Check and help plz

As before  using Heron's formula

s =   [7 + 7 + 6 ]  /2  =  20 /2  =10

Area  =   √[10 * (10 - 7) (10 - 7) (10- 6)  ]  = √[10 * 3 * 3 * 4  ]  = √ 360  = √ [36 * 10 ] = 

6√10  units^2

Second one

Area of sector   =  (1/2) r^2  ( theta in rads)   =  (1/2) (5^2) (5pi/12)      =

125 pi

_____    in^2

  24

Last one

Area =  (1/2)  (XZ)(YZ)  * sin ( Z)   =  (1/2) (9)(4) sin (58°)  =  18 * sin (58°)  ≈  15.3  units^2

1. For this problem, recognize that the given side lengths are part of an isoceles triangle. 

It is advisable to drop the altitude to the base length of six, bisecting that side length.

By the Pythagorean Theorem, we have 7^2-3^2=y^2, y^2=40, \(y=2\sqrt{10}\).

Thus, the area of the triangle is \(\frac{1}{2}*6*2\sqrt{10}=6\sqrt{10}\)