Answer:
0.125 m²
Step-by-step explanation:
The area of a rectangle is the product of its length and breadth. Therefore:
area of the rectangle = length (l) × breadth (b) = 1 m².
l × b = 1
A triangle is cut off from that rectangle along a line connecting the midpoints of two adjacent sides, therefore the base of the triangle = length of rectangle/2 and the height of triangle = breadth of rectangle/2.
Area of triangle = 1/2(base × height) = ![\frac{1}{2}*\frac{l}{2}*\frac{b}{2}=\frac{1}{8}lb](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%2A%5Cfrac%7Bl%7D%7B2%7D%2A%5Cfrac%7Bb%7D%7B2%7D%3D%5Cfrac%7B1%7D%7B8%7Dlb)
since lb = 1:
Area of triangle = ![\frac{1}{8}*1=0.125m^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B8%7D%2A1%3D0.125m%5E2)
Answer:
C (11.3) = 165
P (3,3) = 6
Step-by-step explanation:
We want to select 3 players out of 11 regardless of the order. That is, there is no difference between selecting the players {2,5,7} or {7,2,5}
Then we use the formula of combinations:
![C(n, r) = \frac{n!}{r!(n-r)!}\\\\C(11, 3) = \frac{n!}{r!(n-r)!}\\\\C(11, 3) = 165](https://tex.z-dn.net/?f=C%28n%2C%20r%29%20%3D%20%5Cfrac%7Bn%21%7D%7Br%21%28n-r%29%21%7D%5C%5C%5C%5CC%2811%2C%203%29%20%3D%20%5Cfrac%7Bn%21%7D%7Br%21%28n-r%29%21%7D%5C%5C%5C%5CC%2811%2C%203%29%20%3D%20165)
There are 165 ways to choose 3 players out of 11.
Now we want to know how many ways you can designate those 3 players as first, second and third. Now if we care about the order of selection. Then we use permutations.
![P(n, r) = \frac{n!}{(n-r)!}\\\\P(3,3) = \frac{3!}{(3-3)!}\\\\P(3,3) = 6](https://tex.z-dn.net/?f=P%28n%2C%20r%29%20%3D%20%5Cfrac%7Bn%21%7D%7B%28n-r%29%21%7D%5C%5C%5C%5CP%283%2C3%29%20%3D%20%5Cfrac%7B3%21%7D%7B%283-3%29%21%7D%5C%5C%5C%5CP%283%2C3%29%20%3D%206)
They can be designated in 6 different ways
Answer:
The answer to your question is At = 48 + 36![\sqrt{3}](https://tex.z-dn.net/?f=%5Csqrt%7B3%7D)
Step-by-step explanation:
Area = Area of the base + Lateral area
a) Area of the base
Area = length x height / 2
length = 12
height² = 12² - 6²
= 144 - 36
= 108
height =
= 6![\sqrt{3}](https://tex.z-dn.net/?f=%5Csqrt%7B3%7D)
Area = 12 x 6![\sqrt{3} / 2](https://tex.z-dn.net/?f=%5Csqrt%7B3%7D%20%2F%202)
= 36![\sqrt{3}](https://tex.z-dn.net/?f=%5Csqrt%7B3%7D)
b) Lateral area
height² = 10² - 6²
= 100 - 36
= 64
height = 8
Area = 12 x 8/2
= 48
c) Total area
At = 48 + 36![\sqrt{3}](https://tex.z-dn.net/?f=%5Csqrt%7B3%7D)