Aside of a triangle is divided into three congruent parts. two lines, parallel to another side of the triangle, are drawn through each dividing point. find the area of the quadrilaterals formed by those lines if the area of the original triangle is 24.

Step-by-step explanation:

Suppose the base and height of the given triangle are represented by "b" and "h". Then the area of that triangle is ...

  A = (1/2)bh = 24 . . . . . . given

  bh = 48 . . . . . . . . . . . . . multiply by 2

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Now the larger quadrilateral is a trapezoid with height h/3 and bases b and (2/3)b. Its are will then be ...

  A = (1/2)(b1 +b2)h = (1/2)(b +(2/3)b)(1/3h) = 1/6·5/3·bh

Substituting the above value of bh, we find ...

  area of larger quadrilateral = (5/18)(48) = 40/3 = 13 1/3

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In similar fashion, the smaller quadrilateral is a trapezoid with bases 2/3b and 1/3b and height 1/3h. Its area is then ...

  A = (1/2)(2/3b +1/3b)(1/3h) = 1/6·bh

As before, substituting the value of bh, we find ...

  area of smaller quadrilateral = (1/6)(48) = 8

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A geometry application confirms these values.

y/x of point b= y/x of point a

step-by-step explanation:

(i got the same question, this was the correct answer.)