Question

$please \: solve \: this$

ᴘ ᴀɴᴅ ϙ ᴀʀᴇ ᴀɴʏ ᴛᴡᴏ ᴘᴏɪɴᴛs ʟʏɪɴɢ ᴏɴ ᴛʜᴇ sɪᴅᴇs ᴅᴄ ᴀɴᴅ ᴀᴅ ʀᴇsᴘᴇᴄᴛɪᴠᴇʟʏ ᴏғ ᴀ ᴘᴀʀʀᴀʟᴇɢʀᴀᴍ ᴀʙᴄᴅ.sʜᴏᴡ ᴛʜᴀᴛ ᴀʀ( ᴀᴘʙ)=ᴀʀ(ʙϙᴄ).






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Answer 1

Answer:

If a parallelogram and a triangle are on the same base and between the same parallels then the area of the triangle is half the area of the parallelogram.

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Given, let p and q be the points on the sides DC and AD of the parallelogram ABCD.

To pove: ar(APB) =ar(BQC)

Proof : here in triangle APB and parallelogram ;

which have the same base AB and are between same parallels AB and DC

Therefore ; ar (APB) = 1/2 ar(ABCD) ____[1]

Similarly,

In Triangle BQC and parallelogram; which have the same base BC and are between the same parallels BC and AD.

Therefore; ar(BQC) =1/2 ar(ABCD) ____[2]

FROM EQUATION 1 AND 2; WE GET

ar (APB) =ar (BQC)

Hope it helps you!

Answer 2

Answer:

A ( △APB ) = A ( △BQC )

Step-by-step-explanation:

NOTE: Refer to the attachment for the diagram.

In figure,

□ABCD is a parallelogram.

Points P and Q are on the sides DC & AD respectively.

Join AP & BP.

Draw QM ⊥ BC and PN ⊥ AB. - - [ Construction ]

Now, we know that,

Area of parallelogram = Base * Height

⇒ A ( □ABCD ) = BC * QM - - ( 1 )

Also, by taking base AB and height PN,

A ( □ABCD ) = AB * PN - - ( 2 )

Now, in △BQC,

QM ⊥ BC - - [ Construction ]

∴ A ( △BQC ) = ½ * BC * QM

⇒ A ( △BQC ) = ½ * ( BC * QM )

⇒ A ( △BQC ) = ½ * A ( □ABCD ) - - ( 3 ) [ From ( 1 ) ]

Now, in △APB,

PN ⊥ AB - - [ Construction ]

∴ A ( △APB ) = ½ * AB * PN

⇒ A ( △APB ) = ½ * ( AB * PN )

⇒ A ( △APB ) = ½ * A ( □ABCD ) - - ( 4 ) [ From ( 2 ) ]

Now,

A ( △BQC ) = A ( △APB ) - - [ From ( 3 ) & ( 4 ) ]

∴ A ( △APB ) = A ( △BQC )

Hence shown!