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

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

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प्रश्न

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

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उत्तर

Let the corresponding sides of similar triangles be s1 and s2.

Let A1 and A2 be their corresponding areas.

s1 : s2 = 4 : 7   ...[Given]

∴ `(s_1)/(s_2) = 4/7`   ...(i)

`(A_1)/(A_2) = (s_1^2)/(s_2^2)`   ...[Theorem of areas of similar triangles]

`(A_1)/(A_2) = (s_1/s_2)^2`

`(A_1)/(A_2) = (4/7)^2`   ...[From (i)]

`(A_1)/(A_2) = 16/49`

∴ Ratio of areas of similar triangles = 16 : 49

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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.


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


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.


In the given figure, AD is the bisector of the exterior ∠A of ∆ABC. Seg AD intersects the side BC produced in D. Prove that:

\[\frac{BD}{CD} = \frac{AB}{AC}\]

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)"`


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 the following ratio.

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


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))` = ?


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


In fig., AB ⊥ BC and DC ⊥ BC, AB = 6, DC = 4 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))`


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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