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In the adjoining figure, a parallelogram ABCD is shown. If DM ⊥ AB, DN ⊥ BC, then area of ΔABD is equal to: - Mathematics

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Question

In the adjoining figure, a parallelogram ABCD is shown. If DM ⊥ AB, DN ⊥ BC, then area of ΔABD is equal to:

Options

  • `1/2 xx AB xx DN`

  • `1/2 xx BC xx DN`

  • `1/2 xx BC xx DM`

  • `1/2 xx AD xx DM`

MCQ
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Solution

`bb(1/2 xx BC xx DN)`

Explanation:

Diagonal BD splits parallelogram ABCD into two equal-area triangles.

So, area (△ABD) = area (△CBD).

In △CBD, the base is BC and the height from D to BC is DN (given DN ⟂ BC).

So, area = `1/2` × BC × DN.

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Chapter 13: Theorems on Area - Exercise 13B [Page 260]

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Nootan Mathematics [English] Class 9 ICSE
Chapter 13 Theorems on Area
Exercise 13B | Q 2. | Page 260
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