# Isosceles trapezoid

Calculate the area of an isosceles trapezoid whose bases are in the ratio of 4:3; leg b = 13 cm and height = 12 cm.

Correct result:

S =  420 cm2

#### Solution:

$S = (a+c)h/2 \ \\ \ \\ a_1^2 = b^2 - h^2 \ \\ a_1 = \sqrt{b^2 - h^2} = \sqrt{ 13^2 - 12^2} = 5 \ cm \ \\ a = c+2a_1 \ \\ a/c = 4/3 \ \\ (c+2a_1)c = 4/3 \ \\ 3(c+2a_1) = 4c \ \\ 6a_1 = c \ \\ c = 6a_1 = 6 \cdot 5 = 30 \ cm \ \\ a = c + 2a_1= 30 + 2\cdot 5 = 40 \ cm \ \\ \ \\ S = (a+c)h/2 = (40 + 30) \cdot 12 /2 = 420 \ \text{cm}^2 \ \\$

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