# Isosceles trapezoid

In an isosceles trapezoid KLMN intersection of the diagonals is marked by the letter S. Calculate the area of trapezoid if /KS/: /SM/ = 2:1 and a triangle KSN is 14 cm2.

Result

S =  63 cm2

#### Solution:

$S_{ 1 } = 14 \ cm^2 \ \\ S_{ 2 } = S_{ 1 } = 14 = 14 \ cm^2 \ \\ S_{ 1 } = 2x v_{ 1 }/2 \ \\ S_{ 3 } = x v_{ 1 } /2 \ \\ S_{ 3 } = S_{ 1 }/2 = 14/2 = 7 \ cm^2 \ \\ S_{ 1 } = 1 \ y v_{ 2 } /2 \ \\ S_{ 4 } = 2 \ y v_{ 2 } /2 \ \\ S_{ 4 } = y v_{ 2 } = 2 \cdot \ S_{ 1 } \ \\ \ \\ S_{ 4 } = 2 \cdot \ S_{ 1 } = 2 \cdot \ 14 = 28 \ cm^2 \ \\ S = S_{ 1 }+S_{ 2 }+S_{ 3 }+S_{ 4 } = 14+14+7+28 = 63 = 63 \ cm^2$

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