# 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:

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

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