Question

If in the given figure (refer attachment) ∠PQR = 90° , O is the centroid of ΔPQR , PQ = 5 cm and QR = 12 cm , then OQ is equal to

A. 3 1/2 cm

B. 4 1/3 cm

C. 4 1/2 cm

D. 5 1/3 cm​

Answer 1

Answer:

Terms to know:

Centroid: Intersection point of all the three medians of a triangle which divides the median in the ratio 2:1.

Coming to the question,

Since it is a right angles triangle, the third side can be found out by using Pythagoras Formula. Applying Pythagoras Formula we get,

⇒ PQ² + QR² = PR²

⇒ 5² + 12² = PR²

⇒ 25 + 144 = PR²

⇒ 169 = PR²

⇒ PR = √169 = 13 cm.

Therefore the third side is 13 cm.

Now since O is the centroid, the line passing through the centroid should be a median. Therefore, QM is the median which dives PR as PM and RM which are equal in length. Therefore, PM = RM. The value of their length is:

⇒ PM + RM = PR

⇒ 2 PM = PR

⇒ PM = PR / 2 = 13/2 cm

Now we will do some construction. Refer Attachment for diagram.

Construction:

• Draw a perpendicular MP to the line QR.

As MP passes through the mid-point M and is parallel to PQ , it divides the side QR into two equal parts i.e. PQ = PR

Now Consider Δ MPQ and Δ MPR.

⇒ MP = MP ( Common Side )

⇒ ∠ MPQ = ∠ MPR ( 90° By Construction )

⇒ PQ = PR ( Proved Above )

Therefore by SAS Congruence,

⇒ Δ MPQ ≅ Δ MPR

⇒ By CPCT, MQ = MR = 13/2 cm

Now we know that,

⇒ OQ + OM = MQ = MR

⇒ 2x + x = MR  [ Since Ratio is 2:1 ]

⇒ 3x = 13/2

⇒ x = 13/2 ÷ 3 = 13/6 cm

Therefore OQ = 2x

⇒ OQ = 2 × 13/6 = 13/3 cm

⇒ OQ = 4 ( 1/3 ) cm.

Therefore Option (B) is correct.

Hope it helped !!

Answer 2

Answer:

OQ = $4\frac{1}{3} \: cm$

Step-by-step explanation:

PR² = PQ² + QR²  (Pythagoras theorem)

=> PR² = 5² + 12²

=> PR² = 25 + 144

=> PR² = 169

=> PR² = 13²

=> PR = 13

O is the centroid.

Centroid is where Median intersects

=> QM is Median  to hypotenuse

Hence PM = RM = PR/2 = 13/2

Now We can use Either

in a right triangle , the length of median to hypotenuse is half the length of the hypotenuse.

=> QM = 13/2

or We can use in Δ QMR    or ΔPQM using formula

c² = a² + b² - 2abCos∠C

QM² = QR² + RM² - 2*QR*RM Cos∠R = 12² + (13/2)² - 2(12)(13/2)(12/13)  = 12² + (13/2)² - 12² = (13/2)² => QM = 13/2

QM² = PQ² + PM² - 2*PQ*PM Cos∠P = 5² + (13/2)² - 2(5)(13/2)(5/13)  = 5² + (13/2)² - 5² =(13/2)² => QM = 13/2

Median intersects them selves in 2: 1 ratio  ( from Vertex to opposite side)

=> OQ  = (2/3) QM

=> OQ = (2/3) (13/2) = 13/3

=> OQ = 4  1/3   cm