QUESTION 5
The side length of the first tra-pezoid and the second tra-pezoid are in the ratio [tex]5:7[/tex].
To find the corresponding ratio of their area, we square each term in the ratio.
The ratio of the area of the first figure, to the ratio of the area of the second is
[tex]5^2:7^2=25:49[/tex]
QUESTION 6
The side length of the first figure and the second figure are in the ratio [tex]12:7[/tex].
To find the corresponding ratio of their area, we square each term in the ratio.
The ratio of the area of the first figure, to the ratio of the area of the second figure is
[tex]12^2:7^2=144:49[/tex]
QUESTION 7
Let [tex]h[/tex] represent the height of this triangle.
[tex]\sin 30\degree=\frac{h}{6}[/tex]
This implies that;
[tex]h=6\sin 30\degree[/tex]
[tex]h=3ft[/tex]
The area of the triangle is [tex]=\frac{1}{2}bh[/tex]
The base is 9 ft and the height is 3 ft.
We substitute these values to obtain;
[tex]A=\frac{1}{2}(9)(3)[/tex]
[tex]A=13.5ft^2[/tex] to the nearest tenth.
QUESTION 8
The area of a parallelogram is [tex]base\times\:height[/tex]
Let h represent the height of this parallelogram.
[tex]\sin60\degree=\frac{h}{6}[/tex]
[tex]h=6\sin60\degree[/tex]
[tex]h=3\sqrt{3}m[/tex]
The area of the parallelogram is [tex]=10\times3\sqrt{3}m^2[/tex]
[tex]=30\sqrt{3}m^2[/tex]
[tex]=52.0m^2[/tex] to the nearest tenth.
