# PYTHAGOREAN THEOREM WORD PROBLEMS FOR GRADE 10

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

Now,

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, P = 12

P = 12 cm

Again,

P - 5 = 0 => 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.] Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

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