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

In fig., TP = 10 cm, PS = 6 cm. (A(ΔRTP))/(A(ΔRPS)) = ?

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प्रश्न

In fig., TP = 10 cm, PS = 6 cm. `(A(ΔRTP))/(A(ΔRPS))` = ?

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उत्तर


Draw RE ⊥ TS, T-E-S

ΔRTP and ΔRPS have same height RE.

`(A(ΔRTP))/(A(ΔRPS)) = (TP)/(PS)`   ...[Triangles having equal height]

`(A(ΔRTP))/(A(ΔRPS)) = 10/6`   ...[Given]

∴ `(A(ΔRTP))/(A(ΔRPS)) = 5/3`

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  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 1: Similarity - Q.1 (B)

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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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`(A(ΔPQS))/(A(ΔQRS)) = (square)/(NR)`,

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


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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