Question

How do you find the perimeter of an isosceles triangle whose base is 16 dm and whose height is 15 dm?

Answer 1

Hint: Here in this question, we have to find the perimeter of an isosceles triangle by using the given base is 16 dm and height is 15 dm. first we have to find the length of another two sides of a isosceles triangle by using the concept of Pythagoras theorem i.e., \[hy{p^2} = ad{j^2} + op{p^2}\] and by using formula of a perimeter \[p = a + b + c\] later on simplification we get the required solution.

Complete step-by-step solution:
Finding the perimeter of a triangle means finding the distance around the triangle. The simplest way to find the perimeter of a triangle is to add up the length of all of its sides
The perimeter of the triangle with sides a, b and c is \[p = a + b + c\].

Consider a triangle \[\Delta \,ABC\] whose Base is 16dm i.e., \[BC = 16\]dm
Draw a line AD perpendicular to the line BC i.e., \[AD \bot BC\].
Where the height of the \[\Delta \,ABC\] is \[AD = 15\,dm\] and \[BC = BD + CD\]
\[BD = CD\]
\[BC = BD + BD\]
\[BC = 2BD\]
\[BD = \dfrac{{BC}}{2} = \dfrac{{16}}{2}\]
\[\therefore BD = 8\]dm
By the use of Pythagoras theorem and get the slant height or height of another two lines
i.e., \[hy{p^2} = op{p^2} + ad{j^2}\]
for \[\Delta \,ABD\]
\[ \Rightarrow \,\,\,A{B^2} = A{D^2} + B{D^2}\]
\[ \Rightarrow \,\,\,A{B^2} = A{D^2} + B{D^2}\]
\[ \Rightarrow \,\,\,A{B^2} = {15^2} + {8^2}\]
\[ \Rightarrow \,\,\,A{B^2} = {15^2} + {8^2}\]
\[ \Rightarrow \,\,\,A{B^2} = 225 + 64\]
\[ \Rightarrow \,\,\,A{B^2} = 289\]
Taking square root on both the side, then
\[ \Rightarrow \,\,\,AB = \sqrt {289} \]
289 is the square number of 17
\[ \Rightarrow \,\,\,AB = \sqrt {{{17}^2}} \]
\[ \Rightarrow \,\,\,AB = 17\]dm
The given triangle is an isosceles triangle that has two equal sides with two equal angles opposite of those sides.
Given the length of base line \[AC = 16\]dm and another two which are having same length
\[AB = AC = 17\]dm
Perimeter of \[\Delta \,ABC\] is
\[P = 17 + 17 + 16\]
\[P = 50\]dm

Hence, the perimeter of give isosceles triangle \[\Delta \,ABC\] is 50dm

Note: While determining the perimeter we use the formula. The unit for the perimeter will be the same as the unit of the length of a side or triangle. Whereas the unit for the area will be the square of the unit of the length of a triangle. We should not forget to write the unit. we should also know about the Pythagoras theorem.