<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
There is a website on that
Answer:
(2,10) (4,20) (6,30) (8, 40)
Step-by-step explanation:
every minute, the pool fills up with 5 gallons.
if you look at the points, you will see a pattern
Answer:
Step-by-step explanation:
85
Given: PQ=2x-3. if QR=5x+7 and PR=10x-11
To Find: PR
Solution: From the attached diagram, point P, Q and R are co-linear.
Hence, PR = PQ + QR
or, 10x-11 = 2x-3 + 5x+7
or, 10x - 7x = 15
or, x = 5
Now, we shall calculate the value of PR
PR = 10x-11
or, PR = 10× 5 - 11
or, PR = 39
Hence, the required value of PR will be 39.