Please help on the area for 7 8 9 please all i need help is

The area of the equilateral, isosceles and right angled triangle are 12.6mm², 9.61in² and 16.81yds² respectively.
Note that:
The area of an Equilateral triangle is expressed as A = ((√3)/4)a²
Where a is the dimension of the side.
The area of an Isosceles triangle is expressed as A = (ah)/2
Where a is the dimension of the base and h is the height.
The area of a Right angled triangle is expressed as A = (ab)/2
Where a and b is the dimension of the two sides other than the hypotenuse.
For the Equilateral triangle.
Given that;
A = ((√3)/4)(5.4mm)²
A = ((√3)/4)( 29.16mm² )
A = 12.6mm²
Area of the Equilateral triangle is 12.6mm²
For the Isosceles triangle.
Given that;
The height h is the imaginary line drawn upward from the center of a.
First, we calculate the height using Pythagorean theorem
x² = y² + z²
Where x = b = 5.9in, y = a/2 = 3.4in/2 = 1.7in, and z = h
(5.9in)² = (1.7in)² + h²
34.81in² = 2.89in² + h²
h² = 34.81in² - 2.89in²
h² = 31.92in²
h = √31.92in²
h = 5.65in
Now, the area will be;
A = (ah)/2
A = (3.4in × 5.65in )/2
A = 19.21in²/2
A = 9.61in²
Area of the Isosceles triangle is 9.61in².
For the Right angled triangle
Given that;
A = (ab)/2
A = ( 8.2yds × 4.1yds)/2
A = ( 33.62yds²)/2
A = 16.81yds²
Area of the Right angled triangle is 16.81yds²
Therefore, the area of the equilateral, isosceles and right angled triangle are 12.6mm², 9.61in² and 16.81yds² respectively.
Learn more about Pythagorean theorem here: brainly.com/question/343682
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