In trapezoid ABCD, AB is the longer base. Diagonals AC and BD intersect at E.

a. Prove triangle AEB ~(similar) triangle CED

b. If the bases of the trapezoid are 5 inches and 15 inches fid the ratio of corresponding altitudes of (triangle) CED and (triangle) AEB

c. If the altitude of the trapezoid is 8 inches, find the number of square inches in the area of AEB(triangle)

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That's what my teacher gave us to work on and its exactly what she gave us.

Wait give me 5 - 10 min I will check out some how if it happens....

Ok thank you. Because when my tutor and I did it we were having trouble figuring it out

Solution:

a) Since DC || AB in a trapezoid shown we have,

Angle BDC = Angle DBA (Alternate Angles)

Angle DCA = Angle CAB (Alternate Angles)

Angle DEC = Angle AEB (Vertically Opposite Angles)

Hence, triangle AEB ~ triangle CED

b) Ratio will be in accordance with the sides of the Triangles.

Hence Ratio of altitudes for triangle CED : triangle AEB will be 5/15 = 1 / 3

c) Altitude = 8 inch

Altitude for triangle AEB = 3/4 * 8 = 6 Inch

Area = ½ Base * Height = ½ x 15 x 6 = 45 inch^2

Ari.docx ..............Solution along with diagram for your reference.

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Thank you so much ! I'll put it into a traditional proof form. Haha when you did it it made a lot more sense

Ok.

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