Draw a figure.
Rohan walks a distance of 3 km towards North.
Then turns to his left and walks for 2 km.
He again turns left and walks for 3 km.
At this point he turns to his left and walks for 3 km.
First move to north direction, then turn right, again turn right without moving from the position than turn left and move.
Draw a figure first when man is moving in clockwise direction then for anticlockwise. The man will be finally standing in North East position as explain in below figure.
Let us assume her house is situated at point A.
Draw a figure of movement of Radhika.
Draw the figure of movement with the help of given below direction graph.
A person starts from a point A and travels 3 km East - Wards to B. Then he turns left and travels 9 km to reach C.
He again turns left and travels 15 km and reaches his destination D.
The movements of the person are as shown in Figure
Clearly, AB = 3 km, BC = 3AB = (3 × 3) km = 9 km, CD = 5AB = (5 × 3) km = 15 km,
Draw AE ? CD.
Then, CE = AB = 3 km
AE = BC = 9 km
DE = (CD ? CE) = (15 ? 3) km = 12 km.
In triangle AED, AD2 = AE2 + DE2
or AD2 = 92 + 122
or AD2 = 81+ 144
or AD2 = 225
or AD2 = 152
AD = 15
Required distance = AD = 15 km
The movements of Amit are as shown in Fig (P to Q, Q to R and R to S).
Clearly, his final position is S which is to the South ? East from the starting point P.
The movements of Maya from T to R are as shown in Figure.
Distance between T and R = TR = TU + UR = TU + PW + QV = ( 4 + 3 + 1) ft = 8 ft.
The movements of Sanjeev from A to F are as shown in Figure Clearly,
Sanjeev's distance from starting point A
= AF = (AB + BF) = AB + ( BE ? EF ) = AB + (CD ? EF)
= [10 + (20 ? 10)] m = (10 + 10) m = 20 m.
Also, F lies to the South of A. So, Sanjeev is 20 meters to the south of his starting point.
Drawing a figure for the movement of a man as below.
Draw a figure for 1 km towards East.
Then Draw another figure for 5 km in south.
Again draw a line when he turns to East and walks 2 km,
after this draw a line when he turns to North and walks 9 km.
The movements of A are as shown in Figure (O to P, P to Q, Q to R, R to S and S to T)
Since TS = OP + QR, so T lies in line with O. A's distance from the starting point O = OT = (RS ? PQ) = (15 ? 10)m = 5 m.
The movements of the man from A to F are as shown in Figure Clearly,
DC = AB + EF
Therefore, F is in line with A
Also, AF = (BC ? DE) = 5 m.
So, the man is 5 meters away from his initial position.
Comments
There are no comments.Copyright ©CuriousTab. All rights reserved.