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Maharashtra State BoardSSC (English Medium) 10th Standard

In the following figure RP : PK= 3 : 2, then find the value of A(ΔTRP) : A(ΔTPK).

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Question

In the following figure RP : PK= 3 : 2, then find the value of A(ΔTRP) : A(ΔTPK).

Sum
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Solution

Ratio of the areas of two triangles with common or equal heights is equal to the ratio of their corresponding bases.

`(A(triangleTRP))/(A(triangleTPK))`

`="RP"/"PK"`

`=3/2`

A(△TRP) : A(△TPK) = 3 : 2​
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2013-2014 (March)

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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`

A(ΔPBC)  = `1/2 xx square xx square`

Therefore, 

`(A(ΔPQB))/(A(ΔPBC)) = (1/2 xx square xx square)/(1/2 xx square xx square)`

= `square/square`


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