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# In Fig.9.29, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums

**Solution:**

We have to prove quadrilaterals ABCD and DCPR are trapeziums.

It is given that Area (ΔDRC) = Area (ΔDPC)

As ΔDRC and ΔDPC are lying on the same base DC having equal areas, therefore, they must lie between the same parallel lines.

According to Theorem 9.3: Two triangles having the same base (or equal bases) and equal areas lie between the same parallels.

∴ DC || RP

Therefore, DCPR is a trapezium.

It is also given that Area (ΔBDP) = Area (ΔARC)

Now, subtract ar (ΔDPC) form ar (ΔBDP) and ar (ΔDRC) from ar (ΔARC)

ar (ΔBDP) - ar (ΔDPC) = ar (ΔARC) - ar (ΔDRC) [Since, ar (ΔDPC) = ar (ΔDRC)]

ar (ΔBDC) = ar (ΔADC)

Since ΔBDC and ΔADC are on the same base CD having equal areas, they must lie between the same parallel lines. (According to Theorem 9.3)

∴ AB || CD

Therefore, ABCD is a trapezium.

**☛ Check: **NCERT Solutions for Class 9 Maths Chapter 9

**Video Solution:**

## In Fig.9.29, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums

Maths NCERT Solutions Class 9 - Chapter 9 Exercise 9.3 Question 16:

**Summary:**

In the given figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC), then we have proved that ABCD and DCPR are trapeziums.

**☛ Related Questions:**

- In Fig.9.23, E is any point on median AD of a ∆ ABC. Show that ar (ABE) = ar (ACE).
- In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) = 1/4 ar(ABC).
- Show that the diagonals of a parallelogram divide it into four triangles of equal area.
- In Fig. 9.24, ABC and ABD are two triangles on the same base AB. If line- segment CD is bisected by AB at O, show that ar(ABC) = ar (ABD).

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