Question

what is the area of this figure?
7 mm
8 mm
9 mm
3 mm 1 mm
4 mm
□ square millimeters
submit
work it out

Explanation:

Step1: Split the figure into two rectangles.

First rectangle: length = 7 mm, height = 8 mm. Area = \(7 \times 8 = 56\) square mm.
Second rectangle: length = 4 mm, height = \(9 - 8 = 1\) mm? Wait, no. Wait, the lower part: the horizontal length is 4 mm, vertical height is 1 mm? Wait, no, let's re - examine. The total height is 9 mm, the upper part is 8 mm, so the lower part's height is \(9 - 8=1\) mm? Wait, no, the small rectangle that is missing? Wait, no, the figure can be considered as a large rectangle minus a small rectangle. The large rectangle: length = 7 mm, height = 9 mm. Area = \(7\times9 = 63\) square mm. The small rectangle that is "cut out": length = 3 mm, height = 1 mm. Area = \(3\times1=3\) square mm. Then the area of the figure is \(63 - 3=60\) square mm. Wait, let's check another way. Split into two rectangles: upper rectangle: length = 7 mm, height = 8 mm, area = \(7\times8 = 56\). Lower rectangle: length = 4 mm, height = 1 mm (since \(9 - 8 = 1\)), area = \(4\times1=4\). Total area = \(56 + 4=60\). Or, another split: left part: length = 3 mm, height = 9 mm? No, that's not right. Wait, the horizontal length of the bottom part: 3 mm + 4 mm=7 mm, which matches the top length. The vertical height: 9 mm. The missing part is a rectangle with length 3 mm and height 1 mm (since \(9 - 8 = 1\)). So area of large rectangle (7x9) minus area of missing rectangle (3x1): \(7\times9-3\times1 = 63 - 3=60\).

Step2: Calculate the area.

Using the large rectangle minus the small rectangle method:
Large rectangle area: \(A_{1}=7\times9 = 63\)
Small rectangle area: \(A_{2}=3\times1 = 3\)
Figure area: \(A = A_{1}-A_{2}=63 - 3=60\)

Answer:

60