Advertisements
Advertisements
प्रश्न
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
उत्तर
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
संबंधित प्रश्न
In ∆PQR, point S is the midpoint of side QR. If PQ = 11, PR = 17, PS = 13, find QR.
In ∆ABC, ∠BAC = 90°, seg BL and seg CM are medians of ∆ABC. Then prove that:
4(BL2 + CM2) = 5 BC2
In right angle ΔABC, if ∠B = 90°, AB = 6, BC = 8, then find AC.
In the following figure, AD is perpendicular to BC and D divides BC in the ratio 1: 3.
Prove that : 2AC2 = 2AB2 + BC2
In the given figure, angle ACP = ∠BDP = 90°, AC = 12 m, BD = 9 m and PA= PB = 15 m. Find:
(i) CP
(ii) PD
(iii) CD
In the given figure, angle ACB = 90° = angle ACD. If AB = 10 m, BC = 6 cm and AD = 17 cm, find :
(i) AC
(ii) CD
The sides of the triangle are given below. Find out which one is the right-angled triangle?
11, 12, 15
From a point O in the interior of aΔABC, perpendicular OD, OE and OF are drawn to the sides BC, CA and AB respectively. Prove that: AF2 + BD2 + CE2 = AE2 + CD2 + BF2
In ∆PQR, PD ⊥ QR such that D lies on QR. If PQ = a, PR = b, QD = c and DR = d, prove that (a + b)(a – b) = (c + d)(c – d).
If two legs of a right triangle are equal to two legs of another right triangle, then the right triangles are congruent.
