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3 years ago

What is the area of ABDC? Assume ABDC lies in the center of the larger rectangle.

Mathematics

2 answers:

nadezda [96]3 years ago

8 0

Answer:

Area of ABDC is 24 square feet.

Step-by-step explanation:

Given rectangle ABDC which lies in the center of the larger rectangle. We have to find the area of ABDC.

Length of rectangle=outer length-given gap from both sides

=12-3-3=6 ft.

Breadth of rectangle=outer breadth-given gap from both sides

=10-3-3=4 ft.

Now, $\text{area of rectangle ABDC=}Length\times breadth$

$=6\times 4=24ft^2$

hence, area of ABDC is 24 square feet.

Option C is correct.

MrRa [10]3 years ago

7 0

The area of rectangle ABCD should be 24. Since the length is 6 and the height is 4.

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Answer:

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A quadratic equation is in the form: $ax^2+bx+c = 0$ where a, b ,c are the coefficient and a≠0 then the solution is given by :

$x_{1,2} = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$ ......[1]

On comparing with given equation we get;

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$x_{1,2} = \frac{-10\pm \sqrt{10^2-4\cdot 3 \cdot c}}{2 \cdot 3}$

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$x_1 -x_2 = \frac{14}{3}$

Here,

$x_1 = \frac{-10 + \sqrt{100- 12c}}{6}$ , ......[2]

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$\frac{-10 + \sqrt{100- 12c}}{6} - (\frac{-10 + \sqrt{100- 12c}}{6}) = \frac{14}{3}$

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$x_1 = \frac{-10 + \sqrt{4}}{6}$

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$x_2 = \frac{-10 - \sqrt{100- 12\cdot 8}}{6}$

$x_2 = \frac{-10 - \sqrt{100- 96}}{6}$ or

$x_2 = \frac{-10 - \sqrt{4}}{6}$

Simplify:

$x_2 = -2$

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