Question

4. In trapezium ABCD, M and N are mid-points
of AC and BD, and AB II CD. Find
(1) MN if AB = 7.5 cm, CD = 5.5 cm
(ii) AB if CD = 16 cm, MN = 10 cm
(iii) CD if AB = 14 cm, MN = 16.5 cm
​

Answer 1

Given :- . In trapezium ABCD, M and N are mid-points

of AC and BD, and AB II CD. Find

(1) MN if AB = 7.5 cm, CD = 5.5 cm

(ii) AB if CD = 16 cm, MN = 10 cm

(iii) CD if AB = 14 cm, MN = 6.5 cm

Concept used :-

• In Trapezium the line segment joining the midpoints of the diagonals of a trapezium is parallel to the parallel sides and equal to half of their difference.
• MN = (1/2)[AB - CD] .

Solution :-

(i) MN if AB = 7.5 cm, CD = 5.5 cm

since,

→ MN = (1/2)[AB - CD] .

so,

→ MN = (1/2)[7.5 - 5.5]

→ MN = (1/2) * 2

→ MN = 1 cm.

(ii) AB if CD = 16 cm, MN = 10 cm

→ MN = (1/2)[AB - CD] .

→ 10 = (1/2)[AB - 16]

→ 20 = AB - 16

→ AB = 36 cm.

(iii) CD if AB = 14 cm, MN = 6.5 cm .

→ MN = (1/2)[AB - CD] .

→ 6.5 = (1/2)[14 - CD]

→ 13 = 14 - CD .

→ CD = 1 cm.

Learn more :-

3.

In the fig, AB || CD,FIND x.(Hint:Prove that AOB-COD).

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