Perimeter & Area

Problem

Find a missing side, then sum sides as fractions

A rectangle has width 3 515\frac{5}{15} cm. Its height is 1 1115\frac{11}{15} cm longer than the width. I must find the perimeter (the sum of all four side lengths) in cm.
FractionsMeasurement & data
Your answer
cm
How to solve
Strategy Identify Subproblems — First find the height by adding to the width, then apply the rectangle perimeter formula P = 2 × (width + height). Two clean subproblems, both with denominator 15.
1STEP 1

Find the height

Add to the width: 3 515\frac{5}{15} + 1 1115\frac{11}{15} = 4 + 1615\frac{16}{15}, and 1615\frac{16}{15} = 1 115\frac{1}{15}, so height = 5 115\frac{1}{15} cm.

3515+11115=4+1615=51153 \frac{5}{15}+1 \frac{11}{15}=4+\frac{16}{15}=5 \frac{1}{15}
2STEP 2

Add width and height

width + height = 3 515\frac{5}{15} + 5 115\frac{1}{15} = 8 + 615\frac{6}{15} = 8 615\frac{6}{15} cm.

3515+5115=86153 \frac{5}{15}+5 \frac{1}{15}=8 \frac{6}{15}
3STEP 3

Double to get the perimeter

Perimeter = 2 × (width + height) = 2 × 8 615\frac{6}{15} = 16 1215\frac{12}{15} = 16 45\frac{4}{5} cm.

2×8615=161215=16452× 8 \frac{6}{15}=16 \frac{12}{15}=16 \frac{4}{5}
Answer
16 45\frac{4}{5} cm
2 × 8 615\frac{6}{15} = 16 45\frac{4}{5}
Width ≈ 3.33 cm and height ≈ 5.07 cm, so perimeter ≈ 2 × 8.4 ≈ 16.8 cm, matching 16 45\frac{4}{5} = 16.8 cm. Units stay in cm and 1215\frac{12}{15} simplifies neatly to 45\frac{4}{5}.
Takeaway

This only needs Grade 4 fraction adding plus the rectangle perimeter formula — two widths plus two heights!

  • Find the height
  • Add width and height
  • Double to get the perimeter
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