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Help need now Find to the nearest tenth the perimeter of a quadrilateral with vertices A 2 1 B7 3 C12 1 and D 7 -4 and what is shape's most discriptive name Please show work or expln?
Wiki User
∙ 15y ago
A = (2, 1) B = (7, 3) Therefore, AB^2 = (7-2)^2 + (3-1)^2 = 25 + 4 = 29 and so AB = sqrt(29) units.
You can find BC, CD, and DA similarly.
You will find that AB = BC and CD = DA but that BC is not = CD.
Therefore the shape is a kite (two pairs of adjacent sides equal).
And perimeter = 2*sqrt(29) + 2*sqrt(50) = 24.9 units.
Wiki User
∙ 8y ago
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