Question

parallelograms on the same base and having equal areas like between the same parallels​

Answer 1

Step-by-step explanation:

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Answer 2

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$\huge\underline\bold\orange{Question}$

parallelograms on the same base and having equal areas like between the same parallels

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$\huge\underline\bold\orange{Solution}$

$\blue{Given}$

Two ||gms ABCD and ABMN on the same base AB such that ar(||gm ABCD) = ar(||gm ABMN).

$\green{To\:prove}$

DCNM is a straight line parallel to AB, i.e. ,

CN || AB.

$\red{Construction}$

Draw CP ⊥ AB and NQ ⊥ AB.

$\purple{proof}$

Since CP⊥AB and NQ⊥AB,we have

CP || NQ (lines perpendicular to the same line are parallel)

Now, ar(||gm ABCD) = ar(||gm ABMN)

$⇒ AB \times CP = AB \times NQ \: (area \: of \: a \: || gm \: = \: base \times height)$

$⇒ CP = NQ.$

Now, CP || NQ and CP = NQ ⇒ CPQN is a ||gm

$⇒ CN || PQ ⇒ CN || AB.$

Thus, parallelograms ABCD and ABMN lie between the same parallels.

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