Question

In ΔABC, P is point on BC such that BP : PC = 4 : 11. If Q is the midpoint of BP, then ar(ΔABQ): ar(ΔABC) is equal to:
In ΔABC में, P, BC पर बिंदु है जैसे BP: PC = 4: 11. यदि Q , BP का मध्य बिंदु है, तो ar(ΔABQ) : ar(ΔABC) बराबर है:​

Answer 1

Answer:

2/15

Step-by-step explanation:

In ΔABC, P is point on BC such that BP : PC = 4 : 11. If Q is the midpoint of BP, then ar(ΔABQ): ar(ΔABC) is equal to:

Let say BC = 15K

BP = 4K

PC = 11 K

BQ = 4K/2 = 2K

CQ = 15K - 2K = 13K

Area of ΔABC = (1/2)BC * H  =  15KH/2 ( H = Altitude)

Ares of ΔAQC = (1/2)CQ * H  = 13KH/2

Area of ΔABQ = Area of ΔABC - Ares of ΔAQC  = 15KH/2 - 13KH/2  = KH

ar(ΔABQ) : ar(ΔABC) = KH/(15KH/2)  = 2/15

Answer 2

Answer:

Write in english. Please.

Step-by-step explanation: