Area 8.1 Rectangles & Parallelograms 8.2 Triangles,Trapezoids & Kites 8.3 Application Word Problems 8.4 Regular Polygons 8.5 Circles 8.6 Sectors, Segments of Circles, & Annulus 8.7 Surface Area – Prisms, Pyramids, Cylinders, & Cones

Rectangles & Parallelograms 3 Height, h = 3 units The area of a rectangle is determined with the formula, A= bh, where b is the length of the base and h is the height of the rectangle. 6 Base, b= 6 units A = 6  3 = 18 sq. units

Rectangles & Parallelograms 3 Height, h = 3 units The area of a parallelogram is determined with the formula, A= bh, where b is the length of the base and h is the height of the parallelogram. 6 Base, b= 6 units A = 6  3 = 18 sq. units

A Closer Look at Parallelograms Length Height Side length Note that the height of the parallelogram is an altitude, perpendicular to the bases, and that it is used in determining the area and not the side lengths.

Practice Problems for 8.1 12 cm What is the area of the shaded region? A =( 12 x 6) – (4x 6) = 72 – 24 = A = 48 cm 2 6 cm 4 cm 6 cm 9.4 cm What is the area of the shaded region? A = ( 8 x 5) = 20 cm 2 5 cm 8 cm

Homework 8.1 Page 413 problems 1 – 9 Page 414 problems 17,19 Page 415 problems 22 – 24 (honors)

Triangles, Trapezoids, & Kites Note that the triangle is formed from using one-half of a parallelogram or rectangle. As such, the area of a triangle should be one-half of the area of the parallelogram. A=½ bh h b Also note that the height of a triangle is the altitude from a vertex to the opposite side. The opposite side is used as the base length.

Triangles, Trapezoids, & Kites b1 A =½ bh A1 h h A2 b2 Base, b A = A1 + A2 A1 = ½ b1 h ; A2 = ½ b2 h A = ½ h (b1 + b2)

Triangles, Trapezoids,& Kites A1 = ½h1d1 ; A2 = ½h2d1 h1 A = ½h1d1 + ½h2d1 d1 A = ½ d1(h1 + h2) h2 Since d2 = (h1 + h2) d2 A = ½ d1 d2

Practice Problems for 8.2 A= ½bh; A = ½(11)(4) A = ½(44) = 22 cm2 h = 4 cm A = ½(44) = 22 cm2 b = 11 cm b1 = 8 cm A = ½h(b1 + b2) h = 6 cm A = ½(6)(8 + 15) b2 = 15 cm A = (3)(23) = 69 cm2

Homework 8.2 Page 418 problems 1 – 6 Page 419 problems 7 - 12 Page 420 problem 20 (honors)

Areas of Irregular Shapes A = Arectangle + 3(Atriangle) A = Arectangle + ½Acircle

Practice Problems for 8.3 What is the area of the shaded region? 2 cm 4 cm 3 cm 7 cm What is the area of the shaded region? 10 cm A = 58 cm2

Practice Problems for 8.3 A = 3692 ft 2 36.4 ft 10 ft 8 ft 20 ft What is the total interior surface area for all walls and the ceiling? 30 ft 15 ft A = 3692 ft 2 40 ft

Homework 8.3 Page 423 problems 1 – 3 Page 424 problems 4 - 7 Page 425 problem 8 (honors)

Regular Polygons The area of a regular polygon is dependent on the length of the sides, s, the number of sides, n, and the distance from the center to a side and perpendicular to the side, a, called the apothem. a s A = ½ asn or ½ aP This equation will work on any regular polygon.

Regular Polygons Explanation of equation origin. The area of a regular polygon is the sum of the area of all of the triangles, each with a height of a, and a base length of s. Since n = 6, then the area is equal to n(½ as). This simplifies to A = ½asn.

Regular Polygons A = ½asn - ½bh or A = ½(asn – bh) Solving for a partial area, the shaded region, of a regular polygon is done by solving for the area of the entire polygon and the unshaded region followed by subtraction of the unshaded from the shaded area. h b a s A = ½asn - ½bh or A = ½(asn – bh)

Homework 8.4 Page 427 problems 1 – 8 Page 428 problem 12, 13 (honors)

Circles The area of a circle can be approximated by dividing it into very small triangles and then determining the sum of the areas of the triangles.

Circles ½C r The area of a circle can be approximated by using r as the height and ½ C as the base. This becomes A = ½ Cr. By substitution we get, A = ½ (2r)r and this then simplifies to, A = r 2. This is the formula for area of a circle.

Circles When polygons are inscribed in a circle, the area of the shaded region is the difference of the area of the circle and the area of the polygon. A = (r2) – (bh)

Circles When polygons are inscribed in a circle, the area of the shaded region is the difference of the area of the circle and the area of the polygon. r h b A = (r2) – (½bh)

Homework 8.5 Page 435 problems 1 – 10 Page 435 problem 12, 13 (honors)

Sectors, Segments of a Circle, and the Annulus A sector or a circle is the region between two radii of a circle and the included arc. r A = a x r2 360 a o Ex. If a = 60o and the radius = 4 cm, then the area of the sector would be: A = 60 x  42 = 8/3  cm2 360

Sectors, Segments of a Circle, and the Annulus A segment of a circle is the region between a chord of a circle and the included arc. r h b A = a x r2 - ½bh 360 90 o Ex. If a = 90o , the radius = 4 cm, b=5.66 cm, h = 2 cm then the area of the sector would be: A = 90 x  42 - (½*5.66 * 2 2)= 1.25 cm2 360

Sectors, Segments of a Circle, and the Annulus The annulus is the region between two concentric circles (the blue region). The area of the shaded region is determined by solving for the area of both circles and subtracting the smaller one from the larger one. r1 r2 A = (½r12) - (½r22)

Homework 8.6 Page 439 problems 1 – 12 Page 441 problem 17, 18 (honors)

Surface Area of Solids Prism Pyramid Cylinder Cone

Prisms w The prism has n sides called lateral faces and 2 bases (top & bottom). The lateral faces are rectangles that can have their area calculated with bh * n. The bases are rectangles with an area of bw * 2. h This prism has 4 lateral sides and 2 bases, so the surface area will be: A = 4bh + 2bw b

Prisms The prism has n sides called lateral faces and 2 bases (top & bottom). The lateral faces are rectangles that can have their area calculated with bh * n. The bases are polygons with an area of ½ asn* 2 which simplifies to asn. h This prism has 5 lateral sides and 2 bases, so the surface area will be: A = 5bh + asn. a b s

Cylinders r r C or 2r A = 2r h + 2r 2 = 2r (h + r) The cylinder is really a rectangle rolled into a circle with two circular bases. r r h h C or 2r A = 2r h + 2r 2 = 2r (h + r)

Pyramids A = ½snl + ½asn = ½sn(l + a) The pyramid has n number of triangle lateral faces and a polygon base. The surface area is the sum of the areas of the lateral faces and the base. The lateral length is l. l s s s s l a s s s s A = ½snl + ½asn = ½sn(l + a)

Cones When the base is opened and the sides are laid down flat, they form a fan shape with a curved bottom (2r) with a length of l. This lateral area LA = rl. The base is a circle and has an area of A = r 2. Together, the surface area of a cone is A = r(l +r). l h r l l 2r 2r

Homework 8.7 Page 450 problems 1 – 10 Page 451 problem 12, 18 (honors)

Homework Chapter 8 Review Page 455 problems 1 – 16 Page 456 problem 17 - 25 Page 457 problems 29 – 31 Page 458 problems 40 - 44(honors)