WWW.C E M C .U WAT E R LO O.C A | T h e C E N T R E fo r E D U C AT I O N i n M AT H E M AT I C S a n d CO M P U T I N G
Problem of the Week
Problem C and Solution
A Kaleidoscope of Areas
Problem
In the diagram, a 5 cm by 5 cm square is formed by arranging 25 identical 1
cm by 1 cm squares in five rows, each of which contains five squares. The large
square contains four shaded triangles. Find the ratio of the shaded area to the
unshaded area.
Solution
B
We will start by determining the area of the four shaded triangles.
We have labeled the regions A, B, C and D. We will calculate the
area of each triangle using the formula for the area of a triangle:
A
D
base × height
area =
2
The triangle in region A has base 4 cm and height 3 cm.
12
2
The area of this triangle is 4×3
2 = 2 = 6 cm .
The triangle in region B has base 3 cm and height 2 cm.
6
2
The area of this triangle is 3×2
2 = 2 = 3 cm .
The triangle in region C has base 4 cm and height 2 cm.
8
2
The area of this triangle is 4×2
2 = 2 = 4 cm .
The triangle in region D has base 3 cm and height 3 cm.
9
2
The area of this triangle is 3×3
2 = 2 cm .
The total shaded area is therefore 6 + 3 + 4 +
9
2
= 13 +
9
2
=
26
2
+
9
2
=
35
2
cm2 .
We will now determine the area of the unshaded region. The square is 5 cm
long and 5 cm wide. Therefore, the area of the entire square is 5 × 5 = 25 cm2 .
The area of the unshaded region is the area of the entire square minus the area
of the shaded region. So the area of the unshaded region is
50
35
15
2
25 − 35
2 = 2 − 2 = 2 cm .
The ratioof the shaded area to the unshaded area is
35
15
2 : 2 = 35 : 15 = 7 : 3.
C