हिंदी

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 □.

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प्रश्न

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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उत्तर

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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अध्याय 1: Similarity - Q.2 (A)

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