InterviewSolution
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InterviewSolution
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In the adjoining figure D, E and F are the mid-points of the sides BC, CA and AB respectively of `Delta ABC`. Prove that: (i) `square BDEF` is a parallelogram (ii) area of `Delta DEF = (1)/(4) xx " area of " Delta ABC` (iii) area of `square BDEF = (1)/(2) xx " area of " Delta ABC` |
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Answer» (i) In `Delta ABC` Since, F is the mid-point of AB and E is the mid-point of AC `{:( :." "FE ||BC),(rArr" "FE ||BD),("Similarly "ED||FB):}` (mid - point theorem) `:. Square BDEF` is a parallelogram (`:.` pairs of opposite sides are parallel) (ii) Since, `square BDEF` is a parallelogram and FD is its diagonal `:.` area of `Delta FBD = " area of " Delta DEF`..(1) Similarly, we can prove that area of `Delta CDE = " area of " Delta DEF`...(2) and area of `Delta AFE = " area of " Delta DEF`...(3) `:. 4 xx " area of " Delta DEF = " area of "Delta DEF + " area of " Delta DEF + " area of "Delta DEF + " area of "Delta DEF` `= " area of " Delta DEF + " area of "Delta AEF + " area of "Delta FBD + " area of "Delta CDE` = area of `Delta ABC` `rArr " area of " Delta DEF = (1)/(4) xx " area of " Delta ABC` (iii) Now, area of `square BDEF = 2 xx " area of " Delta ABC` (`:.` diagonal divides the parallelogram into two equal areas) `= 2 xx (1)/(4) xx " area of " Delta ABC` `= (1)/(2) xx " area of "Delta ABC` |
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