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

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))

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प्रश्न

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))`

बेरीज
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उत्तर


Draw AE ⊥ BC, B-E-C.

BC = BD + DC   ...[B-D-C]

∴ 20 = 7 + DC

∴ DC = 20 – 7

∴ DC = 13

(i) ΔABD and ΔADC have same height AE.

`(A(ΔABD))/(A(ΔADC)) = (BD)/(DC)`   ...[Triangles having equal height]

∴ `(A(ΔABD))/(A(ΔADC)) = 7/13`

(ii) ΔABD and ΔABC have same height AE.

`(A(ΔABD))/(A(ΔABC)) = (BD)/(BC)`   ...[Triangles having equal height]

∴ `(A(ΔABD))/(A(ΔABC)) = 7/20`

(iii) ΔADC and ΔABC have same height AE.

`(A(ΔADC))/(A(ΔABC)) = (DC)/(BC)`   ...[Triangles having equal height]

∴ `(A(ΔADC))/(A(ΔABC)) = 13/20`

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पाठ 1: Similarity - Exercise

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))`


The ratio of the areas of two triangles with common base is 6:5. Height of the larger triangle of 9 cm, then find the corresponding height of the smaller triangle.


Base of a triangle is 9 and height is 5. Base of another triangle is 10 and height is 6. Find the ratio of areas of these triangles.


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"))`

 In trapezium PQRS, side PQ || side SR, AR = 5AP, AS = 5AQ then prove that, SR = 5PQ 

 

 


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`"A(∆ ABD)"/"A(∆ ADC)"`


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


In the figure, PM = 10 cm, A(∆PQS) = 100 sq.cm, A(∆QRS) = 110 sq. cm, then find NR.


The ratio of the areas of two triangles with the common base is 4 : 3. Height of the larger triangle is 2 cm, then find the corresponding height of the smaller triangle.


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


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

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


In the given, seg BE ⊥ seg AB and seg BA ⊥ seg AD.

if BE = 6 and AD = 9 find `(A(Δ ABE))/(A(Δ BAD))`.


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

`100/110 = (square)/(NR)`,

NR = `square` cm


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