I may be wrong, but im pretty sure its C
<h3>The lateral area for the pyramid with the equilateral base is 144 square units</h3>
<em><u>Solution:</u></em>
The given pyramid has 3 lateral triangular side
The figure is attached below
Base of triangle = 12 unit
<em><u>Find the perpendicular</u></em>
By Pythagoras theorem
![hypotenuse^2 = opposite^2 + adjacent^2](https://tex.z-dn.net/?f=hypotenuse%5E2%20%3D%20opposite%5E2%20%2B%20adjacent%5E2)
Therefore,
![opposite^2 = 10^2 - 6^2\\\\opposite^2 = 100 - 36\\\\opposite^2 = 64\\\\opposite = 8](https://tex.z-dn.net/?f=opposite%5E2%20%3D%2010%5E2%20-%206%5E2%5C%5C%5C%5Copposite%5E2%20%3D%20100%20-%2036%5C%5C%5C%5Copposite%5E2%20%3D%2064%5C%5C%5C%5Copposite%20%3D%208)
<em><u>Find the lateral surface area of 1 triangle</u></em>
![\text{ Area of 1 lateral triangle } = \frac{1}{2} \times opposite \times base](https://tex.z-dn.net/?f=%5Ctext%7B%20Area%20of%201%20lateral%20triangle%20%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20opposite%20%5Ctimes%20base)
![\text{ Area of 1 lateral triangle } = \frac{1}{2} \times 8 \times 12\\\\\text{ Area of 1 lateral triangle } = 48](https://tex.z-dn.net/?f=%5Ctext%7B%20Area%20of%201%20lateral%20triangle%20%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%208%20%5Ctimes%2012%5C%5C%5C%5C%5Ctext%7B%20Area%20of%201%20lateral%20triangle%20%7D%20%3D%2048)
<em><u>Thus, lateral surface area of 3 triangle is:</u></em>
3 x 48 = 144
Thus lateral area for the pyramid with the equilateral base is 144 square units
Answer:
Step-by-step explanation:
74939e+1364
Answer:
360 students
Step-by-step explanation:
We start off with 368.
Then 72 leave the school, so subtract 72 from our total.
368 - 72 = 296
But, then 64 new students join, so add this to our new total.
296 + 64 = 360.
Ask any questions you have in the comments, and if this helps, maybe consider giving a Brainliest :P
Answer:
22.5
Step-by-step explanation:
i dit it