Question

$\rm \red{In \: a \: paralleogram \: ABCD, E \: and \: F} \\ \rm \red{are \: the \: mid-point \: of \: Side \: AB} \\ \rm \red{and \: CD, AF \: and \: CE \: meets} \\ \rm \red{ diagonal \: BD \: of \: length \: 12 \: cm \: at \: P }\\ \rm \red{and \: Q \: Find \: PQ}$​

Answer 1

Answer:

Given :-

• In a parallelogram ABCD, E and F are the mid-point of side AB and CD, AF and CE meets diagonal BD of length 12 cm at P and Q.

To Find :-

• What is the value of PQ.

Solution :-

$\bigstar$ ABCD is a parallelogram.

So,

$\leadsto \bf AB \parallel DC$

$\leadsto \bf AB = DC$

Now,

$\bigstar$ E and F are the mid-point of side AB and CD.

So,

$\implies \bf \dfrac{1}{2}AB =\: \dfrac{1}{2}DC$

$\implies \sf AE =\: FC$

$\implies \sf AE \parallel FC$

Hence, AECF is a parallelogram.

So,

$\bullet \: \sf EC \parallel AF$

$\bullet \: \sf EQ \parallel AP$

$\bullet \: \sf QC \parallel PF$

And, In ∆BPA, EQ is the line joining and E is the mid-point of AB and parallel to PA.

By converse of mid-point theorem Q is mid-point to BP.

$\implies \sf\bold{\purple{BQ =\: QP\: ------\: (Equation\: No\: 1)}}$

Similarly,

$\implies \sf\bold{\purple{DP =\: QP\: ------\: (Equation\: No\: 2)}}$

From, the equation no 1 and equation no 2 we get,

$\implies \sf\bold{\green{BQ =\: QP =\: DP}}$

✫ AF and CE meets diagonal BD of length 12 cm at P and Q.

According to the question,

$\implies \sf PQ =\: \dfrac{1}{3}BD$

$\implies \sf PQ =\: \dfrac{1}{3} \times BD$

$\implies \sf PQ =\: \dfrac{1}{\cancel{3}} \times {\cancel{12}}$

$\implies \sf\bold{\red{PQ =\: 4\: cm}}$

$\therefore$ The value of PQ is 4 cm .

[Note : Please refer that attachment for the diagram. ]

Answer 2

Solution :-

\bigstar★ ABCD is a parallelogram.

So,

\leadsto \bf AB \parallel DC⇝AB∥DC

\leadsto \bf AB = DC⇝AB=DC

Now,

\bigstar★ E and F are the mid-point of side AB and CD.

So,

\implies \bf \dfrac{1}{2}AB =\: \dfrac{1}{2}DC⟹

2

1

AB=

2

1

DC

\implies \sf AE =\: FC⟹AE=FC

\implies \sf AE \parallel FC⟹AE∥FC

Hence, AECF is a parallelogram.

So,

\bullet \: \sf EC \parallel AF∙EC∥AF

\bullet \: \sf EQ \parallel AP∙EQ∥AP

\bullet \: \sf QC \parallel PF∙QC∥PF

And, In ∆BPA, EQ is the line joining and E is the mid-point of AB and parallel to PA.

By converse of mid-point theorem Q is mid-point to BP.

\begin{gathered}\implies \sf\bold{\purple{BQ =\: QP\: ------\: (Equation\: No\: 1)}}\\\end{gathered}

⟹BQ=QP−−−−−−(EquationNo1)

Similarly,

\begin{gathered}\implies \sf\bold{\purple{DP =\: QP\: ------\: (Equation\: No\: 2)}}\\\end{gathered}

⟹DP=QP−−−−−−(EquationNo2)

From, the equation no 1 and equation no 2 we get,

\implies \sf\bold{\green{BQ =\: QP =\: DP}}⟹BQ=QP=DP

✫ AF and CE meets diagonal BD of length 12 cm at P and Q.

According to the question,

\implies \sf PQ =\: \dfrac{1}{3}BD⟹PQ=

3

1

BD

\implies \sf PQ =\: \dfrac{1}{3} \times BD⟹PQ=

3

1

×BD

\implies \sf PQ =\: \dfrac{1}{\cancel{3}} \times {\cancel{12}}⟹PQ=

3

1

×

12

\implies \sf\bold{\red{PQ =\: 4\: cm}}⟹PQ=4cm

\therefore∴ The value of PQ is 4 cm .

[Note : Please refer that attachment for the diagram. ]