"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"
Answers
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.
_________________________________________________________
Hope this will help you...
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....
