2. © 2007 - 09 by S-Squared, Inc. All Rights Reserved. 1.Find the distance, d, and the midpoint, m, between (4, − 2) and (2, 6) Distance Formula: d = (x 2 – x 1 ) 2 + (y 2 – y 1 ) 2 Substitute Simplify (6 – (− 2)) 2 + (2 – 4) 2 (6 + 2) 2 + (2 – 4) 2 (8) 2 + (− 2) 2 64 + 4 68 4 17 2 17

3. ﴾, ﴿ 2 3 1.Find the distance, d, and the midpoint, m, between (4, − 2) and (2, 6) Midpoint Formula: m = ﴾, ﴿ 2 y 1 + y 2 x 1 + x 2 2 Substitute Simplify ﴾, ﴿ 2 − 2 + 6 4 + 2 2 m = ﴾, ﴿ 2 4 6 2 m =

4. 2.Find the perimeter and the area of the given rectangle: 6 cm 14 cm Area = length width Perimeter = 2 length + 2 width Area = 14 6 Substitute Area = 84 cm 2 Perimeter = 2 14 + 2 6 Perimeter = 28 + 12 Perimeter = 40 cm

5. 3.The perimeter of the rectangle below is 54 cm. Find the length. 9 cm l Perimeter = 2 length + 2 width width = 9 length = l Perimeter = 54 Formula Identify variables Substitute 54 = 2 l + 2 9 Simplify 54 = 2 l + 18 – 18 Subtract – 18 36 = 2 l Divide 2 2 18 = l 18cm = length

6. x 2 + 5 2 = (x + 1) 2 x 2 + 25 = (x + 1)(x + 1) 4.Find the two unknown side lengths of the given triangle: x + 1 x 5 ft Hint: Use Pythagorean’s Theorem to find x. a 2 + b 2 = c 2 Note: Identify a, b and c Substitute b = 5 a = x c = x + 1 Simplify Subtract – x 2 – 1 25 = 2x + 1 24 = 2x x 2 + 25 = x 2 + 2x + 1 – x 2 Subtract – 1 Divide 2 2 12 = x Note: Side lengths are x and x + 1 x = 12 ft x + 1 = 13 ft

7. 5.Find the perimeter and the area of the given rectangle: yd Volume = length width height Volume = Substitute Volume = or yds 3 4 3 5 4 2 1 yds. 2 4 3 5 4 2 1 2 Volume = 4 3 5 4 2 5 Note: Turn mixed number to improper fraction 1 1 1 1 Volume = 2 3 2 3 2 1 1 Reduce

8. 6.Find the circumference and the area of the given circle: 7 m Circumference = 2 π r where r is the radius and let π = 3.14 Substitute Circumference = 2 3.14 7 Circumference = 6.28 7 Circumference = 43.96 m Note: Identify r r = 7 Simplify

9. 6.Find the circumference and the area of the given circle: 7 m Area = π r 2 where r is the radius and let π = 3.14 Substitute Area = 3.14 7 2 Area = 3.14 49 Area = 153.86 m 2 Note: Identify r r = 7 Simplify

10. 7.Find the area and perimeter of the given shape: 3 in Since the altitude divides the base of the triangle, which is also a side length of the square, in half each segment is 4 inches. 8 in Hint: The altitude divides the base of the triangle in half 4 in All side lengths of a square are equal so the other side length is 8 inches. 8 in Use the Pythagorean Theorem to find the legs of the triangle. 4 2 + 3 2 = c 2 16 + 9 = c 2 25 = c 2 Substitute Simplify Square Root 5 = c 25 = c 2 5 in

11. 7.Find the area and perimeter of the given shape: 3 in 8 in Hint: The altitude divides the base of the triangle in half 4 in 8 in 5 in Note: Perimeter is distance around Perimeter = 8 + 8 + 8 + 5 + 5 Perimeter = 16 + 8 + 5 + 5 Perimeter = 24 + 5 + 5 Perimeter = 29 + 5 Perimeter = 34 in

12. Note: Area of the shape is the sum of the area of the triangle and the area of the square. 7.Find the area and perimeter of the given shape: 3 in 8 in Hint: The altitude divides the base of the triangle in half 4 in 8 in 5 in Area(square) = 8 2 Area(square) = 64 in 2 Area of the square is s 2 Area of the triangle is b h 2 1 base = 8 height = 3 Area(triangle) = (8)(3) 2 1 Area(triangle) = (4)(3) Area(triangle) = 12 in 2 Area(shape) = 64 + 12 Area(shape) = 76 in 2