Surface Area of a pyramid.....PLEASE

[Kimiko]

Triangular Base measures 14 in. Area of Base 85in. Slant Height is 19.8. I have tried to use the formula A=S sq plus 2 side times slant height. so,,, 196 sq + 2(196)(19.8) Can anyone assist me please?

 

[soroban]

Hello, Kimiko!

Triangular base measures 14 in.[ . Area of base 85 sq.in. .Slant height is 19.8 in.
Find the total surface area.

I assume that the base is an equilateral triangle.
Then its area is: .$\displaystyle \frac{\sqrt{3}}{4}14^2 \:=\:84.87048957 \:\approx\:85\text{ in}^2$


A triangular side looks like this:

            *
           /|\
          / | \
         /  |  \ 19.8
        /   |h  \
       /    |    \
      * - - * - - *
         7     7

We see that: .$\displaystyle h^2 + 7^2 \:=\:19.8^2 \quad\Rightarrow\quad h \:=\:18.52133904 \:\approx\:18.5$

The area of this triangle is: .$\displaystyle \frac{1}{2}(14)(18.5) \:=\:129.5\text{ in}^2$

The area of the three triangles is: .$\displaystyle 3 \times 129.5 \:=\:388.5\text{ in}^2.$


The total surface area of the pyramid is: .$\displaystyle 388.5 + 85 \:=\:473.5\text{ in}^2$

 

[Kimiko]

Thank you soo much for the detailed explanation.

Hello, Kimiko!


I assume that the base is an equilateral triangle.
Then its area is: .$\displaystyle \frac{\sqrt{3}}{4}14^2 \:=\:84.87048957 \:\approx\:85\text{ in}^2$


A triangular side looks like this:

            *
           /|\
          / | \
         /  |  \ 19.8
        /   |h  \
       /    |    \
      * - - * - - *
         7     7

We see that: .$\displaystyle h^2 + 7^2 \:=\:19.8^2 \quad\Rightarrow\quad h \:=\:18.52133904 \:\approx\:18.5$

The area of this triangle is: .$\displaystyle \frac{1}{2}(14)(18.5) \:=\:129.5\text{ in}^2$

The area of the three triangles is: .$\displaystyle 3 \times 129.5 \:=\:388.5\text{ in}^2.$


The total surface area of the pyramid is: .$\displaystyle 388.5 + 85 \:=\:473.5\text{ in}^2$

THE PYRAMID IS A TRIANGULAR PYRAMID. I AM TOTALLY CONFUSED AS THE FORMULA GIVEN IS SURFACE AREA A=1/2bh WHICH IS 1/2 ^14^19.8=138.6in sq. SINCE IT HAS 3 SIDES I TOOK 138.6 X 3=415.8 in sq. THEN, THE SURFACE AREA OF THE BASE IS A=415.8 +85=500.8 IN SQUARED. eNGLISH IS VER DIFFICULT FOR ME SO THE ERROR COULD POSSIBLY BE INTERPRETATION. AGAIN, THANK YOU VERY MUCH FOR YOUR TIME. KIMIKO-SAN

 

[Kimiko]

Thank you Denis

Kimiko, I agree with your 500.8
Soroban's diagram should show h = 19.8 (not the sides).

http://www.google.ca/search?q=slant...QHHuqmLDA&sqi=2&ved=0CHMQsAQ&biw=1014&bih=559

Thank you so much for the confirmation.