Logan created a poster to advertise the Harvest Festival at his school. For the poster’s border, he traced an elm leaf 20 times on an 8-inch by 10-inch sheet of construction paper. After all 20 of the identical leaves were cut from the sheet, only 19% of the original sheet remained. What was the area of each leaf? Express your answer as a decimal to the nearest hundredth.
Initially, the sheet of construction paper has dimensions 8 inches by 10 inches. Therefore, it has an area of 8 × 10 = 80 in2. We are told that all but 19% of the sheet was used for the leaves. It follows, then, that the 20 leaves accounted for 100 – 19 = 81%, or 0.81 × 80 = 64.8 in2, of the sheet of construction paper. Dividing, we see that each leaf had an area of 64.8 ÷ 20 = 3.24 in2.
Logan glued those 20 leaves, with none overlapping, onto the front of the poster. If the leaves covered 7.5% of the front of the poster, what was the poster’s area, in square inches?
From the previous problem, we know that the 20 leaves had a combined area of 64.8 in2. We are told that the leaves covered only 7.5% of the poster. So, the poster had an area of 64.8 ÷ 0.075 = 864 in2.
The leaves, placed completely on the poster, along its perimeter, formed the poster’s border. If the poster’s width was 2/3 its length, what was the perimeter of Logan’s poster, in inches?
We are told that the width of the poster was 2/3 the length, so w = (2/3) × l. From the previous problem, we know that the poster has an area of 864 in2, and w × l = 864. Substituting for w in the second equation, we get (2/3) × l × l = 864. Now we solve for l to determine the length of the poster. We get (2/3)l2 = 864 → l2 = 1296 → l = 36 inches. It follows that w = (2/3) × 36 = 24 inches. Thus, the poster had a perimeter of 2 × w + 2 × l = 2 × 36 + 2 × 24 = 72 + 48 = 120 inches.
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