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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) - Geometry Mathematics 2

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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))`

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

संबंधित प्रश्‍न

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In ∆ABC, B - D - C and BD = 7, BC = 20 then find following ratio. 

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From adjoining figure, ∠ABC = 90°, ∠DCB = 90°, AB = 6, DC = 8, then `("A"(Δ"ABC"))/("A"(Δ"BCD"))` = ?


In ΔABC, B − D − C and BD = 7, BC = 20, then find the following ratio.

(i) `"A(ΔABD)"/"A(ΔADC)"`

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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:

  1. Draw two triangles, give the names of all points, and show heights.
  2. Write 'Given' and 'To prove' from the figure drawn.

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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