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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) - Geometry Mathematics 2

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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

The ratio of the areas of two triangles with the common base is 14 : 9. Height of the larger triangle is 7 cm, then find the corresponding height of the smaller triangle.


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


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

`"A(∆ ABD)"/"A(∆ ADC)"`


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


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

`(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))`.


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


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


Ratio of corresponding sides of two similar triangles is 4:7, then find the ratio of their areas = ?


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 [______]

`("A"("PQS"))/("A"("QRS")) = (["______"])/"NR"`,

`100/110 = (["______"])/"NR"`,

NR = [ ______ ] cm


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:

  1. Draw two triangles, give the names of all points, and show heights.
  2. Write 'Given' and 'To prove' from the figure drawn.

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.


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