हिंदी

In ∆ABC, D is the mid-point of AB and P is any point on BC. If CQ || PD meets AB in Q (Figure), then prove that ar (BPQ) = 12 ar (∆ABC).

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प्रश्न

In ∆ABC, D is the mid-point of AB and P is any point on BC. If CQ || PD meets AB in Q (Figure), then prove that ar (BPQ) = `1/2` ar (∆ABC).

योग
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उत्तर


Given: In ∆ABC, D is the mid-point of AB and P is any point on BC.

CQ || PD means AB in Q.

To prove: ar (∆BPQ) = `1/2` ar (∆ABC)

Construction: Join PQ and CD.

Proof: Since, D is the mid-point of AB. So, CD is the median of ∆ABC.

We know that, a median of a triangle divides it into two triangles of equal areas.

∴ ar (∆BCD) = `1/2` ar (∆ABC)

⇒ ar (∆BPD) + ar (∆DPC) = `1/2` ar (∆ABC)  ...(i)

Now, ∆DPQ and ∆DPC are on the same base DP and between the same parallel lines DP and CQ.

So, ar (∆DPQ) = ar (∆DPC)  ...(ii)

On putting the value from equation (ii) in equation (i), we get

ar (∆BPD) + ar (∆DPQ) = `1/2` ar (∆ABC)

⇒ ar (∆BPQ) = `1/2` ar (∆ABC)

Hence proved.

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अध्याय 9: Areas of Parallelograms & Triangles - Exercise 9.3 [पृष्ठ ९०]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 9
अध्याय 9 Areas of Parallelograms & Triangles
Exercise 9.3 | Q 4. | पृष्ठ ९०

वीडियो ट्यूटोरियलVIEW ALL [1]

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