Midsegment of a Triangle | MathHelp.com @yourteachermathhelp

Uploaded January 2016 | Updated September 2026, 2 weeks ago

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This lesson covers the midsegment of triangle. Students learn the following theorems related to parallel lines. If 3 parallel lines form congruent segments along one transversal, then they form congruent segments along every transversal. If a line intersects the midpoint of one side a triangle and is parallel to another side of the triangle, then it intersects the midpoint of the third side of the triangle. If a segment joins the midpoints of two sides of a triangle, then the segment is (a) parallel to the third side of the triangle, and (b) half the length of the third side. Students are then asked to solve problems related to these theorems using Algebra.

$Dividing Square Roots | MathHelp.com Need a custom math course? Visit https://www.MathHelp.com. This lesson covers dividing square roots. Students learn to divide square roots by dividing the numbers that are inside the radicals. Students also learn that if there is a square root in the denominator of a fraction, the problem can be simplified by multiplying both the numerator and denominator by the square root that is in the denominator.$

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$Work Word Problems | MathHelp.com Need a custom math course? Visit https://www.MathHelp.com Today! McCauley can paint a house in 10 hours, while it takes Clayton 15 hours. If they work together, how long will it take them to paint the house? To solve this kind of a problem, which is called a work problem, it’s important to understand the following idea. Since McCauley can paint a house in 10 hours, we know that in 1 hour, McCauley can paint 1/10 of the house. And in 2 hours, McCauley can paint 2/10 of the house. Therefore, in t hours, McCauley can paint t tenths of the house. And since it takes Clayton 15 hours to paint the house, in t hours, Clayton can paint t fifteenths of the house. Pause the audio for a moment if you need time to understand this idea…Now, to solve the problem, we use the following formula: part of job done by McCauley + part of job done by Clayton = 1 job done. And we’re asked how long will it take them to paint the house, so we’re looking for the time, or t. Remember that in t hours, McCauley can paint t/10 of the house, so the part of the job done by McCauley is t over 10. And in t hours, Clayton can paint t/15 of the house, so the part of the job done by Clayton is t over 15. Now, we have the equation t/10 + t/15 = 1. To solve this equation for t, we first get rid of the fractions by multiplying both sides of the equation by the common denominator of 10 and 15, which is 30. Distributing on the left side, 30 times t over 10 is 30t over 10, which simplifies to 3t, and 30 times positive t over 15 is positive 30t over 15, which simplifies to positive 2t. And on the right, 1 times 30 is 30. So we have 3t + 2t = 30, or 5t = 30, and dividing both sides by 5, t = 6. So if Clayton and McCauley work together, they can paint the house in 6 hours. Finally, it’s a good idea to check your answer. If they work together for 6 hours, then McCauley paints 6/10 of the house, and Clayton paints 6/15 of the house, so we have 6/10 + 6/15 = 1. And reducing on the left side, we have 3/5 + 2/5 = 1, which simplifies to 5/5 = 1, which is a true statement, so our answer checks$

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