QUESTION IMAGE
Question
find the area of the figure. (sides meet at right angles.)
Step1: Divide the figure into two rectangles
We can split the L - shaped figure into two rectangles. One rectangle has dimensions \(8\mathrm{m}\) (height) and \((8 - 5)\mathrm{m}=3\mathrm{m}\) (width). The other rectangle has dimensions \(5\mathrm{m}\) (height) and \(8\mathrm{m}\) (width)? Wait, no, a better way: The first rectangle: length \(8 - 5=3\mathrm{m}\), width \(8\mathrm{m}\)? No, let's re - examine. The vertical side is \(8\mathrm{m}\), the horizontal side is \(8\mathrm{m}\). The notch is \(5\mathrm{m}\) (horizontal) and \(3\mathrm{m}\) (vertical, since \(8 - 5 = 3\)). So we can split it into a rectangle of \(8\mathrm{m}\times(8 - 5)\mathrm{m}\) and a rectangle of \(5\mathrm{m}\times5\mathrm{m}\). Wait, \(8-5 = 3\), so first rectangle: length \(8\mathrm{m}\), width \(3\mathrm{m}\) (because the left part has width \(8 - 5=3\) and height \(8\)). The second rectangle: length \(5\mathrm{m}\), height \(5\mathrm{m}\) (the square - like part on the right).
Step2: Calculate the area of the first rectangle
The area of a rectangle is given by \(A = l\times w\). For the first rectangle, \(l = 8\mathrm{m}\), \(w=(8 - 5)\mathrm{m}=3\mathrm{m}\). So \(A_1=8\times3 = 24\mathrm{m}^2\).
Step3: Calculate the area of the second rectangle
For the second rectangle, \(l = 5\mathrm{m}\), \(w = 5\mathrm{m}\) (since it's a square with side \(5\)). So \(A_2=5\times5=25\mathrm{m}^2\).
Step4: Sum the areas of the two rectangles
The total area \(A=A_1 + A_2\). Substitute \(A_1 = 24\) and \(A_2 = 25\) into the formula: \(A=24 + 25=49\mathrm{m}^2\). Wait, another way: The area of the big square (if there was no notch) would be \(8\times8 = 64\mathrm{m}^2\). The area of the notch (the missing rectangle) is \(5\times3=15\mathrm{m}^2\) (since the notch has length \(5\) and width \(3\) (because \(8 - 5 = 3\))). Then the area of the figure is \(64-15 = 49\mathrm{m}^2\). Let's check with the first method: \(8\times3+5\times5=24 + 25 = 49\). Yes, that's correct.
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\(49\)