In △abc, point p∈ ab is so that ap: bp=1: 3 and point m is the midpoint of segment cp . find the area of △abc if the area of △bmp is equal to 21m2.

56 m^2

Step-by-step explanation:

Area of

Step-by-step explanation:

Given :

In ,

M is the midpoint of segment CP.

Calculation :

BMP and CMB are the triangles whose area are same because there height and base are same. And hence area of   which is due to transitivity.

Area of triangle PCB is equal to the combine area of triangles BMP and CMB which is equal to .

Now, area of triangle ACP

Now we add all the areas, we get the are of triangle ABC

Therefore, area of

For more information, refer the link given below

link

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐

42

step-by-step explanation: