Three problems about fitting congruent shapes together:
- (i) Rectangle ABCD has sides a, b, and rectangle PQRS has sides 2a, 2b. Show that PQRS has 4 times the area of ABCD. Does this mean that 4 copies of rectangle ABCD will fit into rectangle PQRS? Check and see!
- (ii) ∆ABC has sides a, b, c, and ∆PQR has sides 2a, 2b, 2c. Show that ∆PQR has 4 times the area of ∆ABC. Does this mean that 4 copies of ∆ABC will fit into ∆PQR? Check and see!
- (iii) ∆ABC has sides a, b, c, and ∆PQR has sides 3a, 3b, 3c. Show that ∆PQR has 9 times the area of ∆ABC. Does this mean that 9 copies of ∆ABC will fit into ∆PQR? Check and see!
Answer: (i) 2a × 2b = 4ab; yes, 4 copies fit (a 2 × 2 grid). (ii) Heron with s doubled gives √16 = 4 times; yes, joining midpoints gives 4 copies. (iii) √81 = 9 times; yes, trisecting the sides and drawing parallels gives 9 copies.
Step-by-step solution
Idea: Scaling all lengths by k multiplies area by k2. For rectangles and triangles the bigger shape can actually be tiled by k2 copies.
(i) Rectangle ABCD has sides a, b, and rectangle PQRS has sides 2a, 2b. Show that PQRS has 4 times the area of ABCD. Does this mean that 4 copies of rectangle ABCD will fit into rectangle PQRS? Check and see!
- Area(PQRS) = 2a × 2b = 4ab = 4 × area(ABCD).½ mark
- Yes: lines through the midpoints of the sides of PQRS cut it into 4 rectangles, each a × b, i.e. 4 copies of ABCD fit exactly.1 mark
4 times; yes, 4 copies fit
(ii) ∆ABC has sides a, b, c, and ∆PQR has sides 2a, 2b, 2c. Show that ∆PQR has 4 times the area of ∆ABC. Does this mean that 4 copies of ∆ABC will fit into ∆PQR? Check and see!
- Semi-perimeter of PQR = 2s. Heron: area = √(2s × 2(s − a) × 2(s − b) × 2(s − c)) = √16 × √(s(s − a)(s − b)(s − c)) = 4 × area(ABC).1 mark
- Yes: joining the midpoints of the sides of PQR gives 4 triangles, each with sides a, b, c (a midpoint segment is half the parallel side), so 4 copies of ABC fit (the middle one turned upside down).½ mark
4 times; yes, 4 copies fit
(iii) ∆ABC has sides a, b, c, and ∆PQR has sides 3a, 3b, 3c. Show that ∆PQR has 9 times the area of ∆ABC. Does this mean that 9 copies of ∆ABC will fit into ∆PQR? Check and see!
- Semi-perimeter 3s: area = √(3s × 3(s − a) × 3(s − b) × 3(s − c)) = √81 × area(ABC) = 9 × area(ABC).1 mark
- Yes: divide each side of PQR into 3 equal parts and draw lines through the division points parallel to the sides. This makes a triangular grid of 9 small triangles (6 pointing the same way as PQR, 3 upside down), each with sides a, b, c. So 9 copies of ABC fit.1 mark
9 times; yes, 9 copies fit
(i) 4 times, and 4 copies fit. (ii) 4 times, and 4 copies fit (joining midpoints). (iii) 9 times, and 9 copies fit (trisecting the sides).
Answer to write in the exam
(i)
ar(PQRS) = 2a × 2b = 4ab = 4 ar(ABCD)
Join midpoints of opposite sides ⇒ 4 rectangles a × b
∴ Yes, 4 copies fit.
(ii)
s′ = 2s; ar(PQR) = √(2s · 2(s − a) · 2(s − b) · 2(s − c)) = 4 ar(ABC)
Midpoints of sides of PQR joined ⇒ 4 triangles with sides a, b, c
∴ Yes, 4 copies fit.
(iii)
s′ = 3s; ar(PQR) = √(34 × s(s − a)(s − b)(s − c)) = 9 ar(ABC)
Trisect sides, draw parallels ⇒ 9 triangles with sides a, b, c
∴ Yes, 9 copies fit.
Common mistakes that cost marks
- Thinking doubling the sides doubles the area. Area grows by the square of the scale: 22 = 4, 32 = 9.
- In (ii) and (iii), forgetting that some copies must be turned upside down to fit.
- In Heron’s formula, scaling s but not s − a, etc.
How this can come in the exam
MCQ (1 mark)
Each side of a triangle is made 5 times as long. Its area becomes
- 5 times
- 10 times
- 25 times
- 125 times
Show answer
(C) 25 times
Area scales by 52 = 25.
Try one yourself
A triangle has area 12 cm2. What is the area of a triangle whose sides are 4 times as long, and how many copies of the small one fit inside it?
Show answer
42 × 12 = 192 cm2; 16 copies.
More questions like this
- What fraction of the triangle is shaded? What fraction of the square is shaded?
- What fraction of the rectangle is covered by the circles?
- Use the above to make a conjecture about the area occupied by circles fitted into a rectangle in the manner shown. Test your conjecture for particular cases: 10 circles; 20 circles; 50 circles. Then prove your conjecture!
- The figure shows nine identical rectangles fitted together to make a large rectangle whose area is 72 cm2. Find the perimeter of each small rectangle.
- Show that the areas of the shaded blue triangle and the shaded red triangle are equal. Find a way of cutting up the blue triangle into some number of pieces and rearranging the pieces to cover the red triangle.