Diagonal

he rectangular ABCD trapeze, whose AD arm is perpendicular to the AB and CD bases, has area 15cm square. Bases have lengths AB = 6cm, CD = 4cm. Calculate the length of the AC diagonal.

Correct result:

u =  3.606 cm

Solution:

S=15 cm2 a=6 cm c=4 cm  S=(a+c)d/2  d=2⋅ S/(a+c)=2⋅ 15/(6+4)=3 cm u=d2+(a−c)2=32+(6−4)2=13=3.606 cmS=15 \ \text{cm}^2 \ \\ a=6 \ \text{cm} \ \\ c=4 \ \text{cm} \ \\ \ \\ S=(a+c)d/2 \ \\ \ \\ d=2 \cdot \ S/(a+c)=2 \cdot \ 15/(6+4)=3 \ \text{cm} \ \\ u=\sqrt{ d^2+(a-c)^2 }=\sqrt{ 3^2+(6-4)^2 }=\sqrt{ 13 }=3.606 \ \text{cm}

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