मराठी

In the following figure, ABCD and EFGD are two parallelograms and G is the mid-point of CD. Then ar (DPC) = 12 ar (EFGD).

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प्रश्न

In the following figure, ABCD and EFGD are two parallelograms and G is the mid-point of CD. Then ar (DPC) = `1/2` ar (EFGD).

पर्याय

  • True

  • False

MCQ
चूक किंवा बरोबर
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उत्तर

This statement is False.

Explanation:

In the given figure, join PG.

Since, G is the mid-point of CD.

Thus, PG is a median of ΔDPC and it divides the triangle into parts of equal areas.

Then, ar (ΔDPG) = ar (ΔGPC) = `1/2` ar (ΔDPC)  ...(i)

Also, we know that, if a parallelogram and a triangle lie on the same base and between the same parallels, then area of triangle is equal to half of the area of parallelogram.

Here, parallelogram EFGD and ΔDPG lie on the same base DG and between the same parallels DG and EF.

So, ar (ΔDPG) = `1/2` ar (EFGD)  ...(ii)

From equations (i) and (ii),

`1/2` ar (ΔDPG) = `1/2` ar (EFGD)

⇒ ar (ΔDPC) = ar (EFGD)

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  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 9: Areas of Parallelograms & Triangles - Exercise 9.2 [पृष्ठ ८८]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 9
पाठ 9 Areas of Parallelograms & Triangles
Exercise 9.2 | Q 5. | पृष्ठ ८८

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