Question

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

Answer 1

The area of the equilateral, isosceles and right angled triangle are 12.6mm², 9.61in² and 16.81yds² respectively.

What is the area of the equilateral, isosceles and right angle triangle?

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 = 5.4mm
• Area A = ?

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;

• Base a = 3.4in
• Slant height b = 5.9in
• height h = ?
• Area A = ?

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 = 8.2yds
• b = 4.1yds
• c = 9.17yds
• Area A = ?

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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