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Ratio of corresponding sides of two similar triangles is 4:7, then find the ratio of their areas = ? - Geometry Mathematics 2

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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)

by theorem of areas of similar triangles,

`"A"_1/"A"_2 = "s"_1^2/"s"_2^2`  

`"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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अध्याय 1: Similarity - Q.1 (B)

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

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


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In the given, seg BE ⊥ seg AB and seg BA ⊥ seg AD.

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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.

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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`

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Therefore, 

`(A(ΔPQB))/(A(ΔPBC)) = (1/2 xx square xx square)/(1/2 xx square xx square)`

= `square/square`


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