a typical problem involving the area and perimeter of a rectangle gives us the area, perimeter and/r...

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Shape Scrapbook By, LaPreya Simmons Mrs. Togans

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Page 1: A typical problem involving the area and perimeter of a rectangle gives us the area, perimeter and/r length and width of the rectangle. We may also

Shape Scrapbook

By, LaPreya SimmonsMrs. Togans

Page 2: A typical problem involving the area and perimeter of a rectangle gives us the area, perimeter and/r length and width of the rectangle. We may also

Rectangle A typical problem involving the area and perimeter of a

rectangle gives us the area, perimeter and/r length and width of the rectangle. We may also be given a relationship between the area and perimeter or between the length and width of the rectangle.

The length is twice the width l = 2w The area is 32, so I use the area formula for a rectangle. lw =

2w(w) = 2w2 = 32 Solving for w: 2w2 = 32 w2 = 16 w = 4 The length equals  l = 2w l = 8 and the dimensions are 8 x 4. The perimeter is the sum of the lengths of all four sides or 2w

+ 2l = 2(4) +2(8) = 24

Page 3: A typical problem involving the area and perimeter of a rectangle gives us the area, perimeter and/r length and width of the rectangle. We may also

Triangle This triangle has internal angles 'A', 'B' and 'C', and sides of length

'a', 'b' and ‘c‘ If three of these six measurements are known, then it may be

possible to find the other three. This is called 'solving' the triangle, and this topic will show you how

to solve triangles for the three unknown angles when the three side lengths 'a', 'b' and 'c' are known.

1. The sum of the internal angles equals 180º A + B + C = 180º  2. The 'sine rule' ... 3. The 'cosine rule' ... a² = b² + c² - 2bc cosA or b² = a² + c² - 2ac cosB or c² = b² + a² -

2ba cosC

Page 4: A typical problem involving the area and perimeter of a rectangle gives us the area, perimeter and/r length and width of the rectangle. We may also

Parallelogram A quadrangle, the opposite sides of which are parallel, is called a

parallelogram.Each couple of opposite sides in a parallelogram are equal.

Perimeter of a parallelogram: P = 2a + 2b Area of a parallelogram formula: S = b.h = ab.sinA The sum of two contiguous angles is 180° A + B = 180° è A + D = 180° Dependency of diagonals and sides of a parallelogram formula p2 + q2

=2(a2+ b2)

Page 5: A typical problem involving the area and perimeter of a rectangle gives us the area, perimeter and/r length and width of the rectangle. We may also

Circle Since the radius of this this circle is 1, and its center is

the origin, this picture's equation is (Y-0)² +(X-0)² = 1 ² Y² + X² = 1

(y-3)2+(X-1)2 =9 (y-5)2+(X-14)2 =16 (y-1)2+(X-5)2 =25 (X+2)2++(y-12)2 =36 (y+7)2+(X +5)2 =49 (X +8)2+(y+17)2 =49

Answer r = radius (1, 3) r = 3 (14, 5) r = 4 (5, 1) r = 5 (-2, 12) r = 6 (-5, -7) r = 7 (-8, -17) r = 7  

Page 6: A typical problem involving the area and perimeter of a rectangle gives us the area, perimeter and/r length and width of the rectangle. We may also

Irregular Shape

Area square= s2 Area square = 42 Area square = 16 Area circle = pi × r2 Notice that the radius of the circle is 4/2 = 2 Area circle = 3.14 ×

22 Area circle 3.14 × 4 Area circle = 12.56 Since you only have half a circle, you have to multiply the result

by ½ 1/2 × 12.56 = 6.28 Area of this shape = 16 + 6.28 = 22.28 An irregular shaped object cannot be readily so described.

Natural rocks are irregularly shaped. Crystals, however, have regular shapes.