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

In adjoining figure, PQ ⊥ BC, AD ⊥ BC then find following ratios. (i) APQBAPBCA(∆PQB)A(∆PBC) (ii) APBCAABCA(∆PBC)A(∆ABC) (iii) AABCAADCA(∆ABC)A(∆ADC) (iv) AADCAPQCA(∆ADC)A(∆PQC)

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प्रश्न

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"))`
बेरीज
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उत्तर

(i) In ∆PQB and ∆PBC,

ΔPQB and ΔPBC have same height PQ.   ...(Given)

Areas of triangles with equal heights are proportional to their corresponding bases.

∴ `("A"(∆"PQB"))/("A"(∆"PBC")) = "BQ"/"BC"`

(ii) In ∆PBC and ∆ABC,

∆PBC and ∆ABC have same base BC.   ...(Given)

Areas of triangles equal bases are proportional to their corresponding heights.

∴ `("A"(∆"PBC"))/("A"(∆"ABC")) = "PQ"/"AD"`

(iii) In ∆ABC and ∆ADC,

∆ABC and ∆ADC have same height AD.   ...(Given)

Areas of triangles with equal heights are proportional to their corresponding bases.

∴ `("A"(∆"ABC"))/("A"(∆"ADC")) = "BC"/"DC"`

(iv) In ∆ADC and ∆PQC,

Ratio of areas of two triangles is equal to the ratio of the products of their bases and corresponding heights.

∴ `("A"(∆"ADC"))/("A"(∆"PQC")) = "DC × AD"/"QC × PQ"`

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पाठ 1: Similarity - Practice Set 1.1 [पृष्ठ ६]

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))`


In the following figure RP : PK= 3 : 2, then find the value of A(ΔTRP) : A(ΔTPK).


In the given figure, BC ⊥ AB, AD ⊥ AB, BC = 4, AD = 8, then find `("A"(∆"ABC"))/("A"(∆"ADB"))`


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

 

 


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

`"A(∆ ABD)"/"A(∆ ADC)"`


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.


A roller of diameter 0.9 m and the length 1.8 m is used to press the ground. Find the area of the ground pressed by it in 500 revolutions.
`(pi=3.14)`


If ΔXYZ ~ ΔPQR then `(XY)/(PQ) = (YZ)/(QR)` = ?


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


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


In fig., PM = 10 cm, A(ΔPQS) = 100 sq.cm, A(ΔQRS) = 110 sq.cm, then NR?


ΔPQS and ΔQRS having seg QS common base.

Areas of two triangles whose base is common are in proportion of their corresponding `square`.

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

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

NR = `square` cm


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


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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