Question

In a parallelogram ABCD, E and F are the mid-points of the sides AB and CD respectively. AF and CE meet the diagonal BD of length 12 cm at P and Q. Then length of PQ is
a) 6 cm b) 4 cm c) 3 cm d) 5 cm​

Answer 1

Answer:

a) 6 cm is the answer please mark me the brainliest pls mark me the brainliest

Answer 2

$\pink {\large {\sf { Answer:} }}$

b) 4 cm

$\pink {\large {\sf { Step-by-step \: explanation:} }}$

Given:

☛ ABCD is || gm

☛ E and F are the mid-points of the sides AB and CD.

☛ AF and CE meet the diagonal BD

☛ BD = 12 cm

To Find:

The length of PQ.

Solution:

☛ ABCD is || gm

So, AB || DC and AB = DC

☛ E and F are the mid-points of the sides AB and CD.

Now, AB = DC

➯ $\sf\dfrac{1}{2} AB = \dfrac{1}{2} DC$

➯ AE = FC also, AE || FC

∴ AECF is a || gm.

⇰ EC || AF

⇰ EQ || AP

⇰ QC || PF

So, In ∆BPA, EQ is a line joining the midpoint E of AB and parallel to PA.

Then, by converse of midpoint theorem, Q bisects BP.

➨ BQ = QP .......①

Similarly, by taking ∆CQD, we can prove that

➨ DP = QP ........②

From ① and ②

BQ = QP = DP

AF and CE trisect the diagonal BD.

➩ $\sf{PQ = \dfrac{1}{3} BD}$

➩ $\sf{PQ = \dfrac{1}{3} \times 12}$

➩ $\sf{PQ = \dfrac{1}{ \cancel3} \times \cancel {12 }^{4} }$

➩ 4 cm

∴ The length of PQ = 4 cm.

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