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McGraw Hill Integrated II, 2012

MH

McGraw Hill Integrated II, 2012 View details

3. Special Right Triangles

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Exercise 42 Page 646

Start with plotting the given points in a coordinate plane.

≈(-6.47,-4)

Practice makes perfect

Let's begin with plotting the given points in a coordinate plane. We can connect these two points with a segment.

Since we know that △ EFG is a right triangle with a right angle at the vertex F, we can deduct that the vertex E is lying on the horizontal line y=-4.

If △ EFG is a 30^(∘)-60^(∘)-90^(∘) triangle, then the length of GF is sqrt(3) times the length of EF. This means that EF= GFsqrt(3). First let's evaluate the length of GF.

Next we will substitute 6 for GF to find the length of EF.

EF=GF/sqrt(3)

GF= 6

EF=6/sqrt(3)

Calculate quotient

EF=3.468...

Round to 2 decimal place(s)

EF≈3.47

Therefore, the third vertex E is about 3.47 units from F either in a positive or a negative direction. However, as we are asked to find the coordinates of E in Quadrant III we will consider only the second case.

The coordinates of the vertex E are approximately (-6.47,-4).

Subchapter links

Exercises p.644-648