Question

The sides of a triangle are of lengths of 20, 21 & 29. The sum of lengths of altitude will be?
A.\[\dfrac{{1609}}{{29}}\] units
B.40 units
C. \[\dfrac{{1609}}{{21}}\]units
D.70 units

Answer 1

Hint: Try to find, either it is a right-angled triangle or not. Then, find the altitude by equating the area.
Since, \[area = \dfrac{1}{2} \times base \times altitude\]
Given: all the three sides of the triangle (i.e. 20, 21 &29)

Stepwise Solution:
Step 1: First we need to find an approach for the question as it is very difficult to find altitude just by the given data.
Since, we can see,
\[{(29)^2} = {(20)^2} + {(21)^2}\]
being, \[841 = 400 + 441\]
Since, the square of one side is equal to the sum of the square of the other two sides,we can conclude and say it is a right angled triangle
So, we can state that the given triangle is a right angled triangle.
Step 2: Now, let us consider 3 altitudes being a, b & c.
We can easily see that the altitude a & b are the respective sides being 20 & 21.
So,
\[a = 20cm\]& \[b = 21cm\]
Now, for determining c,
Since the area of a fixed shape is fixed.
Hence, the side used does not matter.
Step 3: So, by the above statement,
Either we use 20 as base or 29, the value is the same.
\[area = \dfrac{1}{2} \times base \times altitude\]
Since, no matter which is the altitude taken, the area of a triangle will be the same
\[\dfrac{1}{2} \times 20 \times 21 = \dfrac{1}{2} \times 29 \times c\]
So, we got, \[c = \dfrac{{420}}{{29}}\].
Step 4: Now if we add a, b & c.
We get, \[ = 20 + 21 + \dfrac{{420}}{{29}}\]
\[ = \dfrac{{1609}}{{29}}units\].

So, the correct option is (A).

Note: It is important, the Pythagoras triples (i.e. \[{a^2} + {b^2} = {c^2}\]) so as to identify the right-angled triangle. Otherwise using the other approaches is very difficult. Whenever you have to find altitude, check if the triangle is a right angled-triangle. Otherwise, in other questions, you can find the area of the triangle by other methods and then compute it with half of the multiple of base and height.