Question

triangle PQR is an equilateral triangle in which A, B, and C are the midpoints of PO, OR and PR respectively. The centroid of this triangle is G. If AG - BG + CG = 20 cm, what is the semiperimeter of triangle POR?​

Answer 1

Given :-

• Triangle PQR is an equilateral triangle in which A, B, and C are the midpoints of PO, OR and PR respectively. The centroid of this triangle is G. If AG - BG + CG = 20 cm.

To Find :-

• The semiperimeter of ∆ PQR = ?

Solution :-

• To calculate the semi perimeter of ∆ PQR , at first we have to find the side of ∆. To calculate the side of triangle we have to construct the median.

Construction :-

• Join the point A to R , Point B to P , point C to Q such that at the midpoint point become perpendicular at point A, B and C.

Theorm used :-

• As we the medians of triangle intersect at point G . Here G is the centroid of triangle. Point G divides the median AR , BP and CQ in the ratio of 2 : 1.

Calculation begins :-

⇒AG - BG + CG = 20cm

⇒ AG = BG = CD = x (it's ratio in 2 : 1)

⇒ X - X + X = 20

⇒ 2X - X = 20

⇒ X = 20 cm

Therefore, AG = BG = CG = 20cm

Let's focus on ∆ BQG and BRG :-

• Let's prove congruency :-

⇒ BG = BG (common side)

⇒ QB = BR (PB _|_ QR)

⇒ ∠GBQ = ∠GBR (each 90°)

∴ ∆BQG ≅ ∆BRG ( By S.A.S criterian of congruency)

Now , let's focus on ∆ BQG :-

⇒ QG = 2 × CG (G divides the median in ratio 2 : 1)

⇒ QG = 2 × 20 = 40cm

⇒ GB = 20cm (as above calculated)

• By applying Pythagoras theorem :-

⇒ QG² = BG² + QB²

⇒ 40² = 20² + QB²

⇒ QB² = 1600 - 400

⇒ QB² = 1200

⇒ QB = √400 × 3

⇒ QB = 20√3 cm

⇒ QB = BR (PB _|_ QR)

⇒ QR = QB + BR

⇒ QR = 20√3 + 20√3

⇒ QR = 40√3 cm

Therefore, QR is the side of ∆PQR

• Now calculate Semi perimeter :-

⇒ S = a + b + c/2

• PQ = a , QR = b , PR = c
• PQ = QR = PQ (equilateral ∆ PQR)

⇒ S = (40√3 + 40√3 + 40√3)/2

⇒ S = 120√3/2

⇒ S = 60√3 cm

Hence,

• The semi perimeter of ∆ PQR = 60√3cm

Answer 2

Given :-

Triangle PQR is an equilateral triangle in which A, B, and C are the midpoints of PO, OR and PR respectively. The centroid of this triangle is G. If AG - BG + CG = 20 cm

To Find :-

Semiperimeter

Solution :-

G is the centroid

And

AG - BG + CG = 20

Here

AG = BG = CG

Let the side be a

a - a + a = 20

(a + a) - a = 20

2a - a = 20

a = 20

Now, ATQ

QG/2 = CG  

QG = 2(CG)

• CG = 20(Discussed above)

QG = 2(20)

QG = 40 cm

Now

QB = √(QG² - BG²)

QB = √(40² - 20²)

QB = √(1600 - 400)

QB = √1200

QB = 20√3 cm

QB = BR = 20√3 cm

Length of QR = QB + BR

QR = 20√3 + 20√3

QR = 40√3

Now

Perimeter = PQ + PR + QR

Perimeter = 40√3 + 40√3 + 40√3

Perimeter = 120√3

Now

Semi-perimeter = Perimeter × 1/2

Semi-perimeter = 120√3 × 1/2

Semi-perimeter = 60√3 cm