Question

Fig. 1.14
I BC,
4. In adjoining figure, AP
AD || BC, then find
A(A ABC): A(A BCD).
B
P
11​

Answer 1

Step-by-step explanation:

Area of Triangle = (1/2) * Base * Height

Area of Triangle Δ ABC  = (1/2) BC * AP

Lets draw a perpendicular DQ from point D at BC line

DQ ║ AP

AD║BC

=> DQ = AP

Area of Δ BDQ = (1/2)(BQ) * DQ

= (1/2)(BC + CQ) * AP

Area of ΔCDQ = (1/2) CQ * DQ

= (1/2)CQ * AP

Area of Δ BCD = Area of Δ BDQ - Area of Δ CDQ

= (1/2)(BC + CQ) * AP -  (1/2)CQ * AP

= (1/2)(BC) * AP

= Area of Triangle Δ ABC

=> A(Δ ABC): A(Δ BCD) :: 1: 1

Answer 2

Answer:

Given:

AP ⊥ BC

AD || BC

∴A(△ABC)/A(△BCD)=AP×BC/AP×BC

=1/1

Hence, the ratio of A(∆ABC) and A(∆BCD) is 1 : 1.