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प्रश्न
The ratio of the areas of two triangles with common base is 6:5. Height of the larger triangle of 9 cm, then find the corresponding height of the smaller triangle.
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उत्तर
Let the height of the larger triangle be h1 and that of the smaller triangle be `h_2`
The ratio of the areas of two triangles with common base is equal to the ratio of their corresponding heights.
`(A("Larger"triangle))/(A("Smaller"triangle)) = h_1/h_2`
`:. 6/5 = 9/h_2` ...(Substituting the given values)
`∴ 6 xx h_2 = 9 xx 5`
`:. h_2 = (9 xx 5)/6 = 15/2`
`:. h_2= 7.5 cm`
The corresponding height of the smaller traingles of 7.5 cm.
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संबंधित प्रश्न
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\[\frac{A\left( ∆ ADC \right)}{A\left( ∆ ABC \right)}\]
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In fig., PM = 10 cm, A(ΔPQS) = 100 sq. cm, A(ΔQRS) = 110 sq. cm, then NR?
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In ΔABC, B-D-C and BD = 7, BC = 20, then find the following ratio.
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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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- Write 'Given' and 'To prove' from the figure drawn.
If ΔABC ∼ ΔDEF, length of side AB is 9 cm and length of side DE is 12 cm, then find the ratio of their corresponding areas.
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
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Therefore,
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= `square/square`
