1 Standards 7, 8, 10 Review Classification of Triangles Review Area of Triangle Area of an EQUILATERAL TRIANGLE given the side Area of an EQUILATERAL TRIANGLE given the height Area of an ISOSCELES RIGHT TRIANGLE given the hypotenuse Area of an ISOSCELES TRIANGLE given the vertex and base length Area of an ISOSCELES TRIANGLE given the vertex and one leg Area of a REGULAR OCTAGON using isosceles triangles 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 55° 64°61° 21° 110° 49° RIGHT TRIANGLE ACUTE TRIANGLE OBTUSE TRIANGLE ANGLES CLASSIFYING TRIANGLES BY ANGLES REVIEW Click to continue … Standards 7, 8, 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

4 SIDES CLASSIFYING TRIANGLES BY SIDES SCALENE TRIANGLE ISOSCELES TRIANGLE EQUILATERAL TRIANGLE REVIEW Standards 7, 8, 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

5 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

6 Finding the area of an EQUILATERAL TRIANGLE with side An equilateral triangle is also equiangular, so each angle measures 180° 3 =60° 2. Let’s draw an Altitude that is also an Angle Bisector and Median so each half of the angle is 60° 2 =30° and the side is 20 2 =10 each half So, we have a 30°-60°-90° triangle and we may use it to find the height of the equilateral triangle which is the unknown leg of this right triangle. 2 3S S S 60° 30° 10= = Click to continue the solution… 60° 4. We know that: 10 60° 30° 60° 30° 20 Standards 7, 8, 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

We have now a triangle like the one above, and we know: A = bh 1 2 b= 20 A= ? and h = So, we substitute values in the formula: A = (20) = = square units Reflection: Could you have found the height of the triangle using the Pythagorean Theorem instead? Standards 7, 8, 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

8 Given the height of an EQUILATERAL TRIANGLE find its area: 6 1. The height of the triangle is the Altitude, Angle Bisector and Median, so we have a 30°-60°-90° triangle (see PROBLEM 1). 60° 30° 60° 30° 6 2. And we know that: 2 3S S S 60° 30° = 6 6 = S =S ( ) =S Click to continue… Standards 7, 8, 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

9 60° 30° 60° 30° The short leg of the right triangle is half the side of the equilateral triangle, so we add both halves and draw the resulting triangle. 4. We have now a triangle like the one above at the right, and we know: A = bh 1 2 A= ? and h = 6 5. So, we substitute values in the formula: = square units b= 3 4 A = = Standards 7, 8, 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

10 Find the area of an ISOSCELES RIGHT TRIANGLE given the length of the hypotenuse: 10 45° The base angles of the isosceles triangle are congruent, each is 90° 2 =45° 2. Let’s draw the height which is Altitude, Angle Bisector and Median 3. Therefore we have a new isosceles triangle with both acute angles of 45° and the hypotenuse is bisected with each half 10 2 =5=5 4. Both legs of the new triangle are congruent. 5. Let’s draw the resulting triangle at the right and we apply the Area Formula. A = bh 1 2 b= 10 h= 5 A= 25 square units. A = (10)(5) 1 2 Reflection: Would the case with same triangle and a given height be the same? Standards 7, 8, 10

11 Find the area of an NON-RIGHT ISOSCELES TRIANGLE given the base lenght and vertex angle. 6 40° 70° 3 Tan 70°= 3 h Opposite side Adjacent Tan 70°= 3 h (3) Tan 70°=(3) h h = 3 ( ) h = A = bh 1 2 b= 6 h= 8.2 A = (6)(8.2) A 24.6 square units. 6. Substituting values: 1. Let’s obtain the base angles: 180°-40° 2 =70° 2. Let’s draw the Altitude and Median 3. Let’s draw the formed right triangle 4. Let’s use trigonometry to find the height h 3 5. Let’s redraw the original triangle and use the area formula: Standards 7, 8, 10 h PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

12 Find the area of an NON-RIGHT ISOSCELES TRIANGLE given the one leg and vertex angle. 50° 65° A = bh 1 2 b= 7.4 h= 8.2 A = (7.4)(8.2) A 30.3 square units. 6. Substituting values: 1. Let’s obtain the base angles: 180°-50° 2 =65° 2. Let’s draw the Altitude and Median 3. Let’s draw the formed right triangle 5. Let’s redraw the original triangle and use the area formula: 9 b b 1 Sin( )= h b 1 65° h 9 9 Sin 65°= h 9 (9) h=9 Sin 65° h=9( ) 65° = b Let’s use trigonometry and Pytagoream Theorem to find the h and and b. b 1 81 = b h b 1 = b If b=2 b 1 b=2(3.7) 7.4 h Standards 7, 8, =

13 What would be the Area for the regular octagon below if the sides measure 8? 1. Let’s divide in 8 congruent isosceles triangles the total area Let’s take one of the triangles to get its area The vertex angle for each triangle would be: 4. Let’s obtain the base angles: 180°-45° 2 =67.5° 360° 8 =45° 45° 67.5° 5. Let’s draw the altitude and median which is the height h h h ° 6. Let’s use trigonometry to find “h” Tan67.5°= 4 h (4) Tan 67.5°=(4) h h = 4 ( ) h 9.7 Tan 67.5°= 4 h 7. The area for the triangle is: b=8 h=9.7 A = bh 1 2 A = (8)(9.7) 1 2 A 38.8 square units. 8. There are 8 triangles like this in the regular octagon, so: Area of Regular Octagon= 8 (38.8) square units Standards 7, 8, 10