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Question
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.
Areas of two triangles whose base is common are in proportion of their corresponding `square`.
`(A(ΔPQS))/(A(ΔQRS)) = (square)/(NR)`,
`100/110 = (square)/(NR)`,
NR = `square` cm
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Solution
ΔPQS and ΔQRS having seg QS common base.
Areas of two triangles whose base is common are in proportion of their corresponding \[\boxed{\text{heights}}\].
`(A(ΔPQS))/(A(ΔQRS))` = \[\frac{{\boxed{PM}}}{{{NR}}}\],
`100/110` = \[\frac{{\boxed{10}}}{{{NR}}}\],
∴ NR = `(110 xx 10)/100`
∴ NR = \[\boxed{11}\] cm
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