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Areas of Parallelograms and Triangles
Skill 34
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Objective HSG-MG.1/7: Students are responsible for finding the areas of parallelograms and triangles.
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Definitions A base of a parallelogram can be any one of its sides. The corresponding altitude is a segment perpendicular to the line containing the base, drawn from the side opposite the base. The height is the length of the altitude. The base of a triangle can be any of its sides. The corresponding altitude is the length of the altitude to the line containing the base.
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Thm. 61: Area of a Rectangle
The area of a rectangle is the product of a base and height. Thm. 62: Area of a Parallelogram The area of a parallelogram is the product of a base and the corresponding height. Thm. 63: Area of a Triangle The area of a triangle is the half the product of a base and the corresponding height.
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Example 1; Finding Angle Measures of Trapezoids
a) πΆπ·πΈπΉ is an isosceles trapezoid and πβ πΆ=65α΅. What are πβ π·, πβ πΈ, and πβ πΉ? πβ πͺ+πβ π«=πππ ππ+πβ π«=πππ E F C D πβ π«=πππα΅ πβ πͺ=πβ π=ππα΅ 65α΅ πβ π«=πβ π¬=πππα΅
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Example 1; Finding Area of a Parallelogram
b) 4.6 cm 3.5 cm 2 cm 4.5 in. 4 in. 5 in. π¨=ππ π¨=ππ π¨=π π π¨=π π.π π¨=ππ ππ π π¨=π ππ π c) 10 m 9 m 12 m π¨=ππ π¨=ππ π π¨=πππ π π
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Example 2: Find the Missing Dimension
a) Find DπΈ to the nearest tenth. 9 in. 13 in. 9.4 in. A B C D E π¨=ππ π¨=ππ π π¨=πππ ππ π π¨=ππ πππ=π.π π«π¬ π«π¬=ππ.π ππ.
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Example 2: Find the Missing Dimension
b) A parallelogram has sides 15 cm and 18 cm. The height corresponding to a 15-cm base is 9 cm. What is the height corresponding to an 18-cm base? Sketch the situation to help you. π¨=ππ 15 cm 18 cm A B C D E π¨=ππ π 9 cm π¨=πππ ππ π π¨=ππ πππ=ππ π«π¬ π«π¬=π.π ππ
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Example 3; Finding Area of a Triangle
a) Find the area of the triangle at the left. In square feet. π=ππππβ ππππ πππ +πππ π=ππππβ ππππ πππ +πππ π=πππ ππ π=πππ ππ π¨= π π ππ = π π πππ πππ 12ft 2in. 13ft 4in. π¨=πππππ ππ π πππππ ππ π β πππ ππππ β πππ ππππ ππ π π ππ π
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Example 3; Finding Area of a Triangle
b) Find the area of the triangle at the left. In square inches. π=ππ ππ π=π ππ 1 ft. 1 ft. 1 in. 5 in. π¨= π π ππ = π π ππ π π¨=ππ ππ π
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Example 4; Finding Area of the Irregular Figure
a) What is the area of the figure at the right in square inches? π¨= π π ππ 8 in. 6 in. π¨= π π π π π¨=ππ ππ π π¨=ππ π¨= π π π¨=ππ ππ π π¨=ππππππππ+ππππππ π¨= ππ ππ π¨=ππ ππ π
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Example 4; Finding Area of the Irregular Figure
b) What is the area of the figure at the right in square inches? 7 in. 5 in. π¨= π π ππ π¨= π π π π π¨=ππ.π ππ π π¨=ππ π¨= π π π¨=ππ ππ π π¨=πππππππππ+ππππππ π¨=π ππ.π ππ π¨=ππ ππ π
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#34: Areas of Parallelograms and Triangles
Questions? Summarize Notes Homework Quiz
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