6-10.

Mathematics
Tutor: None Selected Time limit: 1 Day

May 4th, 2015

6.  Area = 1/2 B * H = 1/2 (6)(4) = 24/2 = 12; Perimeter: 6 + 4 (SQRT 2) + 2 (SQRT 5), If you cut the big triangle into two smaller right triangles, you get 2 base of left triangle, 4 base of right side triangle, and they share height of 4, then by Pythagorean theorem, left hypotenuse = 2^2 + 4^2 = c^2 = 4 + 16 = 20 = 4 (SQRT 5), and right hypotenuse is 4^2 + 4^2 = SQRT 32 = 2 (SQRT 5),

7.  Area = 1/2 B * H = 1/2 (3) (4) = 12/2 = 6: Perimeter:  3 +5 + 2 SQRT 3 = 8 + 2 SQRT 3, By Pythagorean theorem, height = 4, base = 3 (of small right triangle), hypotenuse of small right triangle = 5, (which we only use to use Pythagorean theorem again on the obtuse angle where the base = 6, other side, 5 (which we just did by Pythagorean theorem) and larger side 2 (SQRT 3), 6^2 + 4^2 = 36 + 16 = SQRT 52 = 2 SQRT 3

8.  Area = 1/2 B*H (right triangle) + Area of square (B*H) = 1/2 (3)(4) = 12/2 = 6 area of triangle left side, Area of square 4 * 4 = 16, so Total Area = 16 + 6 = 22

Perimeter: 7 + 4 + 4 + 5 = 20 (the two 4s are from the square, the 7 (base) and the 5 (hypotenuse of right triangle)

9.  Area of two right triangles (one on each end) + area square (middle), 1/2 (3) (3) = 9/2 (left right triangle) + 9 (3*3 = 9 for middle square) + 1/2 (1)(3) = 3/2 Total area = 9/2 + 9 + 3/2 = 12/2 + 18/2 = 30/2 = 15

Perimeter: 7 (total base length) + SQRT 13 (left hypotenuse of left right triangle) + 3 (top of square) + SQRT 10 (hypotenuse of right hand side right triangle) = 10 + SQRT 13 + SQR 10

10.  Area circle = PI R^2 = PI * 3^2 = 9(PI), Perimeter of circle = circumference = 2* PI* R = 2 PI * 3 = 6 PI

If you wish you can estimate the square roots and PI, but if you want exact answer, keep the SQRT and PI as is.  Estimate for PI is 3.1415

Best me if I explained it better and went through all of the steps.  Let me know if you have any questions.


May 4th, 2015

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