the breadth (width) of the rectangle is 6 m.

Ex 10.3 Class 6 Maths Question 5.
What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of ₹ 8 per hundred sq. m?
Solution:
Area of rectangular piece of land = 500 x 200 sq. m
= 100000 sq. m
Cost of tiling the land at the rate of ₹ 8 per hundred sq. m
= ₹ (100000×1100) = ₹ 8000

Ex 10.3 Class 6 Maths Question 6.
A table-top measures 2 m by 1 m 50 cm. What is its area in square metres?
Solution:
Length of table’s top = 2 m
Breadth of table’s top = 1 m 50 cm = 1.50 m
∴ Area = Length x Breadth
= (2 x 1.50) sq m = 3 sq m

Ex 10.3 Class 6 Maths Question 7.
A room is 4 m long and 3 m 50 cm wide. How many square metres of carpet is needed to cover the floor of the room?
Solution:
Length of the room = 4 m
Breadth of the room = 3 m 50 cm = 3.50 m
Carpet needed to cover the floor of the room
= Area of the floor of the room
= Length x Breadth
= (4 x 3.50) sq. m = 14 sq. m

Ex 10.3 Class 6 Maths Question 8.
A floor is 5 m long and 4 m wide. A square carpet of sides 3 m is laid on the floor. Find the area of the floor that is not carpeted.
Solution:

Area of the floor of the room = (5 x 4) sq. m
= 20 sq. m
Area of the carpet = (3 x 3) sq. m = 9 sq. m
Area of the floor not carpeted = (20 – 9) sq. m
= 11 sq. m

Ex 10.3 Class 6 Maths Question 9.
Five square flower beds each of sides 1 m are dug on a piece of land 5 m long and 4 m wide. What is the area of the remaining part of the land?
Solution:
Area of the piece of land =(5 x 4) sq. m
= 20 sq. m
Area of square flower bed =(1 x 1) sq. m
= 1 sq. m
Area of 5 such flower beds =(5 x 1) sq. m
= 5 sq. m
∴ Area of the remaining part of land
= (20 – 5) sq. m = 15 sq. m

Ex 10.3 Class 6 Maths Question 10.
By splitting the following figures into rectangles, find their areas (The measures are given in centimetres).

Solution:
(a) Let the figure may be divided into rectangles marked as A, B, C etc. Area of rectangle (A) = (3 x 3) sq. cm = 9 sq. cm

Area of rectangle (B) = (1 x 2) sq. cm = 2 sq. cm
Area of rectangle (C) = (3 x 3) sq. cm = 9 sq. cm
Area of rectangle (D) = (4 x 2) sq. cm = 8 sq. cm
∴ The total area of the figure = (9 + 2 + 9 + 8) sq. cm = 28 sq. cm
(b) Let the figure may be divided into rectangle marked as shown.

Area of rectangle (A) = (2 x 1) sq cm = 2 sq cm
Area of rectangle (B) = (5 x 1) sq cm = 5 sq cm
Area of rectangle (C) = (2 x 1) sq cm = 2 sq cm
∴ Total area of the figure = (2 + 5 + 2) sq cm = 9 sq cm

Ex 10.3 Class 6 Maths Question 11.
Split the following shapes into rectangles and find their areas. (The measures are given in centimetres)

Solution:
(a) Splitting the given shape into rectangles as marked A and B.

Area of the rectangle (A) = (10 x 2) sq cm
= 20 sq cm
Area of the rectangle (B) = (10 x 2) sq cm
= 20 sq cm
∴ Total area of given shape = (20 +20) sq cm
= 40 sq cm
(b) Splitting the given shape into five squares each of side 7 cm.

Area of the given shape
= 5 x Area of one square of side 7 cm = 5 x (7 x 7) sq. cm
= (5 x 49)sq cm 7
= 245 sq cm
(c) Splitting the given shape into two rectangles named A and B. Area of rectangle (A) = (5 x 1) sq cm = 5 sq cm
Area of rectangle (B) = (4 x 1) sq cm = 4 sq cm
∴ Total area of given shape = (5 + 4) sq cm = 9 sq cm

Ex 10.3 Class 6 Maths Question 12.
How many tiles whose length and breadth are 12 cm and 5 cm, respectively will be needed to fit in a rectangular region whose length and breadth are respectively:
(a) 100 cm and 144 cm
(b) 70 cm and 36 cm
Solution:
(a) Area of the rectangular region
= (100 x 144) sq cm
= 14400 sq cm
Area of one tile = (5 x 12) sq cm
= 60 sq cm

(b) Area of the rectangular region
= (70 x 36) sq cm
= 2520 sq cm

Ex 9.1 Class 6 Maths Question 1.
In a Mathematics test, the following marks were obtained by 40 students. Arrange these marks in a table using tally marks.

(a) Find how many students obtained marks equal to or more than 7?
(b) How many students obtained marks below 4?
Solution:
In the first column of the table, we write all the values of marks scored by the students starting from the lowest to the highest. In the second column, a vertical bar (|) called the tally mark is put against the number, whenever it occurs. For our convenience, we shall keep the tally marks in bunches of five, the fifth mark being drawn diagonally across the first four. We continue this process for all the values of the first column. Finally, we count the number of tally marks corresponding to each observation and write the number in the third column to represent the number of students.
Thus, we have the table as under:


From the above, clearly the number of students who obtained marks equal to or more than 7 marks are
5 + 4+3=12
Clearly, from the above table the number of students scoring marks below 4 are
2 + 3 + 3 = 8

Ex 9.1 Class 6 Maths Question 2.
Following is the choice of sweets of 30 students of class VI. Ladoo, Barfi, Ladoo, Jalebi, Ladoo, Rasgulla, Jalebi, Ladoo, Barfi, Rasgulla, Ladoo, Jalebi, Jalebi, Rasgulla, Ladoo, Rasgulla, Jalebi, Ladoo, Rasgulla, Ladoo, Ladoo, Barfi, Rasgulla, Rasgulla, Jalebi, Rasgulla, Ladoo, Rasgulla, Jalebi, Ladoo.
(a) Arrange the names of sweets in a table using tally marks.
(b) Which sweet is preferred by most of the students?
Solution:
(a) The required table is as under:

(b) The sweet Ladoo is preferred by the most of the students.

Ex 9.2 Class 6 Maths Question 1.
Total number of animals in five villages are as follows:
Village A    :   80
Village B    :   120
Village C     :   90
Village D     :  40
Village E     :   60
Prepare a pictograph of these animals using one symbol  to represent 10 animals and answer the following questions:
(a) How many symbols represent animals of village E?
(b) Which village has the maximum number of animals?
(c) Which village has more animals: village A or village C?
Solution:
The given data can be represented by a pictograph as given below:

(i) 6 symbols represent the animals of village E.
(ii) Village B has the maximum number of animals.
(iii) Village C has more animals than that of village A.

Ex 9.2 Class 6 Maths Question 2.
Total number of students of a school in different years is shown in the following table:

A. Prepare a pictograph of students using one symbol A to represent 100 students and answer the following questions:
(a) How many symbols represent total number of students in the year 2002?
(b) How many symbols represent total number of students for the year 1998?
B. Prepare another pictograph of students using any other symbol each representing 50 students. Which pictograph do you find more informative?
Solution:
The given data can be represented by a pictograph as given below:

(a) 6 symbols represent total number of students in the year 2002.
(b) 5 complete symbols and one incomplete symbol of 35 students
B. Let us prepare another pictograph of students using symbol  each representing 50 students.

Second pictograph is more informative as it is accurate.

Ex 9.3 Class 6 Maths Question 1.
The bar graph given alongside shows the amount of wheat purchased by government during the year 1998-2002.
Read the bar graph and write down your observations. In which year was.
(a) the wheat production maximum?
(b) the wheat production minimum?

Solution:
The given bar graph represents the amount of wheat (in thousand tonnes) purchased by government during the year 1998-2002. The amount of wheat purchased during 1998 – 2002 = 15 + 25 + 20 +20 +30 = 110 (in thousand tonnes).
(a) The maximum wheat production was in the year 2002.
(b) The minimum wheat production was in the year. 1998.

Ex 9.3 Class 6 Maths Question 2.
Observe this bar graph which is showing the sale of shirts in a ready made shop from Monday to Saturday.

Now answer the following questions:
(a) What information does the above bar graph given?
(b) What is the scale chosen on the horizontal line representing number of shirts?
(c) On which day were the maximum number of shirts sold? How many shirts were sold on that day?
(d) On which day were the minimum number of shirts sold?
(e) How many shirts were sold on Thursday?
Solution:
(a) The given bar graph represents the number of shirts sold from Monday to Saturday.
(b) Scale: 1 unit length = 5 shirts.
(c) Saturday and 60 in number.
(d) Tuesday.
(e) 35

Ex 9.3 Class 6 Maths Question 3.
Observe this bar graph which shows the marks obtained by Aziz in half yearly examination in different subjects.

Answer the given questionsn.
(a) What information does the bar graph give?
(b) Name the subject in which Aziz scored maximum marks.
(c) Name the subject in which he has scored minimum marks.
(d) State the names of the subjects and marks obtained in each of them.
Solution:
(a) The given bar graph represents the marks obtained by Aziz in half yearly examination in different subjects.
(b) Hindi
(c) Social Studies.
(d) Hindi-80, English-60, Mathematics-70, Science-50 and Social Studies-40.

Ex 9.4 Class 6 Maths Question 1.
A survey of 120 school students was done to find which activity they prefer to do in their free time:

Draw a bar graph to illustrate the above data taking scale of 1 unit length = 5 students.
Which activity is preferred by most of the students other than playing?
Solution:
Here, 5 values of the data are given. So, mark 5 points on the horizontal axis at equal distances and erect rectangles of the same width whose heights are proportional to the values of the numerical data.

The activity of ‘reading story books’ is preferred by most of the students other than playing.

Ex 9.4 Class 6 Maths Question 2.
The number of Mathematics books sold by a shopkeeper on six consecutive days is shown below:

Draw a bar graph to represent the above information choosing the scale of your choicer
Solution:
Here 6 values of the data are given. So, mark 6 points on the horizontal axis at equal distances and erect rectangles of the same width whose heights are proportional to the values of the numerical data.

Ex 9.4 Class 6 Maths Question 3.
Following table shows the number of bicycles manufactured in a factory during the years 1998 to 2002. Illustrate this data using a bar graph. Choose a scale of your choice.

(a) In which year were the maximum number of bicycles manufactured?
(b) In which year were the minimum number of bicycles manufactured.
Solution:
Here, 5 values of the data are given. So, mark 5 points on the horizontal axis at equal distances and erect rectangles of the same width whose heights are proportional to the values of the numerical data.

(a) The maximum number of bicycles were manufactured in the year 2002.
(b) The minimum number of bicycles were manufactured in the year 1999.

Ex 9.4 Class 6 Maths Question 4.
Number of persons in various age groups in a town is given in the following table.

Draw a bar graph to represent , the above information and answer the following questions.
(take 1 unit length = 20 thousands)
(a) Which two age groups have same population?
(b) All persons in the’ age group of 60 and above are called senior citizens. How many senior citizens are there in the town?
Solution:
Here, 6 values of the data are given. So, mark 6 points on the horizontal axis at equal distances and erect rectangles of the same width whose heights are as under:
Age Group

(a) Age groups 30 – 44 and 45 – 59 have the same population.
(b) Number of senior citizens are 80000 + 40000 = 120000.