Question

Solve the following problems:
a) Lengths of sides of a right-angled triangle are in arithmetic sequence with common difference d. If the length of the smallest side of the triangle is x – d, write the length of its other two sides.
b) Show that any right-angled triangle with sides in arithmetic sequence is similar to the right-angled triangle with sides 3, 4 and 5.

Answer 1

Hint: For the problem a) the smallest side of the triangle which is also the first term of the arithmetic progression is given. We just have to suppose the other two sides to be x and x+d by assuming common difference of the arithmetic progression to be d and then we will apply Pythagoras theorem on the sides to find the relation between x and d, thereby representing the length of the sides of triangle in terms of d.
For the problem b) we will assume the sides of the right-angled triangle to be x – d, x, x + d. After that we will apply Pythagoras theorem to find the relation between x and d and we will get the length of all the sides and then we will just compare them with the Pythagorean triplet 3, 4, 5.

Complete step-by-step answer:
Problem a) Given length of the smallest side of the triangle = x – d, where d is the common difference
So, first term = x – d
Second term = x – d + d = x
Third term = x + d
So, the three sides of the triangle = x – d, x, x + d
Now, we have been given that the triangle is a right-angled triangle. So, it will follow the Pythagoras theorem,
$ \Rightarrow {H^2} = {P^2} + {B^2}$ where H = Hypotenuse of the right-angled triangle, P represents the perpendicular of the right-angled triangle and B represents the Base of the right-angled triangle.
Here H = x + d, P = x and B = x – d
On substituting the values in the equation, we get
$ \Rightarrow {(x + d)^2} = {x^2} + {(x - d)^2}$
Using ${(a + b)^2} = {a^2} + 2ab + {b^2}$ and ${(a - b)^2} = {a^2} - 2ab + {b^2}$
$ \Rightarrow {x^2} + 2xd + {d^2} = {x^2} + {x^2} - 2xd + {d^2}$
$ \Rightarrow {x^2} + 2xd + {d^2} = 2{x^2} - 2xd + {d^2}$
$ \Rightarrow {x^2} + 2xd = 2{x^2} - 2xd$
\[ \Rightarrow 2xd + 2xd = 2{x^2} - {x^2}\]
\[ \Rightarrow 4xd = {x^2}\] or \[{x^2} = 4xd\]
\[ \Rightarrow x = 4d\]
Hence, the length of the sides of the right-angled triangle are,
x – d = 4d – d = 3d
x = 4d
x + d = 4d + d = 5d
Hence, the length of the other two sides of the right-angled triangle are 4d and 5d respectively.

Problem b) Since it is given that the sides of the right-angled triangle are in arithmetic progression, let us suppose the sides are x – d, x, x + d where d is our common difference.
Now, we have been given that the triangle is a right-angled triangle. So, it will follow the Pythagoras theorem,
$ \Rightarrow {H^2} = {P^2} + {B^2}$ where H = Hypotenuse of the right-angled triangle, P represents the perpendicular of the right-angled triangle and B represents the Base of the right-angled triangle.
Here H = x + d, P = x and B = x – d
On substituting the values in the equation, we get
$ \Rightarrow {(x + d)^2} = {x^2} + {(x - d)^2}$
Using ${(a + b)^2} = {a^2} + 2ab + {b^2}$ and ${(a - b)^2} = {a^2} - 2ab + {b^2}$
$ \Rightarrow {x^2} + 2xd + {d^2} = {x^2} + {x^2} - 2xd + {d^2}$
$ \Rightarrow {x^2} + 2xd + {d^2} = 2{x^2} - 2xd + {d^2}$
$ \Rightarrow {x^2} + 2xd = 2{x^2} - 2xd$
\[ \Rightarrow 2xd + 2xd = 2{x^2} - {x^2}\]
\[ \Rightarrow 4xd = {x^2}\] or \[{x^2} = 4xd\]
\[ \Rightarrow x = 4d\]
Hence, the length of the sides of the right-angled triangle are,
x – d = 4d – d = 3d
x = 4d
x + d = 4d + d = 5d
And we can clearly see that the sides 3d, 4d, 5d are similar to Pythagorean triplet sides 3, 4, 5.
Hence verified that the sides 3d, 4d, 5d are similar to Pythagorean triplet sides 3, 4, 5.

Note: For both of the problems we just have to keep in mind that we will take the sides of the triangle to be x – d, x, x + d, where d is the common difference, if the sides are given in an Arithmetic progression. Also, we should keep in mind that whenever we have been given a right-angled triangle, we just use the Pythagoras theorem if we have to find the length of the sides of the triangle which simply states that in a right-angled triangle, the square on the hypotenuse of a right-angled triangle is equal to the sum of the squares on the other two sides i.e. ${H^2} = {P^2} + {B^2}$ where H = Hypotenuse of the right-angled triangle, P represents the perpendicular of the right-angled triangle and B represents the Base of the right-angled triangle.