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3 years ago

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Which simplified expression represents the area of the parallelogram

1 answer:

Charra [1.4K]3 years ago

5 0

Answer:

The area of a parallelogram is its base times its height.

Step-by-step explanation:

Area is for 2 dimensional shapes. (Shapes with two dimensions in parallelograms they are often referred to as base times height, or length times width). Area is the space inside a two dimensional object, and to find this use the length and the height to find the space in the 2-D figure.

Side note: Volume is "the area of 3 dimensional figures". Three dimensions meaning Length, Width and Height. So there is no area for 3-D figures. (unless you are using Surface Area)

Hope this Helps!

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If you have 14 feet of fencing, what are the areas of the different rectangles you could enclose with the fencing? Consider only

disa [49]

Answer:

Area of rectangle possible. : 6ft² ; 10 ft², 12 ft²

Step-by-step explanation:

Feets of fencing = 14

Perimeter = 14

Let x = Length y = width

Perimeter = 2(x + y)

For x = 1

2(1 + y) = 14

1 + y = 7

y =6

Area of rectangle ; x *y = 1 * 6 = 6 ft²

For x = 2

2(2 + y) = 14

2 + y = 7

y = 5

Area of rectangle ; x *y = 2 * 5 = 10 ft²

For x = 3

2(3 + y) = 14

3 + y = 7

y =4

Area of rectangle ; x *y = 3 * 4 = 12 ft²

For x = 4

2(1 + y) = 14

4 + y = 7

y = 3

Area of rectangle ; x *y = 4 * 3 =12 ft²

Hence, Area of rectangle possible. : 6ft² ; 10 ft², 12 ft²

8 0

3 years ago

Can you help me with my math questions

GrogVix [38]

$p^2+2(7)p+7^2=(p+7)^2$

Justification:

Given: $p^2+14p+49$

Split the middle term;

$=p^2+7p+7p+49$

Factor:

$=p(p+7)+7(p+7)$

$=(p+7)(p+7)$

$=(p+7)^2$

QUESTION 3a

The given expression for the area of the rectangle is $36x^2-12x+1$ .

This is equal to the indicated area which is $289\;in^2$

This implies that

$36x^2-12x+1=289$

$\Rightarrow (6x-1)^2=289$

$\Rightarrow (6x-1)(6x-1)=289$

$\Rightarrow l=(6x-1),w=(6x-1)$

This implies that, the dimensions of the rectangle are equal;

Using the square root method, we obtain

$\Rightarrow (6x-1)=\pm \sqrt{289}$

$\Rightarrow (6x-1)=\pm 17$

$\Rightarrow 6x=1\pm 17$

$\Rightarrow 6x=18 \:or\:6x=-16$

$\Rightarrow x=3 \:or\:x=-\frac{2}{3}$

We discard the negative value.

The side length of this rectangle is

$\Rightarrow l=(6(3)-1)=17,w=6(3)-1=17$

QUESTION 3b

The given expression for the area of the rectangle is $25x^2-50x+25$ .

This is equal to the indicated area which is $1225\;in^2$

This implies that

$25x^2-50x+25=1225$

$25(x^2-2x+1)=1225$

$(5(x-1))^2=35^2$

$5(x-1)5(x-1)=49$

$\Rightarrow l=5(x-1),w=5(x-1)$

This implies that, the dimensions of the rectangle are equal;

Using the square root method, we obtain

$\Rightarrow 5(x-1)=\pm \sqrt{1225}$

$\Rightarrow 5(x-1)=\pm 35$

$\Rightarrow x=1\pm 7$

$\Rightarrow x=8 \:or\:x=-6$

We discard the negative value.

The side length of this rectangle is

$\Rightarrow l=5(8-1)=35,w=5(8-1)=35$

QUESTION 3c

The given expression for the area of the rectangle is $49x^2-56x+16$ .

This is equal to the indicated area which is $289\;in^2$

This implies that

$49x^2-56x+16=289$

$(7x-4)^2=289$

$(7x-4)(7x-4)=289$

$\Rightarrow l=(7x-4),w=(7x-4)$

Applying the laws of indices gives;

$(7x-4)^2=17^2$

This implies that;

$7x-4=17$

$7x=21$

$x=3in.$

The side length of this rectangle is

$\Rightarrow l=7(3)-4=17\:in.,w=7(3)-4=17\:in.$

Dont forget that the square is also a rectangle.

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