Lesson Example Discussion Quiz: Class Homework |
Example |
Title: Geometry |
Grade: Best-SAT3 Lesson: S3-P1 |
Explanation: The best way to understand SAT-3 is by looking at some examples. Take turns and read each example for easy understanding. |
Examples:
An arc length in a circle is 6 cm, and the central angle corresponding to the arc is 40°. Find the radius of the circle.
Step 1a
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The given arc length is 6 cm and the central angle is 40° To find the radius of the circle, we can use the formula relating the arc length, central angle, and radius: arc length = radius \$times\$ central angle |
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Explanation:
Here we have to find the radius(r) of the circle by using the arc length and central angle |
Step 1b
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Substituting these values into the formula, we have: 6 cm = r × 40° The angle to be in radians rather than°. Since 1 radian is equal to 180° / π, we can convert the angle to radians. ⇒ \$6 "cm" = "r" \times ((40°) times (π / 180°))\$ ⇒ \$6 = "r" \times ((2π) / 9) \$ radians ⇒ \$6 / ((2π) / 9) = "r"\$ radians ⇒ \$"r" = (54 "cm") / π\$ radians ⇒ \$"r" = (54 "cm") / 3.14\$ radians Thus, the radius of the circle is 17.1828 cm. |
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Explanation: Here, we substitute the given values in the formula. To find the circle’s radius, we substitute the π = 3.14. On solving, we get the radius of the circle is 17.1828cm. |
Find the area of an equilateral triangle of side 12 cm (\$\sqrt 3 = 1.732\$).
Step 2a
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Given the side of a triangle = 12cm Area of an equilateral triangle = \$\sqrt 3/4\$ a² square units ⇒ \$\sqrt 3/4 \times 12 \times 12\$ ⇒ \$\sqrt 3/4 \times 144\$ ⇒ \$\sqrt 3 * \cancel144^36/\cancel4^1\$ ⇒ \$1.73 \times 36\$ ⇒ 62.28 cm² |
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Explanation: If the side of an equilateral triangle is 12cm, then the area of an equilateral triangle is 62.28 cm². |
To find the equation of the line passing through the points (1, -2) and (4, -5).
Step 3a
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First, to find the slope (m) of a line passing through two points (x1, y1) and (x2, y2) is given by: \$m = (y2 - y1) / (x2 - x1)\$. Let’s plug in the coordinates of the given points (1, -2) and (4, -5): \$m = (-5 - (-2)) / (4 - 1)\$ then \$m = (-5 + 2) / 3 \$ ⇒ m = -1. |
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Explanation: To find the slope (m) of a line passing through two points (x1, y1) and (x2, y2), you can use this formula: m = (y2 - y1) / (x2 - x1) Using the points (1, -2) and (4, -5), the slope(m) is -1. |
Step 3b
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In this step, the point-slope form of the equation of a line is: Using the point (1,−2) and the slope m=−1: ⇒y - (-2) = -1(x - 1) ⇒ y + 2 = -1 (x) + 1 ⇒ y = -1 (x) + 1 - 2 ⇒ y = -1 (x) - 1 |
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Explanation: To proceed, apply the point-slope form. With the slope (m = -1) and one point (1, -2), we can formulate the equation of the line as follows: y − y1 = m(x − x1). Therefore, substituting the values, we get y = -1x - 1. |
Step 3c
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So, the equation of the line passing through the points (1, -2) and (4, -5) is y = -x - 1. |
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Explanation: The equation for the line that passes through the coordinates (1, -2) and (4, -5) is y = -x - 1. |
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