Chapter 8: Playing with Constructions (Constructions)

All the points that are 4 cm from P together form a circle with centre P and radius 4 cm. We can draw it at once with a compass opened to 4 cm, placing its tip on P.

A circle with centre P and radius 4 cm.

The two half circles together cover AB, so each half circle stands on half of AB: AX = $8÷2=$ 4 cm. A half circle on a 4 cm line has its centre at the middle of the line, so the radius is $4÷2=$ 2 cm.

Radius 2 cm; AX = 4 cm.

For a central line of length $L$: AX = $\frac{L}{2}$ and the radius = $\frac{L}{4}$. For example, for $L=6$ cm: mark X at 3 cm, and draw half circles of radius 1.5 cm with centres at 1.5 cm and 4.5 cm from A, one above and one below the line.

For AB = 6 cm: AX = 3 cm and radius 1.5 cm (in general, radius = AB ÷ 4).

Use a radius larger than half of AX, and place the compass tip away from the line: for the upper wave the centre lies below the line, on the perpendicular through the middle of AX; for the lower wave it lies the same distance above the line, through the middle of XB. Using the same radius and the same distance for both makes the waves identical.

Take a larger radius (e.g. 3 cm on AX = 4 cm) with the centres at equal distances below and above the line, on the perpendiculars through the midpoints of AX and XB.

Going around the square, the corners come in the order S → P → Q → R. A valid name must follow this order (in either direction, starting anywhere).

PQSR is not a valid name: after Q it jumps across to S (along a diagonal), instead of going to R.

1. PQSR

Draw the rectangle with its corners on dots. Then draw four equal small squares, one at each corner, each starting the same number of dots diagonally away from the corner of the rectangle (for example, leave one dot diagonally and then draw a square of side 1 or 2 dot-gaps). Because each square is at the same position relative to its corner, the figure is symmetric.

Draw the rectangle on dots, then place four equal squares at the same diagonal distance (e.g. one dot) from each corner.

Only A is a square (the NCERT key agrees). B and D have unequal neighbouring sides, so they are rectangles, not squares; C has all its sides on dots but its neighbouring sides are of different lengths.

Yes, we can reason it out from the dots. For each side, count how many dots it goes across and up (or down). In A, every side goes "3 across and 3 up" (or 3 across and 3 down), so all four sides are equal, and the turns at the corners match those of a square, so the angles are right angles. In the others, the "across and up" counts of neighbouring sides are different, so the sides are not all equal.

Only A is a square. Yes: counting the dots each side moves across and up shows whether the sides are equal and the angles right.

Rotated squares: from a corner, go "2 right, 1 up", then "1 left, 2 up", then "2 left, 1 down", then "1 right, 2 down". All sides are equal and each turn is a right angle. Others: "3 right, 1 up" with "1 left, 3 up"; "2 right, 2 up" with "2 left, 2 up".

Rotated rectangles: use "4 right, 2 up" for the long sides and "1 left, 2 up" for the short sides (going around: 4 right 2 up, 1 left 2 up, 4 left 2 down, 1 right 2 down). Another: "3 right, 3 up" and "1 left, 1 up".

Verify: measure (or count) that opposite sides are equal (and, for the squares, all four sides), and check each corner with the corner of a page (a right angle).

For example, a square with sides "2 right, 1 up" / "1 left, 2 up", and a rectangle with sides "4 right, 2 up" / "1 left, 2 up"; check equal sides and right angles.

Steps: draw AB = 6 cm. At A and B draw perpendiculars (90° with the protractor or set square). Mark D on the perpendicular at A and C on the perpendicular at B, each 4 cm from AB. Join DC.

Check: opposite sides are equal (AB = DC = 6 cm, AD = BC = 4 cm) and all four angles measure 90°.

Draw AB = 6 cm, perpendiculars of 4 cm at A and B, and join; AB = CD = 6 cm, AD = BC = 4 cm, all angles 90°.

Same steps with PQ = 10 cm and perpendiculars of 2 cm at P and Q.

Check: PQ = SR = 10 cm, PS = QR = 2 cm, and all angles are 90°.

PQ = SR = 10 cm, PS = QR = 2 cm, all angles 90°.

No. If all four angles are 90°, the two sides standing on the bottom side are both perpendicular to it, so they are parallel and the top side meets them at right angles; the top side is therefore exactly as long as the bottom side, and the two upright sides are equal too. Any such figure is a rectangle, with opposite sides equal.

No: a 4-sided figure with four right angles is always a rectangle, so its opposite sides are equal.

• Closest: when X and Y are at the same height (X and Y equally far from A and B), XY is horizontal and equals AB = 7 cm; it can never be shorter than 7 cm.
• Farthest: when X and Y are at opposite corners (X = A and Y = C, or X = D and Y = B); then XY is a diagonal.

XY is shortest (= AB = 7 cm) when X and Y are equally far from A and B, and longest (a diagonal) when they are at opposite corners. Table: about 7.4 cm, 7 cm, 7.3 cm.

i. XY = AB = 7 cm every time.

ii. ABYX is a rectangle (all angles are 90° and XY = AB).

The farthest distance between X and Y is equal to the diagonal AC (or BD), about 8.1 cm; the two diagonals are equal.

XY always equals AB (7 cm) and ABYX is a rectangle; the farthest distance equals the diagonals AC = BD (≈ 8.1 cm).

Plan: the three squares are identical, so the length of the rectangle must be 3 times its breadth. Draw AF (say 3 cm), draw the perpendicular at A, and with the compass opened to AF mark B, E and C along it (AB = BE = EC = AF), so AC = 9 cm. Draw the perpendicular at C, mark D with CD = AF, and join FD. Joining B and E to the bottom side gives three squares.

• Cannot be divided into two identical squares: any rectangle whose length is not twice its breadth, e.g. 4 cm × 2.5 cm (or 5 cm × 3 cm).
• Cannot be divided into three identical squares: any rectangle whose length is not 3 times its breadth, e.g. 7 cm × 2 cm (or 8 cm × 4 cm, which instead splits into two squares).

Make the length 3 times the breadth (e.g. 9 cm × 3 cm), transferring the breadth with a compass. A 4 cm × 2.5 cm rectangle cannot be split into two identical squares, and a 7 cm × 2 cm one cannot be split into three.

The square's side must be the full breadth, 4 cm (its top and bottom lie on the long sides). For the centres to match, the square must be in the middle: the remaining length $8-4=4$ cm is split equally, 2 cm on each side. So mark points 2 cm from each end on both long sides and join them with perpendicular lines.

A 4 cm square placed 2 cm from each short side of the 8 cm × 4 cm rectangle.

Draw the first square of side 4 cm. Its top-right corner becomes the bottom-left corner of the next square: extend the top side of the first square by 4 cm to the right and draw the second square standing on that extension; repeat for the third. Use perpendiculars (90°) at every corner.

For the second figure, start with a 7 cm square; draw the 5 cm square on the extension of its top side starting at its top-right corner, then the 3 cm square in the same way on top of the 5 cm square.

Each new square starts at the top-right corner of the previous one, standing on the extension of its top side; use right angles at all corners.

Take a square of side 8 cm. Divide each side into 4 equal parts of 2 cm and join the opposite marks to get 16 small squares. Draw one diagonal in the squares that are half-shaded, as in the figure, and shade the required triangles with parallel lines. (The bottom-left part is a 4 cm square left unshaded.)

Draw an 8 cm square, divide it into a 4 × 4 grid of 2 cm squares, draw the diagonals shown and shade the half-squares.

At the centre of the square, which is the point where its two diagonals meet. Draw the diagonals, then draw the circle with that point as centre.

At the point where the diagonals of the square meet.

More holes: divide the square into 4 equal squares (join the midpoints of opposite sides); the centre of each hole is the centre of a small square (where its diagonals meet). Draw equal circles there.

Curves: for each side, put the compass tip on the line through the middle of that side, perpendicular to it, at the same distance from the side for all four sides (for example, at the centre of the square or beyond it, on the far side). With the same radius for all four, each arc passes through the two ends of its side and bulges outward by the same amount.

Holes: centres at the centres of the small squares. Curves: compass tip on the perpendicular through the middle of each side, at the same distance for all four sides, with the same radius.

The diagonal splits the corner angles equally (45° and 45°) only when all four sides are equal, i.e. when the rectangle is a square.

Laws observed in any rectangle:

• The two diagonals are equal in length and cut each other into equal halves.
• At each corner, the two parts of the 90° angle made by a diagonal add up to 90°; the same pair of angles appears at the opposite corner (for example, 40° and 50° at A and also at C).
• In a square, every diagonal makes 45° with the sides.

Only in a square does the diagonal divide the corner angles equally (45° each). In every rectangle the diagonals are equal and divide each other into halves.

Steps: draw AB of any length (say 6 cm) and a perpendicular at A. With the protractor, draw a ray from A making 40° with AB (so 50° with the perpendicular). Draw the perpendicular to AB at B; it meets the ray at C. Draw a line through C parallel to AB (perpendicular to BC) to meet the perpendicular at A in D. ABCD is the rectangle and AC divides the angles at A and C into 40° and 50°.

Draw AB, a ray at 40° from A, the perpendicular at B meeting it at C, and complete the rectangle; AC divides the corner angles into 40° and 50°.

Same method, with the ray at 45°. The perpendicular at B meets the ray at C with BC = AB, so all sides come out equal: the rectangle is a square.

The figure turns out to be a square: all four sides are equal.

Steps: draw AD = 4 cm. Draw line $l$ perpendicular to AD at A. With D as centre and radius 8 cm, draw an arc cutting $l$ at B (so DB = 8 cm is the diagonal). At B draw a perpendicular and mark C with BC = 4 cm on the same side as D; join DC. (AB comes out about 6.9 cm.)

Draw AD = 4 cm, the perpendicular at A, an 8 cm arc from D to get B, and complete the rectangle (AB ≈ 6.9 cm).

Same steps with AD = 3 cm and an arc of radius 7 cm from D. (AB comes out about 6.3 cm.)

Draw AD = 3 cm, the perpendicular at A, a 7 cm arc from D to get B, and complete the rectangle (AB ≈ 6.3 cm).

No. A is only needed where the two circles meet, above BC. So it is enough to draw a small arc of each circle (radius 5 cm from B and from C) near where A should be; the point where the arcs cross is A.

No: two short arcs of radius 5 cm from B and C, crossing above BC, are enough.

Steps: construct a square BCED of side 7 cm (draw DE = 7 cm, perpendiculars of 7 cm at D and E to get B and C, join BC). With B as centre and radius 7 cm draw an arc above BC; with C as centre and the same radius draw another arc cutting the first at A. Join AB and AC. Every side of the house is now 7 cm (the roof is an equilateral triangle).

Draw a 7 cm square BCED; arcs of radius 7 cm from B and C meet at A; join AB and AC.

• A Person: construct the body as a rectangle (perpendiculars at the corners). For the curved top (neck) of the body, find the centre of the arc as a point equidistant from both top corners: draw equal arcs from the two corners to cross below the body; with that point as centre, draw the curve through both corners. Draw the head as a circle whose centre is on the perpendicular through the middle of the top.
• Wavy Wave: the centres of the half circles are the points of AX and XB that are equidistant from their ends (their midpoints), found by drawing equal arcs from the two ends.
• Eyes: draw two light horizontal supporting lines. The centres of the arcs forming each eye are points equidistant from the two corners of the eye; find them by crossing equal arcs from the corners, then draw the upper and lower curves through the corners.

Use equal arcs from two points to find the centres of curves (points equidistant from both ends), as for the roof of the house.

Yes: a rhombus. Its sides are all equal but its angles are not 90°.

Construction: draw PQ = 5 cm and a ray from P at 60° to PQ; mark S on it with PS = 5 cm. With Q as centre and with S as centre, draw arcs of radius 5 cm crossing at R. Join QR and SR. PQRS has all sides 5 cm and angles 60° and 120°.

Yes, a rhombus: e.g. PQ = PS = 5 cm at 60°, then arcs of 5 cm from Q and S give R.