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A trapezoid has an area of 20 cm2 and a height z cm. The lengths of the parallel sides are (2z + 3) cm and (6z – 1) cm. Find the height, z, of the trapezoid. In your final answer, include all of the formulas and calculations necessary.

Respuesta :

check the picture below.

[tex]\bf A=\cfrac{h(a+b)}{2}\quad \begin{cases} A=20\\ a=6z-1\\ b=2z+3\\ h=z \end{cases}\implies 20=\cfrac{z[(6z-1)~+~(2z+3)]}{2} \\\\\\ 20=\cfrac{z(8z+2)}{2}\implies 20=\cfrac{2z(4z+1)}{2}\implies 20=z(4z+1) \\\\\\ 20=4z^2+z\implies 0=4z^2+z-20[/tex]

[tex]\bf \qquad \qquad \textit{quadratic formula}\\\\ \begin{array}{llccll} 0=&{{ 4}}z^2&{{ +1}}x&{{ -20}}\\ &\uparrow &\uparrow &\uparrow \\ &a&b&c \end{array} \qquad \qquad z= \cfrac{ - {{ b}} \pm \sqrt { {{ b}}^2 -4{{ a}}{{ c}}}}{2{{ a}}} \\\\\\ z=\cfrac{-1\pm\sqrt{1+320}}{8}\implies z=\cfrac{-1\pm\sqrt{321}}{8}\implies z\approx \begin{cases} \boxed{2.1146}\\ -2.3646 \end{cases}[/tex]

since the height is just a length unit, it can't be -2.3646.
Ver imagen jdoe0001
botellok

Answer:

2.11 cm

Step-by-step explanation:

As you can see in the picture I put the sides in the trapezoid. Now, let's use the area's formula:

A = [tex]\frac{(B+b)*h}{2}[/tex]

where B is the Largest parallel side, b the smallest parallel side and h is the height.

Now,

20 =  [tex]\frac{(6z-1+2z+3)*z}{2}[/tex]

20 =  [tex]\frac{(8z^{2}+2z}{2}[/tex]

20 = [tex]4z^{2}+z[/tex]

[tex]4z^{2}+z-20=0[/tex]

Then, we use the cuadratic formula z= [tex]\frac{-b+-\sqrt{b^{2}-4ac} }{2a}[/tex] where a=4, b=1 and c=-20.

z= [tex]\frac{-1+-\sqrt{1-4(4)(-20)} }{8}[/tex]

z =  [tex]\frac{-1+-\sqrt{1+320} }{8}[/tex]

z =  [tex]\frac{-1+-\sqrt{1+320} }{8}[/tex]

z =  [tex]\frac{-1+-17.91}{8}[/tex]

As we are searching lenght we choose the positive z value

z =  [tex]\frac{-1+17.91}{8}[/tex]

z=  [tex]\frac{16.91}{8}[/tex]

z =  2.11 cm.

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