from a handpicked tutor in LIVE 1-to-1 classes

# ABCD is a trapezium in which AB || DC, DC = 30 cm and AB = 50 cm. If X and Y are, respectively the mid-points of AD and BC, prove that ar (DCYX) = 7/9 ar (XYBA)

**Solution:**

Given, ABCD is a trapezium

AB || DC

Given, AB = 50 cm

DC = 30 cm

X and Y are the midpoints of AD and BC

We have to prove that ar(DCYX) = 7/9 ar(XYBA)

AB is produced to P

Join DY and extend it to meet P

Considering triangles DCY and PBY,

Since Y is the midpoint of BC

CY = BY

We know that the alternate interior angles are equal

∠DCY = ∠PBY

We know that the vertically opposite angles are equal.

∠2 = ∠3

By ASA criteria, the triangles DCY and PBY are congruent.

By CPCTC,

DC = BP

Given, DC = 30 cm

So, BP = 30 cm

Now, AP = AB + BP

AP = 50 + 30

AP = 80 cm

Considering triangle ADP,

By midpoint theorem,

XY = 1/2 AP

XY = 1/2 (80)

XY = 40 cm

Let the distance between AB, XY and DC be h cm.

Area of trapezium = 1/2 × (sum of parallel sides) × (distance between parallel sides)

Area of trapezium DCYX = 1/2 × (30 + 40) × h

= 1/2 × 70 × h

= 35h square cm

Similarly, area of trapezium XYBA = 1/2 × (40 + 50) × h

= 1/2 × 90 × h

= 45h square cm

Now, ar(DCYX)/ar(XYBA) = 35h/45h

ar(DCYX)/ar(XYBA) = 7/9

Therefore, ar(DCYX) = 7/9 ar(XYBA)

**✦ Try This: **Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar(△AOD)=ar(△BOC). Prove that ABCD is a trapezium

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

**NCERT Exemplar Class 9 Maths Exercise 9.4 Problem 5**

## ABCD is a trapezium in which AB || DC, DC = 30 cm and AB = 50 cm. If X and Y are, respectively the mid-points of AD and BC, prove that ar (DCYX) = 7/9 ar (XYBA)

**Summary:**

ABCD is a trapezium in which AB || DC, DC = 30 cm and AB = 50 cm. If X and Y are, respectively, the mid-points of AD and BC, it is proven that ar (DCYX) = 7/9 ar (XYBA)

**☛ Related Questions:**

- In ∆ ABC, if L and M are the points on AB and AC, respectively such that LM || BC. Prove that ar (LO . . . .
- In Fig. 9.25, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn par . . . .
- If the medians of a ∆ ABC intersect at G, show that ar(AGB) = ar(AGC) = ar(BGC) = 1/3 ar(ABC)

visual curriculum