Question

Each diagram below shows a right angle with the altitude drawn to the hypotenuse. How do I find the value of x, y and z?

Answer 1

Hint: According to the question, we have to apply the Pythagoras’ theorem, first in the smaller triangle, then in the bigger triangle, and then in the outer triangle. After that we will substitute the values and find out the values of x, y and z.

Formula used: \[{p^2} + {b^2} = {h^2}\]

Complete step by step solution:
From the above diagram, we can see that there are three right angled triangles.
In the first diagram, we will check for the smaller triangle. We will apply the Pythagoras’ theorem in the smaller triangle. The formula of a Pythagoras’ triangle is:
\[{p^2} + {b^2} = {h^2}\]
Here, \[p = perpendicular\], \[b = base\], \[h = hypotenuse\]
When we apply this formula, on the smaller triangle of the first diagram, then we get:
\[{x^2} + {(2)^2} = {y^2}\]
\[ \Rightarrow {x^2} + 4 = {y^2}\]
We will consider this equation as equation (1).
Now, we will apply the same Pythagoras’ formula in the bigger triangle of the first diagram, and we get:
\[ \Rightarrow {x^2} + {(8)^2} = {z^2}\]
\[ \Rightarrow {x^2} + 64 = {z^2}\]
We will consider this equation as equation (2).
When we combine both the right angles in the first diagram, then:
The hypotenuse is \[10\], the perpendicular is \[z\] and the base is \[y\]. When we apply the Pythagoras’ theorem in the outer bigger triangle, we get:
\[ \Rightarrow {y^2} + {z^2} = 100\]
We will consider this equation as equation (3).
Now, we will try to substitute the equation (1) in the equation (3), and we get:
\[ \Rightarrow {x^2} + 4 + {z^2} = 100\] \[(because\,\,{x^2} + 4 = {y^2})\]
Now, we will substitute this result into the equation (2), and we get:
\[ \Rightarrow {x^2} + 4 + {x^2} + 64 = 100\]
Now, we will solve this, and we get:
\[ \Rightarrow 2{x^2} + 68 = 100\]
\[ \Rightarrow 2{x^2} = 32\]
We will try to solve \[x\]. For that we have to make \[x\] alone:
\[ \Rightarrow {x^2} = \dfrac{{32}}{2}\]
By solving this, we get:
\[ \Rightarrow {x^2} = 16\]
\[ \Rightarrow x = \sqrt {16} \]
\[ \Rightarrow x = 4\]
Now, we will put the value of \[x\] in equation (1), and we get:
\[ \Rightarrow {y^2} = 16 + 4\]
\[ \Rightarrow {y^2} = 20\]
\[ \Rightarrow y = 2\sqrt 5 \]
Therefore, \[y = 2\sqrt 5 \].
Now, we will try to find the value of \[z\]. We will put the value of \[x\] in equation (2), and we get:
\[ \Rightarrow {z^2} = 16 + 64\]
\[ \Rightarrow {z^2} = 80\]
\[ \Rightarrow z = 4\sqrt 5 \]

Note: If we apply Pythagoras’ theorem in the second figure as well, we will find the same values of x, y and z. We will apply the same method, first by putting Pythagoras’ theorem in the smaller circle, then in the bigger circle, and then in the outer circle, and then substitute the values.