What is the area of the figure? figure in attachment below

Mathematics

1 answer:

damaskus [11]3 years ago

3 0

Since we cannot solve irregular shapes in one formula, we will have to divide the irregular form into shapes that have area formulas. We will do the triangle first. The formula for triangle is A=1/2BH. The base is the 6 feet and the height is 2 feet, so we will multiply it to get 12 then divide it by 2, which will get us 6. The formula for the rectangle is A=LW, so we just multiply 5 and 6 to get 30, then add the triangle area, which was 6 So the final answer is that the area of that figure is 36 ft. Squared. Don't forget the squared!

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Given the vectors A⃗ and B⃗ shown in the figure ((Figure 1) ), determine the magnitude of B⃗ −A⃗. A is 28 degrees above the posi

Vlad [161]

This problem is represented in the Figure below. So, we can find the components of each vector as follows:

$\bullet \ cos(28^{\circ})=\frac{Adjacent}{Hypotenuse}=\frac{A_{x}}{44} \\ \\ \therefore A_{x}=44cos(28^{\circ})=38.85m \\ \\ \\ \bullet \ sin(28^{\circ})=\frac{Opposite}{Hypotenuse}=\frac{A_{y}}{44} \\ \\ \therefore A_{y}=44sin(28^{\circ})=20.65m$

$\bullet \ cos(56^{\circ})=\frac{Adjacent}{Hypotenuse}=\frac{-B_{x}}{26.5} \\ \\ \therefore B_{x}=-26.5cos(56^{\circ})=-14.81m \\ \\ \\ \bullet \ sin(56^{\circ})=\frac{Opposite}{Hypotenuse}=\frac{B_{y}}{26.5} \\ \\ \therefore B_{y}=26.5sin(56^{\circ})=21.97m$