Q. 74.8( 8 Votes )
A. Draw the given quadrilateral ABCD.
B. Draw the diagonal DB of quadrilateral ABCD.
C. Draw a parallel line through point C to diagonal DB of quadrilateral ABCD which intersects at Q produced AD.
D. CQ||DB, ΔABQ is the required triangle.
∆DCB = ∆DBQ (on same base DB and between same parallels DB and CQ)
∴ ∆DCB = ∆DBQ
∆ABD + ∆DCB = ∆DBQ + ∆ABD (adding area of ∆ABD on both sides)
∴ quadrilateral ABCD = ∆ABQ
Taking BQ as a base, we draw another ∆BCE with one angle 30⁰ and between the parallels DQ and BC.
∴ ∆BCE = quadrilateral ABCD
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