Question

The length of a hypotenuse of a right triangle exceeds the length of its base by 2 cm and exceeds twice the length of the altitude by 1 cm. Find the length of each side of the triangle (in cm).
A) \[6,{\text{ }}8,{\text{ }}10\]
B) \[7,{\text{ }}24,{\text{ }}25\]
C) \[\;8,{\text{ }}15,{\text{ }}17\]
D) \[7,{\text{ }}40,{\text{ }}41\]

Answer 1

Hint:
In this question, we will get the two values of Hypotenuse. First value will be in terms of Base and second value will be in the terms of perpendicular. Now, we can equate both the values of Hypotenuse. After solving the expression, we will get the value of Base in terms of Perpendicular. Then we will put the value of Base and Hypotenuse in the Pythagoras Theorem. In this way we will solve the question further.

Complete step by step solution:
As it is given that the Length of a hypotenuse exceeds the length of its base by 2 cm and exceeds twice the length of the altitude by 1 cm.
Let us assume, Hypotenuse = H, Base = B and Perpendicular = P
So, from the given information we can say that:
The Length of \[H{\text{ }} = {\text{ }}B{\text{ }} + {\text{ }}2\] (as it is given in the question that length of hypotenuse of right triangle exceeds the length of its base by 2 cm.)
Again, from the given information we can say that:
\[H{\text{ }} = {\text{ }}2P{\text{ }} + {\text{ }}1\] (As it is given in the question that length of hypotenuse exceeds the length of altitude (perpendicular) by 1 cm.)
Also, we can say that,
\[H{\text{ }} = {\text{ }}B{\text{ }} + {\text{ }}2\] And \[H{\text{ }} = {\text{ }}2P{\text{ }} + {\text{ }}1\] both are equal to each other because both are the value of Hypotenuse.
So, from the above two equations we can calculate the value of Hypotenuse:
\[ \Rightarrow B{\text{ }} + {\text{ }}2{\text{ }} = {\text{ }}2P{\text{ }} + {\text{ }}1\]
Here, we will solve for B in terms of P.
\[B{\text{ }} = {\text{ }}2P{\text{ }} + 1{\text{ }}-{\text{ }}2\]
After, solving we get
\[B{\text{ }} = {\text{ }}2P{\text{ }}-{\text{ }}1\]
Now, we know the Pythagoras theorem \[{H^2} = {P^2} + {B^2}\]
Putting the values of the \[B{\text{ }} = {\text{ }}2P{\text{ }}-{\text{ }}1\] and \[H{\text{ }} = {\text{ }}2P{\text{ }} + {\text{ }}1\] in the Pythagoras theorem.
On substituting the values we get,
\[ \Rightarrow \]\[{(2P + 1)^2} = {P^2} + {(2P - 1)^2}\]
Solving the equation with the help of algebraic formulas \[{(a + b)^2} = {a^2} + {b^2} + 2ab\] and \[{a^2} - {b^2} = {a^2} + {b^2} - 2ab\]
\[ \Rightarrow \]\[4{P^2} + 1 + 4P = {P^2} + 4{P^2} + 1 - 4P\]
\[ \Rightarrow 4P = {P^2} - 4P\]
\[ \Rightarrow {P^2} - 4P - 4P = 0\]
\[ \Rightarrow {P^2} - 8P = 0\]
\[ \Rightarrow P(P - 8) = 0\]
From the above final equation we get two values of P:
\[P{\text{ }} = {\text{ }}0\] and \[P{\text{ }} = {\text{ }}8\]
So, we will not take \[P{\text{ }} = {\text{ }}0\] because it is not possible to construct the triangle with 0 cm altitude.
Now, putting the values of \[P{\text{ }} = {\text{ }}8\]cm in the equation \[B{\text{ }} = {\text{ }}2P{\text{ }}-{\text{ }}1\]
\[B = 2 \times 8 - 1 = 16 - 1\]
\[\therefore B{\text{ }} = {\text{ }}15\] cm
Again, putting the value of \[P{\text{ }} = {\text{ }}8\]cm in the equation \[H{\text{ }} = {\text{ }}2P{\text{ }} + {\text{ }}1\]
\[H = 2 \times 8 + 1 = 16 + 1\]
\[\therefore H{\text{ }} = {\text{ }}17\] cm
Thus, Hypotenuse = 17 cm, Perpendicular = 8 cm and Base = 15 cm

Option (C) \[\;8,{\text{ }}15,{\text{ }}17\] is the correct answer.

Note:
In these types of questions, you need to take care while putting the value of Hypotenuse, Base and Perpendicular. All the values above are in terms of ‘P’. So, we could easily find the value of Perpendicular. According to this question we got two values of Perpendicular that are 0 and 8. We will ignore 0 because perpendicular cannot be 0 as we are not able to draw a right triangle with this value. So, we will consider \[P{\text{ }} = {\text{ }}8{\text{ }}cm\]