Question

what is the area of this figure? square miles submit work it out not feeling ready yet? these can area of rectangles and squares (100) lesson: area of compound shapes

Explanation:

Step1: Divide the figure into two rectangles

We can split the compound shape into a larger rectangle (top - right) and a smaller rectangle (bottom - left). The larger rectangle has length \( l_1 = 8\) mi and height \( h_1=9\) mi? Wait, no, let's re - examine. Wait, another way: the bottom rectangle has length \( 11\) mi and height \( 3\) mi, and the top rectangle has length \( 8\) mi and height \( 6\) mi? Wait, no, let's calculate the dimensions correctly.

Wait, the total length of the base is \( 11\) mi, and the top part has length \( 8\) mi, so the length of the bottom - left extension is \( 11 - 8=3\) mi. The height of the bottom rectangle is \( 3\) mi, and the height of the top rectangle is \( 9 - 3 = 6\) mi (since the total height is \( 9\) mi).

So, the area of the bottom rectangle: length \( L_1=11\) mi, height \( H_1 = 3\) mi. Area \( A_1=L_1\times H_1=11\times3 = 33\) square miles.

Step2: Calculate the area of the top rectangle

The top rectangle has length \( L_2 = 8\) mi, height \( H_2=6\) mi. Area \( A_2=L_2\times H_2 = 8\times6=48\) square miles.

Step3: Sum the areas of the two rectangles

Total area \( A = A_1+A_2=33 + 48=81\) square miles.

Wait, another way: Let's check the dimensions again. The figure can also be considered as a large rectangle of \( 11\times9\) minus a smaller rectangle of \( 3\times6\) (the missing part). Let's verify:

Large rectangle area: \( 11\times9 = 99\) square miles.

Missing rectangle: length \( 3\) mi (since \( 11 - 8 = 3\)) and height \( 6\) mi (since \( 9 - 3=6\)). Area of missing rectangle: \( 3\times6 = 18\) square miles.

Then total area \(=99 - 18=81\) square miles.

Answer:

\(81\)