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saw5 [17]
2 years ago
15

If v is the circumcenter of pqr ,pr =46 ,tv=15,and vr=25 find each measure

Mathematics
1 answer:
timofeeve [1]2 years ago
7 0

Answer:

1. a) SR = 23

b) QV = 25

c) QT = 20

d) PQ = 40

e) VS = 4·√6

2. a) LH = 16

b) EL = 2·√185

c) JG = 30

d) EK = 22

e) KG = 30

3. a) XT = 37

b) TZ = 34

c) ZW = 17

d) XZ = 21

e) SY = 69

Step-by-step explanation:

The circumcenter ΔPQR is the center of the circle that circumscribes ΔPQR

The length of the radius of the circle ≡ VR = VP = QV = 25

a) Given that VR ≅ VP - Radius of circumcircle

VS ≅ VS Reflective property

∠VPS ≅ ∠VRS - Base angles of an isosceles triangle

Right triangle VPS ≅ Right triangle VRS -Hypotenuse and one Leg HL congruency

Therefore, SR ≅ PS -Corresponding parts of congruent triangles are congruent CPCTC

SR + PS = PR = 46

SR + PS = SR + SR = 2·SR = 46

∴ SR = 46/2 = 23

b) QV = VR = 25 = Radius of circumcircle of ΔPQR -Given V = center and Q = vertices of the triangle circumscribed by the circle referred to in the question

c) QT = √(QV² - TV²) = √(25² - 15²) = √400 = 20

d) TV ≅ TV - Reflexive property of congruency

ΔTQV ≅ ΔTVP - Hypotenuse and one Leg (HL) congruency

QT ≅ TP -Corresponding parts of congruent triangles are congruent CPCTC

PQ = QT + TP Given

∴ PQ = QT + QT since QT = TP

PQ = 2·QT = 2 × 20 = 40

e) VS = √(VR² - SR²) = √(25² - 23²) = √96 = 4·√6

2.  The incenter is the center of the incircle of ΔEFG

 a) LH = LK = JL = 16 -Radius of incircle of ΔEFG

b) EL = Hypotenuse of right triangle LHE = √(LH² + EH²) = √(16² + 22²) = √740 = 2·√185

c) JG = Leg length of right triangle JGL = √(LG² - JL²) = √(34² - 16²) = √900 = 30

d) EK = Leg length of right triangle LKE = √(EL² - LK²) = √(740 - 256) = 22

e) KG = Leg length of right triangle LKG = √(LG² - LK²) = √(34²- 16²) = √900 = 30

3. Point Z id the centroid of ΔRST

a) XT = XS - point X on ST bisected by median line RX

ST = XT + XS = XT + XT = 2.XT = 74

XT = 74/2 = 37

b) TZ = 2/3×TW  - Length from a vertex to the centroid on a median line is equal to two third the length of the median line

TZ = 2/3×51 = 34

c) TZ + ZW = TW

∴ ZW = TW - TZ = 51 - 34 = 17

d) RZ = 42 = 2/3×RX - Length from a vertex to the centroid on a median line is equal to two third the length of the median line

∴ RX = 3/2×42 = 63

RZ + XZ = RX - Given

XZ = RX - RZ = 63 - 42 = 21

e) SZ = 2/3×SY - Length from a vertex to the centroid on a median line is equal to two third the length of the median line

SZ + ZY = SY

∴ ZY = SY - SZ = SY - 2/3×SY = 1/3×SY = 23

Which gives;

SY = 3 × 23 = 69.

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Answer:

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Step-by-step explanation:

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2x+14=−4x+14

Step 2: Add 4x to both sides.

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The line on the graph passes through to points A (1, 3) B (7, 1)
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Answer:

a. Gradient of line AB is -\frac{1}{3}

b. The gradient of a line perpendicular to line AB is 3

c. The equation of a line passing through point (4,2) and perpendicular to AB is y = 3x - 10

Step-by-step explanation:

a.

Given

Point A (1, 3) B (7, 1)

Required

Gradient of AB

Gradient of a line is represented by m

m is calculated using the following formula

m = \frac{y_{2} - y_{1} }{x_{2} - x_{1}}

Where the general representation of the coordinates are A(x_1,y_1) and B(x_2,y_2)

<em>From the given data, we have that</em>

A(x_1,y_1) = A(1,3)

B(x_2,y_2) = A(7,1)

<em>So, from there we know that</em>

x_1 = 1;y_1 =3; x_2 = 7;y_2 =1

m = \frac{y_{2} - y_{1} }{x_{2} - x_{1}} becomes

m = \frac{1 - 3}{7 - 1}

m = \frac{-2}{6}

m = -\frac{1}{3}

b.

Required

Find the gradient of a line perpendicular to AB

<em>Recall that gradient of a line is represented by m</em>

The condition for perpendicularity is that m_1.m_2 = -1

In (a) above, we solved the gradient of line AB to be -\frac{1}{3}

Let m_1 represent gradient of line AB

Hence, m_1 = -\frac{1}{3}

Substitute -\frac{1}{3} for m_1 in  m_1.m_2 = -1

<em>This will give</em>

\frac{-1}{3} * m_2 = -1

Multiply both sides by -3

-3 * \frac{-1}{3} * m_2 = -1 * -3

m_2 = -1 * -3

m_2 = 3

Hence, the gradient of a line perpendicular to line AB is 3

c.

Required

Find the equation of a line passing through point (4,2) and perpendicular to AB

Equation is calculated using the gradient formula

m = \frac{y_{2} - y_{1} }{x_{2} - x_{1}}

Since only one point is known, the formula is represented as follows

m = \frac{y - y_{1} }{x - x_{1}}

Where x_1 = 4; y_1 = 2

Since, the line is perpendicular to line AB, then its gradient m is equal to 3 (as calculated in b above)

So, we have x_1 = 4; y_1 = 2; m = 3

By substitution

m = \frac{y - y_{1} }{x - x_{1}} becomes

3 = \frac{y - 2}{x - 4}

Multiply both sides by x - 4

3 * (x - 4) = \frac{y - 2}{x - 4} * (x - 4)

3(x - 4) = {y - 2}

Open brackets

3x - 12 = y - 2

Make y the subject of formula

3x - 12 + 2= y

3x - 10 = y

Reorder

y = 3x - 10

Hence, the equation of a line passing through point (4,2) and perpendicular to AB is y = 3x - 10

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Answer:

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Step-by-step explanation:

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