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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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In fig., PM = 10 cm, A(ΔPQS) = 100 sq. cm, A(ΔQRS) = 110 sq. cm, then NR?
ΔPQS and ΔQRS having seg QS common base.
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`(A(ΔPQS))/(A(ΔQRS)) = (square)/(NR)`,
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(ii) `(A(ΔABD))/(A(ΔABC))`
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Prove that, The areas of two triangles with the same height are in proportion to their corresponding bases. To prove this theorem start as follows:
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In the figure, PQ ⊥ BC, AD ⊥ BC. To find the ratio of A(ΔPQB) and A(ΔPBC), complete the following activity.
Given: PQ ⊥ BC, AD ⊥ BC
Now, A(ΔPQB) = `1/2 xx square xx square`
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Therefore,
`(A(ΔPQB))/(A(ΔPBC)) = (1/2 xx square xx square)/(1/2 xx square xx square)`
= `square/square`
