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3 years ago

Find area of a parallelogram with a height of 8 inches and a base of 14 inches

Mathematics

1 answer:

12345 [234]3 years ago

4 0

Answer:

$\boxed{ \bold{ \huge{ \boxed{ \sf{112 \: {inches} \: ^{2} }}}}}$

Step-by-step explanation:

Given,

Height of a parallelogram ( h ) = 8 inches

Base of a parallelogram ( b ) = 14 inches

Area of a parallelogram ( A ) = ?

Finding the area of a parallelogram

$\boxed{ \sf{area \: of \: parallelogram \: = \: base \: \times \: height}}$

⇒ $\sf{area \: of \: parallelogram \: = \: 14 \: \times 8}$

⇒ $\sf{area \: of \: parallelogram = 112 \: {inches}^{2} }$

Hope I helped!

Best regards! :D

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A norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular. Find the dimensions of a norman

Yanka [14]

Answer:

$W\approx 8.72$ and $L\approx 15.57$ .

Step-by-step explanation:

Please find the attachment.

We have been given that a norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular. The total perimeter is 38 feet.

The perimeter of the window will be equal to three sides of rectangle plus half the perimeter of circle. We can represent our given information in an equation as:

$2L+W+\frac{1}{2}(2\pi r)=38$

We can see that diameter of semicircle is W. We know that diameter is twice the radius, so we will get:

$2L+W+\frac{1}{2}(2r\pi)=38$

$2L+W+\frac{\pi}{2}W=38$

Let us find area of window equation as:

$\text{Area}=W\cdot L+\frac{1}{2}(\pi r^2)$

$\text{Area}=W\cdot L+\frac{1}{2}(\pi (\frac{W}{2})^2)$

$\text{Area}=W\cdot L+\frac{\pi}{2}(\frac{W}{2})^2)$

$\text{Area}=W\cdot L+\frac{\pi}{2}(\frac{W^2}{4})$

$\text{Area}=W\cdot L+\frac{\pi}{8}W^2$

Now, we will solve for L is terms W from perimeter equation as:

$L=38-(W+\frac{\pi }{2}W)$

Substitute this value in area equation:

$A=W\cdot (38-W-\frac{\pi }{2}W)+\frac{\pi}{8}W^2$

Since we need the area of window to maximize, so we need to optimize area equation.

$A=W\cdot (38-W-\frac{\pi }{2}W)+\frac{\pi}{8}W^2$

$A=38W-W^2-\frac{\pi }{2}W^2+\frac{\pi}{8}W^2$

Let us find derivative of area equation as:

$A'=38-2W-\frac{2\pi }{2}W+\frac{2\pi}{8}W$

$A'=38-2W-\pi W+\frac{\pi}{4}W$

$A'=38-2W-\frac{4\pi W}{4}+\frac{\pi}{4}W$

$A'=38-2W-\frac{3\pi W}{4}$

To find maxima, we will equate first derivative equal to 0 as:

$38-2W-\frac{3\pi W}{4}=0$

$-2W-\frac{3\pi W}{4}=-38$

$\frac{-8W-3\pi W}{4}=-38$

$\frac{-8W-3\pi W}{4}*4=-38*4$

$-8W-3\pi W=-152$

$8W+3\pi W=152$

$W(8+3\pi)=152$

$W=\frac{152}{8+3\pi}$

$W=8.723210$

$W\approx 8.72$

Upon substituting $W=8.723210$ in equation $L=38-(W+\frac{\pi }{2}W)$ , we will get:

$L=38-(8.723210+\frac{\pi }{2}8.723210)$

$L=38-(8.723210+\frac{8.723210\pi }{2})$

$L=38-(8.723210+\frac{27.40477245}{2})$

$L=38-(8.723210+13.70238622)$

$L=38-(22.42559622)$

$L=15.57440378$

$L\approx 15.57$

Therefore, the dimensions of the window that will maximize the area would be $W\approx 8.72$ and $L\approx 15.57$ .

8 0

3 years ago

AB = x + 8

vivado [14]

Answer: The length of BC is 7

Step-by-step explanation: Assuming the lengths of the opposite sides of the quadrilateral are congruent, then

AB=DC and

AD=BC

Inputting the values of AB, DC and AD as given in the question:

x + 8 = 3x ...(1)

x + 3=? ...(2)

We have to solve for the value of x to get the actual lengths and thus ascertain BD.

From equation (1):

8 = 3x - x

8 = 2x

8/2 = x

Therefore, x = 4.

If x = 4 then equation(2) would be

4 + 3= 7.

Hence, the actual lengths of the quadrilateral are:

AB = 4 + 8. DC = 3(4)

=12. =12.

AD = 4 + 3. AD = BC

= 7. Therefore, BC = 7.

Hence, it is confirmed that quadrilateral ABCD is a parallelogram since both the opposite sides are proven to be congruent.

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3 years ago

Find area of a parallelogram with a height of 8 inches and a base of 14 inches￼

Find area of a parallelogram with a height of 8 inches and a base of 14 inches