Trapezium diagonals

It is given trapezium ABCD with bases | AB | = 12 cm, |CD| = 8 cm. Point S is the intersection of the diagonals for which |AS| is 6 cm long. Calculate the length of the full diagonal AC.

Result

x =  10 cm

Solution:

a=12 cm c=8 cm f1=6 cm  f1:a=f2:c  f2=f1 ca=6 812=4 cm  x=f1+f2=6+4=10 cma=12 \ \text{cm} \ \\ c=8 \ \text{cm} \ \\ f_{ 1 }=6 \ \text{cm} \ \\ \ \\ f_{ 1 }:a=f_{ 2 }:c \ \\ \ \\ f_{ 2 }=f_{ 1 } \cdot \ \dfrac{ c }{ a }=6 \cdot \ \dfrac{ 8 }{ 12 }=4 \ \text{cm} \ \\ \ \\ x=f_{ 1 }+f_{ 2 }=6+4=10 \ \text{cm}



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