+0  
 
+2
77
5
avatar+154 

 

 

 

is that correct

 

 

 

 

 

 

help on this 

 Mar 25, 2020
edited by whatisthis1010  Mar 25, 2020
 #1
avatar+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

 

cool cool cool

 Mar 25, 2020
 #2
avatar+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

 

 

 

cool cool cool

 Mar 25, 2020
 #3
avatar+111330 
+3

Last one

 

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

 

 

 

cool cool cool

 Mar 25, 2020
 #4
avatar+1970 
+3

That's impressive, Chris!!!!

CalTheGreat  Mar 25, 2020
 #5
avatar+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

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