find the perimeter of triangle Abc whose sides are 26/8,9/4,7/2 and a square whose side is 7/2 find whose perimeter is greater?
Answers
Given:
A triangle ABC with lengths 26/8, 9/4 and 7/2
A square with a side of 7/2
To Find:
1. The Perimeter of triangle and square.
2. Whose perimeter is greater.
Solution:
1. Let, the sides of ∆ABC be AB, BC, AC
therefore,
AB=26/8 ; BC=9/4 & AC=7/2
The perimeter of a triangle ABC
= Sum of all sides
= AB+BC+AC
= 26/8 + 9/4 + 7/2
(Simplifying a bit, this expression can also be written as):
= 3.25 + 3.5 + 2.25
= 9.0
Therefore,
The Perimeter of Given Triangle ABC is 9 units.
now,
The perimeter of a square
= Sum of all sides
= 4 x ( 7/2 )
(Simplifying a bit, this expression can also be written as):
= 4 x 3.5
= 14.0
Therefore,
The Perimeter of Given Square is 14 units.
2. Now we have to find whose perimeter is greater,
The perimeter of Triangle is 9 and
The perimeter of Square is 14.
so, we need to compare 9 and 14.
Here, 14 is greater than 9
Thus, the Perimeter of a Square is Greater.
Step-by-step explanation:
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