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The figure, given below, shows a trapezium ABCD. M and N are the mid-points of the non-parallel sides AD and BC respectively. Find :

(i) MN, if AB = 11 cm and DC = 8 cm.

(ii) AB, if DC = 20 cm and MN = 27 cm.

(iii) DC, if MN = 15 cm and AB = 23 cm.

The figure, given below, shows a trapezium ABCD. M and N are the mid-points of the non-parallel sides AD and BC respectively. Find : Mid-point Theorem, Concise Mathematics Solutions ICSE Class 9.

Mid-point Theorem

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Answer

The figure, given below, shows a trapezium ABCD. M and N are the mid-points of the non-parallel sides AD and BC respectively. Find : Mid-point Theorem, Concise Mathematics Solutions ICSE Class 9.

Join BD. Let BD intersects MN at point O.

We know that,

In trapezium the line joining the mid-points of non-parallel sides are parallel to the parallel sides of trapezium.

∴ MN || AB || DC.

By mid-point theorem,

The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and is equal to half of it.

By converse of mid-point theorem,

The straight line drawn through the mid-point of one side of a triangle parallel to another, bisects the third side.

Given,

⇒ MN || AB

⇒ MO || AB

In △ ABD,

M is mid-point of AD and MO || AB.

∴ O is mid-point of BD. (By converse of mid-point theorem)

∴ MO = 12AB\dfrac{1}{2}AB (By mid-point theorem) ……….(1)

Given,

⇒ MN || DC

⇒ ON || DC

In △ BCD,

O is mid-point of BD and N is mid-point of BC.

∴ ON = 12CD\dfrac{1}{2}CD (By mid-point theorem) ……….(2)

Adding equations (1) and (2), we get :

⇒ MO + ON = 12AB+12CD\dfrac{1}{2}AB + \dfrac{1}{2}CD

⇒ MN = 12(AB+CD)\dfrac{1}{2}(AB + CD) ……….(3)

(i) Given,

⇒ AB = 11 cm

⇒ DC = 8 cm

Substituting values in equation (3), we get :

MN=12(AB+CD)MN=12×(11+8)MN=12×19MN=9.5 cm.\Rightarrow MN = \dfrac{1}{2}(AB + CD) \\[1em] \Rightarrow MN = \dfrac{1}{2} \times (11 + 8) \\[1em] \Rightarrow MN = \dfrac{1}{2} \times 19 \\[1em] \Rightarrow MN = 9.5 \text{ cm}.

Hence, MN = 9.5 cm

(ii) Given,

⇒ MN = 27 cm

⇒ DC = 20 cm

Substituting values in equation (3), we get :

MN=12(AB+CD)27=12×(AB+20)27×2=AB+20AB+20=54AB=5420=34 cm.\Rightarrow MN = \dfrac{1}{2}(AB + CD) \\[1em] \Rightarrow 27 = \dfrac{1}{2} \times (AB + 20) \\[1em] \Rightarrow 27 \times 2 = AB + 20 \\[1em] \Rightarrow AB + 20 = 54 \\[1em] \Rightarrow AB = 54 - 20 = 34 \text{ cm}.

Hence, AB = 34 cm.

(iii) Given,

⇒ MN = 15 cm

⇒ AB = 23 cm

Substituting values in equation (3), we get :

MN=12(AB+CD)15=12×(23+DC)15×2=23+DC23+DC=30DC=3023=7 cm.\Rightarrow MN = \dfrac{1}{2}(AB + CD) \\[1em] \Rightarrow 15 = \dfrac{1}{2} \times (23 + DC) \\[1em] \Rightarrow 15 \times 2 = 23 + DC \\[1em] \Rightarrow 23 + DC = 30 \\[1em] \Rightarrow DC = 30 - 23 = 7 \text{ cm}.

Hence, DC = 7 cm.

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