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For the next two (2) items that follow :
ABCD is a trapezium in which AB is parallel to CD. Let M be the midpoint of BC.
68. Consider the following statements :
1. 'Area of triangle ADM + Area of triangle DCM' is equal to three-fourth of the area of trapezium ABCD, if AB = CD.
2. 'Area of triangle DCM + Area of triangle ABM' is always greater than half of the area of trapezium ABCD.
Which of the above statements is/are correct?


A) 1 only

B) 2 only

C) Both 1 and 2

D) Neither 1 nor 2

Correct Answer:
B) 2 only

Description for Correct answer:


Let M be the midpoint of BC and let AD = 2 cm and BC = 4 cm

Now area of \( \Large \vartriangle ADM = 1cm^{2} \)

and area of \( \Large \vartriangle DCM = \frac{1}{2} \times 2 \times 1 = 1 cm^{2} \)

Area of trapezium = \( \Large \frac{1}{2} x 1 x (2 + 4) = 3 cm^{2}\)

Now, \( \Large Area of \vartriangle ADM + Area of \vartriangle DCM = 26 cm^{2}\) Area of trapezium ABCD = \( \Large 3 cm^{2} \)

Statement (1) is wrong

For Statement 2

\( \Large Area of \vartriangle DCM + Area of \vartriangle ABM = \frac{1}{2} \times 2 \times 1 + \frac{1}{2} \times 2 \times 1 \)

= 1 + 1

=2 \( \Large cm^{2} \)

Now \( \Large \frac{1}{2} \times Area of trapezium = \frac{3}{2} cm^{2} \)

\( \Large = 1.5 cm^{2} \)

2 > 1.5

Statement (2) is correct.

Option (B)is correct.

Part of solved CDS Maths(1) questions and answers : Exams >> CDSE >> CDS Maths(1)




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