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

(c) 7 cm
It is given that △ABC~△PQR.
ar△ABC=25 cm2 and ar△PQR=49 cm2 Also, QR = 9.8 cm. We have to find BC.We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

∴ ar△ABCar△PQR=BC2QR2

⇒BCQR2=ar△ABCar△PQR=2549=572

⇒BCQR=57⇒BC9.8=57

⇒BC=5×9.87=497=7
Hence, BC = 7 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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