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What is the area of an isosceles trapezoid that has bases of length 22 and 9 centimeters and legs of length 14 centimeters?
Wiki User
∙ 10y ago
Vertical height = sqrt{142 - [(22-9)/2]2} = sqrt{142 - 6.52} = sqrt(153.75) = 12.40 cm (approx).
So area = 1/2*(22+9)*12.40 = 192.19 cm2.
Vertical height = sqrt{142 - [(22-9)/2]2} = sqrt{142 - 6.52} = sqrt(153.75) = 12.40 cm (approx).
So area = 1/2*(22+9)*12.40 = 192.19 cm2.
Vertical height = sqrt{142 - [(22-9)/2]2} = sqrt{142 - 6.52} = sqrt(153.75) = 12.40 cm (approx).
So area = 1/2*(22+9)*12.40 = 192.19 cm2.
Vertical height = sqrt{142 - [(22-9)/2]2} = sqrt{142 - 6.52} = sqrt(153.75) = 12.40 cm (approx).
So area = 1/2*(22+9)*12.40 = 192.19 cm2.
Wiki User
∙ 10y ago
Wiki User
∙ 10y agoVertical height = sqrt{142 - [(22-9)/2]2} = sqrt{142 - 6.52} = sqrt(153.75) = 12.40 cm (approx).
So area = 1/2*(22+9)*12.40 = 192.19 cm2.
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