Answer:
Length of the rectangle is 15.0325 km and width is 0.3765 km.
Explanation:
Given:
Perimeter of a rectangle = 30.8 km
Length of diagonal of rectangle = 11 km
To find:
The length and width of rectangle=?
Solution:
Lets assume length of the rectangle = x km
And assume width of the rectangle = y km
Lets first create equation using given perimeter
perimeter of rectangle = 2 ( length + width )
=> 30.8 km = 2 ( x + y )
=>![x + y = \frac{30.8}{2}](https://tex.z-dn.net/?f=x%20%2B%20y%20%3D%20%5Cfrac%7B30.8%7D%7B2%7D)
=> y = 15.4 – x ------(1)
As diagonal and two sides of rectangle forms right angle triangle whose hypoteneus is diagonal ,
=> ![length^2 + width^2 = diagonal^2](https://tex.z-dn.net/?f=length%5E2%20%2B%20width%5E2%20%3D%20diagonal%5E2)
=> ![x^2 + y^2 = 11^2](https://tex.z-dn.net/?f=x%5E2%20%2B%20y%5E2%20%3D%2011%5E2)
=> ![x^2 + y^2 = 121](https://tex.z-dn.net/?f=x%5E2%20%2B%20y%5E2%20%3D%20121)
On substituting value of y from (1) in above equation we get
=> ![x^2 + (15.4-x)^2 = 121](https://tex.z-dn.net/?f=x%5E2%20%2B%20%2815.4-x%29%5E2%20%3D%20121)
=>![x^2 + (15.4)^2 + x^2 – 2 x 15.4 \times x = 121](https://tex.z-dn.net/?f=x%5E2%20%2B%20%2815.4%29%5E2%20%2B%20x%5E2%20%E2%80%93%202%20x%2015.4%20%5Ctimes%20x%20%20%20%3D%20121)
=> ![2x^2-30.8x + 237.16 -121 = 0](https://tex.z-dn.net/?f=2x%5E2-30.8x%20%2B%20237.16%20-121%20%20%3D%200)
=> ![2x^2-30.8x + 116.16 = 0](https://tex.z-dn.net/?f=2x%5E2-30.8x%20%2B%20116.16%20%3D%200)
Solving above quadratic equation using quadratic formula
General form of quadratic equation is
![ax^2 +bx +c = 0](https://tex.z-dn.net/?f=ax%5E2%20%2Bbx%20%2Bc%20%3D%200)
And quadratic formula for getting roots of quadratic equation is
![x= \frac{ -b\pm\sqrt{(b^2-4ac)}}{2a}](https://tex.z-dn.net/?f=x%3D%20%5Cfrac%7B%20-b%5Cpm%5Csqrt%7B%28b%5E2-4ac%29%7D%7D%7B2a%7D)
As equation is
, in our case
a = 2 , b = -30.8 and c = 116.16
Calculating roots of the equation we get
![x=\frac{ -(-30.8)\pm\sqrt{(-30.8)^2-4(2)( 11)} } {(2\times2)}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B%20-%28-30.8%29%5Cpm%5Csqrt%7B%28-30.8%29%5E2-4%282%29%28%2011%29%7D%20%7D%20%7B%282%5Ctimes2%29%7D)
![x=\frac{30.8\pm\sqrt{(948.64-88)}}{4}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B30.8%5Cpm%5Csqrt%7B%28948.64-88%29%7D%7D%7B4%7D)
![x=\frac{30.8\pm\sqrt{860.64}}{4}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B30.8%5Cpm%5Csqrt%7B860.64%7D%7D%7B4%7D)
![x=\frac{30.8\pm\sqrt{860.64}}{4}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B30.8%5Cpm%5Csqrt%7B860.64%7D%7D%7B4%7D)
![x=\frac{(30.8\pm29.33)}4](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B%2830.8%5Cpm29.33%29%7D4)
![x=\frac{(30.8+29.33)}{4}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B%2830.8%2B29.33%29%7D%7B4%7D)
![x=\frac{(30.8-29.33)}{4}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B%2830.8-29.33%29%7D%7B4%7D)
=> x = 15.0325 or x = 0.3675
As generally length is longer one ,
So x = 1.0325
From equation (1) y = 15.4 – x = 0.3765
Hence length of the rectangle is 15.0325 km and width is 0.3765 km.