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

Areas of two similar triangles are in the ratio 144 : 49. Find the ratio of their corresponding sides.

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प्रश्न

Areas of two similar triangles are in the ratio 144 : 49. Find the ratio of their corresponding sides.

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उत्तर

Let the areas of two similar triangles be A1, A2 and their corresponding sides be S1, S2 respectively.

∴ `(A_1)/(A_2) = 144/49`    ...(i) [Given]

∴ `(A_1)/(A_2) = (S_1^2)/(S_2^2)`   ...[Theorem of areas of similar triangles]

∴ `144/49 = (S_1^2)/(S_2^2)`    ...[From (i)]

∴ `(S_1)/(S_2) = 12/7`   ...[Taking square root of both sides]

∴ The ratio of the corresponding sides of the given triangles is 12 : 7.

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पाठ 1: Similarity - Exercise

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.


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 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)"`


Ratio of areas of two triangles with equal heights is 2 : 3. If base of the smaller triangle is 6 cm then what is the corresponding base of the bigger triangle ?


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.


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))`.


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


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


From adjoining figure, ∠ABC = 90°, ∠DCB = 90°, AB = 6, DC = 8, then `(A(ΔABC))/(A(ΔBCD))` = ?


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.

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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