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

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

## six.step one and you will six.step three Test

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

Explanation: 5x – cuatro = 4x + three times = eight ?JGK = 4(7) + 3 = 30 m?GJK = 180 – (31 + 90) = 180 – 121 = 59

Explanation: Keep in mind the circumcentre of an effective triangle is equidistant in the vertices out-of a good triangle. Then PA = PB = Desktop computer PA? = PB? = PC? PA? = PB? (x + 4)? + (y – 2)? = (x + 4)? + (y + 4)? x? + 8x + sixteen + y? – 4y + 4 = x? + 8x + sixteen + y? + 8y + 16 12y = -a dozen y = -step 1 PB? = PC? (x + 4)? + (y + 4)? = (x – 0)? + (y + 4)? x? + 8x + sixteen + y? + 8y + 16 = x? + y? + 8y + sixteen 8x = -sixteen x = -2 The circumcenter are (-dos, -1)

Explanation: Recall that the circumcentre away from a triangle are equidistant on vertices out-of a triangle. Let D(3, 5), E(seven, 9), F(11, 5) function as the vertices of your own offered triangle and you can help P(x,y) be the circumcentre associated with triangle. Up coming PD = PE = PF PD? = PE? = PF? PD? = PE? (x – 3)? + (y – 5)? = (x – 7)? + (y – 9)? x? – 6x + 9 + y? – 10y + twenty-five = x? – 14x + 49 + 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 + forty-two + y? – 18y + 81 = x? – 22x + 121 + y? – 10y + twenty five -14x + 22x – 18y + 10y = 146 – 130 8x – 8y = 16 x – y = 2 – (ii) Include (i) (ii) x + y + x – y = twelve + dos 2x = 14 x = eight Lay x = eight from inside the (i) eight + y = 12 y = 5 The fresh circumcenter are (7, 5)

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