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

In fig., AB ⊥ BC and DC ⊥ BC, AB = 6, DC = 4 then (A(ΔABC))/(A(ΔBCD)) = ?

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प्रश्न

In fig., AB ⊥ BC and DC ⊥ BC, AB = 6, DC = 4 then `(A(ΔABC))/(A(ΔBCD))` = ?

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उत्तर

ΔABC and ΔBCD have same base BC.

∴ `(A(ΔABC))/(A(ΔBCD)) = (AB)/(DC)`   ...[Triangles having equal base]

∴ `(A(ΔABC))/(A(ΔBCD)) = 6/4`   ...[Given]

∴ `(A(ΔABC))/(A(ΔBCD)) = 3/2`

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पाठ 1: Similarity - Q.2 (B)

संबंधित 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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In the given figure, AD is the bisector of the exterior ∠A of ∆ABC. Seg AD intersects the side BC produced in D. Prove that:

\[\frac{BD}{CD} = \frac{AB}{AC}\]

In adjoining figure, PQ ⊥ BC, AD ⊥ BC then find following ratios.

  1. `("A"(∆"PQB"))/("A"(∆"PBC"))`
  2. `("A"(∆"PBC"))/("A"(∆"ABC"))`
  3. `("A"(∆"ABC"))/("A"(∆"ADC"))`
  4. `("A"(∆"ADC"))/("A"(∆"PQC"))`

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`"A(∆ ABD)"/"A(∆ ADC)"`


In the given figure, ∠ABC = ∠DCB = 90° AB = 6, DC = 8 then `(A(Δ ABC))/(A(Δ DCB))` = ?


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`(A(triangleABD))/(A(triangleABC))`


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\[\frac{A\left( ∆ ADC \right)}{A\left( ∆ ABC \right)}\] 


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


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In fig. BD = 8, BC = 12, B-D-C, then `(A(ΔABC))/(A(ΔABD))` = ?


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

(i) `(A(ΔABD))/(A(ΔADC))`

(ii) `(A(ΔABD))/(A(ΔABC))`

(iii) `(A(ΔADC))/(A(ΔABC))`


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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