Find the area of the unshaded reason for these two problems

Mathematics

1 answer:

Taya2010 [7]2 years ago

7 0

9514 1404 393

Answer:

a) 96 m²

b) 26,000 ft²

Step-by-step explanation:

The area of the unshaded region is the difference between the overall area of the rectangle and the area of the shaded region. It can also be found directly using the dimensions of the unshaded region.

a) We have the height of the right triangle, but need to know its base. The Pythagorean theorem can help with that. Let the base of the triangle be represented by x. Then ...

x² +16² = 20²

x² = 400 -256 = 144

x = √144 = 12

The area of the unshaded triangle is given by the formula ...

A = (1/2)bh

A = (1/2)(12)(16) = 96 . . . . square meters

b) The shaded area at upper left is a square 100 ft on a side. Its area is ...

A = s² = (100 ft)² = 10,000 ft²

The shaded area at upper right is a triangle with base and height 120 ft and 150 ft. Its area is ...

A = (1/2)bh = (1/2)(120 ft)(150 ft) = 9,000 ft²

The area of the overall rectangle is ...

A = LW = (300 ft)(150 ft) = 45,000 ft²

So, the unshaded area is ...

unshaded area = overall area - left shaded area - right shaded area

unshaded area = 45,000 ft² -10,000 ft² -9,000 ft² = 26,000 ft²

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2. Solve the equation using the quadratic formula. -4x2 + 40x + 16 = 0 Page 1 of 2

kumpel [21]

The solution to given quadratic equation is x = -0.3851 or x = 10.3851

Solution:

Given equation is:

$-4x^2 + 40x + 16 = 0$

To find: solve by quadratic equation formula

$\text {For a quadratic equation } a x^{2}+b x+c=0, \text { where } a \neq 0\\\\x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Using the above formula,

$\text{ For} -4x^2 + 40x + 16 = 0 \text{ we have } a = -4 ; b = 40 ; c = 16$

Substituting the values of a = -4 ; b = 40 ; c = 16 in above quadratic formula we get,

$\begin{aligned}&x=\frac{-40 \pm \sqrt{40^{2}-4(-4)(16)}}{2 \times-4}\\\\&x=\frac{-40 \pm \sqrt{1600+256}}{-8}\\\\&x=\frac{-40 \pm 43.081}{-8}\\\\&x=\frac{-40+43.081}{-8} \text { or } x=\frac{-40-43.081}{-8}\\\\&x=-0.3851 \text { or } x=10.3851\end{aligned}$