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Geometric Constructions — Class 10 Maharashtra SSC Mathematics revision notes

A 4-page visual revision sheet for Geometric Constructions, written for Class 10 Maharashtra SSC Mathematics. Formulas, diagrams and the traps that cost marks, laid out to be read the night before rather than the term before. Free, and it opens inside CapCompass.

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Geometric Constructions in one look

compass and straight edge only, with the arcs left showing because they are the working

The rules of a construction

  • only two instruments are allowed: a compass for arcs and circles, and an ungraduated straight edge for lines
  • the ruler may be used to measure a given length when setting the compass, but never to mark a point by measuring along a line
  • every arc must be left visible, because the arcs are the evidence that the figure was constructed and not merely drawn
  • a protractor is not allowed, so an angle must be built from arcs, not measured out
  • label every point as it is made, and write the steps beside the figure, since both carry marks
  • a rough sketch of the finished figure drawn first is worth the thirty seconds: it tells you what to aim at

Dividing a segment in a given ratio

  1. draw the segment AB of the given length
  2. from A draw a ray AX making any acute angle with AB, on either side
  3. along AX mark off m + n equal arcs of any convenient radius, giving points A₁, Aā‚‚ and so on
  4. for a ratio m:n, join the last point Aā‚m+nā‚Ž to B
  5. through the point A_m draw a line parallel to that join, meeting AB at P
  6. then P divides AB in the ratio m:n, which follows from the basic proportionality theorem

Why that construction works

  • the arcs along the ray are equal by construction, so AX is divided into m + n equal parts
  • the line through A_m is parallel to the join from the last point to B
  • by the basic proportionality theorem a line parallel to one side of a triangle divides the other two sides in the same ratio
  • the ray is divided in the ratio m:n at A_m, so AB must be divided in that same ratio at P
  • the angle at A does not matter, and neither does the size of the arcs, which is why any convenient choice is allowed

Constructing a similar triangle

  1. draw the given triangle ABC, and from A draw a ray AX at an acute angle to AB, on the side away from C
  2. mark as many equal arcs along AX as the larger of m and n, giving A₁ up to that point
  3. join A_n to B, where n is the denominator of the scale factor
  4. through A_m draw a line parallel to A_nB, meeting the line AB at B'
  5. through B' draw a line parallel to BC, meeting the line AC at C'
  6. then triangle AB'C' is similar to ABC with the ratio m:n, and it is larger when m exceeds n and smaller when it does not

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