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Maharashtra State BoardSSC (English Medium) 10th Standard

In the figure, PM = 10 cm, A(∆PQS) = 100 sq.cm, A(∆QRS) = 110 sq. cm, then find NR. - Geometry Mathematics 2

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Question

In the figure, PM = 10 cm, A(∆PQS) = 100 sq.cm, A(∆QRS) = 110 sq. cm, then find NR.

Sum
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Solution

`{:("PM = 10 cm"), ("A(∆PQS) = 100 sq.cm"), ("A(∆QRS) = 110 sq.cm"):}   }"Given"`

Now,

In ∆PQS and ∆QRS,

seg QS is common base of ∆PQS and ∆QRS.

∴ `"A(∆PQS)"/"A(∆QRS)" = "PM"/"NR"`   ...(Triangles having equal base)

∴ `100/110 = 10/"NR"`

∴ `"NR" = (110 × 10)/100`

∴ NR = 11 cm

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Chapter 1: Similarity - Problem Set 1 [Page 27]

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Balbharati Mathematics 2 [English] Standard 10 Maharashtra State Board
Chapter 1 Similarity
Problem Set 1 | Q 5 | Page 27

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