मराठी

In the given figure, BD : DC = 3 : 5, then area of ΔABD : area of ΔACD is:

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प्रश्न

In the given figure, BD : DC = 3 : 5, then area of ΔABD : area of ΔACD is:

पर्याय

  • 5 : 3

  • 3 : 5

  • 25 : 9

  • 9 : 25

MCQ
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उत्तर

3 : 5

Explanation:

Both triangles ΔABD and ΔACD share the same vertex A and have their bases BD and DC along the same straight line BC. Therefore, they share a common height from vertex A to the base BC.

The area of a triangle is given by `1/2` × base} × height. Taking the ratio of their areas:

`\frac{\text{area of }\triangle ABD}{\text{area of }\triangle ACD} = \frac{\frac{1}{2} \times BD \times h}{\frac{1}{2} \times DC \times h} = \frac{BD}{DC} = \frac{3}{5}`

Thus, the ratio of the area of ΔABD to the area of ΔACD is 3 : 5.

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  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 15: Area Theorems [Proof and Use] - Exercise 15 [पृष्ठ २२१]

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सेलिना Concise Mathematics [English] Class 9 ICSE
पाठ 15 Area Theorems [Proof and Use]
Exercise 15 | Q 1. (a) | पृष्ठ २२१
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