Problem 1 :
A man goes 18 m due east and then 24 m due north. Find the distance of his current position from the starting point?
the distance of his current position from the starting point = √182 + 242
= √(324 + 576)
= 30 m
So, the required distance is 30 m.
Problem 2 :
There are two paths that one can choose to go from Sarah’s house to James house. One way is to take C street, and the other way requires to take A street and then B street. How much shorter is the direct path along C street? (Using figure).
By choosing the C street, he has to cover the distance,
= √22 + 1.52
= √(4 + 2.25)
= 2.5 miles
By choosing the alternative way, he has to cover the distance = 2 + 1.5
= 3.5 miles
The difference between these two paths = 3.5 - 2.5
= 1 mile
So, by choosing the direct path, he may save 1 miles faster than other way.
Problem 3 :
To get from point A to point B you must avoid walking through a pond. You must walk 34 m south and 41 m east. To the nearest meter, how many meters would be saved if it were possible to make a way through the pond?
By drawing the rough picture using the given information, we get
AC = √342 + 412
= √1156 + 1681
Miles saved = (34 + 41) - 53.26
= 75 - 53.26
= 21.74 m
Problem 4 :
In the rectangle WXYZ, XY + YZ = 17 cm, and XZ + YW = 26 cm. Calculate the length and breadth of the rectangle?
XY + YZ = 17 cm
XZ + YW = 26 cm
To calculate : - Length and breadth of the rectangle.
We know that,
Diagonals of a rectangle are equal.
So, XZ = YW
Then, XZ = YW = 26/2 = 13 cm
In ∆XYZ ,
Let YZ = P , Then, XY = (17 - P).
Then, by Pythagoras theorem,
(P)2 + (17 - P)² = (13)²
P2 + 289 - 34P + P² = 169
2P2 - 34P = 169 - 289
2(P2 - 17P) = - 120
P2 - 17P = - 120/2
P2 - 17P = - 60
P2 - 17P + 60 = 0
P2 - 12P - 5P + 60 = 0
P(P - 12) - 5(P - 12) = 0
(P - 12)(P - 5) = 0
P - 12 = 0 or P = 12
P = 12 cm or P = 5 cm
YZ = P = 12 cm [Because , YZ is the length of the rectangle ,so we will assign it the greatest value of P]
Again, XY = (17 - P) = (17 - 12) cm = 5 cm
[Because , XY is thee breadth]
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