Question

Prove that parallelogram on the same base and between the same parallelogram lines are equal in area?

Plz help me​

Answer 1

• ᴛʜᴇᴏʀᴇᴍ: ᴘᴀʀᴀʟʟᴇʟᴏɢʀᴀᴍs ᴏɴ ᴛʜᴇ sᴀᴍᴇ ʙᴀsᴇ ᴀɴᴅ ʙᴇᴛᴡᴇᴇɴ ᴛʜᴇ sᴀᴍᴇ ᴘᴀʀᴀʟʟᴇʟs ᴀʀᴇ ᴇϙᴜᴀʟ ɪɴ ᴀʀᴇᴀ. ... ʙᴄ = ᴀᴅ (ᴏᴘᴘᴏsɪᴛᴇ sɪᴅᴇs ᴏғ ᴀ ᴘᴀʀᴀʟʟᴇʟᴏɢʀᴀᴍ ᴀʀᴇ ᴇϙᴜᴀʟ) ∠ʙᴄᴇ ∠ ʙ ᴄ ᴇ = ∠ᴀᴅғ ∠ ᴀ ᴅ ғ (ᴄᴏʀʀᴇsᴘᴏɴᴅɪɴɢ ᴀɴɢʟᴇs) ∠ʙᴇᴄ ∠ ʙ ᴇ ᴄ = ∠ᴀғᴅ ∠ ᴀ ғ ᴅ (ᴄᴏʀʀᴇsᴘᴏɴᴅɪɴɢ ᴀɴɢʟᴇs)

Answer 2

Answer:

Step-by-step explanation:

Let ABCD and EFCD be two parallelograms lying on the same base CD and between the same parallels AF and CD.

We have to prove that area of the parallelograms ABCD and EFCD are equal.

In ADE and BCF,

DAE = CBF (Corresponding angles)

AED = BFC (Corresponding angles)

AD = BC (Opposite sides of the parallelogram ABCD)

ADE  BCF (By ASA congruence rule)

We know that congruent figures have the same area.

area (ADE) = area (BCF) … (1)

Now, we have:

area (ABCD) = area(ADE) + area (EDCB)

= area (BCF) + area (EDCB) [Using (1)]

= area (EDCF)

Thus, the areas of the two parallelograms ABCD and EFCD are the same.