Question

The length of the hypotenuse of a right angled triangle exceeds the length of the base by 2cm and exceeds twice the length of the altitude by $ 1 $ cm. The perimeter of the triangle is:
(A) $ 18 $ cm
(B) $ 17 $ cm
(C) $ 25 $ cm
(D) $ 40 $ cm

Answer 1

Hint: It is given that the hypotenuse exceeds from both base and altitude. Therefore we first let hypotenuse be (x) and use the given conditions to find respective values of base and altitude in terms of ‘x’ and finally use all the three in Pythagoras theorem to get the value of ‘x’ and hence required value of the perimeter.
Pythagoras theorem which states that $ {\left( {Hpy.} \right)^2} = {\left( {Base} \right)^2} + {\left( {altitude} \right)^2} $ in a right angle triangle.

Complete step-by-step answer:
Let the length of the hypotenuse of a given right angle triangle be ‘x’.
Then, it is given that the length of hypotenuse exceeds the length of the base by $ 2 $ cm.
Therefore we have
Length of hypotenuse = length of base + $ 2 $
 $
   \Rightarrow x = length\,\,of\,base\,\, + 2 \\
   \Rightarrow length\,of\,base = x - 2 \\
  $

Also, it is given that the length of hypotenuse exceeds twice the length of the altitude by $ 1 $ cm.
 $
  \Rightarrow Length\,\,of\,\,hypotenuse\,\, = 2\left( {length\,\,of\,\,altitude\,} \right)\, + 1 \\
   \Rightarrow x = 2\left( {length\,\,of\,\,altitude} \right) + 1 \\
   \Rightarrow x - 1 = 2\left( {length\,\,of\,\,altitude} \right) \\
   \Rightarrow length\,\,of\,\,altitude = \left( {\dfrac{{x - 1}}{2}} \right) \;
  $
In the right angle triangle we know that Pythagoras theorem holds true. Which states that:
 $ {\left( {Hyp.} \right)^2} = {\left( {Perp.} \right)^2} + {\left( {Base} \right)^2} $
On substituting values we have
 $ {\left( x \right)^2} = {\left( {\dfrac{{x - 1}}{2}} \right)^2} + {\left( {x - 2} \right)^2} $
On simplifying brackets
 $ {x^2} = \dfrac{{{x^2} + 1 - 2x}}{4} + {x^2} + 4 - 4x $
Taking L.C.M. we have
 $ {x^2} = \dfrac{{{x^2} + 1 - 2x + 4{x^2} + 16 - 16x}}{4} $
 $ {x^2} = \dfrac{{5{x^2} - 18x + 17}}{4} $ On cross multiplication
 $
  4{x^2} = 5{x^2} - 18x + 17 \\
   \Rightarrow 5{x^2} - 18x + 17 - 4{x^2} = 0 \\
   \Rightarrow {x^2} - 18x + 17 = 0 \;
  $
Now, using the middle term splitting method to solve the above quadratic equation.
 $
  {x^2} - 17x - x + 17 = 0 \\
   \Rightarrow x\left( {x - 17} \right) - 1\left( {x - 17} \right) = 0 \\
   \Rightarrow \left( {x - 1} \right)\left( {x - 17} \right) = 0 \\
   \Rightarrow x - 1 = 0\,\,or\,\,x - 17 = 0 \\
   \Rightarrow x = 1\,\,or\,\,x = 17 \;
  $
But, $ x = 1 $ is not possible as if we take $ x = 1 $ the altitude $ \left( {\dfrac{{x - 1}}{2}} \right) $ of triangle will become zero, which is not possible.
Hence, length of hypotenuse is $ 17cm $
Therefore length of altitude will be $ \dfrac{{x - 1}}{2} = \dfrac{{17 - 1}}{2} $ = $ 8 $ cm.
Length of base is $ \left( {x - 2} \right) = \left( {17 - 2} \right) $ = $ 15 $ cm.
We know that the perimeter of the triangle will be given as the sum of the three sides.
Therefore, perimeter of right angle triangle = length of hypotenuse + length of base + length of altitude
 $ Perimeter = 17 + 15 + 8 $
 $ Perimeter = 40 cm $
Therefore, the required perimeter of the triangle is $ 40 $ cm.
So, the correct answer is “Option D”.

Note: In mathematics, the Pythagorean theorem, also known as Pythagoras's theorem, is a fundamental relation in Euclidean geometry among the three sides of a right triangle. It states that the area of the square whose side is the hypotenuse is equal to the sum of the areas of the squares on the other two sides.