Solve for the area using heron´s formula

Given =>

A figure ABCD with two adjacent triangles i.e ABC &ACD

To Acquire =>

The Area of the figure

Points to know while solving this problem=>

Step-by-step explanation:

In ∆ABC,

Perimeter of the ∆

= mAB +mBC+mAC

Where,

mAB =20cm

mAC = 16cm

mBC= ?

Also

mBC

= Base of the ∆ ABC where AB = perpendicular & AC = Hypotenuse

So

mBC = √AC²-AB².......(Using Pythagoras theorem)

mBC = √20²-16²

mBC =√400-256

mBC =√144

mBC = 12.....( The square root of 144 = 12)

Hence,

The perimeter of ∆

= mAB +mBC+mAC

= 16+12+20

= 48cm

Semiperimeter = 48/2

= > 24 cm

.°.

The area of ∆ ABC according to heron's formula

=>

Now,

In ∆ACD,

The perimeter = mAC+mCD+mAD

where,

mAC=16cm

mCD=23cm

mAD=20cm

=>

So...

Semiperimeter = 59/2cm

=> 29.5

Area of the Same∆ using Heron's formula

= 156 .97cm²

Hence,

The area of the figure

=>96cm²+156.97cm²

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: