Question

Find the area of the shape below list the area of each part of the total area.

Answer 1

Step 1

Shape A is parallelogram

Shape B is Trapezium

We are going to find the area of each shapes and sum them to have the total area.

Step 2

`\begin{gathered} \text{Area of shape A =bh} \\ where\text{ b is the base } \\ \text{and h is the height} \\ b=5 \\ h=3 \\ \text{Area of shape A =5}*3 \\ \text{Area of shape A =}15 \end{gathered}`

Step 3

To find the area of shape B

we need to know the distance between the points

`\begin{gathered} \text{Distance betwe}en\text{ BC} \\ D=\sqrt[]{(x_2-x_1)^2+(y_2-y_(1)^2)} \\ p\text{ oint B (2,-4)} \\ p\text{ oint }C\text{ (5,-3)} \\ D=\sqrt[]{(5_{}-2_{})^2+(-3_{}-(-4)^2} \\ \\ D=\sqrt[]{3^2+1^2} \\ D=\sqrt[]{9+1} \\ D=\sqrt[]{10} \\ D=3.1622776601684 \\ \therefore BC\text{ =3.16} \end{gathered}`
`\begin{gathered} \text{Distance betwe}en\text{ DC} \\ D=\sqrt[]{(x_2-x_1)^2+(y_2-y_(1)^2)} \\ P\text{ oint D (4,0)} \\ P\text{ oint C (5,-3)} \\ D=\sqrt[]{(5_{}-4_{})^2+(-3_{}-0)^2} \\ \\ D=\sqrt[]{1^2+(-3_{})^2} \\ D=\sqrt[]{1+9} \\ D=\sqrt[]{10} \\ DC=\sqrt[]{10} \end{gathered}`
`\begin{gathered} \text{Distance betw}eent\text{ AD} \\ D=\sqrt[]{(x_2-x_1)^2+(y_2-y_{1^{}}})^2 \\ p\text{ oint A(-5,-3)} \\ p\text{ oint D (4,0)} \\ D=\sqrt[]{(4_{}-(-5)_{})^2+(0_{}-(-3)})^2 \\ D=\sqrt[]{(9_{})^2+(-3})^2 \\ D=\sqrt[]{81+9} \\ D=\sqrt[]{90} \\ AD=\sqrt[]{90} \\ AD=3\sqrt[]{10} \end{gathered}`

Now

We can find now find the area of shape B

`\begin{gathered} \text{Area of shape B (Trapeziod)=}(a+b)/(2)h \\ \text{where a=}\sqrt[]{10} \\ b=3\sqrt[]{10} \\ h=\sqrt[]{10} \\ \text{Area of shape B (Trapeziod)=}\frac{\sqrt[]{10}+3\sqrt[]{10}}{2}*\sqrt[]{10} \\ \\ \text{Area of shape B (Trapeziod)=}\frac{4\sqrt[]{1}0}{2}*\sqrt[]{10} \\ \\ \text{Area of shape B (Trapeziod)}=20 \end{gathered}`

Finally

We add area of shape A and Shape B

Area of shape A=15

Area of shape B=20

The total area of the shape =35 square units

`\text{The total area = 35}`

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