QUESTION IMAGE
Question
- what is the perimeter of an equilateral triangle if two of its vertices have coordinates (0,2) and (0,5)? a) 3 b) 6 c) 9 d) 12 9. what kind of triangle is formed when its vertices (0,4), (-4,-2) and (4,-2) are plotted on the cartesian plane? a) equilateral b) isosceles c) right d) scalene 10. what type of quadrilateral is formed by the given vertices c (0,0), a (a,0), r (a,a) and e (0,a)? a) kite b) parallelogram c) rectangle d) square
Question 8
Step1: Calculate the distance between two vertices
The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). For points \((0,2)\) and \((0,5)\), \(x_1 = 0,x_2=0,y_1 = 2,y_2 = 5\). Then \(d=\sqrt{(0 - 0)^2+(5 - 2)^2}=\sqrt{0 + 9}=3\).
Step2: Find the perimeter of the equilateral triangle
Since it is an equilateral triangle, all sides are equal. The perimeter \(P\) of a triangle is \(P=a + b + c\). For an equilateral triangle \(a=b=c\). Here \(a = 3\), so \(P=3\times3=9\).
Step1: Calculate the lengths of the sides
Let \(A=(0,4)\), \(B=(-4,-2)\), \(C=(4,-2)\)
The distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)
\(AB=\sqrt{(-4 - 0)^2+(-2 - 4)^2}=\sqrt{16 + 36}=\sqrt{52}=2\sqrt{13}\)
\(AC=\sqrt{(4 - 0)^2+(-2 - 4)^2}=\sqrt{16+36}=\sqrt{52}=2\sqrt{13}\)
\(BC=\sqrt{(4+4)^2+(-2 + 2)^2}=\sqrt{64+0}=8\)
Step2: Classify the triangle
Since \(AB = AC
eq BC\), the triangle is isosceles.
Step1: Calculate the lengths of the sides
Using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)
\(CA=\sqrt{(a - 0)^2+(0 - 0)^2}=a\)
\(AR=\sqrt{(a - a)^2+(a - 0)^2}=a\)
\(RE=\sqrt{(0 - a)^2+(a - a)^2}=a\)
\(EC=\sqrt{(0 - 0)^2+(a - 0)^2}=a\)
Step2: Calculate the slopes of the sides
The slope formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\)
Slope of \(CA\): \(m_{CA}=\frac{0 - 0}{a - 0}=0\)
Slope of \(AR\): \(m_{AR}=\frac{a - 0}{a - a}\), undefined (vertical line)
Since all sides are equal (\(CA = AR=RE = EC=a\)) and adjacent sides are perpendicular (slope of \(CA = 0\) and slope of \(AR\) is undefined), it is a square.
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