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Maharashtra State BoardSSC (English Medium) 10th Standard

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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Question

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"))`
Sum
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Solution

(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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Chapter 1: Similarity - Practice Set 1.1 [Page 6]

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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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