The length of one side of a triangle is one less than twice the smallest side and the length of
other side is three more than the smallest side. If the perimeter of the triangle is 162 cm.
Find the area of the triangle.
Answers
Answer :
Given :
The length of one side of a triangle is one less than twice the smallest side and the length of other side is three more than the smallest side. If the perimeter of the triangle is 162 cm.
To find :
Find the area of the triangle.
Solution :
Let the smallest side of the triangle be
X cm
The length of one side = one less than twice the smallest side
=> (2X-1) cm
And
The length of the other side = three more than the smallest side
=> (X+3) cm
The three sides are X cm , (2X-1) cm and (X+3) cm
Perimeter of a triangle = Sum of all sides
=> P = X+2X-1+X+3
=> P = (4X+2 ) cm
According to the given problem
the perimeter of the triangle is 162 cm.
=> 4X+2 = 162
=> 4X = 162-2
=> 4X = 160
=> X = 160/4
=> X = 40 cm
The smallest side = 40 cm
Now 2X-1 = 2(40)-1=80-1=79 cm
and
X+3 = 40+3 = 43 cm
The three sides are 40 cm ,43 cm , 79 cm
Let a= 40 cm
b = 43 cm
c = 79 cm
P = 162 cm
We know that
S = (a+b+c)/2
=> S = 162/2 = 81 cm
We know that
Heron's formula for finding area of the triangle =√[S(S-a)(S-b)(S-c)] sq.units
On Substituting these values in the above formula then
=> A = √[81(81-40)(81-43)(81-79)]
=> A =√[(81)(41)(38)(2)]
=> A = √(498636)
=> A = 706.14 sq.cm ( approximately)
Answer:-
The area of the given trianagle is 706.14 sq.cm
Used formulae:-
Heron's formula for finding area of the triangle =√[S(S-a)(S-b)(S-c)] sq.units
Where S = (a+b+c)/2
Sum of all the sides is the perimeter of the given figure
Answer:
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