Math  /  Geometry

QuestionHome > CCA2 > Chapter 2>2> Lesson 2.1.2 > Problem 2-21
2-21. Plot each pair of points and find the distance between them. Give answers in both square-root form and as decimal approximations. \square \square Hint: Draw a right triangle with the hypotenuse segment connecting the given points. Recall the Pythagorean Theorem. Find the difference between the xx - and yy coordinates, respectively. Square these values, add them, and find the square root of this value. a. (3,6)(3,-6) and (2,5)(-2,5) b. (5,8)(5,-8) and (3,1)(-3,1) c. (0,5)(0,5) and (5,0)(5,0) d. Write the distance you found in \square DAnswer (a): part (c) in simplified square-root 14612.1\sqrt{146} \approx 12.1 form. \square (2Hint (d): Rewrite 50\sqrt{50} as 252\sqrt{25} \cdot \sqrt{2}. Which factor can be simplified further?

Studdy Solution

STEP 1

What is this asking? We need to find the distance between several pairs of points, showing both the exact square-root answer and a decimal approximation. Watch out! Remember the Pythagorean Theorem: a2+b2=c2a^2 + b^2 = c^2.
Don't mix up your x and y coordinates when finding the differences.

STEP 2

1. Distance between (3, -6) and (-2, 5)
2. Distance between (5, -8) and (-3, 1)
3. Distance between (0, 5) and (5, 0)
4. Simplify the square root from part (c)

STEP 3

Let's **find** the difference in the x-coordinates: 3(2)=3+2=53 - (-2) = 3 + 2 = \mathbf{5}.

STEP 4

Now, the difference in the y-coordinates: 65=11-6 - 5 = \mathbf{-11}.

STEP 5

Using the Pythagorean Theorem, the distance squared is 52+(11)2=25+121=1465^2 + (-11)^2 = 25 + 121 = \mathbf{146}.

STEP 6

So, the distance is 14612.1\sqrt{146} \approx \mathbf{12.1}.

STEP 7

Difference in x-coordinates: 5(3)=5+3=85 - (-3) = 5 + 3 = \mathbf{8}.

STEP 8

Difference in y-coordinates: 81=9-8 - 1 = \mathbf{-9}.

STEP 9

Distance squared: 82+(9)2=64+81=1458^2 + (-9)^2 = 64 + 81 = \mathbf{145}.

STEP 10

Distance: 14512.0\sqrt{145} \approx \mathbf{12.0}.

STEP 11

Difference in x-coordinates: 05=50 - 5 = \mathbf{-5}.

STEP 12

Difference in y-coordinates: 50=55 - 0 = \mathbf{5}.

STEP 13

Distance squared: (5)2+52=25+25=50(-5)^2 + 5^2 = 25 + 25 = \mathbf{50}.

STEP 14

Distance: 507.1\sqrt{50} \approx \mathbf{7.1}.

STEP 15

We found the distance to be 50\sqrt{50}.
We can rewrite this as 252\sqrt{25 \cdot 2}.

STEP 16

Since 25=5\sqrt{25} = 5, we can simplify 50\sqrt{50} to 525\sqrt{2}.

STEP 17

a. 14612.1\sqrt{146} \approx 12.1 b. 14512.0\sqrt{145} \approx 12.0 c. 507.1\sqrt{50} \approx 7.1 d. 525\sqrt{2}

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord