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3 years ago

11

I completely lost!! SOMEONE PLEASE HELP!

2 answers:

borishaifa [10]3 years ago

8 0

To determine the area of the shaded region, you must
1: find the area of the whole rectangle
2: find the area of the non-shaded rectangle inside
3: subtract the non-shaded rectangle from the whole rectangle.

densk [106]3 years ago

8 0

This is easier than you think.

The area of a rectangle is l x w, right?
Well, you have two rectangles: The rectangle that is the whole of the picture, and the tiny rectangle inside the shaded area.
So the big rectangle:

A = 10 x 14 5/8
= 146.25
The insdie rectangle:
A = 5 x 2 1/2
= 12.5

Now take the small one away from the large one:
146.25 - 12.5 = 133.75
Or, to put it in the same way as the question:
133 3/4 inches²

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3 years ago

The dimensions of a rectangle are such that its length is 3 in. more than its width. If the length were doubled and if the width

Kay [80]

Answer:

The length of the rectangle is 11 inch and its width is 8 inch.

Step-by-step explanation:

Given:

Rectangle $(R_1)$ and rectangle $(R_2)$ .

Let the width of the rectangle be 'x' inch.

Its length = '(x+3)' inch

Area of the rectangle = $(A_1)$

⇒ $A_1$ = $(x)(x+3)$

Now for rectangle $(R_2)$ its width is decreased by 1 inch and length is doubled,the area $(A_2)$ will increased by 66 sq-inch.

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$A_2=A_1+66$

According to the question.

⇒ Area of rectangle $(R_2)$ with width '(x-1)' and length '2(x+3)'='(2x+6)' .

⇒ $A_2=(x-1)(2x+6)$

Plugging the values of $A_2$ in terms of $A_1$ .

⇒ $A_1+66= (x-1)(2x+6)$

Plugging $A_1$ = $(x)(x+3)$ into the equation.

⇒ $(x)(x+3)+66=(x-1)(2x+6)$

⇒ $x^2+3x+66=2x^2+6x-2x-6$

⇒ $x^2-2x^2+3x-6x+2x+6+66=0$

⇒ $-x^2-x+72=0$

Now using quadratic formula we can find the value of 'x'.

Quadratic formula, (x)= $\frac{\pm b+\sqrt{b^2-4ac} }{2a}$

x= -9 and x= 8

Discarding the negative value.

Our width = 'x' = 8 inches.

So the length of the rectangle = 11 inches and width of the rectangle = 8 inches.

7 0

3 years ago