Two similar triangles abc and pqr have their areas 25 cm2 and 49 cm2 respectively. if qr = 9.8 cm

3. The areas of two similar triangles ABC and ΔΡQR are 25 cm2 and 49 cm2 respectively and QR 9.8 cm. Then BC-?

The area of two similar triangles ABC and PQR are $$25\ cm^{2}\ \& \  49\ cm^{2}$$, respectively. If QR $$=9.8$$ cm, then BC is:

• A

  9.8 cm

• B

  7 cm

• C

  49 cm

• D

  25 cm

$$\dfrac { ar(ABC) }{ ar(PQR) } =\dfrac { 25 }{ 49 } $$

In two similar triangles, the ratio of their areas is the square of the ratio of their sides

$$\Rightarrow { \left( \dfrac { BC }{ QR }  \right)  }^{ 2 }=\dfrac { 25 }{ 49 } \\ \Rightarrow \dfrac { BC }{ QR } =\dfrac { 5 }{ 7 } \\ \Rightarrow \dfrac { BC }{ 9.8 } =\dfrac { 5 }{ 7 } \\ \Rightarrow BC=\dfrac { 5 }{ 7 } \times 9.8=7$$

The areas of two similar triangles ∆ABC and ∆PQR are 25 cm2 and 49 cm2 respectively. If QR = 9.8 cm, find BC

It is being given that ∆ABC ~ ∆PQR, ar (∆ABC) = 25 cm2 and ar (∆PQR) = 49 cm2 .

We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

`\therefore \text{ }\frac{ar\ (\Delta ABC)}{ar\ (\DeltaPQR)}=\frac{BC^{2}}{QR^{2}}`

`\Rightarrow \frac{25}{49}=\frac{x^{2}}{(9.8)^{2}}`

`\Rightarrowx^{2}=( \frac{25}{49}\times 9.8\times 9.8)`

`\Rightarrow x=\sqrt{\frac{25}{49}\times 9.8\times 9.8}=(5/7xx9.8)=(5xx1.4)=7`

Hence BC = 7 cm.

Concept: Areas of Similar Triangles

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