In triangle PQR, C is the centroid. PQ = 30 cm, QR = 36 cm and PR = 50 cm. If D is the midpoint of QR, then what is the length (in cm) of CD ?
A. $$\frac{{4\sqrt {86} }}{3}$$
B. $$\frac{{2\sqrt {86} }}{3}$$
C. $$\frac{{5\sqrt {86} }}{3}$$
D. $$\frac{{5\sqrt {86} }}{2}$$
Answer: Option A
Solution (By Examveda Team)

(PQ)2 + (PR)2 = 2(QD2 + PD2)
(30)2 + (50)2 = 2[(18)2 + PD2]
900 + 2500 = 2[324 + PD2]
3400 = 2[324 + PD2]
1700 = 324 + PD2
PD2 = 1376
PD = $$4\sqrt {86} $$
CD = $$\frac{1}{3}$$PD
CD = $$\frac{{4\sqrt {86} }}{3}$$
Related Questions on Geometry
A. $$\frac{{23\sqrt {21} }}{4}$$
B. $$\frac{{15\sqrt {21} }}{4}$$
C. $$\frac{{17\sqrt {21} }}{5}$$
D. $$\frac{{23\sqrt {21} }}{5}$$
In the given figure, ∠ONY = 50° and ∠OMY = 15°. Then the value of the ∠MON is

A. 30°
B. 40°
C. 20°
D. 70°


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