+0

# Check and help plz

+2
77
5
+154

is that correct

help on this

Mar 25, 2020
edited by whatisthis1010  Mar 25, 2020

#1
+111330
+3

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

Mar 25, 2020
#2
+111330
+3

Second one

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

125 pi

_____    in^2

24

Mar 25, 2020
#3
+111330
+3

Last one

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

Mar 25, 2020
#4
+1970
+3

That's impressive, Chris!!!!

CalTheGreat  Mar 25, 2020
#5
+4568
+1

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 $$\frac{1}{2}*6*2\sqrt{10}=6\sqrt{10}$$

.
Mar 25, 2020