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Gre4nikov [31]
3 years ago
10

Original scale: 1 inch = 8 feet. A parallelogram with a side length of 24 feet. A larger parallelogram with a side length of 72

feet.
Noah planned a parallelogram-shaped sidewalk to lead from the street to the front door of his house. He realized the walkway was too short and he needs to change the length to 72 feet. He wrote a ratio to find the change in the scale factor.

24 ft
72 ft
=


The change in the scale factor is
✔ 1/3
.
The change of scale means that 1 inch represented
✔ 8
feet, but now 1 inch represents
✔ 24
feet.
Mathematics
1 answer:
musickatia [10]3 years ago
4 0

Answer:

The change in the scale factor is

✔ 1/3

The change of scale means that 1 inch represented

✔ 8

feet, but now 1 inch represents

✔ 24

feet.

Step-by-step explanation:

If you have a rectangle or square, each of the angles measures 90°. If you have a parallelogram or rhombus, the opposite angles are the same and the consecutive angles are supplementary. For other types of quadrilaterals you may need to add the given angles and then subtract from 360°.

[Also Pic with proof]

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Solve the following equation: *<br> -9 (t – 2) = 4(t – 15)
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5 0
3 years ago
Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.
melamori03 [73]

Answer:

6+2\sqrt{21}\:\mathrm{cm^2}\approx 15.17\:\mathrm{cm^2}

Step-by-step explanation:

The quadrilateral ABCD consists of two triangles. By adding the area of the two triangles, we get the area of the entire quadrilateral.

Vertices A, B, and C form a right triangle with legs AB=3, BC=4, and AC=5. The two legs, 3 and 4, represent the triangle's height and base, respectively.

The area of a triangle with base b and height h is given by A=\frac{1}{2}bh. Therefore, the area of this right triangle is:

A=\frac{1}{2}\cdot 3\cdot 4=\frac{1}{2}\cdot 12=6\:\mathrm{cm^2}

The other triangle is a bit trickier. Triangle \triangle ADC is an isosceles triangles with sides 5, 5, and 4. To find its area, we can use Heron's Formula, given by:

A=\sqrt{s(s-a)(s-b)(s-c)}, where a, b, and c are three sides of the triangle and s is the semi-perimeter (s=\frac{a+b+c}{2}).

The semi-perimeter, s, is:

s=\frac{5+5+4}{2}=\frac{14}{2}=7

Therefore, the area of the isosceles triangle is:

A=\sqrt{7(7-5)(7-5)(7-4)},\\A=\sqrt{7\cdot 2\cdot 2\cdot 3},\\A=\sqrt{84}, \\A=2\sqrt{21}\:\mathrm{cm^2}

Thus, the area of the quadrilateral is:

6\:\mathrm{cm^2}+2\sqrt{21}\:\mathrm{cm^2}=\boxed{6+2\sqrt{21}\:\mathrm{cm^2}}

4 0
3 years ago
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