Answer :
The area of the trapezoid is maximum when the fourth side be 20 inches.
What is meant by trapezium?
Trapezoids are quadrilaterals with two parallel and two oblique sides. It is also known as a trapezium. A trapezoid is a four-sided closed shape or figure with a perimeter that covers some area.
By applying Pythagorean theorem,
[tex]c^2=a^2+b^2[/tex]
[tex]10^2=h^2+x^2[/tex]
[tex]100=h^2+x^2[/tex]
[tex]h^2=100-x^2[/tex]
[tex]h=\sqrt{100-x^2}[/tex]
Area of the trapezoid is,
[tex]A=\frac12 \times h\times (b_1+b_2)[/tex]
By substituting the values of h, b1, and b2 in the above equation,
[tex]A=\frac12 \times \sqrt{100-x^2} \times (10+2\times x+10)[/tex]
[tex]A=\sqrt{100-x^2} \times (x+10)[/tex]
If we take the derivative of both sides of the equation in relation to x, we get
[tex]\frac{dA}{dx}=\frac{d}{dx}[\sqrt{100-x^2} \times (x+10)][/tex]
[tex]\frac{dA}{dx}=\sqrt{100-x^2} \ \frac{d}{dx} (x+10) + (x+10) \ \frac{d}{dx} \sqrt{100-x^2}[/tex]
[tex]\frac{dA}{dx}=\sqrt{100-x^2} - \frac{x^2+10 x}{\sqrt{100-x^2}}[/tex]
[tex]\frac {dA}{dx}=0[/tex]
[tex]0=\sqrt{100-x^2} - \frac{x^2+10 x}{\sqrt{100-x^2}}[/tex]
[tex]\sqrt{100-x^2} = \frac{x^2+10 x}{\sqrt{100-x^2}}[/tex]
100-x²=x²+10x
2x²+10x-100=0
x²+5x-50=0
(x-5)(x+10)=0
If, x+10=0
x=-10
The above value is not accepted because it is a negative value,
If, x-5=0
x= 5
The value is accepted because it is positive value
The length of the trapezium which should be fourth is,
b₂=2x+10
b₂= 2(5)+10
b₂=20
The area of the trapezoid is maximum when the fourth side be 20 inches
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