Explanation: QP = QR 6x = 3x + 9 3x = nine x = step three QP = 6(3) = 18

six.step 1 and 6.step 3 Test

Explanation: SV = VU 2x + eleven = 8x – 1 8x – 2x = eleven + step one 6x = several x = 2 Ultraviolet = 8(2) – step 1 = fifteen

Explanation: Remember that circumcentre away from an effective triangle was equidistant in the vertices regarding good triangle. Assist A(- cuatro, 2), B(- cuatro, – 4), C(0, – 4) function as vertices of one’s considering triangle and let P(x,y) function as circumcentre for the triangle. Then PA = PB = Pc PA? = PB? = PC? PA? = PB? (x + 4)? + (y – 2)? = (x + 4)? + (y + 4)? x? + 8x + 16 + y? – 4y + 4 = x? + 8x + 16 + y? + 8y + 16 12y = -twelve y = -1 PB? = PC? (x + 4)? + (y + 4)? = (x – 0)? + (y + 4)? x? + 8x + sixteen + y? + 8y + 16 = x? + y? + 8y + sixteen 8x = -16 x = -2 The brand new circumcenter is actually (-dos, -1)

Explanation: Keep in mind that circumcentre off a beneficial triangle is actually equidistant about vertices of an excellent triangle. Let D(3, 5), E(7, 9), F(11, 5) function as vertices of the offered triangle and you can help P(x,y) function as circumcentre on the triangle. After that PD = PE = PF PD? = PE? = PF? PD? = PE? (x – 3)? + (y – 5)? = (x – 7)? + (y – 9)? x? – 6x + nine + y? – 10y + twenty-five = x? – 14x + forty two + y? – 18y + 81 -6x + 14x – 10y + 18y = 130 – 34 8x + 8y = 96 x + y = twelve – (i) PE? = PF? (x – 7)? + (y – 9)? = (x – 11)? + (y – 5)? x? – 14x + 49 ceny xmatch + y? – 18y + 81 = x? – 22x + 121 + y? – 10y + 25 -14x + 22x – 18y + 10y = 146 – 130 8x – 8y = sixteen x – y = dos – (ii) Put (i) (ii) x + y + x – y = a dozen + 2 2x = fourteen x = seven Lay x = seven in the (i) seven + y = twelve y = 5 New circumcenter try (7, 5)

Explanation: NQ = NR = NS 2x + 1 = 4x – 9 4x – 2x = ten 2x = 10 x = 5 NQ = 10 + step 1 = eleven NS = 11

Explanation: NU = NV = NT -3x + six = -5x -3x + 5x = -6 2x = -6 x = -step three NT = -5(-3) = 15

Explanation: NZ = New york = NW 4x – 10 = 3x – step one x = 9 NZ = 4(9) – 10 = 36 – ten = 26 NW = twenty-six

Explanation: 5x – 4 = 4x + three times = 7 ?JGK = 4(7) + step three = 30 meters?GJK = 180 – (29 + 90) = 180 – 121 = 59

Find the coordinates of your own centroid of your own triangle wilt the newest offered vertices. Question nine. J(- 1, 2), K(5, 6), L(5, – 2)

Explanation: The slope of TU = \(\frac < 1> < 0>\) = -2 The slope of the perpendicular line is \(\frac < 1> < 2>\) The perpendicular line is y – 5 = \(\frac < 1> < 2>\)(x – 2) 2y – 10 = x – 2 x – 2y + 8 = 0 The slope of UV = \(\frac < 5> < 2>\) = 2 The slope of the perpendicular line is \(\frac < -1> < 2>\) The perpendicular line is y – 5 = \(\frac < -1> < 2>\)(x + 2) 2y – 10 = -x – 2 x + 2y – 8 = 0 equate both equations x – 2y + 8 = x + 2y – 8 -4y = -16 y = 4 x – 2(4) + 8 = 0 x = 0 So, the orthocenter is (0, 4) The orthocenter lies inside the triangle TUV

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