J Jason76 Senior Member Joined Oct 19, 2012 Messages 1,180 Jan 20, 2014 #1 Shouldn't the area formula be \(\displaystyle A = \dfrac{1}{2}bh\) ? Here it is only \(\displaystyle A= bh\) How would this differ from this problem, where \(\displaystyle A= \dfrac{1}{2} bh\) Last edited: Jan 20, 2014
Shouldn't the area formula be \(\displaystyle A = \dfrac{1}{2}bh\) ? Here it is only \(\displaystyle A= bh\) How would this differ from this problem, where \(\displaystyle A= \dfrac{1}{2} bh\)
D daon2 Full Member Joined Aug 17, 2011 Messages 999 Jan 20, 2014 #2 Jason76 said: View attachment 3689 Shouldn't the area formula be \(\displaystyle A = \dfrac{1}{2}bh\) ? Here it is only \(\displaystyle A= bh\) How would this differ from this problem, where \(\displaystyle A= \dfrac{1}{2} bh\) View attachment 3690 Click to expand... The area of a rectangle is not 1/2(b)(h)...
Jason76 said: View attachment 3689 Shouldn't the area formula be \(\displaystyle A = \dfrac{1}{2}bh\) ? Here it is only \(\displaystyle A= bh\) How would this differ from this problem, where \(\displaystyle A= \dfrac{1}{2} bh\) View attachment 3690 Click to expand... The area of a rectangle is not 1/2(b)(h)...
J Jason76 Senior Member Joined Oct 19, 2012 Messages 1,180 Jan 20, 2014 #3 Right, I realized that a little later. Two different shapes.