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:

$$\begin{gathered} S=(a+c)h\mathrm{/}2 \\ {a}_1^2=b^2-h^2 \\ {a}_1=\sqrt{b^2-h^2}=\sqrt{13^2-12^2}=5cm \\ a=c+2a_1 \\ a\mathrm{/}c=4\mathrm{/}3 \\ (c+2a_1)c=4\mathrm{/}3 \\ 3(c+2a_1)=4c \\ 6a_1=c \\ c=6a_1=6\cdot 5=30cm \\ a=c+2a_1=30+2\cdot 5=40cm \\ S=(a+c)h\mathrm{/}2=(40+30)\cdot 12\mathrm{/}2=420{\text{cm}}^2 \end{gathered}$$

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