Question

a triangle ABC, AD is the median and P is the point on AD such that AP:PD=1:3, then what is the area of triangle ABP?

Answer 1

ABC is a triangle, and AD is a median,

P is a point on AD, And AP:PD=1:3,

There fore ,

BD = DC,

AD median divides triangle ABC into two equal triangles, they are

a) Triangle ABD ,

b) Triangle ADC,

Here ,Area of triangle ABD = Area of triangle ADC,

Area of triangle ABP = 1/4 of area of triangle ABD =1/4 × 1/2 area of triangle ABC,

Hence, area of Triangle ABP = 1/8 area of triangle of ABC,

Ans:-- Area of triangle is , 1/8 area of triangle ABC,

Answer 2

$\bold {Area\ of \Delta ABP=\frac{1}{8}\times ( Area\ of\ \Delta ABC)}$

Step-by-step explanation:

Given,ΔABC where AD is a median and AP:PD=1:3

Ratio of their area is equal to the ratio of their respective base .

Since AD is a median

⇒ BD=CD

∴ , area o(ΔABD)= area(ΔACD) ( both the triangle have same height and same base )

$Area\ of \Delta ABD =\frac{1}{2}( Area\ of\triangle ABC)$............(1)

Also, $area\ of\ \Delta ABP=\frac{1}{4}(Area\ of\ \Delta ABD)=\frac{1}{4}\times \frac{1}{2}(area of \Delta ABC)$)

Hence,$\bold{Area\ of \Delta ABP=\frac{1}{8}\times ( Area\ of\ \Delta ABC)}$