Up next in 10
Find the Sides of a Right Triangle with Perimeter and Area of 30
0 views
Jun 28, 2025
In this video, I solve a right triangle problem where both the perimeter and area are 30. Using the Pythagorean theorem and algebra, I break down the steps to find the sides of the triangle. Watch to learn how to solve this challenging geometry problem!
View Video Transcript
0:00
find the sides of a right triangle whose
0:02
perimeter and area are equal to 30 as
0:05
usual try to solve the problem on your
0:07
own the perimeter of the triangle by
0:10
definition is equal to the sum of the
0:12
lengths of all its sides and the area of
0:14
the right triangle in this case is equal
0:16
to half the product of its legs
0:18
additionally we will use the Pythagorean
0:20
theorem for this right triangle which
0:22
states that the sum of the squares of
0:24
the legs is equal to the square of the
0:27
hypotenuse to solve the system we will
0:29
try to express the sum of A and B
0:32
through the sum of their squares for
0:34
this we will need to add to the sum of
0:36
the squares of A and B their double
0:38
product which we can find from the
0:40
second equation so from the first
0:42
equation we express the sum of A and B
0:45
and from the second equation we express
0:47
the product of A and B for this both
0:50
sides of the equation need to be
0:51
multiplied by two and to each side of
0:53
the third equation we will add the
0:55
double product of A and B as a result in
0:58
the left part of the Third thir equation
1:00
we get the sum of A and
1:03
B let's solve the third equation of the
1:06
system separately for this we substitute
1:09
from the first equation instead of the
1:11
sum of a and b 30- c and instead of the
1:15
product of A and B we substitute the
1:18
number 60 from the second equation then
1:20
30 - c will be equal to C of 2qu + 2 *
1:24
60 we open the parentheses and subtract
1:27
from both sides of the equation we
1:29
express press and find C then return to
1:32
the system and find A and B from the
1:34
first two equations then the sum of A
1:37
and B will be equal to 30 minus C that
1:40
is 13 the product of A and B is equal to
1:43
60 the system can be solved in different
1:46
ways for example you can express one of
1:49
the variables from the first equation
1:50
and substitute it into the second
1:52
equation or you can recall vieta's
1:55
theorem According to which the numbers A
1:58
and B are the roots of the quadratic
2:00
equation x^ 2 minus the sum of A and B *
2:05
X plus the product of the roots that is
2:08
60 60 is the product of 5 and 12 and 17
2:12
is the sum of the roots 5 and 12 with a
2:15
negative sign so either a = 5 and B = 12
2:20
or a = 12 and B = 5 thus the sides of
2:25
the right triangle are 5 12 and 13 if
2:29
you unod give a like write comments and
2:33
subscribe to the channel
#Science