Question

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. Find the area of the middle part if the top and bottom parts have area S1 and S3.

Answer 1

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

`b_1 +(b_2-b_1)/(3) =(2_b1+b_2)/(3) and\\b_1 +2(b_2-b_1)/(3) =(_b1+2b_2)/(3)`

So S2 will have area as averge of the top and bottom.

Area of middle part =
`(S_1+S_3)/(2)`

Answer 2

`(S_1+S_3)/(2)\ un^2.`

Explanation:

Consider trapezoid ABCD. Let BC=a, AD=b and the height of trapezoid be h. Then segment

`EF=(a+b)/(3),\\ \\GH=(2(a+b))/(3)`

and trapezoids EBCF, GEFH, AGHD have the sqme height
`(h)/(3).`

The area of trapezoid EBCF is

`S_1=(a+(a+b)/(3))/(2)\cdot (h)/(3)=((4a+b)h)/(18)\ un^2.,`

the area of trapezoid GEFH is

`S_2=((a+b)/(3)+(2(a+b))/(3))/(2)\cdot (h)/(3)=((a+b)h)/(6)\ un^2.,`

the area of trapezoid AGHD is

`S_3=((2(a+b))/(3)+b)/(3)}{2}\cdot (h)/(3)=((2a+5b)h)/(18)\ un^2.`

Now,

`(4a+b)h=18S_1,\\ \\(2a+5b)h=18S_3.`

Multiplying the second equation by 2 and subtracting the first equation, you get

`9bh=36S_3-18S_1,\\ \\bh=4S_3-2S_1.`

Multiplying the first equation by 5 and subtracting the second equation, you get

`18ah=90S_1-18S_3,\\ \\ah=5S_1-S_3.`

Hence,

`S_2=((a+b)h)/(6)=(ah+bh)/(6)=(5S_1-S_3+4S_3-2S_1)/(6)=(3S_1+3S_3)/(6)=(S_1+S_3)/(2)\ un^2.`