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InterviewSolution
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A rectangular yard contains two flower beds in the shape of congruent isosceles right triangles. The remaining portion is a yard of trapezoidal shape whose parallel sides have length 15m and 25m. What fraction of the yard is occupied by the flower bed? |
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Answer» in the fig AB = 25 m and FE = 15 m ∴ DF + EC = 25 – 15= 10m But DF = EC ∴DF = EC = 5m also AD = BC = 5m Area of rectangle ∆BCD = AB x BC = 1 x b = 25 x 5 = 125. m2 Area of ∆ADF = \(\frac{1}{2}\)base × height = \(\frac{1}{2}\)5 × 5 = \(\frac{25}{2}\)m2 ∆ADF ≅ ABCE [data] Area of ∆BCE = \(\frac{25}{2}\)m2 Sum of Area of the two triangles = \(\frac{25}{2}\) + \(\frac{25}{2}\) = \(\frac{50}{2}\) = 25m2 \(\frac{Area\; of\; 2\; triangles}{Area\; of\; rectangles}\) = \(\frac{25}{125} = \frac{1}{5}\) ∴ \(\frac{1}{5}\) of the field is occupied by the flower bed. |
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