Question

PLEASE HELPPP!!!!!!!!!!!!!
1. Find the area of a triangle with sides of 20, 23, and 27 mm (millimeters).
2. Calculate the area of a triangle with a base of 20 m (meters) and an altitude of 18 m.
3. Laura bought a rectangular rug for her living room. It measures 9 feet long and 8 feet wide. What is the area of her rug?
4. One can of paint will cover an area of 400 square feet. Will one can of paint be enough to cover an exterior brick wall that is 48 feet long and 9 feet high?
5. A regular hexagon has a base of 10 cm and a height of 5 cm. What's the area of the hexagon?

Answer 1

For this case we have:

Question 1:

According to Heron's formula, the area of ​​a triangle is given by:

$A=\sqrt{(s(s-a)(s-b)(s-c))}$

Where a, b and c are the sides of the triangle and s is the semi-perimeter of the triangle given by:

$s=\frac{(a+b+c)}{2}$

Let:

$a = 20mm\\b = 23mm\\c = 27mm$

We have s:

$s=\frac{(20+23+27)}{2}\\s=\frac{70}{2}\\s=35$

Substituting in the formula of the area:

$A=\sqrt{(35(35-20)(35-23)(35-27))}\\A=\sqrt{(35(15)(12)(8))}\\A=\sqrt{(50400)}\\A=224.5mm^2$

Thus, the area of ​​the triangle is $A = 224.5mm ^ 2$

Answer:

$A = 224.5mm ^ 2$

Question 2:

The area of ​​a traingule can be expressed as:

$A=\frac{(b*h)}{2}$

Where:

b: Base of the triangle

h: Triangle height

Let:

$b = 20m\\h = 18m$

Substituting the values ​​in the expression we have:

$A=\frac{(20*18)}{2}\\A=\frac{360}{2}\\A=180m^2$

Thus, the area of ​​the triangle is $A = 180m ^ 2$

Answer:

$A = 180m ^ 2$

Question 3:

It is known that the area of ​​a rectangle is given by:

$A = l * w$

Where:

l: It is the length of the rectangle

w: It is the width of the rectangle

So:

$l = 9 feet\\w = 8 feet$

Substituting:

$A = (9 * 8) feet ^ 2\\A = 72 feet ^ 2$

Thus, the area of ​​Laura's carpet is given by $A = 72 feet ^ 2$

Answer:

$A = 72 feet ^ 2$

Question 4:

We must take out the area of ​​the outer wall, given by a rectangle, then:

$A = l * a$

Where:

l: It is the length of the rectangle

a: It is the height of the rectangle

We have:

$l = 48 feet\\a = 9 feet$

Substituting in the formula we have:

$A = (48 * 9) feet ^ 2\\A = 432ft ^ 2$

Thus, the area of ​​the exterior wall is given by: $A = 432 feet ^ 2$

$432ft ^ 2> 400ft ^ 2$

So, a can of paint is not enough to cover the exterior wall.

Answer:

A can of paint is not enough to cover the exterior wall.

Question 5:

A regular hexagon is formed by 6 equal triangles. The area of ​​the hexagon is given by:

$A=\frac{(perimeter* apothem)}{2}$

Where the perimeter is given by the sum of the sides, that is:

$perimeter = (10 + 10 + 10 + 10 + 10 + 10) cm\\perimeter = 60cm$

And the apothem is the height of each of the triangles that make up the hexagon, that is:

$apothem = 5cm$

Substituting in the formula:

$A=\frac{(60*5)}{2}\\A=\frac{300}{2}\\A=150cm^2$

Thus, the area of ​​the hexagon is $A = 150cm ^ 2$

Answer:

$A = 150cm ^ 2$