Question

In fig. , \[ar(DRC)=ar(DPC)\] and \[ar(BDP)=ar(ARC)\]; show that both the quadrilaterals ABCD and DCPR are trapeziums.

Answer 1

Hint: In geometry, a quadrilateral can be defined as a closed, two dimensional shape which has four straight lines. Here basically the quadrilateral is trapezoid whose opposite sides are parallel and adjacent angles add up to \[\text{180}{}^\circ \].

Complete step-by-step solution -
A trapezoid, also known as a trapezium, is a flat closed shape having four parallel sides. The parallel sides of a trapezium are known as the bases, and its non – parallel sides are called rays.
A trapezium can also have parallel rays.
it is given that, Area of triangle \[\text{( }\!\!\Delta\!\!\text{ DRC)=}\]Area of triangle \[\text{( }\!\!\Delta\!\!\text{ DPC)}\] and also given that
Area of triangle \[\text{( }\!\!\Delta\!\!\text{ BDP)=}\]Area of triangle \[\text{( }\!\!\Delta\!\!\text{ ARC)}\]
Basically triangle area in terms of base and triangle and height
The area of triangle is:
Area \[\text{=}\dfrac{1}{2}(b\times h)\], where b is the base of the triangle and h is the height of the triangle.
The base of the triangle is any one of the sides, and the height of the triangles is the length of the altitude from the opposite vertex to that base.
As, Area of triangle \[\text{( }\!\!\Delta\!\!\text{ DRC)=}\]Area of triangle \[\text{( }\!\!\Delta\!\!\text{ DPC)}\] then triangle \[\Delta DRC\] and the triangle \[\Delta DRC\] lies on the same base DC and have equal areas,
Therefore, they must lie between the same parallel lines.
\[\text{DC}~\left| \text{ } \right|~RP\]
Therefore, DCPR is a trapezium.
[Two triangles using the same base and equal areas lie between the same parallels]
Here DCPR is a trapezium (prove)
Therefore, it is given that,
Area of triangle \[\text{( }\!\!\Delta\!\!\text{ BDP)=}\]Area of triangle \[\text{( }\!\!\Delta\!\!\text{ ARC)}\]
As Area of triangle \[\text{( }\!\!\Delta\!\!\text{ DRC)=}\]Area of triangle \[\text{( }\!\!\Delta\!\!\text{ DPC)}\]
Then we assume that,
Area of triangle \[\text{( }\!\!\Delta\!\!\text{ BDP)=}\] Area of triangle \[\text{( }\!\!\Delta\!\!\text{ ARC)}\] …. 1
And Area of triangle \[\text{( }\!\!\Delta\!\!\text{ DRC)=}\] Area of triangle \[\text{( }\!\!\Delta\!\!\text{ DPC)}\] …. 2
If we subtracting 1 and 2 such that
Then we get,
Area of triangle \[\text{( }\!\!\Delta\!\!\text{ BDP)}-\] Area of triangle \[\text{( }\!\!\Delta\!\!\text{ DPC)=}\] Area of triangle \[\text{( }\!\!\Delta\!\!\text{ ARC)}-\] Area of triangle \[\text{( }\!\!\Delta\!\!\text{ DRC)}\]
\[\Rightarrow \] Area of \[\text{ }\!\!\Delta\!\!\text{ BDP=}\] Area of \[\text{ }\!\!\Delta\!\!\text{ ADC}\]
[area of \[\text{ }\!\!\Delta\!\!\text{ BDP}-\] area of triangle \[\text{( }\!\!\Delta\!\!\text{ DPC)=}\] Area of triangle \[\text{( }\!\!\Delta\!\!\text{ ADC)}\]]
[Area of triangle \[\text{( }\!\!\Delta\!\!\text{ ARC)}-\] Area of triangle \[\text{( }\!\!\Delta\!\!\text{ DRC)=}\] area of \[\text{( }\!\!\Delta\!\!\text{ ADC)}\]]
As \[\text{( }\!\!\Delta\!\!\text{ BDC)}\] and \[\text{( }\!\!\Delta\!\!\text{ ADC)}\] are on the same base CD and have equal areas, they must lie between the same parallel lines.
\[AB~\left| \text{ } \right|~CD\]
Therefore, ABCD is a trapezium.
[Two triangles having the same base and equal area lie between the same parallels.]
Since one pair of opposite sides of quadrilateral ABCD are parallel. Hence, ABCD is a trapezium.

Note: In general another way to prove this problem that,
1. In fig. \[ar(DRC)=ar(DPC)\] and \[ar(BDP)=ar(ARC)\] ……. 2
\[\therefore \text{ar( }\!\!\Delta\!\!\text{ DRC)+ar( }\!\!\Delta\!\!\text{ DPC)=}\,\text{ar( }\!\!\Delta\!\!\text{ BDP)+ar( }\!\!\Delta\!\!\text{ ARC)}\]
But \[\text{ }\!\!\Delta\!\!\text{ DRC= }\!\!\Delta\!\!\text{ DPC}\]
\[\text{ar( }\Delta \text{ DCB)=ar( }\Delta \text{ DCA)}\]
They are on the same base DC and in between straight lines DC and AB. As \[\text{DC}~\left| \text{ } \right|~AB\] then ABCD is trapezium.
2. \[\text{ }\!\!\Delta\!\!\text{ DRC}\]and \[\text{ }\!\!\Delta\!\!\text{ DPC}\] are on the same base DC and in between DC and RP and they are equal in area.
\[\therefore \text{DC}~\left| \text{ } \right|~RP\]
Hence, DCPR is a trapezium.
3. Always we have to mind it that two triangles having the same base and equal areas lie between the same parallels.