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sammy [17]
2 years ago
12

Triangle MRN is created when an equilateral triangle is folded in half. Triangle N R M is shown. Angle N R M is a right angle. A

n altitude is drawn from point R to point S on side M N to form a right angle. The length of N S is 6, the length of S M is 2, the length of R M is x, the length of N R is y, and the length of R S is z. What is the value of x? 2 StartRoot 3 EndRoot units 4 units 4 StartRoot 3 EndRoot units 8 units

Mathematics
2 answers:
Fed [463]2 years ago
7 0

Answer:

The value of x is 4.

Step-by-step explanation:

It is given that triangle MRN is created when an equilateral triangle is folded in half.

It means original equilateral is triangle MNO and NR is a perpendicular bisector (<em>A line which cuts a line segment into two equal parts at 90°</em>).

The side length of the triangle is

NO = NS + SM = 6 + 2 = 8

Since an equilateral triangle is a triangle in which all three sides are equal and NR is a perpendicular bisector, therefore

RM = MO/2 = 8/2 = 4

The value of x is 4.

Ilya [14]2 years ago
7 0

Answer:

He is right i just took the test and its 4 units

Step-by-step explanation:

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so we have three points, A, B and C, if indeed AC is the diameter of the circle, then half the distance of AC is its radius, and the midpoint of AC is the center of the circle, morever, since B is also on the circle, the distance from B to the center must be the same radius distance.

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\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{7}~,~\stackrel{y_1}{4})\qquad C(\stackrel{x_2}{12}~,~\stackrel{y_2}{3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{12+7}{2}~~,~~\cfrac{3+4}{2} \right)\implies \left( \cfrac{19}{2}~~,~~\cfrac{7}{2} \right)=M\impliedby \textit{center of the circle}

now, let's check the distance from say A to the center, and check the distance of B to the center, if it's indeed the center, they'll be the same and thus AC its diameter.

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{7}~,~\stackrel{y_1}{4})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ AM=\sqrt{\left( \frac{19}{2}-7 \right)^2+\left( \frac{7}{2}-4 \right)^2} \\\\\\ AM=\sqrt{\left( \frac{5}{2}\right)^2+\left( -\frac{1}{2} \right)^2}\implies \boxed{AM\approx 2.549509756796392} \\\\[-0.35em] ~\dotfill

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ B(\stackrel{x_1}{10}~,~\stackrel{y_1}{6})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}}) \\\\\\ BM=\sqrt{\left( \frac{19}{2}-10 \right)^2+\left( \frac{7}{2}-6 \right)^2} \\\\\\ BM=\sqrt{\left( -\frac{1}{2}\right)^2+\left( -\frac{5}{2} \right)^2}\implies \boxed{BM\approx 2.549509756796392}

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