If Q is the center of the line PR and PR=10cm, then PQ=how much?
Answers
Answer:
Step-by-step explanation:
Let the RQ be x cm,
In right angle triangle PQR,
h^2\ =\ b^2+p^2h
2
= b
2
+p
2
QR^2\ =\ PQ^2\ +PR^2\QR
2
= PQ
2
+PR
2
x^2=\ \left(10\right)^2+\left(24\right)^2x
2
= (10)
2
+(24)
2
x^2\ =\ 100\ +576x
2
= 100 +576
x\ =\ \sqrt{100+576}x =
100+576
x\ =\ \sqrt{676}x =
676
x = 26 cm
Thus, the length of QR is 26cm.
Answer:
PQ = 5 cm
Step-by-step explanation:
PR is the line of 10 cm
centre line is a real or imaginary line that is equidistant from the surface or sides of something.
Q is the centre of this line and centre is the mid point of any line and it divides the line in two equal parts.
So, PQ = QR
PQ + QR = PR
but we know that PQ = PR
so,
PQ + PQ = PR
2PQ = PR
PQ = PR ÷ 2
PQ = 10 ÷ 2
PQ = 5 cm
Hence, the length of PQ will be 5 cm.