Respuesta :
Answer:
[tex]\dfrac{S_1+S_3}{2}\ un^2.[/tex]
Step-by-step explanation:
Consider trapezoid ABCD. Let BC=a, AD=b and the height of trapezoid be h. Then segment
[tex]EF=\dfrac{a+b}{3},\\ \\GH=\dfrac{2(a+b)}{3}[/tex]
and trapezoids EBCF, GEFH, AGHD have the sqme height [tex]\dfrac{h}{3}.[/tex]
The area of trapezoid EBCF is
[tex]S_1=\dfrac{a+\frac{a+b}{3}}{2}\cdot \dfrac{h}{3}=\dfrac{(4a+b)h}{18}\ un^2.,[/tex]
the area of trapezoid GEFH is
[tex]S_2=\dfrac{\frac{a+b}{3}+\frac{2(a+b)}{3}}{2}\cdot \dfrac{h}{3}=\dfrac{(a+b)h}{6}\ un^2.,[/tex]
the area of trapezoid AGHD is
[tex]S_3=\dfrac{\frac{2(a+b)}{3}+b}{3}}{2}\cdot \dfrac{h}{3}=\dfrac{(2a+5b)h}{18}\ un^2.[/tex]
Now,
[tex](4a+b)h=18S_1,\\ \\(2a+5b)h=18S_3.[/tex]
Multiplying the second equation by 2 and subtracting the first equation, you get
[tex]9bh=36S_3-18S_1,\\ \\bh=4S_3-2S_1.[/tex]
Multiplying the first equation by 5 and subtracting the second equation, you get
[tex]18ah=90S_1-18S_3,\\ \\ah=5S_1-S_3.[/tex]
Hence,
[tex]S_2=\dfrac{(a+b)h}{6}=\dfrac{ah+bh}{6}=\dfrac{5S_1-S_3+4S_3-2S_1}{6}=\dfrac{3S_1+3S_3}{6}=\dfrac{S_1+S_3}{2}\ un^2.[/tex]
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Answer:
Step-by-step explanation:
Given that Two lines, parallel to the bases of a trapezoid divide each leg into three congruent parts and divide the trapezoid into three parts as well.
We find that trapezium area is
1/2 h (b1+b2) where h is the altitude and b1 and b2 are bases of trapezium.
Here we have 3 trapeziums as top, middle and bottom.
Since parallel lines making equal intercepts are drawn we have altitude of each divided trapezium would be the same.
If original trapezium has base 1 as a and base 2 as b, then the middle lines would have length as
[tex]b_1 +\frac{b_2-b_1}{3} =\frac{2_b1+b_2}{3} and\\b_1 +2\frac{b_2-b_1}{3} =\frac{_b1+2b_2}{3}[/tex]
So S2 will have area as averge of the top and bottom.
Area of middle part = [tex]\frac{S_1+S_3}{2}[/tex]