Problem 1 :
The hypotenuse of a right triangle is 6 m more than twice of the shortest side. If the third side is 2 m less than the hypotenuse, find the sides of the triangle.
Solution :
Let x be the length of shorter side,
length of hypotenuse = 2x + 6
length of third side = 2x + 6 - 2
= 2x + 4
(2x + 6)2 = x2 + (2x + 4)2
4x2 + 24x + 36 = x2 + 4x2 + 16x + 16
x2 + 16x - 24x + 16 - 36 = 0
x2 - 8x - 20 = 0
(x - 10) (x + 2) = 0
x = 10 and x = -2
Hence the length of shorter side is 10
hypotenuse = 2(10) + 6 = 26
length of third side = 20 + 4 = 24
AC = √342 + 412
= √1156 + 1681
= √2837
= 53.26
Miles saved = (34 + 41) - 53.26
= 75 - 53.26
= 21.74 m
Problem 2 :
In the rectangle WXYZ, XY + YZ = 17 cm, and XZ + YW = 26 cm. Calculate the length and breadth of the rectangle?
Solution :
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
Now,
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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