Advertisements
Advertisements
Question
Two trains leave a railway station at the same time. The first train travels due west and the second train due north. The first train travels at a speed of `(20 "km")/"hr"` and the second train travels at `(30 "km")/"hr"`. After 2 hours, what is the distance between them?
Advertisements
Solution
A is the position of the 1st train.
B is the position of the 2nd train.
Distance Covered in 2 hours
OA = 2 × 20 = 40 km
OB = 2 × 30 = 60 km
Distance between the train after 2 hours
AB = `sqrt("OA"^2 + "OB"^2)`
= `sqrt(40^2 + 60^2)`
= `sqrt(1600 + 3600)`
= `sqrt(5200)` or `sqrt(52 xx 100)`
= `10sqrt(4 xx 13)`
= `20sqrt(13)`
= 72.11 km
Distance between the two train = 72.11 km or `20sqrt(13) "km"`
APPEARS IN
RELATED QUESTIONS
The points A(4, 7), B(p, 3) and C(7, 3) are the vertices of a right traingle ,right-angled at B. Find the values of p.
The diagonal of a rectangular field is 16 metres more than the shorter side. If the longer side is 14 metres more than the shorter side, then find the lengths of the sides of the field.
From a point O in the interior of a ∆ABC, perpendicular OD, OE and OF are drawn to the sides BC, CA and AB respectively. Prove
that :
`(i) AF^2 + BD^2 + CE^2 = OA^2 + OB^2 + OC^2 – OD^2 – OE^2 – OF^2`
`(ii) AF^2 + BD^2 + CE^2 = AE^2 + CD^2 + BF^2`
A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.
For finding AB and BC with the help of information given in the figure, complete following activity.
AB = BC .......... 
∴ ∠BAC =
∴ AB = BC = × AC
= × `sqrt8`
= × `2sqrt2`
=
A ladder 13 m long rests against a vertical wall. If the foot of the ladder is 5 m from the foot of the wall, find the distance of the other end of the ladder from the ground.
In an isosceles triangle ABC; AB = AC and D is the point on BC produced.
Prove that: AD2 = AC2 + BD.CD.
The sides of a certain triangle is given below. Find, which of them is right-triangle
6 m, 9 m, and 13 m
In ΔABC, AD is perpendicular to BC. Prove that: AB2 + CD2 = AC2 + BD2
If in a ΔPQR, PR2 = PQ2 + QR2, then the right angle of ∆PQR is at the vertex ________
