मराठी
महाराष्ट्र राज्य शिक्षण मंडळएस.एस.सी (इंग्रजी माध्यम) इयत्ता १० वी

In the following figure seg AB ⊥ seg BC, seg DC ⊥ seg BC. If AB = 2 and DC = 3, find A(△ABC)/A(△DCB)

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प्रश्न

In the following figure seg AB ⊥ seg BC, seg DC ⊥ seg BC. If AB = 2 and DC = 3, find `(A(triangleABC))/(A(triangleDCB))`

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उत्तर

In the following figure ΔABC and ΔDCB have a comman base BC.

`therefore(A(triangleABC))/(A(triangleDCB))=(AB)/(DC)`

(∵The ratio of areas of two triangles with the same base is equal to the ratio of their corresponding heights.)

`therefore(A(triangleABC))/(A(triangleDCB))=2/3`

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2014-2015 (March) Set B

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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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