In two similar triangles ABC and PQR, if their corresponding altitudes AD and PS are in the ratio 4 : 9, find the ratio of the areas of $△$ ABC and $△$ PQR.

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Solution

Since the areas of two similar triangles are in the ratio of the squares of the corresponding altitudes

∴ $\frac{A r e a (△ A B C)}{A r e a (△ P Q R)} = \frac{A D^{2}}{P S^{2}}$

$\Rightarrow$ $\frac{A r e a (△ A B C)}{A r e a (△ P Q R)} = \frac{4^{2}}{9^{2}}$ = $\frac{16}{81}$

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If $△$ ABC and $△$ PQR are two similar triangles shown in the figure. AM and AN are the medians on $△$ ABC and $△$ PQR respectively. The ratio of areas of $△$ ABC and $△$ PQR is 9:25. If AM = PO = 5 cm. Find the value of 3ON.

Q. The areas of two similar triangles ABC and PQR are in the ratio 9 : 16. If BC = 4.5 cm, find the length of QR.

Q. If

Δ

ABC and

Δ

Δ

ABC and

Δ

PQR respectively. The ratio of areas of

Δ

ABC and

Δ

PQR is 9:25. If AM = PO = 5 cm. Find the value of 3(ON) in cm ___

$△ A B C$ is similar to $△ P Q R$ , in $△ A B C, A B = 3.6 c m, B C = 4 c m$ and $A C = 4.2 c m$ . The corresponding sides of $△ A B C$ and $△ P Q R$ are in the ratio $2 : 3$ . Construct $△ P Q R$ .

Find $m ∠ P R Q$ (in deg.)

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In two similar triangles ABC and PQR, if their corresponding altitudes AD and PS are in the ratio 4 : 9, find the ratio of the areas of △ABC and △PQR.

Since the areas of two similar triangles are in the ratio of the squares of the corresponding altitudes ∴ Area (△ABC)Area (△PQR)=AD2PS2 ⇒ Area (△ABC)Area (△PQR)=4292 = 1681

In two similar triangles ABC and PQR, if their corresponding altitudes AD and PS are in the ratio 4 : 9, find the ratio of the areas of $△$ ABC and $△$ PQR.

Since the areas of two similar triangles are in the ratio of the squares of the corresponding altitudes

∴ $\frac{A r e a (△ A B C)}{A r e a (△ P Q R)} = \frac{A D^{2}}{P S^{2}}$

$\Rightarrow$ $\frac{A r e a (△ A B C)}{A r e a (△ P Q R)} = \frac{4^{2}}{9^{2}}$ = $\frac{16}{81}$