Question

If in a triangle XYZ,P,Q are the
points on XY,YZ respectively such that XP=2PY,XQ=2QZ then find the ratio of the∆XPQ and ∆XYZ?

Answer 1

Answer:

ratio of the∆XPQ and ∆XYZ?

sides 2:3

Area 4:9

Step-by-step explanation:

If in a triangle XYZ,P,Q are the

points on XY,XZ respectively such that XP=2PY,XQ=2QZ then find the ratio of the∆XPQ and ∆XYZ?

XP = 2PY

Let say PY = A then XP = 2A

XY = XP + PY = 2A + A = 3A

XQ = 2QZ

Let say QZ = B then XQ = 2B

XZ = XQ + QZ = 2B + B = 3B

in ∆XPQ and ∆XYZ

XP/XY = 2A/3A = 2/3

XQ/XZ = 2B/3B = 2/3

=> XP/XY = XQ/XZ

∠X is common

=> ∆XPQ ≅ ∆XYZ

Ration of  sides of ∆XPQ ≅ ∆XYZ  = 2/3

Ratio of Area  = Side²

=> Ration of  Areas of ∆XPQ ≅ ∆XYZ  = (2/3)² = 4/9

Answer 2

Answer:

ratio of the∆XPQ and ∆XYZ?

sides 2:3

Area 4:9

Step-by-step explanation:

If in a triangle XYZ,P,Q are the

points on XY,XZ respectively such that XP=2PY,XQ=2QZ then find the ratio of the∆XPQ and ∆XYZ?

XP = 2PY

Let say PY = A then XP = 2A

XY = XP + PY = 2A + A = 3A

XQ = 2QZ

Let say QZ = B then XQ = 2B

XZ = XQ + QZ = 2B + B = 3B

in ∆XPQ and ∆XYZ

XP/XY = 2A/3A = 2/3

XQ/XZ = 2B/3B = 2/3

=> XP/XY = XQ/XZ

∠X is common

=> ∆XPQ ≅ ∆XYZ

Ration of sides of ∆XPQ ≅ ∆XYZ = 2/3

Ratio of Area = Side²

=> Ration of Areas of ∆XPQ ≅ ∆XYZ = (2/3)² = 4/9