QUESTION 1
The figure is a tra-pezoid.
The area of a trapezoid is given by the formula;
[tex]A=\frac{1}{2}(Sum\:of\:parallel\:sides)\times h[/tex]
We substitute the values to obtain;
[tex]Area=\frac{1}{2}(13.2cm+8.4cm)\times 6cm[/tex]
[tex]Area=\frac{1}{2}(21.6cm)\times 6cm[/tex]
[tex]Area=64.8cm^2[/tex]
QUESTION 2
The approximate length of the ribbon is equal to the circumference of the circular table cloth.
We can find this by using the formula for calculating the circumference of a circle.
[tex]C=2\pi r[/tex]
where [tex]r=3.5ft[/tex] is the radius of the circular table cloth.
We substitute this value and [tex]\pi=3.14[/tex] into the formula to get;
[tex]C=2(3.14)(3.5)ft[/tex]
[tex]C=21.98ft[/tex]
The approximate length of the ribbon is 22.0ft to the nearest tenth.
QUESTION 3
Type of quadrilateral:Rectangle
Explanation: A rectangle has 4 angles that are right angles.
The two pairs of opposite sides of a rectangle are also congruent.
QUESTION 4.
The circumference of a semi circle is calculated using the formula;
[tex]C=\pi r[/tex]
where [tex]r=12.5cm[/tex] is the radius of the semicircle.
[tex]C=3.14\times 12.5cm[/tex]
[tex]C=39.25cm[/tex]
QUESTION 5
The area of the entire figure is the area of the semicircle plus the area of the isosceles triangle.
[tex]Area=\frac{1}{2}\pi r^2+\frac{1}{2}bh[/tex]
[tex]Area=\frac{1}{2}\times3.14\times12.5^2+\frac{1}{2}\times25\times 24[/tex]
[tex]Area=345.3125+300[/tex]
[tex]Area=645.3125cm^2[/tex]
[tex]Area\approx645cm^2[/tex]
QUESTION 6
The area of the parallelogram is given by the formula;
[tex]A=bh[/tex]
From the diagram, [tex]b=y\:in.,h=3in.[/tex].
Given, A=23.7 inches squared.
We substitute the values to obtain;
[tex]23.7=3y[/tex]
[tex]y=\frac{23.7}{3}[/tex]
[tex]y=7.9in.[/tex]
QUESTION 7
The area of a rectangle is given by the formula;
[tex]Area=l\times b[/tex]
The area of the bigger rectangle [tex]=9\times15=135in^2[/tex]
The area of the smaller rectangle [tex]=8\times 13=104in^2[/tex]
The area of the shaded region [tex]=135-104=31in^2[/tex]
