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 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.

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Presentation on theme: " 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."— Presentation transcript:

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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 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) = 2w 2 = 32  Solving for w: 2w 2 = 32 w 2 = 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

3  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

4  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 p 2 + q 2 =2(a 2 + b 2 )

5  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 

6  Area square= s 2 Area square = 4 2 Area square = 16  Area circle = pi × r 2  Notice that the radius of the circle is 4/2 = 2 Area circle = 3.14 × 2 2  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.


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