Advertisements
Advertisements
प्रश्न
An aeroplane leaves an airport and flies due north at a speed of 1000km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after` 1 1/2` hours?
उत्तर
Let A be the first aeroplane flied due north at a speed of 1000 km/hr and B be the second aeroplane flied due west at a speed of 1200 km/hr
Distance covered by plane A in` 1 1/2 hours=1000xx3/2=1500km`
Distance covered by plane B in `1 1/2 hours = 1200xx3/2=1800 km`
Now, In right triangle ABC
By using Pythagoras theorem, we have
`AB^2=BC^2+CA^2`
=(1800)^2+(1500)^2
=3240000+2250000
=5490000
∴ `AB^2=5490000 `
⇒` AB=300sqrt61m`
Hence, the distance between two planes after `1 1/2 house is 300 sqrt61m`
APPEARS IN
संबंधित प्रश्न
In a ΔABC, D and E are points on the sides AB and AC respectively. For the following case show that DE || BC
AB = 2cm, AD = 8cm, AE = 12 cm and AC = l8cm.
In a ΔABC, D and E are points on the sides AB and AC respectively. For the following case show that DE || BC
AB = 5.6cm, AD = 1.4cm, AC= 7.2 cm and AE = 1.8 cm.
In a ΔABC, D and E are points on AB and AC respectively such that DE || BC. If AD = 2.4cm, AE = 3.2 cm, DE = 2cm and BC = 5 cm, find BD and CE.
D and E are points on the sides AB and AC respectively of a ΔABC such that DE║BC.
If AD = 3.6cm, AB = 10cm and AE = 4.5cm, find EC and AC.
D and E are points on the sides AB and AC respectively of a ΔABC such that DE║BC. Find the value of x, when
AD = 4cm, DB = (x – 4) cm, AE = 8cm and EC = (3x – 19) cm.
Show that the line segment which joins the midpoints of the oblique sides of a trapezium is parallel sides
Each of the equal sides of an isosceles triangle is 25 cm. Find the length of its altitude if the base is 14 cm.
In the adjoining figure,
seg XY || seg AC, If 3AX = 2BX
and XY = 9 then find the length of AC.
From fig., seg PQ || side BC, AP = x + 3, PB = x – 3, AQ = x + 5, QC = x – 2, then complete the activity to find the value of x.
In ΔPQB, PQ || side BC
`"AP"/"PB" = "AQ"/(["______"])` ...[______]
`(x + 3)/(x - 3) = (x + 5)/(["______"])`
(x + 3) [______] = (x + 5)(x – 3)
x2 + x – [______] = x2 + 2x – 15
x = [______]
In fig., PS = 2, SQ = 6, QR = 5, PT = x and TR = y. Then find the pair of value of x and y such that ST || side QR.
