Question

Prove that :- a) Parallelograms on the same base AB and between the

same parallels are equal in area.

b) Two triangles having the same base (or equal bases)

and equal areas lie between the same parallels.​

Answer 1

Answer:

Hyy

Step-by-step explanation:

(a) Given two parallelogram ABCD and EFCD that have the same base CD and lie between same parallel AF and CD.

We have prove that ar(ABCD)=ar(EFCD)

Since opposite sides of ∥gm are parallel AB∥CD and ED∥FC with transversal AB

⇒∠DAB=∠CBF [ Corresponding angles ]

with transversal EF

⇒∠DEA=∠CFE [ Corresponding angles ]

⇒AD=BC [ Opposite sides of parallelogram are equal ]

In △AED ξ △BFC

⇒∠DEA=∠CFE

∠DAB=∠CBF

∴AD=BC

⇒△AED≅△BFC [ AAS congruency ]

Hence, ar(△AED)=ar(△BFC)

( Areas of congruent figures are equal )

⇒ar(ABCD)=ar(△ADE)+ar(EBCD)

=ar(△BFC)+ar(EBCD)

=ar(EBCD)

∴ar(ABCD)=ar(EBCD)

Hence, the answer is proved.

ANSWER

(b) Given:△ABC and △ABD are two trianles on same base AB such that area(△ABC)=area(△ABD).

Join CD

To prove:CD∥AB

Construction:Draw the altitutdes of △ABC and △ABD throught C and D and meeting base at E and F respectively.

Proof:Since CE⊥AB and DF⊥AB(by construction)

Since, lines perpendicular to same line are parallel to each other.

∴CE∥DF .......(1)

Now, area(△ABC)= 21×AB×CE

area(△ABD)= 21 ×AB×DF

Since area(△ABC)=area(△ABD)(Given)

∴CF=DF .........(2)

Now, In CDFE,CF∥DF and CE∥DF

Since one pair of opposite sides are equal and parallel

Hence CDFE is a parallelogram.

So, CD∥EF (opposite sides of parallelogram are parallel)

∴CD∥AB

Hence, proved.