Three circles of radius 5 cm each touch each other. If the points of contact are P, Q and R, then what is the area of the triangle PQR in sq. cm?

Correct — B, 25√3 / 4. Two equal circles that touch externally meet at a single point on the line joining their centres, exactly halfway along it. So join the three centres: each side of that triangle is 5 + 5 = 10 cm, the triangle is equilateral, and P, Q and R are the midpoints of its three sides. The triangle formed by the midpoints of a triangle has sides half as long, so PQR is equilateral with side 5 cm. Its area is (√3/4) × 5² = 25√3/4 sq. cm.

• (a)25√3 / 6 — This comes from dividing the centre triangle's area by 6. Joining the three midpoints cuts a triangle into four congruent pieces, not six, so the divisor is 4.
• (c)25√3 / 2 — Halving the centre triangle. Halving the sides quarters the area, because area scales with the square of the length ratio.
• (d)25√3 — That is the area of the triangle joining the three centres, side 10 cm. It is the figure you draw on the way to the answer, not the triangle the question asks about.

When two circles touch externally the distance between their centres equals the sum of the radii, and the point of contact lies on the segment joining the centres. With equal radii that point is the midpoint of the segment. Three mutually touching equal circles therefore give an equilateral triangle of centres whose side is twice the radius, and the three contact points are the midpoints of its sides — the medial triangle.

The figure is never printed, so the whole item is a test of whether you can build it. Draw the centres first, not the circles: the moment the centre triangle appears with side 10 cm, the contact points sit visibly at the midpoints. The medial triangle divides the parent into four congruent small triangles, which is the cleanest way to see the factor of 4 without any algebra. Every wrong option here is a different divisor applied to 25√3, so a candidate who reaches the centre triangle and then guesses has a one-in-four chance — worth remembering that the last step is the one being tested.

• Externally touching circles: distance between centres = r₁ + r₂, and the contact point lies on the centre line.
• Equal radii put the contact point exactly at the midpoint of the centre-to-centre segment.
• Joining the midpoints of a triangle's three sides gives a triangle with half the side lengths and one quarter of the area.
• Area of an equilateral triangle of side a is (√3/4)a².
• Here the centre triangle has side 10 cm and area 25√3 sq. cm; the contact triangle has side 5 cm and area 25√3/4 sq. cm.

• Answering with the centre triangle's area because it is the figure you actually constructed.
• Halving the area when the sides are halved.
• Assuming the contact points are the feet of the altitudes rather than the midpoints.

Asked without a figure, so the examiner is really testing whether you construct the triangle of centres before reaching for a formula.

• practice — not a real PYQ

  Two circles of radii 7 cm and 4 cm touch each other externally. What is the distance between their centres?

  • (a)3 cm
  • (b)5.5 cm
  • (c)11 cm
  • (d)14 cm

  Answer(c) 11 cm — for external contact the centre distance is the sum of the radii, 7 + 4.

• practice — not a real PYQ

  The midpoints of the sides of an equilateral triangle of side 12 cm are joined. What is the area of the inner triangle so formed?

  • (a)9√3 sq. cm
  • (b)18√3 sq. cm
  • (c)36√3 sq. cm
  • (d)12√3 sq. cm

  Answer(a) 9√3 sq. cm — the inner triangle has side 6 cm, so its area is (√3/4) × 36 = 9√3.