JEE Mathematics Geometrical Constructions Notes Set 01

Geometrical Constructions of Triangles

The constructions below all rest on one fact: any point on the perpendicular bisector of a segment is at equal distances from its two end points.

Type 1: Base, Sum of the Other Two Sides, and One Base Angle

To construct triangle ABC with BC = a, angle B = α and AB + AC = x (with x > a):

1. Draw BC = a.
2. At B draw a ray BY making angle CBY = α.
3. On ray BY mark D with BD = x.
4. Join CD and draw its perpendicular bisector, meeting BD at A.
5. Join AC. Triangle ABC is the required triangle.

Why it works: A lies on the perpendicular bisector of CD, so AC = AD. Therefore AB + AC = AB + AD = BD = x.

Example 1: Construct a triangle ABC with BC = 6 cm, angle B = 40° and AB + AC = 7.5 cm.
Solution: Draw BC = 6 cm, make angle CBY = 40°, cut BD = 7.5 cm on BY, join DC and draw the perpendicular bisector of DC to meet BD at A. Join AC. (Calculation check: AB is about 3.49 cm and AC about 4.01 cm.)

Type 2: Base, Difference of the Other Two Sides, and One Base Angle

Let BC = a, angle B = α and the difference of the other two sides = d (with d < a).

1. Case AB > AC: draw BC and ray BY with angle CBY = α. Mark D on BY with BD = d. Join CD, draw its perpendicular bisector and let it meet BY at A. Then AD = AC and AB − AC = AB − AD = BD = d.
2. Case AC > AB: draw BC and ray BY as before, and extend YB backwards to Y'. Mark D' on BY' with BD' = d. Join CD', draw its perpendicular bisector and let it meet BY at A. Then AC = AD' = AB + BD', so AC − AB = d.

Type 3: Perimeter and Both Base Angles

To construct a triangle of perimeter p with base angles α and β:

1. Draw PQ = p.
2. At P draw a ray making angle α/2 with PQ and at Q draw a ray making angle β/2 with QP on the same side; let the rays meet at A.
3. Draw perpendicular bisectors of AP and AQ, meeting PQ at B and C.
4. Join AB and AC. Triangle ABC is the required triangle.

Why it works: BA = BP and CA = CQ, so AB + BC + CA = PB + BC + CQ = PQ = p. In triangle ABP, angle BAP = angle BPA = α/2, and the exterior angle ABC = α/2 + α/2 = α. In the same way angle ACB = β.

Example 2: Construct a triangle with perimeter 10.5 cm and base angles 60° and 45°.
Solution: Draw PQ = 10.5 cm. At P draw a ray at 30° to PQ, and at Q a ray at 22.5° to QP. They meet at A. Draw the perpendicular bisectors of AP and AQ meeting PQ at B and C. Triangle ABC has angles B = 60°, C = 45° and A = 75°, and perimeter 10.5 cm.

Type 4: Two Sides and a Median

To construct triangle ABC with BC = 4 cm, AB = 5 cm and median AE = 3.5 cm to side BC: draw BC = 4 cm and bisect it at E. With centre E draw an arc of radius 3.5 cm and with centre B an arc of radius 5 cm; they meet at A. Join AB and AC. The construction is possible only if the three lengths BE, AB and AE can form a triangle. Here BE = 2, AE = 3.5, AB = 5 and 2 + 3.5 > 5, so it works. By the median formula AB² + AC² = 2(AE² + BE²), giving AC² = 7.5.

Type 5: A Triangle Equal in Area to a Quadrilateral

1. Join the diagonal BD of the quadrilateral ABCD.
2. Through C draw a line parallel to BD, meeting AB produced at E.
3. Join DE. Triangle AED is the required triangle.

Why it works: triangles DBC and DBE lie on the same base DB between the same parallels DB and CE, so their areas are equal. Then Area(ABCD) = Area(ABD) + Area(DBC) = Area(ABD) + Area(DBE) = Area(AED).

The triangle obtained is not unique: choosing a different pair of adjacent sides gives a different triangle of the same area. The same idea reduces an n-sided polygon to an (n − 1)-sided polygon of equal area, and by repeating it any polygon can be turned into a triangle of equal area.