Exercises 1 & 2 – Animated Solutions സമഭാഗം · അസമഭാഗങ്ങൾ
Six constructions and one proof. Every construction here starts the same way: draw the perimeter as one straight line, cut it into the right parts with a slanted ray and parallels, then swing compass arcs. Read each question, try it on paper (use the hint if you are stuck), then press Next ▶ and draw along.
- Equal parts: ray from one end, n equal compass steps, join the last mark to the other end, parallels through the other marks (page 5).
- Parts in a ratio m : n: the same, with steps of m units and n units (page 6, Key idea 4).
- Triangle from three sides: one side on the line, arcs of the other two lengths from its ends (page 4). Square: perpendiculars at both ends, arcs of the side length. Regular hexagon: side = radius of its circle; step the compass six times round the circle.
- Pieces: 11 ÷ 3 = 3⅔, 15 ÷ 4 = 3¾, 20 ÷ 6 = 3⅓, 7½ × 3/7 = 3 3/14, 13 × 5/12 = 5 5/12 … (mixed numbers, page 3).
Exercise 1 · Question 1 – Equilateral triangle, perimeter 11 cm
(11 സെന്റിമീറ്റർ ചുറ്റളവുള്ള സമഭുജത്രികോണം വരയ്ക്കുക.)
Each side is 11 ÷ 3 = 3⅔ cm. You cannot measure that with a ruler, but you can divide an 11 cm line into three equal parts with parallels and use one part as the side.
Exercise 1 · Question 2 – Square, perimeter 15 cm
Side = 15 ÷ 4 = 3¾ cm. Divide a 15 cm line into four equal parts (four equal steps on the ray) and build the square on the first part.
Exercise 1 · Question 3 – Regular hexagon, perimeter 20 cm
Side = 20 ÷ 6 = 3⅓ cm. A regular hexagon is six equilateral triangles around a centre, so its side equals the radius of the circle through its corners: draw a circle of radius 3⅓ cm and step the same compass opening round it.
Joining the centre of a regular hexagon to its six corners makes six triangles with angle 360° ÷ 6 = 60° at the centre and two equal sides (radii). Equal sides give equal base angles, (180° − 60°) ÷ 2 = 60° each – so each triangle is equilateral and the side of the hexagon equals the radius.
Exercise 2 · Question 1 – Rectangle, perimeter 15 cm, sides 3 : 4
(ചുറ്റളവ് 15 സെന്റിമീറ്ററും വീതിയും നീളവും 3 : 4 എന്ന അംശബന്ധത്തിലുമായ ചതുരം വരയ്ക്കുക.)
Length + breadth = half the perimeter = 7½ cm. Divide a 7½ cm line in the ratio 3 : 4 (steps of 3 cm and 4 cm on the ray). The two pieces are the breadth and the length.
Exercise 2 · Question 2 – Triangles of perimeter 13 cm
Straighten the perimeter into a 13 cm line and cut it in the ratio of the sides: (i) 1 : 1 : 1, (ii) 3 : 4 : 5, (iii) base : side : side = 1 : 1½ : 1½ = 2 : 3 : 3. Then keep the first piece as the base and swing arcs equal to the other two pieces.
(i) Equilateral: sides 1 : 1 : 1
(ii) Sides 3 : 4 : 5
(iii) Isosceles, lateral sides 1½ × base: sides 2 : 3 : 3
Builders are often given a total length and a ratio – a 13 m roll of fencing for a plot whose sides must be 3 : 4 : 5 (a right-angled corner for free), a frame of fixed perimeter with sides 3 : 4 for a photo. "Divide the total in the ratio, then build" is exactly what you have just practised.
Exercise 2 · Question 3 – Diagonals of a trapezium
You need three parallel lines to use Key idea 4. The two parallel sides of the trapezium are two of them. Where could the third one go so that it passes through the crossing point of the diagonals?
Move A and B along the bottom line and C and D along the top line. AO/OC and BO/OD are always the same number – and that number is AB/DC. Make AB = DC and the figure becomes a parallelogram: the diagonals bisect each other (ratio 1).
Check yourself
- Any regular figure with a given perimeter: divide the perimeter into equal parts first (3 for a triangle, 4 for a square, 6 for a hexagon).
- Any figure with given perimeter and side ratio: divide the perimeter (or half of it, for a rectangle) in that ratio, then construct with arcs and perpendiculars.
- To prove a ratio statement, look for – or draw – a third parallel line, and then use "parallels cut any two lines in the same ratio".
Next: Sections 3 and 4 – a parallel inside a triangle, and the midpoint theorem ▶