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3 years ago

What is the length of segment CD question mark

Mathematics

1 answer:

blagie [28]3 years ago

8 0

Answer:

CD = two square root of 10 end square root

Step-by-step explanation:

To find the length of a segment, use the distance formula. Substitute the order pairs for the endpoints of the segment. CD has the end points (-7, -4) and (-1, -2).

$d = \sqrt{(y_2-y_1)^2 + (x_2-x_1)^2} \\\\d = \sqrt{(-4--2)^2 + (-7--1)^2} \\\\d = \sqrt{-2^2 + -6^2}\\\\d = \sqrt{4 + 36}\\\\d=\sqrt {40} = 2\sqrt{10}$

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polygon A i simllar to polgon B. find the perimeer of polygon B if one side of polygon A is 24 A sde to polygon B is 15 and the

Gemiola [76]

Answer:

Perimeter of polygon B = 80 units

Step-by-step explanation:

Since both polygons are similar, their corresponding sides and perimeters are proportional. Knowing this we can setup a proportion to find the perimeter of polygon B.

$\frac{side-polygonA}{side-polygonB} =\frac{perimeter-polygonA}{perimeter-polygonB}$

Let $x$ be the perimeter of polygon B. We know from our problem that the side of polygon A is 24, the side of polygon B is 15, and the perimeter of polygon A is 128.

Let's replace those value sin our proportion and solve for $x$ :

$\frac{side-polygonA}{side-polygonB} =\frac{perimeter-polygonA}{perimeter-polygonB}$

$\frac{24}{15} =\frac{128}{x}$

$x=\frac{128*15}{24}$

$x=\frac{1920}{24}$

$x=80$

We can conclude that the perimeter of polygon B is 80 units.

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2 years ago

I need to know How to solve it

dezoksy [38]

SohCahToa
Heard of it?
Sine- opposite over hypotenuse
Cosine-adjacent over hypotenuse
Tangent- opposite over adjacent

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3 years ago

Express the area of the triangle shown below in terms of b and θ only.

Paul [167]

Given a right triangle with base, b and height, h with the angle opposide side h as θ.

The area of a triangle is given by
$A= \frac{1}{2} bh$

But,
$\tan\theta= \frac{h}{b} \\ \\ \Rightarrow h=b\tan\theta$

Therefore, the area of the triangle in terms of <span>b and θ only is given by
$A= \frac{1}{2} b(b\tan\theta)= \frac{1}{2} b^2\tan\theta$ </span>

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2 years ago