Question

$\bf \huge {\underline {\underline \red{QuEsTiOn}}}$

P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (APB) = ar (BQC).​

Answer 1

Step-by-step explanation:

Since △APB and parallelogram ABCD are on the same base AB and between the same parallels.

Therefore,

ar(△APB)=21ar(∥gmABCD)       .......(1)

Similarly,

ar(△BQC)=21ar(∥gmABCD)     ......(2)

From 1 and 2, we get,

ar(△APB)=ar(△BQC)

have a nice day sister

Answer 2

Answer:

$\huge{ \underline{ \underline{ \boxed{ \boxed{ \boxed{ \pink{answer}}}}}}}$

$\huge \underline \mathfrak{ \red{given}}$

➡ ABCD is a parallelogram

➡P and Q are any two points lying on the sides DC and AD.

$\huge \underline \mathfrak{ \blue{to \: prove}}$

➡ ar(APB) = ar (BQC).

$\huge \underline \mathfrak{ \green{proof}}$

➡ ABCD is a parallelogram .

So, AB||CD and AD || BC ( opposite sides of parallelogram are equal)

$➡ar \: (∆APB) \: = \frac{1}{2} ar(||gm ABCD) -----(1) \\ ➡Similarly \: (∆BQC) = \frac{1}{2} ar(||gm ABCD) ----(2) \\ ➡from (1) and (2) we \: have \\ ➡ ar(∆APB) = ar (∆BQC) \: (Hence \: proved)$

Step-by-step explanation:

Hope I help you!!