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2 years ago

How do you prove each of the following theorems using either a two-column paragraph or flow chart proof?

Mathematics

1 answer:

anzhelika [568]2 years ago

4 0

Answer:

Step-by-step explanation:

1) The Triangle Sum Theorem states that the sum of the angles in a triangle = 180°

2) The triangle inequality theorem states that the sum of any two sides of a triangle is larger than the third side

3) Isosceles triangle theorem states that the angles opposite the equal sides of an isosceles triangle are congruent

4) Converse of the Isosceles theorem states that the sides opposite the equal angles of an isosceles triangle are congruent

5) Midsegment of a triangle theorem states that the midsegment of two sides of a triangle is equal to half of the side it is parallel to

6) Concurrency of medians theorem states that the medians of a triangle intersect at a point within the triangle

Step-by-step explanation:

1) The Triangle Sum Theorem states that the sum of the angles in a triangle = 180°

Proof: To draw a triangle ABC starting from the point A we move 180° - ∠A to get to ∠B

From ∠B we turn 180° - ∠B to get to ∠C and from ∠C we turn 180° - ∠C to get back to A we therefore have turned 360° to get to A which gives;

180° - ∠A + 180° - ∠B + 180° - ∠C = 360°

Hence;

- ∠A - ∠B - ∠C = 360° - (180°+ 180°+ 180°) = -180°

-(∠A + ∠B + ∠C) = -180°

∴ ∠A + ∠B + ∠C = 180°

2) The triangle inequality theorem states that the sum of any two sides of a triangle is larger than the third side

Proof: Given ΔABC with height h from B to D along AC, then

AC = AB×cos∠A + CB×cos∠C

Since ∠A and ∠C are < 90 the cos∠A and cos∠C are < 1

∴ AC < AB + CB

3) Isosceles triangle theorem

Where we have an isosceles triangle ΔABC with AB = CB, we have by sine rule;

Therefore;

sin(C) = sin(A) hence ∠A = ∠C

4) Converse of the Isosceles theorem

Where we have an isosceles triangle ΔABC with ∠A = ∠C, we have by sine rule;

Therefore;

sin(C) = sin(A) hence AB = CB

5) Midsegment of a triangle theorem states that the midsegment of two sides of a triangle is equal to half of the side it is parallel to

Given triangle ABC with midsegment at DF between BA and BC respectively, we have;

in ΔABC and ΔADF

∠A ≅ ∠A

BA = 2 × DA, BC = 2 × FA

Hence;

ΔABC ~ ΔADF (SAS similarity)

Therefore,

BA/DA = BC/FA = DF/AC = 2

Hence AC = 2×DF

6) Concurrency of Medians Theorem

By Ceva's theorem we have that the point of intersection of the segments from the angles in ΔABC is concurrent when the result of multiplying ratio the ratios of the segment formed on each of the triangle = 1

Since the medians bisect the segment AB into AZ + ZB

BC into BX + XB

AC into AY + YC

Where:

AZ = ZB

BX = XB

AY = YC

We have;

AZ/ZB = BX/XB = AY/YC = 1

∴ AZ/ZB × BX/XB × AY/YC = 1 and the median segments AX, BY, and CZ are concurrent (meet at point within the triangle).

PLZ MARK ME BRAINLY

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3 years ago

Suppose the roots of the polynomial $x^2 - mx + n$ are positive prime integers (not necessarily distinct). Given that $m < 20

Vsevolod [243]

Answer:

18 values for n are possible.

Step-by-step explanation:

Given the quadratic polynomial:

$x^2 - mx + n$

such that:

Roots are positive prime integers and

$m < 20$

To find:

How many possible values of $n$ are there ?

Solution:

First of all, let us have a look at the sum and product of a quadratic equation.

If the quadratic equation is:

$Ax^{2} +Bx+C$

and the roots are: $\alpha$ and $\beta$

Then sum of roots, $\alpha+\beta = -\frac{B}{A}$

Product of roots, $\alpha \beta = \frac{C}{A}$

Comparing the given equation with standard equation, we get:

A = 1, B = -m and C = n

Sum of roots, $\alpha+\beta = -\frac{-m}{1} = m$

Product of roots, $\alpha \beta = \frac{n}{1} = n$

We are given that $m$

$\alpha$ and $\beta$ are positive prime integers such that their sum is less than 20.

Let us have a look at some of the positive prime integers:

2, 3, 5, 7, 11, 13, 17, 23, 29, .....

Now, we have to choose two such prime integers from above list such that their sum is less than 20 and the roots can be repetitive as well.

So, possible combinations and possible value of $n (= \alpha \times \beta)$ are:

$1.\ 2, 2\Rightarrow n = 2\times 2 = 4\\2.\ 2, 3 \Rightarrow n = 6\\3.\ 2, 5 \Rightarrow n = 10\\4.\ 2, 7\Rightarrow n = 14\\5.\ 2, 11 \Rightarrow n = 22\\6.\ 2, 13 \Rightarrow n = 26\\7.\ 2, 17 \Rightarrow n = 34\\8.\ 3, 3\Rightarrow n = 3\times 3 = 9\\9.\ 3, 5 \Rightarrow n = 15\\10.\ 3, 7 \Rightarrow n = 21\\$

$11.\ 3, 11\Rightarrow n = 33\\12.\ 3, 13 \Rightarrow n = 39\\13.\ 5, 5 \Rightarrow n = 25\\14.\ 5, 7 \Rightarrow n = 35\\15.\ 5, 11 \Rightarrow n = 55\\16.\ 5, 13 \Rightarrow n = 65\\17.\ 7, 7 \Rightarrow n = 49\\18.\ 7, 11 \Rightarrow n = 77$

So,as shown above 18 values for n are possible.

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3 years ago

A 14 inch pizza (diameter is 14) was cut in to 8 pieces. What is the angle measure of each slice? *

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Answer:

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Step-by-step explanation:

In order to calculate the angle of each slice, we first need to calculate the total area of the pizza, because we will use that to find the area of each slice and as a result it's angle. To calculate the area of the pizza we must use:

pizza area = pi*r²

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pizza area = pi*(7)² = 153.86 square inches

Since the pizza was divided in 8 pieces, the area of each piece is given by the area of the pizza divided by the number of slices. We have:

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Each slice is a circle sector, therefore it's area is given by:

slice area = (angle*pi*r²)/360

Therefore we can solve for angle:

19.2325 = (angle*pi*7²)/360

angle*pi*49 = 6923.7

angle = 6923.7 / pi*49 = 45 degrees

The angle for each slice is 45 degrees.

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