What is the distance from QR?
Q
6n + 3
s
4n + 11
R

The distance of QR is 54 units

Step-by-step explanation:

In the isosceles triangle, The height drawn from the vertex angle to the base bisects the base

In ΔPQR

∵ PQ = PR

∴ Δ PQR is an isosceles triangle

∵ P is the vertex angle

∵ QR is the base

∵ PS ⊥ QR

→ By using the fact above

∴ PS bisects QR

∴ S is the midpoint of QR

→ That means S divided QR into 2 equal parts QS and SR

∵ QS = SR

∵ QS = 6n + 3

∵ SR = 4n + 11

→ Equate them to find n

∴ 6n + 3 = 4n + 11

→ Subtract 4 n from both sides

∵ 6n - 4n + 3 = 4n - 4n + 11

∴ 2n + 3 = 11

→ Subtract 3 from both sides

∵ 2n + 3 - 3 = 11 - 3

∴ 2n = 8

→ Divide both sides by 2

∴ n = 4

→ Find the length of QS by substitute x in its expression by 4

∵ QS = 6(4) + 3 = 24 + 3

∴ QS = 27

∵ QS = SR

∴ SR = 27

∵ QR = QS + SR

∴ QR = 27 + 27

∴ QR = 54 units

∴ The distance of QR is 54 units

the answer is 31.27mph

hi

step-by-step explanation: