ABCD is a trapezium in which AB ∥ DC, M is mid-point of AD and N is

1.	ABCD is a trapezium in which AB ∥ DC, M is mid-point of AD and N is mid-point of BC. If the length of AB is 71.5 cm and the length of CD is 52 cm, then find the length of MN.1. 59.50 cm2. 61.75 cm3. 63.50 cm4. 55.25 cm
Answer» Correct Answer - Option 2 : 61.75 cm Given: ABCD is a trapezium, AB ∥ DC M is mid-point of AD N is mid-point of BC AB = 71.5 cm CD = 52 cm Formula Used: MN = 1/2 × (AB + CD) In a trapezium AB ∥ DC And M and N are mid-points of AD and BC respectively. Calculation: MN = 1/2 × (AB + CD) ⇒ MN = 1/2 × (71.5 + 52) ⇒ MN = 1/2 × 123.5 ⇒ MN = 61.75 cm ∴ The length of MN is 61.75 cm.

ABCD is a trapezium in which AB ∥ DC, M is mid-point of AD and N is mid-point of BC. If the length of AB is 71.5 cm and the length of CD is 52 cm, then find the length of MN.1. 59.50 cm2. 61.75 cm3. 63.50 cm4. 55.25 cm

Answer» Correct Answer - Option 2 : 61.75 cm

Given:

ABCD is a trapezium, AB ∥ DC

M is mid-point of AD

N is mid-point of BC

AB = 71.5 cm

CD = 52 cm

Formula Used:

MN = 1/2 × (AB + CD)

In a trapezium AB ∥ DC

And M and N are mid-points of AD and BC respectively.

Calculation:

MN = 1/2 × (AB + CD)

⇒ MN = 1/2 × (71.5 + 52)

⇒ MN = 1/2 × 123.5

⇒ MN = 61.75 cm

∴ The length of MN is 61.75 cm.

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ABCD is a trapezium in which AB ∥ DC, M is mid-point of AD and N is mid-point of BC. If the length of AB is 71.5 cm and the length of CD is 52 cm, then find the length of MN.1. 59.50 cm2. 61.75 cm3. 63.50 cm4. 55.25 cm

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