All categories

3 years ago

The perimeters of similar triangles are in the same ratio as the corresponding sides

Mathematics

1 answer:

wariber [46]3 years ago

3 0

Answer:

Always

Step-by-step explanation:

Suppose you have triangle ABC with side lengths a, b, c. Suppose that is similar to triangle DEF with side lengths d, e, f.

Now, let k be the ratio of corresponding sides ...

k = d/a

Because the same factor applies to all sides, we also have ...

k = e/b = f/c

That is, if we multiply by the denominators of each of these fractions, we get ...

d = a·k
e =b·k
f = c·k

The perimeter of ΔABC is ...

perimeter(ABC) = a + b + c

The perimeter of ΔDEF is ...

perimeter(DEF) = d + e + f = a·k + b·k + c·k

perimeter(DEF) = k(a + b + c) = k·perimeter(ABC)

k = perimeter(DEF)/perimeter(ABC)

That is, the perimeters are in the same ratio as corresponding sides.

You might be interested in

1.Find the area of a circle with the radius of 48mm

OLga [1]

Answer:

hope it helps you see the attachment for further information....

3 0

3 years ago

A field bordering a straight stream is to be enclosed. The side bordering the stream is not to be fenced. If 1000 yards of fenci

sveta [45]

Answer:

x = 500 yd

y = 250 yd

A(max) = 125000 yd²

Step-by-step explanation:

Let´s call x the side parallel to the stream ( only one side to be fenced )

y the other side of the rectangular area

Then the perimeter of the rectangle is p = 2*x + 2* y ( but only 1 x will be fenced)

p = x + 2*y

1000 = x + 2 * y ⇒ y = (1000 - x )/ 2

And A(r) = x * y

Are as fuction of x

A(x) = x * ( 1000 - x ) / 2

A(x) = 1000*x / 2 - x² / 2

A´(x) = 500 - 2*x/2

A´(x) = 0 500 - x = 0

x = 500 yd

To find out if this value will bring function A to a maximum value we get the second derivative

C´´(x) = -1 C´´(x) < 0 then efectevly we got a maximum at x = 500

The side y = ( 1000 - x ) / 2

y = 500/ 2

y = 250 yd

A(max) = 250 * 500

A(max) = 125000 yd²

3 0

3 years ago

How do I find the scale factor

iris [78.8K]

You write it as a ratio eg. SF(large→small)=20:7

4 0

4 years ago

Q.6. The equation of the ellipse whose centre is at the origin and the x-axis, the major axis, which passes

azamat

<h3>Answer:</h3>

Equation of the ellipse = 3x² + 5y² = 32

<h3>Step-by-step explanation:</h3>

<h2>Given:</h2>

The centre of the ellipse is at the origin and the X axis is the major axis

It passes through the points (-3, 1) and (2, -2)

The equation of the ellipse

<h2>Solution:</h2>

The equation of an ellipse is given by,

$\sf \dfrac{x^2}{a^2} +\dfrac{y^2}{b^2} =1$

Given that the ellipse passes through the point (-3, 1)

Hence,

$\sf \dfrac{(-3)^2}{a^2} +\dfrac{1^2}{b^2} =1$

Cross multiplying we get,

9b² + a² = 1 ²× a²b²
a²b² = 9b² + a²

Multiply by 4 on both sides,

4a²b² = 36b² + 4a²------(1)

Also by given the ellipse passes through the point (2, -2)

Substituting this,

$\sf \dfrac{2^2}{a^2} +\dfrac{(-2)^2}{b^2} =1$

Cross multiply,

4b² + 4a² = 1 × a²b²
a²b² = 4b² + 4a²-------(2)

Subtracting equations 2 and 1,

3a²b² = 32b²
3a² = 32
a² = 32/3----(3)

Substituting in 2,

32/3 × b² = 4b² + 4 × 32/3
32/3 b² = 4b² + 128/3
32/3 b² = (12b² + 128)/3
32b² = 12b² + 128
20b² = 128
b² = 128/20 = 32/5

Substituting the values in the equation for ellipse,

$\sf \dfrac{x^2}{32/3} +\dfrac{y^2}{32/5} =1$

$\sf \dfrac{3x^2}{32} +\dfrac{5y^2}{32} =1$

Multiplying whole equation by 32 we get,

3x² + 5y² = 32

<h3>Hence equation of the ellipse is 3x² + 5y² = 32</h3>

8 0

3 years ago

Find the sum.

Strike441 [17]

You are gonna use the distributive property.

4(2.5g - 4) + 3(1.2g - 2)

So you are going to multiply 4 with everything inside the first set of parenthesis and your gonna multiply 3 with everything insides the second set of parenthesis.

4 x 2.5g - 4 x 4 + 3 x 1.2g - 3 x 2

10g - 16 + 3.6g - 6

You wanna combine like terms:

10 - 3.6g + 16 - 6

6.4g + 10

You answer is 6.4g + 10.

Hope this Helps!!

6 0

3 years ago