Area of iso-trap

Find the area of an isosceles trapezoid if the lengths of its bases are 16 cm and 30 cm, and the diagonals are perpendicular to each other.

Correct result:

A =  529 cm2

Solution:

c=16 cm a=30 cm  a2=x2+x2 x=a/2=30/2≐15 2 cm≐21.2132 cm  c2=y2+y2 y=c/2=16/2≐8 2 cm≐11.3137 cm  d=x+y=21.2132+11.3137≐23 2 cm≐32.5269 cm  h1=x2−(a/2)2=21.21322−(30/2)2=15 cm h2=y2−(c/2)2=11.31372−(16/2)2=8 cm  h=h1+h2=15+8=23 cm  A1=a+c2⋅ h=30+162⋅ 23=529 cm2  A=d⋅ d2=32.5269⋅ 32.52692=529 cm2c=16 \ \text{cm} \ \\ a=30 \ \text{cm} \ \\ \ \\ a^2=x^2 + x^2 \ \\ x=a / \sqrt{ 2 }=30 / \sqrt{ 2 } \doteq 15 \ \sqrt{ 2 } \ \text{cm} \doteq 21.2132 \ \text{cm} \ \\ \ \\ c^2=y^2 + y^2 \ \\ y=c / \sqrt{ 2 }=16 / \sqrt{ 2 } \doteq 8 \ \sqrt{ 2 } \ \text{cm} \doteq 11.3137 \ \text{cm} \ \\ \ \\ d=x+y=21.2132+11.3137 \doteq 23 \ \sqrt{ 2 } \ \text{cm} \doteq 32.5269 \ \text{cm} \ \\ \ \\ h_{1}=\sqrt{ x^2-(a/2)^2 }=\sqrt{ 21.2132^2-(30/2)^2 }=15 \ \text{cm} \ \\ h_{2}=\sqrt{ y^2-(c/2)^2 }=\sqrt{ 11.3137^2-(16/2)^2 }=8 \ \text{cm} \ \\ \ \\ h=h_{1}+h_{2}=15+8=23 \ \text{cm} \ \\ \ \\ A_{1}=\dfrac{ a+c }{ 2 } \cdot \ h=\dfrac{ 30+16 }{ 2 } \cdot \ 23=529 \ \text{cm}^2 \ \\ \ \\ A=\dfrac{ d \cdot \ d }{ 2 }=\dfrac{ 32.5269 \cdot \ 32.5269 }{ 2 }=529 \ \text{cm}^2

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