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Areas Of Parallelograms And Triangles — Maths STD 9 — Question

Maharashtra BoardEnglish MediumSTD 9MathsAreas Of Parallelograms And Triangles5 Marks
Question
In figure, ABCD and AEFD are two parallelograms. Prove that:
  1. $\text{PE}=\text{FQ}$
  2. $\text{ar}(\triangle\text{APE}) : \text{ar}(\triangle\text{PFA})\\=\text{ar}(\triangle\text{QFD}): \text{ar}(\triangle\text{PFD})$
  3. $\text{ar}(\triangle\text{PEA})=\text{ar}(\triangle\text{QFD})$
✓

Answer

Given that, ABCD and AEFD are two parallelograms
  1. In triangles, EPA and FQD
$\angle\text{PEA}=\angle\text{QFD}$ [corresponding angles]

$\angle\text{EPA}=\angle\text{FQD}$ [corresponding angles]

PA = QD [opposite sides of parallelogram]

Then, $\triangle\text{EPA}\cong\triangle\text{FQD}$ [By AAS condition]

Therefore, EP = FQ [C.P.C.T]
  1. Since triangles, PEA and QFD stand on equal bases PE and FQ lies between the same parallels EQ and AD
Therefore, $\text{ar}(\triangle\text{PEA})=\text{ar}(\triangle\text{QFD})\ ...(1)$

Since, triangles PEA and PFD stand on the same base PF and between same parallels PF and AD

Therefore, $\text{ar}(\triangle\text{PFA})=\text{ar}(\triangle\text{PFD})\ ...(2)$

Divide the equation 1 by equation 2

$\text{ar}(\triangle\text{PEA})\text{ ar}(\triangle\text{PFA})=\text{ar}(\triangle\text{QFD})\text{ ar}(\triangle\text{PFD})$
  1. From part (i),
$\triangle\text{EPA}\cong\triangle\text{FQD}$

Then, $\text{ar}(\triangle\text{PEA})=\text{ar}(\triangle\text{QFD}).$.

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