In the given figure, $△ A B C - △ P Q R, A M$ and $P N$ are altitudes, where as $A X$ and $P Y$ are medians. $\frac{A M}{P N} = \frac{A X}{P Y}$

True

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

False

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A True
To prove
$\begin{matrix} Δ A B C \sim Δ P Q R {\frac{A B}{P Q} = \frac{B C}{Q R} = \frac{A C}{P R} \to (i) I n Δ A B M a n d Δ P Q N ∠ A B M = ∠ P Q N {Δ A B C \sim Δ P Q R} ∠ B M A = ∠ Q N P {e a c h 90^{\circ}} B y A A, Δ A B M \sim Δ P Q N B y C . P . S . T = \frac{A B}{P Q} = \frac{B M}{Q N} = \frac{A M}{P N} \to (i i) Δ A X C a n d Δ P Y R ∠ C = ∠ R {g i v e n} \frac{A C}{P R} = \frac{B C}{Q R} {G i v e n} \frac{A C}{P R} = \frac{\frac{1}{2} B C}{\frac{1}{2} Q R} \frac{A C}{P R} = \frac{X C}{Y R} B Y S A S, Δ A X C \sim Δ P Y R \frac{A X}{P Y} = \frac{A C}{P R} {B y C . P . S . T} F r o m (1), (2) a n d (3) \frac{A M}{P N} = \frac{A X}{P Y} \end{matrix}$

Suggest Corrections

Similar questions

In the given figure, $△ A B C \sim △$ PQR and AM and PN are medians of $△$ ABC and $△$ PQR respectively. Then $\frac{A r e a o f △ A B C}{A r e a o f △ P Q R}$ = _____

$△$ ABC and $△$ PQR are two similar triangles as shown in the figure such that $\frac{A r e a o f A B C}{A r e a o f P Q R}$ = $\frac{9}{25}$ . AM and PN are the medians on $△$ ABC and $△$ PQR respectively. If AM = PO = 5 cm, find the value of 3ON.

Find the ratio of the areas of two similar triangles ABC and PQR shown in the figure where AM and PN are the medians of the $△$ ABC and $△$ PQR respectively.

If $△$ ABC and $△$ PQR are two similar triangles shown in the figure. AM and PN 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.