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In-s [12.5K]
3 years ago
13

Given: The coordinates of triangle PQR are P(0, 0), Q(2a, 0), and R(2b, 2c).

Mathematics
1 answer:
masha68 [24]3 years ago
6 0

Answer:

The line containing the midpoints of two sides of a triangle is parallel to the third side ⇒ proved down

Step-by-step explanation:

* Lets revise the rules of the midpoint and the slope to prove the

  problem

- The slope of a line whose endpoints are (x1 , y1) and (x2 , y2) is

  m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

- The mid-point of a line whose endpoints are (x1 , y1) and (x2 , y2) is

  (\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})

* Lets solve the problem

- PQR is a triangle of vertices P (0 , 0) , Q (2a , 0) , R (2b , 2c)

- Lets find the mid-poits of PQ called A

∵ Point P is (x1 , y1) and point Q is (x2 , y2)

∴ x1 = 0 , x2 = 2a and y1 = 0 , y2 = 0

∵ A is the mid-point of PQ

∴ A=(\frac{0+2a}{2},\frac{0+0}{2})=(\frac{2a}{2},\frac{0}{2})=(a,0)

- Lets find the mid-poits of PR which called B

∵ Point P is (x1 , y1) and point R is (x2 , y2)

∴ x1 = 0 , x2 = 2b and y1 = 0 , y2 = 2c

∵ B is the mid-point of PR

∴ B=(\frac{0+2b}{2},\frac{0+2c}{2})=(\frac{2b}{2},\frac{2c}{2})=(b,c)

- The parallel line have equal slopes, so lets find the slopes of AB and

  QR to prove that they have same slopes then they are parallel

# Slope of AB

∵ Point A is (x1 , y1) and point B is (x2 , y2)

∵ Point A = (a , 0) and point B = (b , c)

∴ x1 = a , x2 = b and y1 = 0 and y2 = c

∴ The slope of AB is m=\frac{c-0}{b-a}=\frac{c}{b-a}

# Slope of QR

∵ Point Q is (x1 , y1) and point R is (x2 , y2)

∵ Point Q = (2a , 0) and point R = (2b , 2c)

∴ x1 = 2a , x2 = 2b and y1 = 0 and y2 = 2c

∴ The slope of AB is m=\frac{2c-0}{2b-2a}=\frac{2c}{2(b-c)}=\frac{c}{b-a}

∵ The slopes of AB and QR are equal

∴ AB // QR

∵ AB is the line containing the midpoints of PQ and PR of Δ PQR

∵ QR is the third side of the triangle

∴ The line containing the midpoints of two sides of a triangle is parallel

  to the third side

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A building engineer analyzes a concrete column with a circular cross section. The circumference of the column is 18 \pi18π18, pi
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Step-by-step explanation:

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5 0
3 years ago
A) 0.158<br> B) 15.8<br> C) 1.58<br> D) 1.625
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6 0
3 years ago
Will mark brainliest for the correct answer!
romanna [79]

Part (a)

Focus on triangle PSQ. We have

angle P = 52

side PQ = 6.8

side SQ = 5.4

Use of the law of sines to determine angle S

sin(S)/PQ = sin(P)/SQ

sin(S)/(6.8) = sin(52)/(5.4)

sin(S) = 6.8*sin(52)/(5.4)

sin(S) = 0.99230983787513

S = arcsin(0.99230983787513)

S = 82.889762826274

Which is approximate

------------

Use this to find angle Q. Again we're only focusing on triangle PSQ.

P+S+Q = 180

Q = 180-P-S

Q = 180-52-82.889762826274

Q = 45.110237173726

Which is also approximate.

A more specific name for this angle is angle PQS, which will be useful later in part (b).

------------

Now find the area of triangle PSQ

area of triangle = 0.5*(side1)*(side2)*sin(included angle)

area of triangle PSQ = 0.5*(PQ)*(SQ)*sin(angle Q)

area of triangle PSQ = 0.5*(6.8)*(5.4)*sin(45.110237173726)

area of triangle PSQ = 13.0074347717966

------------

Next we'll use the fact that RS:SP is 2:1.

This means RS is twice as long as SP. Consequently, this means the area of triangle RSQ is twice that of the area of triangle PSQ. It might help to rotate the diagram so that line PSR is horizontal and Q is above this horizontal line.

We found

area of triangle PSQ = 13.0074347717966

So,

area of triangle RSQ = 2*(area of triangle PSQ)

area of triangle RSQ = 2*13.0074347717966

area of triangle RSQ = 26.0148695435932

------------

We're onto the last step. Add up the smaller triangular areas we found

area of triangle PQR = (area of triangle PSQ)+(area of triangle RSQ)

area of triangle PQR = (13.0074347717966)+(26.0148695435932)

area of triangle PQR = 39.0223043153899

------------

<h3>Answer: 39.0223043153899</h3>

This value is approximate. Round however you need to.

===========================================

Part (b)

Focus on triangle PSQ. Let's find the length of PS.

We'll use the value of angle Q to determine this length.

We'll use the law of sines

sin(Q)/(PS) = sin(P)/(SQ)

sin(45.110237173726)/(PS) = sin(52)/(5.4)

5.4*sin(45.110237173726) = PS*sin(52)

PS = 5.4*sin(45.110237173726)/sin(52)

PS = 4.8549034284642

Because RS is twice as long as PS, we know that

RS = 2*PS = 2*4.8549034284642 = 9.7098068569284

So,

PR = RS+PS

PR = 9.7098068569284 + 4.8549034284642

PR = 14.5647102853927

-------------

Next we use the law of cosines to find RQ

Focus on triangle PQR

c^2 = a^2 + b^2 - 2ab*cos(C)

(RQ)^2 = (PR)^2 + (PQ)^2 - 2(PR)*(PQ)*cos(P)

(RQ)^2 = (14.5647102853927)^2 + (6.8)^2 - 2(14.5647102853927)*(6.8)*cos(52)

(RQ)^2 = 136.420523798282

RQ = sqrt(136.420523798282)

RQ = 11.6799196828694

--------------

We'll use the law of sines to find angle R of triangle PQR

sin(R)/PQ = sin(P)/RQ

sin(R)/6.8 = sin(52)/11.6799196828694

sin(R) = 6.8*sin(52)/11.6799196828694

sin(R) = 0.4587765387107

R = arcsin(0.4587765387107)

R = 27.3081879220073

--------------

This leads to

P+Q+R = 180

Q = 180-P-R

Q = 180-52-27.3081879220073

Q = 100.691812077992

This is the measure of angle PQR

subtract off angle PQS found back in part (a)

angle SQR = (anglePQR) - (anglePQS)

angle SQR = (100.691812077992) - (45.110237173726)

angle SQR = 55.581574904266

--------------

<h3>Answer: 55.581574904266</h3>

This value is approximate. Round however you need to.

8 0
3 years ago
QUESTION IS DOW BELOW 5 POINTS EACH PLEASE HELP PLEASE HELP PLEASE HELP
masha68 [24]

The central angle in the circle is ∠DAC,major arc is BED, minor arc is ADC and BC=(5π*BD)/18.

Given that BD is diameter of the circle and angle BAC is 100°.

We are required to find the central angle, major arc, minor arc, m BEC, BC.

Angle is basically finding out the intensity of inclination of something on the surface.

In the circle central angles are many like BAC and CAD. We can write CAD as DAC also.

Major arc of a circle is that arc whose length is larger than all other arcs in the circle.

In our circle the major arc is arc BED.

Minor arc of a circle is that arc whose length is smaller.

In our circle the minor arc is arc ADC.

We know that arc's length is 2πr(Θ/360)

In this way BC=2π*(BD/2)*100/360

=(5π*BD)/18

We cannot find angle BEC.

Hence the central angle in the circle is ∠DAC,major arc is BED, minor arc is ADC and BC=(5π*BA)/18.

Learn more about angles at brainly.com/question/25716982

#SPJ1

7 0
2 years ago
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