QUESTION 1
The given figure is made up of a parallelogram and a rectangle.
The area of the parallelogram is
[tex]=base\times height[/tex]
The base of the parallelogram is 6 units (Just count the boxes).
The height is also 1 unit.
This implies that the area of the parallelogram is
[tex]=6\times 1 \:units^2[/tex]
[tex]=6\:units^2[/tex]
To find the area of the rectangle, we need to find the width and the length using the distance formula or the Pythagoras Theorem.
Using the Pythagoras Theorem,
[tex]l^2=a^2+b^2[/tex]
[tex]\Rightarrow l^2=2^2+6^2[/tex]
[tex]\Rightarrow l^2=4+36[/tex]
[tex]\Rightarrow l^2=40[/tex]
[tex]\Rightarrow l=\sqrt{40}[/tex]
[tex]\Rightarrow l=2\sqrt{10}[/tex]
Similarly,
[tex]w^2=1^2+3^2[/tex]
[tex]\Rightarrow w^2=1+9[/tex]
[tex]\Rightarrow w^2=10[/tex]
[tex]\Rightarrow w=\sqrt{10}[/tex]
The area of the rectangle is
[tex]Area=l\times w[/tex]
We substitute the values into the formula to obtain;
[tex]Area=2\sqrt{10}\times \sqrt{10}[/tex]
[tex]Area=2\times10[/tex]
[tex]Area=20\:units^2[/tex]
The area of the figure is
[tex]=20+6[/tex]
[tex]=26\:units^2[/tex]
QUESTION 2
We can divide the given polygon into two parts to obtain a triangle and a rectangle. See diagram in attachment.
The area of the triangular portion is
[tex]=\frac{1}{2}\times base\times height.[/tex]
[tex]=\frac{1}{2}\times 9\times 6.[/tex]
[tex]=9\times 3.[/tex]
[tex]=27\:units^2.[/tex]
The area of the rectangle is
[tex]=l\times w[/tex]
[tex]=9\times 2[/tex]
[tex]=18\:units^2[/tex]
The area of the polygon is
[tex]=27+18\:units^2[/tex]
[tex]=45\:units^2[/tex]
