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Maharashtra State BoardSSC (English Medium) 10th Standard

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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Question

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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Solution

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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2017-2018 (March) Set A

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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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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

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`


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