The area and perimeter of an isosceles trapezoid with a 60° base angles and bases 9 and 13 is 38.105 squared units and 24 units respectively.
The area of the isosceles trapezoid is the space occupied by it. It can be find out using the following formula,
[tex]A=\dfrac{a+b}{2}\times h[/tex]
Permiter of the isosceles trapezoid is the total length of the boundary by which it is enclosed. It can given as,
[tex]P=a+b+2c[/tex]
Here (a,b) are the base side (c) is the side of leg and (h) is the height.
The image of the given isosceles trapezoid is attached below. Let the value of leg is x units. Thus using right angle property the cos theta is,
[tex]\cos60=\dfrac{2}{x}\\x=1[/tex]
And the height of this trapezoid is,
[tex]\tan60\dfrac{h}{2}\\h=3.46[/tex]
Thus the area of the solid is,
[tex]A=\dfrac{9+13}{2}\times 3.46\\A=38.105\rm\; units^2[/tex]
The perimeter of the solid is,
[tex]P=9+13+2\times1\\P=24\rm\; units[/tex]
Thus, the area and perimeter of an isosceles trapezoid with a 60° base angles and bases 9 and 13 is 38.105 squared units and 24 units respectively.
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