An aeroplane leaves an airport and flies due | vedclass

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(B) Distance travelled by the first aeroplane in $1 \frac{1}{2}$ hours (or $1.5$ hours) $= 300 	imes 1.5 = 450 \, km$.Distance travelled by the secon

Vedclass · Accepted Answer

$750$

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(B) Distance travelled by the first aeroplane in $1 \frac{1}{2}$ hours (or $1.5$ hours) $= 300 	imes 1.5 = 450 \, km$.Distance travelled by the second aeroplane in $1.5$ hours $= 400 	imes 1.5 = 600 \, km$.Since one aeroplane flies North and the other flies West,the paths are perpendicular to each other,forming a right-angled triangle with the airport at the vertex of the right angle.Let $O$ be the airport,$A$ be the position of the first aeroplane,and $B$ be the position of the second aeroplane.Then $OA = 450 \, km$ and $OB = 600 \, km$.Using the Pythagoras theorem in $	riangle AOB$:$AB^2 = OA^2 + OB^2$$AB^2 = (450)^2 + (600)^2$$AB^2 = 202500 + 360000$$AB^2 = 562500$$AB = \sqrt{562500} = 750 \, km$.Thus,the two aeroplanes will be $750 \, km$ apart.

$1$. $\square ABCD$ is a rhombus.	$a$. $\overline{AC}$ and $\overline{BD}$ bisect each other.
$2$. $\square ABCD$ is a parallelogram.	$b$. $\overline{AC}$ and $\overline{BD}$ bisect each other at right angles.
$3$. $\square ABCD$ is a rectangle.	$c$. $\overline{AC}$ and $\overline{BD}$ are congruent and bisect each other at right angles.
$4$. $\square ABCD$ is a square.	$d$. $\overline{AC}$ and $\overline{BD}$ are congruent and bisect each other.

An aeroplane leaves an airport and flies due North at $300 \, km/h$. At the same time,another aeroplane leaves the same airport and flies due West at $400 \, km/h$. How far apart will the two aeroplanes be after $1 \frac{1}{2}$ hours (in $, km$)?

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