What is the area of an equilateral triangle with vertices at (–1, –3),

Question

What is the area of an equilateral triangle with vertices at (–1, –3), (9, –3), and (m, n) where m and n are both positive numbers?

A.  25 \sqrt{2}

B.  50 \sqrt{2}

C.  10 \sqrt{3}

D.  25 \sqrt{3}

E.  50 \sqrt{3}

Bunuel · Answer

Similar question: https://gmatclub.com/forum/what-is-the- ... 25983.html

SSM700plus · Answer

Area of Equilaterla Trailnbgel is sqrt(3)*a. a - side of eq triangle.

distance between vertices at (–1, –3), (9, –3) is 10. Thus side of Triangle is also 10.

Area = side *side * sqrt(3) / 4 = 100 * sqrt(3)/4

Area - 25*Sqrt(3)
Ans - D.

apurva908 · Answer

Dis tance between (–1, –3), (9, –3) is Sq root(9-(-1))2+(-3-(-3))2=sq root(10)2+ (0)2=10

area of equilateral triangle Sq root 3/4(a)2=Sq root3/4(100)= 25 Sq root(3)
D option is correct.

generis · Answer

Formula Area of equilateral triangle = \frac{s^3 \sqrt{3}}{4} Side length: The given points have the same y-coordinate. To find side length, s , subtract x_1 from x_2 : s = 9 - (-1) = 10 Area = \frac{10^3 \sqrt{3}}{4}= 25\sqrt{3} Answer D Area without the formula - diagram below If you do not recall the formula for the area of an equilateral triangle, derive the area by finding height. 1) Find length of a side from coordinates Given points have the same y-coordinates Length: subtract one x-coordinate from the other Length of side AC = (x_2 - x_1)=(9 - (-1) = 10 = base 2) Find the midpoint of AC Midpoint: \frac{-1+9}{2}=(4,-3) Midpoint is important only for the length of one leg of a right triangle 3) Height, h = BX: Sketch a vertical line up from (4,-3) to make a vertex Just guess at how high to go; we are trying to get a visual of the right triangle 4) height is derived from 30-60-90 ∆ BCX Median (= altitude) is a perpendicular bisector of vertex and opposite side -- creates two congruent 30-60-90 triangles with corresponding opposite sides in ratio x : x\sqrt{3} : 2x For right ∆ BCX CX = x = 5, so BX = x\sqrt{3} = 5\sqrt{3} (BC = 2x = 10 ) AC = base = 10 Area = \frac{1}{2} b*h = \frac{1}{2} 10*5\sqrt{3} = 25\sqrt{3} OR 5) Use Pythagorean theorem to find BX for right ∆ BCX One leg = 5 (CX, half of 10) Hypotenuse BC = 10 Let height, h = BX = z 5^2+ z^2= 10^2 z^2=75 z =\sqrt{25*3}=5\sqrt{3}=h= BX b = 10 (AC) Area = \frac{b*h}{2}=\frac{10*5\sqrt{3}}{2}=25\sqrt{3} Answer D 2018-06-19 equitri.png

What is the area of an equilateral triangle with vertices at (–1, –3),

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What is the area of an equilateral triangle with vertices at (–1, –3),

Prep Toolkit

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