Question

Find the length of QP given Q is the midpoint of XF, PQ =2X+1, XF=7X-4, PF=X

Answer 1

Answer:

Therefore the length of QP = 3.4 units

Step-by-step explanation:

Given:

PQ = 2x + 1

XF = 7x - 4

PF = x

Q is the mid poimt of XF

∴ XQ = QF

QF = PQ - PF  ..........( Q - F - P )

     = 2x + 1 - x

∴ QF      = x + 1

∴ XQ = QF = x + 1

TO Find:

QP = ?

Solution:

By Addition Property we have

$XP = XQ + QF+PF ..........(X-Q-F-P)$

$XF + PF =XQ + QF+PF ..........(X-Q-F-P)$

Substituting the given values in above equation we get

(7x - 4) + x = (x +1) + (x +1) + x

8x -4 = 3x +2

8x - 3x + 4 + 2

5x = 6

∴ $x = \frac{6}{5}$

Now we require

QP = (2x + 1)

∴ $QP = 2\times \frac{6}{5} +1\\\\QP = \frac{12+5}{5} \\\\QP =\frac{17}{5} \\\\\therefore QP = 3.4\ unit$

Therefore the length of QP = 3.4 units