CBSE Class 6 Mathematics Playing with Constructions Worksheet Set 01

Long Answer Questions

Question 1. A square PQRS of side 7 cm has to be drawn.
(i) Write the complete steps of construction, naming the instrument used in each step.
(ii) Explain how a compass can be used, instead of measuring with the ruler, to mark the corner R.
(iii) After the figure is drawn, how will you check that it is really a square?
Answer:
(i) Steps of construction:
1. Ruler: draw a line segment PQ = 7 cm.
2. Protractor: place its centre at P with the base along PQ, mark 90° and draw the line through P perpendicular to PQ.
3. Ruler: on this perpendicular mark S with PS = 7 cm.
4. Protractor: in the same way draw a perpendicular to PQ at Q, on the same side as S.
5. Ruler: mark R on it with QR = 7 cm.
6. Ruler: join S and R. PQRS is the required square.
(ii) Open the compass with its tip at P and its pencil at Q; this opening is the side PQ. Now place the tip at Q and cut the perpendicular at Q: the cut is the corner R. The compass carries the length across, which is quicker and more exact than reading the ruler again.
(iii) Measure the four sides with the ruler; each should be 7 cm (S1). Measure the four angles with the protractor; each should be 90° (S2). Both properties hold, so it is really a square.
In simple words: Draw one side, raise right angles at both ends, copy the side length with a compass, join, then check S1 and S2.

Teacher's Note:
1. Name the instrument at every step.
2. The compass transfer avoids small reading errors.

Teacher's Note:
1. Use a proper table with clear headings.
2. Every square is a rectangle, but not the other way round.

Question 3. A rectangle ABCD with AB = 8 cm and BC = 4.5 cm has to be constructed.
(i) Write the steps of construction, naming the instrument used in each step.
(ii) After drawing, check the two properties of a rectangle and write what you find.
(iii) Write two more correct names for this rectangle and one name that would be wrong.
Answer:
(i) Steps of construction:
1. Ruler: draw AB = 8 cm.
2. Set square (or protractor): draw a line perpendicular to AB at B.
3. Ruler: mark C on it with BC = 4.5 cm.
4. Set square: draw a perpendicular to AB at A, on the same side.
5. Compass: open it to the length BC and, with the tip at A, cut this perpendicular at D, so AD = 4.5 cm.
6. Ruler: join D to C. ABCD is the required rectangle.
(ii) On measuring, AB = DC = 8 cm and BC = AD = 4.5 cm, so R1 holds. With the protractor, ∠A = ∠B = ∠C = ∠D = 90°, so R2 holds. It is a rectangle.
(iii) Correct names: BCDA and DCBA, as the corners come in the order of going round. A wrong name: ACBD, since it jumps from A to C and skips B.
In simple words: Draw AB, raise right angles at both ends, mark 4.5 cm on each, join, and check R1 and R2.

Teacher's Note:
1. The compass copies BC on to AD.
2. A name is wrong if two neighbouring letters are opposite corners.

Question 4. Using only a ruler and a compass, explain how you will draw each piece of artwork given below. Name the instrument used in each step.
(i) Two circles of radius 2.5 cm that just touch each other.
(ii) A ring made of two circles of radii 2 cm and 3 cm drawn about the same centre.
(iii) A row of three circles of radius 2 cm, in which each circle just touches the next one.
Answer:
(i) Ruler: draw a line and mark P and Q on it with PQ = 2.5 + 2.5 = 5 cm.
Compass: open it to 2.5 cm against the ruler; draw one circle with the tip at P and another with the tip at Q. They meet at exactly one point, the mid-point of PQ.
(ii) Mark a point O. Compass and ruler: open the compass to 2 cm and draw a circle with the tip at O.
Without moving the tip from O, open the compass to 3 cm and draw a second circle. Both circles have the same centre, and the space between them is the ring.
(iii) Ruler: draw a straight line and mark P, Q, R on it with PQ = QR = 2 + 2 = 4 cm.
Compass: with an opening of 2 cm, draw a circle with the tip at each point. Each circle just touches the next, because the distance between two centres equals the sum of the two radii.
In simple words: Touching circles need centres one 'radius + radius' apart; a ring needs one centre and two radii.

Teacher's Note:
1. Distance between centres = sum of radii for touching circles.
2. Keep the tip fixed for circles with the same centre.

Question 5. A rectangle ABCD has to be constructed in which one of its diagonals divides the opposite right angles into 55° and 35°.
(i) Draw a rough diagram of the rectangle, mark the diagonal and show where the angle of 55° lies.
(ii) Write the steps of construction in order, naming the instrument used in each step.
(iii) Into which two parts does this diagonal divide the right angle at the opposite corner? Write their measures.
Answer:
(i) Draw rectangle ABCD with A at the bottom left, B bottom right, C top right and D top left, and draw the diagonal AC. At A the diagonal makes 55° with AB and the remaining 35° with AD, since the two parts add up to 90°.

(ii) Steps of construction:
1. Ruler: draw AB of any convenient length, say 7 cm.
2. Protractor: at A mark 55° above AB and draw a long ray; the diagonal lies along it.
3. Set square: draw a perpendicular to AB at B. It meets the ray at C, and AC is the diagonal.
4. Set square: draw a perpendicular to AB at A. The right angle at A is now split into 55° and 90° − 55° = 35°.
5. Compass: open it to the length BC and, with the tip at A, cut this perpendicular at D, so AD = BC.
6. Ruler: join DC. ABCD is the required rectangle; check R1 and R2 by measuring.
(iii) At C the right angle is split into 35° and 55°: the part next to CB is 35° and the part next to CD is 55°. The same pair appears again, with places interchanged.
In simple words: Draw AB, a 55° ray at A, and a right angle at B to find C; then complete the rectangle.

Teacher's Note:
1. The 55° angle fixes the direction of the diagonal.
2. With AB = 7 cm the side BC comes out about 10 cm.

Question 6. A rectangle has to be constructed that can be divided into three identical squares. No measurement is given in the question.
(i) Draw a rough diagram, mark its corners and write down which sides must be equal.
(ii) Taking the shorter side as 3 cm, write the steps of construction and name the instrument used in each step.
(iii) Give the lengths of the sides of one rectangle that cannot be divided into three identical squares, and say why.
Answer:
(i) Let the rectangle be ADEH, with the top side AD divided at B and C, and the bottom side HE divided at G and F. The squares are ABGH, BCFG and CDEF; equal sides are marked with small strokes.

The squares are identical and all sides of a square are equal, so AB = BC = CD = AH = BG = CF = DE. Hence AD = 3 × AH.
(ii) Steps of construction (shorter side 3 cm):
1. Ruler: draw AD = 3 × 3 = 9 cm.
2. Compass: open it to 3 cm and, starting from A, cut off this length twice on AD to fix B and C.
3. Set square: draw perpendiculars to AD at A, B, C and D.
4. Compass: with the same 3 cm opening cut these perpendiculars at H, G, F and E.
5. Ruler: join H to E. The rectangle ADEH is complete and cut into three identical squares.
(iii) A rectangle of sides 3 cm and 7 cm cannot be divided so, because the longer side would have to be 3 × 3 = 9 cm, not 7 cm.
In simple words: The long side must be three times the short side: 3 cm and 9 cm works, 3 cm and 7 cm does not.

Teacher's Note:
1. Small strokes in a rough diagram show equal sides.
2. The compass marks equal pieces without measuring each one.

Question 7. A rectangle ABCD is to be constructed in which the side DC = 6 cm and the diagonal DB = 10 cm.
(i) Write the steps of construction in order, naming the instrument used in each step.
(ii) Why is drawing an arc of radius 10 cm with centre D better than moving a ruler about to find the point B?
(iii) Measure the side CB of your figure and write what you get.
Answer:
(i) Steps of construction:
1. Ruler: draw the side DC = 6 cm.
2. Set square: draw a line l perpendicular to DC at C. The corner B lies somewhere on l.
3. Compass (opened to 10 cm against a ruler): with the tip at D, draw an arc cutting l. The cutting point is B, and DB = 10 cm is the diagonal.
4. Set square: draw a perpendicular to DC at D and a perpendicular to CB at B. They meet at the fourth corner A.
5. Ruler: go over DA and AB to complete rectangle ABCD, and check R1 and R2.
(ii) Every point of the arc is 10 cm from D, so we draw all such points at one stroke. Where the arc crosses l is the only point on l that is 10 cm from D. Moving a ruler about would be endless trial and error.
(iii) On measuring, CB comes out to be 8 cm.
In simple words: Draw DC, a right angle at C, and a 10 cm arc from D to fix B; CB measures 8 cm.

Teacher's Note:
1. The arc gives B exactly, without guessing.
2. 6, 8 and 10 cm is a handy set to remember.

Question 8. In a rectangle PQRS, both the diagonals PR and QS are drawn, as shown.
(i) What do you find when you measure the two diagonals?
(ii) The diagonal PR divides the right angles at P and at R into smaller parts. Write what you notice about these smaller angles.
(iii) What new things happen when the rectangle is a square?
Answer:
(i) On measuring PR and QS with a ruler, both have the same length. The two diagonals of a rectangle are always equal.
(ii) PR cuts the right angle at P into two parts and the right angle at R into two parts. At each corner the two parts add up to 90°.
Usually the two parts are not equal: the part next to the longer side is the smaller one. The same two sizes appear again at the opposite corner, with their places interchanged.
(iii) In a square each part becomes 45°, so a diagonal cuts every right angle into two equal halves. This happens in no other rectangle. The two diagonals still remain equal in length.
In simple words: Diagonals are equal; they split each corner into two parts adding to 90°, which are 45° each only in a square.

Teacher's Note:
1. Measure carefully with a protractor to see the parts.
2. Equal halves are the special mark of a square.

Question 9. The ‘House’ figure is to be drawn with every line of its border 6 cm long. The body of the house is the square BDEC (D and E at the bottom), the roof is made of the two lines BA and CA, and a curve joins B to C. A small door of width 1.5 cm and height 2.5 cm stands on DE.
(i) Write the order in which you will draw the parts of the figure, with the steps for the square and the door.
(ii) How will you fix the top point A of the roof without any trial and error?
(iii) How will you draw the curve that joins B to C?
Answer:
(i) Draw the body first, then the door, then the roof lines, and the curve last, because the curve needs the roof point A.
1. Ruler: draw the base DE = 6 cm.
2. Set square: draw perpendiculars to DE at D and at E.
3. Compass (opened to 6 cm): cut these perpendiculars at B and C, so DB = EC = 6 cm. The top BC is not drawn, as the curve comes there.
4. Door: 6 − 1.5 = 4.5 and 4.5 ÷ 2 = 2.25, so mark a point 2.25 cm from D on DE, cut off 1.5 cm from it, raise two perpendiculars of 2.5 cm with the set square and join their tops.
(ii) A must be 6 cm from B and from C. Open the compass to 6 cm; with the tip at B draw an arc above the body, and with the tip at C draw another. The crossing point is A. Join BA and CA with the ruler.
(iii) A is 6 cm from both B and C, so keep the 6 cm opening, place the tip at A and draw the arc from B to C. The house is complete.
In simple words: Square first, door centred on the base, two 6 cm arcs to fix A, and finally an arc centred at A from B to C.

Teacher's Note:
1. The door is centred: equal gaps of 2.25 cm on both sides.
2. Every border line is 6 cm, so one compass opening serves throughout.

Question 10. A rectangle of sides 10 cm and 6 cm has to be drawn, and inside it a square whose centre is the same as the centre of the rectangle, with the corners of the square lying on the two longer sides.
(i) Find the side of this square and the distance of the square from each shorter side of the rectangle. Explain how you got these.
(ii) Write the steps of construction, naming the instrument used in each step.
(iii) How will you check, after drawing, that the square and the rectangle really have the same centre?
Answer:
(i) The corners of the square lie on the two longer sides, which are 6 cm apart, so the side of the square is 6 cm.
Along the length 10 − 6 = 4 cm is left over, shared equally at the two ends because the centres are the same. So the square is 4 ÷ 2 = 2 cm from each shorter side.
(ii) Steps of construction:
1. Ruler and set square: draw rectangle ABCD with AB = 10 cm and BC = 6 cm.
2. Ruler: on AB mark P with AP = 2 cm and Q with QB = 2 cm, so PQ = 6 cm.
3. Compass: with the same 2 cm opening, mark S on DC with DS = 2 cm and R with RC = 2 cm.
4. Ruler: join PS and QR. PQRS is the required square of side 6 cm.
(iii) Measure the gap at both ends: each must be 2 cm, and the square must touch both long sides. Then lightly draw the diagonals of the rectangle and of the square; all four pass through one point, the common centre.
In simple words: The square's side equals the breadth, 6 cm; the leftover 4 cm is split into 2 cm at each end.

Teacher's Note:
1. Same centre means equal gaps at both ends.
2. The diagonals meeting at one point is a neat check.

Case Study Based Questions

Question 11. A carpenter is making a rectangular window frame for a classroom. On his drawing the frame is 90 cm long and 60 cm wide. He cuts four wooden strips, joins them at the corners, and then checks every corner with his set square. Next he stretches a measuring tape from one corner of the frame right across to the corner opposite it, and finds this distance to be about 108 cm. He measures the other corner-to-corner distance in the same way. The carpenter says that the frame will slide into the window opening only if the joined strips really make a rectangle, and not a pushed-over slanting shape.
(i) Write the lengths of the other two sides of the frame.
(ii) What should the reading be when he checks a corner with the set square?
(iii) What should the second corner-to-corner measurement be? Give a reason.
(iv) If the second measurement had come out as 101 cm, what would the carpenter decide about his frame?
Answer:
(i) 90 cm and 60 cm, because opposite sides of a rectangle are equal.
(ii) Each corner should be 90°, since all angles of a rectangle are right angles.
(iii) Also about 108 cm. The corner-to-corner lines are the diagonals, and the diagonals of a rectangle are equal.
(iv) The frame is not a rectangle. It has been pushed into a slanting shape and its corners are no longer 90°, so he must straighten it before fitting it.
In simple words: A true rectangle frame has equal opposite sides, 90° corners and equal diagonals.

Teacher's Note:
1. Unequal diagonals show a slanting frame.
2. Carpenters really use this diagonal check.

Question 12. A tailor has to cut square napkins of side 30 cm from a long piece of cloth. For the first napkin she measures 30 cm along the straight edge of the cloth, turns a right angle with her set square, measures 30 cm again, and goes on in this way until the fourth corner meets the first one. Before stitching the border she measures all the four sides once more and also checks all the four corners. While cutting the next napkin she is in a hurry. One side of it comes out 30 cm and the next side 32 cm, although all its four corners are correct right angles.
(i) How many sides of the first napkin are 30 cm long?
(ii) Which instrument did she use to get the corners right?
(iii) Name the shape of the second napkin and give a reason.
(iv) Which property of a square fails in the second napkin, and which property still holds?
Answer:
(i) All four sides are 30 cm long.
(ii) The set square, which gives a right angle at each corner (a protractor could also be used).
(iii) A rectangle. Its four angles are 90° and its opposite sides are equal (30, 32, 30, 32 cm), so R1 and R2 hold.
(iv) S1 fails, since the four sides are not all equal. S2 still holds, since all four angles are 90°.
In simple words: Four right angles with sides 30 and 32 cm make a rectangle, not a square.

Teacher's Note:
1. Four right angles force opposite sides to be equal.
2. S1 is about sides, S2 about angles.

Question 13. For the wall magazine, Nisha is drawing a border pattern of circles along a straight line. She opens her compass against a ruler so that the distance between the compass tip and the pencil point is 3 cm. On the straight line she marks points that are 6 cm apart, and then draws one circle by keeping the compass tip at each marked point, one after the other. Her friend Sameer likes the pattern and copies it, but he marks his points only 5 cm apart with the same compass opening. A third friend marks his points 8 cm apart and draws circles with the same opening.
(i) What is the radius of each circle that Nisha draws?
(ii) Which two instruments does she use together to set this radius?
(iii) In Nisha’s pattern, do two circles next to each other touch, cut each other or stay apart? Why?
(iv) Describe how Sameer’s pattern and the third friend’s pattern look.
Answer:
(i) 3 cm, the opening between the compass tip and the pencil point.
(ii) The compass and the ruler: the compass is opened against the ruler's scale to exactly 3 cm.
(iii) They just touch. The centres are 6 cm apart, which is exactly 3 + 3, the two radii together, so they meet at only one point.
(iv) Sameer's centres are 5 cm apart, less than 6 cm, so his circles cut each other and overlap. The third friend's centres are 8 cm apart, more than 6 cm, so his circles stay apart with a gap of 8 − 6 = 2 cm.
In simple words: Centres 6 cm apart touch, closer ones overlap, farther ones leave a gap.

Teacher's Note:
1. Compare the gap between centres with radius + radius.
2. Draw all three patterns to see the difference.

Question 14. A chessboard is a square board of side 32 cm. Its playing area is divided into 8 rows and 8 columns of small cells of the same size, coloured dark and light one after the other. Rohit wants to draw such a board in his notebook. He first draws the big square with a ruler and a set square. Then he marks equal lengths along two sides of the square and joins the opposite marks with straight lines to get all the lines of the grid. When the drawing is over, his friend tells him that the big board may also be called a rectangle, and Rohit is surprised.
(i) How many small cells are there on the board?
(ii) What is the side of each small cell?
(iii) What is the measure of each angle of a small cell? Which property gives you this?
(iv) Is Rohit’s friend right in calling the board a rectangle? Explain.
Answer:
(i) 8 × 8 = 64 small cells.
(ii) Each side holds 8 equal cells, so a cell's side is 32 ÷ 8 = 4 cm.
(iii) 90°. The grid lines are perpendicular to the sides, and each cell is a square, whose angles are all 90° (property S2).
(iv) Yes. The board has opposite sides equal and all angles 90°, so it satisfies R1 and R2. Every square is also a rectangle, though not every rectangle is a square.
In simple words: 64 cells of 4 cm each with 90° corners; a square board is also a rectangle.

Teacher's Note:
1. Rows × columns gives the number of cells.
2. A square is a special rectangle.

Question 15. The back of a rectangular photo frame is 60 cm long and 45 cm wide. To stop the frame from being pushed out of shape, a carpenter nails two thin wooden strips across the back, each one going from a corner to the corner opposite it. He cuts the first strip to exactly 75 cm and finds that it fits the two opposite corners perfectly. He then cuts the second strip, fixes it in the same way, and hammers one small nail at the point where the two strips cross each other.
(i) What are these two crossing strips called in the rectangle?
(ii) To what length should the second strip be cut? Give a reason.
(iii) At one corner of the frame, a strip divides the right angle into two parts and one part measures 37°. What does the other part measure?
(iv) How does this pair of strips help to keep the frame a rectangle?
Answer:
(i) They are the two diagonals of the rectangle.
(ii) Also 75 cm, because the two diagonals of a rectangle are equal in length.
(iii) The two parts add up to 90°, so the other part is 90° − 37° = 53°.
(iv) If the frame is pushed sideways, its corners stop being 90° and the corner-to-corner distances change. The fixed strips do not let these distances change, so the corners stay at 90° and the frame stays a rectangle.
In simple words: The equal diagonal strips lock the corners at 90°.

Teacher's Note:
1. Equal diagonals are a test for a rectangle.
2. Parts of a corner angle always add to 90°.

Question 16. A gardener is marking a rectangular flower bed with pegs and string. The bed is 5 m wide. He wants to divide it into two identical square plots by stretching just one more string across the bed, so that roses can be grown in one plot and marigolds in the other. Later the owner asks him to plan another bed, also 5 m wide, that can be divided in the same way into three identical square plots. The owner then points to an old bed in the corner of the garden, which is 5 m wide and 12 m long, and asks whether that one can also be split into three identical squares.
(i) What will be the side of each square plot of the first bed?
(ii) What must be the length of the first bed?
(iii) What length should the new bed have, so that it gives three identical square plots?
(iv) Can the old 5 m by 12 m bed be divided into three identical squares? Give a reason.
Answer:
(i) 5 m, the same as the width of the bed.
(ii) 2 × 5 = 10 m, as the two squares lie side by side.
(iii) 3 × 5 = 15 m.
(iv) No. For three identical squares the length must be 3 × 5 = 15 m, but the old bed is only 12 m long.
In simple words: Square side = width 5 m; two squares need 10 m, three need 15 m, so 12 m does not work.

Teacher's Note:
1. Length = number of squares × width.
2. Always check the rule before saying yes.

Question 17. Ananya is copying a design called ‘Falling Squares’ into her drawing notebook. The design is made of three squares, each of side 5 cm. Every square is placed a little above and to the right of the one below it, in such a way that the top right corner of one square is exactly the bottom left corner of the next square, as shown in the picture. Ananya has a ruler, a set square and a compass in her geometry box. She draws the lowest square first and then goes on upwards.
(i) How many squares are there in the design and what is the side of each?
(ii) Name the instrument that helps her to make each corner 90°.
(iii) What is the total height of the design, measured from the bottom line of the lowest square to the top line of the highest square?
(iv) At how many points does one square touch another square, and what is special about such a point?
Answer:
(i) Three squares, each of side 5 cm.
(ii) The set square, which gives a right angle at once (a protractor may also be used).
(iii) Each square starts exactly one square higher than the one below, so the height is 3 × 5 = 15 cm. (The width of the design is also 15 cm.)
(iv) Two neighbouring squares touch at only one point, so the whole design has two such touching points. Each one is a corner of two squares at the same time: the top right corner of the lower square is the bottom left corner of the square above it.
In simple words: Three 5 cm squares stacked corner to corner make a 15 cm tall design.

Teacher's Note:
1. Count heights square by square.
2. The touching points are shared corners.

Teacher's Note:
1. Check both sides and angles for each figure.
2. Rotation never changes the name of a shape.

Question 19. A craft worker makes square wooden coasters of side 12 cm. In the middle of each coaster he cuts a round hole of radius 3 cm, so that a small cup can rest in it. To find the right place for the hole, he first draws two light lines on the wood, each line joining one corner of the coaster to the corner opposite it. He keeps the tip of his compass at the point where these two lines cross, opens the compass to the correct radius and draws the circle. A new worker in the workshop wants to cut a hole of radius 7 cm in a coaster of the same size.
(i) What are the two light lines called?
(ii) Which instrument draws the outline of the hole, and how is it set to a radius of exactly 3 cm?
(iii) How far is the edge of the 3 cm hole from each side of the coaster?
(iv) Explain why a hole of radius 7 cm cannot be cut in this coaster, and give the largest radius that can be cut.
Answer:
(i) They are the diagonals of the square coaster.
(ii) The outline is a circle, so it is drawn with a compass, opened against the ruler's scale until the gap between the tip and the pencil point is exactly 3 cm.
(iii) The centre is 12 ÷ 2 = 6 cm from each side and the hole reaches 3 cm out, so the edge is 6 − 3 = 3 cm from each side.
(iv) A 7 cm hole would be 7 + 7 = 14 cm across, more than the 12 cm side, so it would run past the edges. The largest radius is 6 cm, when the circle just touches all four sides.
In simple words: The diagonals find the centre; the hole can be at most 6 cm in radius on a 12 cm coaster.

Teacher's Note:
1. The diagonals of a square meet at its centre.
2. Diameter = 2 × radius must fit inside the side.

Question 20. A long banner has to be hung on the wall of the school hall. Two hooks P and Q are already fixed on the wall in a straight line, 60 cm apart. The banner is held by three strings. Two of them are tied to the hooks and the third one is tied to a nail that the peon has to drive into the wall. The nail must be exactly 50 cm away from the hook P and also exactly 50 cm away from the hook Q, so that the banner hangs straight. The peon keeps moving his measuring tape about on the wall and is not able to find the correct spot.
(i) All the points of the wall that are 50 cm away from P lie on one figure. Name it.
(ii) Describe how two arcs give the correct place for the nail at once.
(iii) Is there more than one such point on the wall? Explain.
(iv) The headmaster now says that the nail must be only 25 cm from each hook. Explain why this cannot be done.
Answer:
(i) A circle with centre P and radius 50 cm.
(ii) Open the compass (or a string pinned at one end) to 50 cm. With the tip at P draw an arc, and with the same opening and the tip at Q draw another. The crossing point is 50 cm from both hooks, so the nail goes there, with no trial and error.
(iii) Yes, two points: the arcs cross once above the line PQ and once below it. Both are 50 cm from each hook; the peon will choose the one above.
(iv) With 25 cm, the two arcs together reach only 25 + 25 = 50 cm, but the hooks are 60 cm apart. The arcs never meet, so no point is 25 cm from both hooks.
In simple words: Two 50 cm arcs from P and Q cross at the nail's place; 25 cm arcs are too short to meet.

Teacher's Note:
1. Arcs meet only if radius + radius is more than the gap.
2. There are two crossing points, one on each side.