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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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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 1: Similarity - Q.1 (B)

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  1. `("A"(∆"PQB"))/("A"(∆"PBC"))`
  2. `("A"(∆"PBC"))/("A"(∆"ABC"))`
  3. `("A"(∆"ABC"))/("A"(∆"ADC"))`
  4. `("A"(∆"ADC"))/("A"(∆"PQC"))`

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if BE = 6 and AD = 9 find `(A(Δ ABE))/(A(Δ BAD))`.


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


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(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:

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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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