1 Standards 7, 8 Review Rhombus Properties Review Isosceles Trapezoid Properties Review Trapezoid Properties Finding the Area of a Triangle PROBLEM 1 PROBLEM 2 Finding the Area of a Trapezoid PROBLEM 3 PROBLEM 4 PROBLEM 5 Finding the Area of a Rhombus PROBLEM 7 PROBLEM 8 PROBLEM 11 Standards 7, 8, 10 PROBLEM 6 PROBLEM 9 PROBLEM 10 END SHOW PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

2 Standard 7: Students prove and use theorems involving the properties of parallel lines cut by a transversal, the properties of quadrilaterals, and the properties of circles. Estándar 7: Los estudiantes prueban y usan teoremas involucrando las propiedades de líneas paralelas cortadas por una transversal, las propiedades de cuadriláteros, y las propiedades de círculos. Standard 8: Students know, derive, and solve problems involving perimeter, circumference, area, volume, lateral area, and surface area of common geometric figures. Estándar 8: Los estudiantes saben, derivan, y resuelven problemas involucrando perímetros, circunferencia, área, volumen, área lateral, y superficie de área de figuras geométricas comunes. Standard 10: Students compute areas of polygons including rectangles, scalene triangles, equilateral triangles, rhombi, parallelograms, and trapezoids. Estándar 10: Los estudiantes calculan áreas de polígonos incluyendo rectángulos, triángulos escalenos, triángulos equiláteros, rombos, paralelogramos, y trapezoides. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

3 1. Two pairs of parallel sides. 2. All sides are congruent. 4. Diagonals bisect each other 6. Opposite angles are congruent and bisected by diagonals. 7. Consecutive angles are supplementary. mAmB+=180° mBmC+= mCmD+= mDmA+= A D B C 3. Diagonals are NOT congruent 5. Diagonals form a right angle RHOMBUS Standards 7, 8, 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

4 1. Exactly one pair of parallel sides. 2. One pair of congruent sides. 4. Diagonals DO NOT bisect each other 5. Base angles are congruent. 6. Opposite angles are supplementary. mAmC+=180° mBmD+= A D B C 3. Diagonals are congruent ISOSCELES TRAPEZOID 7. Line connecting midpoints of congruent sides is called MEDIAN. M b 1 b 2 b 1 b 2 and are bases of trapezoid Mis the median M b 1 b 2 2 = + M = b 1 b Standards 7, 8, 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

5 M b 1 b 2 2 = + M = b 1 b Exactly one pair of parallel sides. TRAPEZOID 2. Line connecting midpoints of congruent sides is called MEDIAN. b 1 b 2 and are bases of trapezoid Mis the median M b 1 b 2 Standards 7, 8, 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

6 Area of a Triangle A = bh 1 2 h b b where: b= base h= height h Standards 7, 8, 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

7 20 cm 14 cm Find the area: A = bh 1 2 b= 20 cm h= 14 cm A = (20cm)(14cm) 1 2 This is a triangle so: where: A =(10 cm )(14cm) A=140 cm 2 Standards 7, 8, 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

8 Find the lenght of the base of the triangle below if its area is 50 cm 2 7 cm A = bh 1 2 The area is: b= ? h= 7 cm where: A= 50 cm 2 50 cm = b(7cm) (2) 100 cm = b(7 cm) 2 (7 cm) b 14.3 cm Substituting and solving for b: Standards 7, 8, 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

9 A B C D Area of Trapezoid Area of Trapezoid = Area of + Area of ABC ADC b 1 b 2 h h Area of Trapezoid = 1 2 b 1 h 1 2 b 2 h b 1 hb 2 + h b 1 b 2 Can you define the Area in function of the Median? M = b 1 b Click to find out… Area of ABC = 1 2 b 1 h Area of ADC = 1 2 b 2 h Standards 7, 8, 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

10 h b 1 b 2 Area of a Trapezoid Area of Trapezoid = 1 2 b 1 hb 2 + b 1 b 2 M h M M b 1 b 2 2 = + M = b 1 b OR where: b 1 b 2 and are bases of trapezoid Mis the median his the height Standards 7, 8, 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

11 h = 7 Find the area of the figure below: This is a trapezoid so: A = 1 2 b 1 hb 2 + where: b 1 =30 b 2 =50 A = (7) = 7 2 (80) = = 280 square units. Substituting: Standards 7, 8, 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

12 40 cm 10 cm Find the length of the missing base of the trapezoid below if the area is 340 cm : 2 The formula for the Area is: A = 1 2 b 1 hb 2 + where: h = 10 b 1 =40 b 2 =? A=340 cm 2 We substitute and solve for : b = 1 2 (10)b = (5)b = (5)(40)b = 200b = b b 2 = 28 cm Which means the trapezoid is: 28 cm 10 cm 40 cm Standards 7, 8, 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

Find the Median, using both bases, for the following trapezoid if its area is 840 square units. The formula for the Area is: A = 1 2 b 1 hb 2 + where: h = 20 b 1 =? b 2 =60 A= 840 We substitute and solve for : b = 1 2 (20)b = (10)b = (10)(60) b = 600b = b 1 10 b 1 = 24 now the Median is: M b 1 b 2 2 = + substituting values: M 2 = M 2 = 84 M= 42 lineal units. Standards 7, 8, 10 Is there a shorter way? Click to find out… PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

14 Area of Trapezoid = h M 840=20M 20 M= 42 Standards 7, 8, 10 If we don’t have to use the bases, then: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

15 The area of an isosceles trapezoid is 88 square in. Its height is 3 in and its sides are 5 in. Find the lenght of the bases. 88 in 2 5 in 3 in Formula for the area: A = 1 2 b 1 hb 2 + b 1 88 = 1 2 ( 3)b 2 + b 1 b = b b 1 b First Equation Z From the figure calculating Z using Pythagorean Theorem: 5 = Z = Z = Z 2 4 = Z Z = 4 Z b 2 Now from the figure: + b 1 b 2 = Z + Z b 2 = 2Z + b 1 b 2 = 2( ) + b 1 4 b 2 = 8 + b 1 Second Equation Solving both equations: b = b b 2 = 8 + b 1 b = ( ) 8 + b 1 b = b 1 3 b = = 6 b b 1 = 25.3 Sustituting in Second Equation: b 2 = 8 + b 1 = b 2 = 33.3 Standards 7, 8, 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved 2

16 The area of an isosceles trapezoid is 88 square in. Its height is 3 in and its sides are 5 in. Find the lenght of the bases. 88 in 2 5 in 3 in Z From the figure calculating Z using Pythagorean Theorem: 5 = Z = Z = Z 2 4 = Z Z = 4 Z Now from the figure: + b 1 b 2 = Z + Z b 2 = 2Z + b 1 b 2 = 2( ) + b 1 4 b 2 = 8 + b 1 Standards 7, 8, A = bh 1 2 A = (4)(3) 1 2 A= 6 Now area of central rectangle is Area of trapezoid minus Area of triangles: A rec = 88-2(6) = 76 Area of triangles: b 1 A rec = (3) b 2 b 1 76 = (3) b 1 3 b 1 = 25.3 b 2 = 8 + ( ) 25.3 b 2 = 33.3 Alternate solution: Area of central rectangle is also: Which solution is easier for you? PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

17 d 2 d 1 Area of a rhombus A = 1 2 d 1 d 2 where: d 1 = small diagonal d 2 =large diagonal Standards 7, 8, 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

18 Find the area of the figure below: A = 1 2 d 1 d 2 where: d 1 = small diagonal d 2 =large diagonal Calculating value for diagonals: d 1 = d 2 = d 2 =60 d 1 = 30 This is a rhombus so: Then substituting values: A = 1 2 (30)(60) A = 15(60) A= 900 Standards 7, 8, 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

19 Giving that the figure below has an area of 300 square units, find the value for both diagonals: A = 1 2 d 1 d 2 where: d 1 = small diagonal d 2 =large diagonal 45 d 2 = d 2 = 90 This is a rhombus so: Then substituting values and solving for : d 1 Calculating value for : d = 1 2 (90) d = 45 d 1 45 d A = 300 Standards 7, 8, 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

20 The area of a rhombus is 300 cm. If one of its diagonals is 1 more than twice the other, find the length of each diagonal. 2 d 2 d 1 A = 1 2 d 1 d 2 A= 300 cm 2 300= 1 2 d 1 d d 1 d = d 1 d 2 First Equation Second Equation Solving both equations together: 600= d 1 d 2 d 1 Standards 7, 8, 10 d 2 =2=2 d d 2 =2=2 d 1 2 d 1 600= 2 d 1 2 d = 2 d 1 2 d Let’s solve this quadratic equation and then return to the problem: =2=2 d 1 +1 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

21 Using the Quadratic Formula: X= -b b - 4ac 2a2a 2 +_ where:0 = aX +bX +c 2 Our equation is: a= 2 b= 1 c= -600 We substitute values: Standards 7, 8, 10 0= 2 d 1 2 d = 4 + d 1 = -( ) ( ) - 4( )( ) 2( ) 2 +_ d 1 = _ d = 4 +_ d = 4 +_ d = =17 d = 4 - d = =-17.6 d 1 No sense in problem 17 d d 1 Let’s go back to problem! PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

22 The area of a rhombus is 300 cm. If one of its diagonals is 1 more than twice the other, find the length of each diagonal. 2 d 1 A = 1 2 d 1 d 2 A= 300 cm 2 300= 1 2 d 1 d d 1 d = d 1 d 2 First Equation Second Equation Solving both equations together: 600= d 1 d 2 Using second equation: 17 d 2 35 Standards 7, 8, 10 d 2 =2=2 d d 2 =2=2 d 1 600= d 1 2 d = 2 d 1 2 d = 2 d 1 2 d d 1 =2=2 d 1 +1 d 2 =2=2 d 1 d 2 =2=2 d 1 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

23 The perimeter of a rhombus is 40 units and one of its diagonals is 5 units. Find the area of the rhombus. Perimeter: P= X X X + X X X P = 4X 40 = 4X 4 4 X= 10 =10 5 To find the area we apply the formula: A = 1 2 d 1 d 2 d 1 = From the figure: d 1 = 5 10 = Z = Z = Z 2 Z 9.7 From the figure: d 2 =2=2 Z d 2 =2=2 ( ) 9.7 A = 1 2 ( )( ) A 48.5 Square units d d 2 Using Area Formula: Standards 7, 8, 10 We need to find the second diagonal using the Pythagorean Theorem; we know that so let’s find Z. d 2 =2=2 Z 2.5 Z PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

24 Calculate the non-shaded area of the figure below: A 1 A 2 Total Area =A 1 A 2 - is a trapezoid:A 1 A = 1 2 b 1 hb where: b 1 =50 b 2 = =80 h= =50 So the trapezoid is: A = 1 2 (50) A = (25) A = A = 1 2 (20) 2 25 A 2 is a triangle: A = bh b= 20 h= 25 Calculating Total Area: A = Total Area = Total Area =3000 The Total non-shaded Area is 3000 square units Standards 7, 8, 10 A = (10) PRESENTATION CREATED BY SIMON PEREZ. All rights reserved