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)
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