The area of a trapezium with equal non-parallel sides is 168m². If the lengths of the parallel sides are 36 m and 20 m, find the length of the non-parallel sides.
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Let ABCD be the isosceles trapezium with AB = 20 m and CD = 36 m. The area of the figure, ABCD is 168 sq m.
Draw AF and BE perpendicular to CD from the vertices A and B.
The area of the figure = (AB+CD)*h/2 = 168 [h = AF = BE], or
(20+36)*h = 2*168 = 336, or
h = 336/56 = 6 m.
Now ADF and BCE are congruent right angle triangle, whose altitude is 6 m and the base = (36–20)/2 = 16/2 = 8 m.
The equal sides of the isosceles trapezium is the same as the hypotenuse of ADF and BCF, which is [6^2 + 8^2]^0.5 = [36+64]^0.5 = 100^0.5 = 10 m.
The equal sides of the isosceles trapezium = 10 m.
Draw AF and BE perpendicular to CD from the vertices A and B.
The area of the figure = (AB+CD)*h/2 = 168 [h = AF = BE], or
(20+36)*h = 2*168 = 336, or
h = 336/56 = 6 m.
Now ADF and BCE are congruent right angle triangle, whose altitude is 6 m and the base = (36–20)/2 = 16/2 = 8 m.
The equal sides of the isosceles trapezium is the same as the hypotenuse of ADF and BCF, which is [6^2 + 8^2]^0.5 = [36+64]^0.5 = 100^0.5 = 10 m.
The equal sides of the isosceles trapezium = 10 m.
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