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This chapter explores fundamental concepts in geometry, focusing on the areas of different shapes, including equilateral triangles, squares, regular hexagons, parallelograms, right triangles, rhombuses, and trapezoids. It also covers the Pythagorean Theorem and how to form right triangles using given segment lengths. The chapter emphasizes formulas for calculating areas, circumferences of circles, and relationships among triangle dimensions, facilitating a deep understanding of geometric properties.
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SIMPLIFY: 20 12 3
20 = 4 5 = 2 512312 3 3 3 3 = = 4 3
COMPLETE:( ___ )2 + ( ___ )2 = ( ___ )2 hyp leg1 leg2
(leg1)2 + (leg2)2 = (hyp)2 • Pythagorean Theorem
Complete to form RIGHT triangles: 3, 4, ____ 5, 12, ____ 6, 8, ____ 8, 15, ____
Given segment lengths a, b, c longest Right c2 a2 + b2 Obtuse c2 a2 + b2 Acute c2 a2 + b2
COMPLETE: • L = ____ • H = ____ 45 45
SOLVE: 45 • x = _____ • y = _____ y 8 45 x
COMPLETE: • LL = _____ • H = _____ 60 Ls 30
SOLVE: y x = _____ y = _____ 60 4 30 x
AREA OF EQUILATERAL TRIANGLES: A = ½ ( ____ )( ____ ) A = ½ ( ____)( ____ ) A = ( ____ )2 ( ____ )
A = ½ bh A = ½ ap A = s2 4
Find the area: 6 (Equilateral Triangle)
18 6 3 30 9 A = s2 4 A = (18)2 4 A = 81 A = ½ bh A = ½ (18)(9 ) A = 81 A = ½ ap A = ½ (3 )(54) A = 81
AREA OF SQUARES: • A = ( ____ )2 • A = ½ ( ____ )( ____ ) • A = ½ ( ____ )( ____ )
A = s2 A = ½ ap A = ½ d1d2 a side (s)
Find the area: 10 (Square)
20 10 10 45 10 A = s2 A = 202 A = 400 A = ½ ap A = ½ (10)(80) A = 400 A = ½ d1d2 A = ½ (20 )(20 ) A = 400
A = ½ ap 120 a 60
Find the area: 8 (Regular Hexagon)
8 4 3 60 4 4 A = ½ ap A = ½ (4 3)(48) A = 96 3
AREA OF PARALLELOGRAMS: • A = ( ____ )( ____ )
A = bh Height (h) Base (b)
A = ½bh height (h) base (b)
Find the height: 6 6 4
6 6 h 2 2 4 h2 +22 =62h2 = 32h = 32 = 4 2
A = ½ (l1)(l2) leg (l1) leg (l2)
AREA OF RHOMBUSES: • A = ½ ( ____ ) ( ____ )
AREA OF TRAPEZOIDS: A = ½ ( ____ ) ( ____ + ____ )
A = ½(h)(b1+ b2) base (b1) height (h) base(b2)
FOR CIRCLES: Circumference = ( __ )( __ )( __ )
AREA OF CIRCLES: A = ( ____ )( ____ )2