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प्रश्न
In the figure, PM = 10 cm, A(∆PQS) = 100 sq.cm, A(∆QRS) = 110 sq. cm, then find NR.
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उत्तर
`{:("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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