Problem 17 worked answer

Corner Courtyard

Designed for Grade 3 · Uses Middle Grade 3 math · About 2–6 minutes

Earn one jewelCalcite RhombLevel 2

At a glance

Part 1: A valid two-rectangle split gives a total area of 57 square meters.

1. Split the courtyard and add the areas

Draw a horizontal segment from the inside corner to the left side. This divides the L-shape into two nonoverlapping rectangles:

a top rectangle measuring 8 m by 4 m, with area $8 \times 4 = 32 m 2$ ; and
a lower-left rectangle measuring 5 m by 5 m, with area $5 \times 5 = 25 m 2$ .

The rectangles meet along an edge and together cover the entire courtyard without a gap or overlap. Add their areas:

$(8 \times 4) + (5 \times 5) = 32 + 25 = 57 m 2$

The courtyard’s total area is 57 square meters.

Another valid split

A vertical segment from the inside corner to the top edge also makes two rectangles: 5 m by 9 m and 3 m by 4 m.

$(5 \times 9) + (3 \times 4) = 45 + 12 = 57 m 2$

Either addition equation satisfies the requested two-rectangle method.

Check

The full 8 m by 9 m bounding rectangle has area $8 \times 9 = 72 m 2$ . Its missing upper-right corner measures 3 m by 5 m, with area $3 \times 5 = 15 m 2$ . Subtracting gives $72 - 15 = 57 m 2$ , confirming the addition result. A subtraction-only response is a useful check but does not replace the requested two-rectangle addition equation.

Technical fit and rating

MJ 3.6 · C2 · W1 Standard challenge · Light workload

Geometry · Measurement

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