Sounds as tho' you have an isosceles triangle (a triangle with 2 equal sides). If this triangle is also a right triangle (with one 90-degree angle), then the side lengths MUST satisfy the Pythagorean Theorem.
Let's see whether they do.
8^2 + 8^2 = 11^2 ???
64 + 64 = 121? NO. This is not a right triangle.
If you really do have 2 sides that are both of length 8, and you really do have a right triangle, then:
8^2 + 8^2 = d^2, where d=hypotenuse. Then 64+64 = d^2, and
d = sqrt(128) = sqrt(8*16) = 4sqrt(8) = 4*2*sqrt(2) = 8sqrt(2) = 11.3.
11 is close to 11.3, but still, this triangle cannot really have 2 sides of length 8 and one side of length 11.
Answer:
The answer is 36.
Step-by-step explanation:
36
/\
6 6
/\ /\
3 2 3 2
The answer is 168 because 11 x 6 is 66 but on a triangle it would be half of that cause the formula of a triangle a = bh/2 so thats 33 and since theres 4 sides you would do 33 x 4 = 132 then 6 x 6 cause the base of the triangle = 36 and 132 plus 36 is 168
Answer:
marke me as brainly please
Answer:
The answer is 5 1/6
Step-by-step explanation:
1 2/3 ÷ 10 + 5
5/3 ÷ 10/1 + 5/1
From the rule of BODMAS, we solve for division before subtraction
to change division to multiplication, 10/1 will become 1/10, then will we have
5/3 × 1/10 + 5/1
1/3 ×1/2 + 5/1
1/6 + 5/1
The common multiple of 6 and 1 is 6, so if we solve for the fraction, we will have
1/6 + 30/6 = 31/6
= 5 1/6