IMC 2014 Q12
12. The diagram shows an isosceles right-angled triangle divided into strips of equal width. Four of the strips are shaded.
What fraction of the area of the triangle is shaded?
- A \(\dfrac{11}{32}\)
- B \(\dfrac{3}{8}\)
- C \(\dfrac{13}{32}\)
- D \(\dfrac{7}{16}\)
- E \(\dfrac{15}{32}\)
Intermediate Mathematical Challenge 2014, Question 12. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: D \(\dfrac{7}{16}\)
Let the width of each strip be 1 unit. Then the triangle has base 8 and perpendicular height 8. So its area is equal to \(\tfrac{1}{2} \times 8 \times 8 = 32\). Looking from the right, the area of the first shaded strip is 1 unit of area less than the first unshaded strip. This difference of 1 unit also applies to the other three pairs of strips in the triangle, which means that the shaded area is 4 less than the unshaded area. So the total shaded area is \(\tfrac{1}{2}(32 - 4) = 14\). Therefore the required fraction is \(\tfrac{14}{32} = \tfrac{7}{16}\).
Solution © UK Mathematics Trust, from the IMC 2014 solutions. Longer versions with investigations: Solutions and investigations (PDF).