Elementary Geometry For College Students, 7e
Elementary Geometry For College Students, 7e
7th Edition
ISBN: 9781337614085
Author: Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher: Cengage,
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### Calculating the Area of Triangle ABC In this problem, we are asked to find the area of triangle ABC. The triangle is an isosceles triangle with two equal angles of 60 degrees, which classifies it as an equilateral triangle. The length of one of the sides, AC, is 12 units. #### Diagram analysis: - Triangle ABC has a vertex C, two other vertices A and B, and D is the midpoint of AB. - Angles at A and B are given as 60 degrees each. - The length of the altitude (height) from point C to side AB is 12 units, and this altitude intersects AB at point D, forming two 30-60-90 right triangles (ADC and BDC). #### Options for the area are: A. \( 18\sqrt{3} \) square units B. \( 72\sqrt{3} \) square units C. \( 6\sqrt{3} \) square units D. \( 36\sqrt{3} \) square units To solve for the area of triangle ABC, we use the formula for the area of a triangle: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \] Given that: - Height (CD) = 12 units - Because it's a 30-60-90 right triangle, and sides of a 30-60-90 triangle are in the ratio 1 : \(\sqrt{3}\) : 2. - The height (opposite 60 degrees) in the 30-60-90 triangle is \( \frac{\sqrt{3}}{2} \times \text{base}\). Therefore, in triangle ABC: \[ AD = DB = \text{base} / 2 \] Given height CD = 12, then \( \text{base} = \frac{12 \times 2}{\sqrt{3}} = 8\sqrt{3} \) units. AB = 2 * 8\sqrt{3} = 16\sqrt{3} units Now, calculate the area: \[ \text{Area} = \frac{1}{2} \times 16\sqrt{3} \times 12 = 96\sqrt{3} \] Given the correct option as: None of
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Transcribed Image Text:### Calculating the Area of Triangle ABC In this problem, we are asked to find the area of triangle ABC. The triangle is an isosceles triangle with two equal angles of 60 degrees, which classifies it as an equilateral triangle. The length of one of the sides, AC, is 12 units. #### Diagram analysis: - Triangle ABC has a vertex C, two other vertices A and B, and D is the midpoint of AB. - Angles at A and B are given as 60 degrees each. - The length of the altitude (height) from point C to side AB is 12 units, and this altitude intersects AB at point D, forming two 30-60-90 right triangles (ADC and BDC). #### Options for the area are: A. \( 18\sqrt{3} \) square units B. \( 72\sqrt{3} \) square units C. \( 6\sqrt{3} \) square units D. \( 36\sqrt{3} \) square units To solve for the area of triangle ABC, we use the formula for the area of a triangle: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \] Given that: - Height (CD) = 12 units - Because it's a 30-60-90 right triangle, and sides of a 30-60-90 triangle are in the ratio 1 : \(\sqrt{3}\) : 2. - The height (opposite 60 degrees) in the 30-60-90 triangle is \( \frac{\sqrt{3}}{2} \times \text{base}\). Therefore, in triangle ABC: \[ AD = DB = \text{base} / 2 \] Given height CD = 12, then \( \text{base} = \frac{12 \times 2}{\sqrt{3}} = 8\sqrt{3} \) units. AB = 2 * 8\sqrt{3} = 16\sqrt{3} units Now, calculate the area: \[ \text{Area} = \frac{1}{2} \times 16\sqrt{3} \times 12 = 96\sqrt{3} \] Given the correct option as: None of
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