Reasoning · Grade 6-2 Area of the Shaded Region

Problem

Area of the shaded part (1)

A circle sits inside a big equilateral triangle, touching all three sides. A smaller equilateral triangle sits in that circle with its corners on it. Both triangles point the same way. Find the area of the small triangle.
A B C D E F
Your answer
How to solve
Strategy Create a Physical Representation — With no measurements on the figure, no formula can be started. But the little triangle is free to be picked up and turned: cut DEF out of card, pin it through the centre of the circle, and give it a half turn. Turning does not change its size, so whatever its area was before, it is the same after - and after the turn it lands in a position where the picture answers the question by itself. That turns a hard 'find the area' question into the much easier related question 'what happens when you join the midpoints of a triangle's sides?'
1STEP 1

Notice that nothing can be measured, so look for a relationship instead

With nothing to measure, hunt for a relationship.

2STEP 2

Locate the three points where the circle touches the big triangle

The circle touches at the midpoints of the sides.

3STEP 3

Give the little triangle a half turn about the centre

A half turn lands the little triangle on those points.

4STEP 4

Solve the easier problem: join the midpoints of a triangle's sides

The midpoint triangle is a quarter of the whole.

(midpoint triangle) = 1/4 × (whole triangle)
5STEP 5

Read off the area of the turned triangle, then untangle the turn

Turning keeps the area, so it is 3.

12 × 1/4 = 3 in²
6STEP 6

Check that no circle arithmetic was needed

The radius was never used.

Answer
3 in²
12 ÷ 4 = 3
The answer 3 in² is an area in square inches, matching the units of the 12 in² that was given, and it is smaller than 12, which it must be since DEF fits inside the circle, which fits inside ABC. A quarter also looks right by eye: in the figure the shaded triangle takes up somewhere near a quarter of the big triangle, clearly more than a tenth and clearly less than half. There is a second check available from lengths. The circle's radius is the distance from O out to a vertex of DEF, and the distance from O out to a vertex of ABC is twice that (in an equilateral triangle the centre sits two thirds of the way along each median, while the inscribed circle reaches only one third of the way, to the opposite side). Doubling every distance from the centre doubles a figure's side lengths and so multiplies its area by 2 x 2 = 4, so ABC is 4 times DEF, giving 12 / 4 = 3 in². Finally, pi was correctly never used - the answer would be 3 in² whatever value pi took.
Takeaway

Give the little triangle a half turn and it lands on the midpoints of the big one, cutting it into four equal pieces - so it is one quarter, every time.

  • Notice that nothing can be measured, so look for a relationship instead
  • Locate the three points where the circle touches the big triangle
  • Give the little triangle a half turn about the centre
  • Solve the easier problem: join the midpoints of a triangle's sides
  • Read off the area of the turned triangle, then untangle the turn
  • Check that no circle arithmetic was needed