Question

"Question 16 In the given figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

Class 9 - Math - Areas of Parallelograms and Triangles Page 164"

Answer 1

Trapezium:

A quadrilateral in which one pair of opposite sides are parallel is called a trapezium.

Two Triangles having the same base and equal areas lie between the same parallels.

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Given,

ar(DRC) = ar(DPC) .....(i)

& ar(BDP) = ar(ARC).....(ii)

To Prove,

Quadrilaterals ABCD and DCPR are trapeziums.

Proof:

On subtracting eq (I) from eq(ii),

 ar(△BDP) – ar(△DPC) = ar(△ARC)- ar(△DRC)

 ar(△BDC) = ar(△ADC)

ar(△BDC) = ar(△ADC).

Since , these two triangles have the same base DC.

 Therefore, they must lying between the same parallel lines.

Thus, AB ∥ CD

So, ABCD is a trapezium.

also,

ar(DRC) = ar(DPC).

Since , these two triangles have the same base DC.

 Therefore, they must lying between the same parallel lines.
Thus, DC ∥ PR

Hence, DCPR is a trapezium.

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Hope this will help you...

Answer 2

Given,

ar(DRC) =

ar(DPC) .....(i)

&

ar(BDP) = ar(ARC).....(ii)

To Prove,

Quadrilaterals

ABCD and DCPR are trapeziums.

Proof:

On

subtracting eq (I) from eq(ii),

 ar(△BDP) – ar(△DPC) = ar(△ARC)- ar(△DRC)

 ar(△BDC) = ar(△ADC)

ar(△BDC) = ar(△ADC).

Since ,

these two triangles have the same base DC.

 Therefore, they must lying between the same

parallel lines.

Thus, AB ∥ CD

So, ABCD

is a trapezium.

also,

ar(DRC) =

ar(DPC).

Since ,

these two triangles have the same base DC.

 Therefore, they must lying between the same

parallel lines.

Thus, DC || PR

Hence, DCPR is a trapezium.

Hope

this will help you...

jai hind....