InterviewSolution
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InterviewSolution
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ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. If `ar (DeltaDFB) = 3 cm^(2)`, then find the area of the parallelogram ABCD. |
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Answer» Given, ABCD is a parallelogram and CE = BC i.e., C is the mid-point of BE. Also, `" " ar (DeltaDFB) = 3 cm^(2)` Now, `DeltaADF` and `DeltaDFB` are on the same base DF and between parallels CD and AB. Then, `ar(DeltaADF) = ar (DeltaDFB) = 3 cm^(2)" "` ...(i) In `DeltaABE`, by the converse of mid-point theorem, `EF = AF" "` [since, C is mid-point of BE] ...(ii) In `DeltaADF` and `DeltaECF`, `angleAFD = angleCFE" "` [Vertically opposite angles] `AF = EF" "` [from Eq. (ii)] and `" " angleDAF = angleCEF` [since, `BE || AD` and AE is transversal, then alternate interior angles are equal] `therefore" "` `DeltaADF ~=DeltaECF" "` [by ASA congruence rule] Then, `" " ar (DeltaADF) = ar (DeltaCFE)" "` [since, congruent figures have equal area] `therefore" "` `ar (DeltaCFE) = ar (DeltaADF) = 3 cm^(2)" "` [from Eq. (i)] ...(iii) Now, in `DeltaBFE`, C is the mid-point of BE then CF is median of `DeltaBFE`, `therefore" "` `ar (DeltaCEF) = ar (DeltaBFC)" "` [since, median of a triangle divides it into two triangles of equal area] `rArr" "` `ar (DeltaBFC) = 3 cm^(2)" "` ...(iv) Now, `" " ar (DeltaBDC) = ar (DeltaDFB) + ar (DeltaBFC)" "` [from Eqs. (i) and (iv)] We know that, diagonal of a parallelogram divides it into two congruent triangles of equal areas. `therefore" "` Area of parallelogram `ABCD = 2 xx "Area of " DeltaBDC` `= 2 xx 6 = 12 cm^(2)` Hence, the area of parallelogram ABCD is `12 cm^(2)` . |
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