Log In Start studying!

Select your language

Suggested languages for you:

Americas

English (US)

Europe

Q6.

Expert-verified

Geometry

Found in: Page 206

Book edition Student Edition

Author(s) Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen

Pages 227 pages

ISBN 9780395977279

Answers without the blur.

Just sign up for free and you're in.

Register for Free I’ll do it later

Short Answer

Write the reasons that justify the statements.
Given: $△ A B C ≅ △ R S T$ . Prove: localid="1637922009427" $A K > R S$

Statement	Proof
1. $△ A B C ≅ △ R S T$	Given
2. $\bar{A B} ≅ \bar{R S}$ , or $A B = R S$	Corresponding parts of congruent triangles are congruent.
3. $A K = A B + B K$	Segment Addition Postulate.
4. $A K > A B$	Property of Inequality If $a = b + c$ and $c > 0$ then $c > 0$ .
5. $A K > R S$	Property of Inequality If $a = b + c$ and $c > 0$ then $a > c$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Draw the given diagram and define the congruent property of triangles

Two triangles ABC and RST as shown below are the given congruent triangles.

The corresponding parts of the congruent triangles are congruent.

Since $△ A B C ≅ △ R S T$ therefore, role="math" localid="1637922313136" $\bar{A B} ≅ \bar{R S}$ , or $A B = R S$

Hence, the reason for 2nd statement is Corresponding parts of congruent triangles are congruent.

From Segment Addition Postulate. B is a point between A and K. Therefore, $A K = A B + B K$

Hence, the reason for 3rd statement is Segment Addition Postulate.

Step 2. Property of inequality

If $a = b + c$ and $c > 0$ then $a > c$ .

Since $A K = A B + B K$ therefore, $A K > A B$ .

Hence, the answer to 4 is Property of Inequality If $a = b + c$ and $c > 0$ then $a > c$ .

Step 3. Prove statement 5

Consider that

From 3rd statement;

$A K = A B + B K$

From 2nd statement;

$\bar{A B} ≅ \bar{R S}$ , or $A B = R S$

Therefore, $A K = R S + B K$

Hence from the property of inequality, If $a = b + c$ and $c > 0$ then role="math" localid="1637921863353" $a > c$ .

Therefore, $A K > R S$

Hence, the answer to 5 is Property of Inequality.

Most popular questions for Math Textbooks

For the following conditional statement, classify the statement as true or false.

If $d > e$ and $f > e$ , then $d > f$ .

For the following conditional, write the contrapositive of the statement and find the inverse is true or false.

If $Δ A B C$ is acute, then $m ∠ C \neq 90$ .

Write the letters (a)-(d) in an order that completes an indirect proof of the statement: Through a point outside a line, there is at most one line perpendicular to the given line.

Given: Point P not on the line k.

Prove: There is at most one line through P perpendicular to k.

(a) But this contradicts corollary3 of theorem 3-11: In a triangle, there can be at most one right angle or obtuse angle.

(b) Then $∠ P A B$ and $∠ P B A$ are right angles, and $Δ P A B$ has two right angles.

(c) Thus, our temporary assumption is false, and there is at most one line through P perpendicular to k.

(d) Assume temporarily that there are two lines through P perpendicular to k at A, B.

For the following conditional statement, classify the statement as true or false.

Given: Points $R, S, T,$ and $W; \overset{\leftrightarrow}{R S}$ and $\vec{T W}$ are skew.

Prove: $\overset{\leftrightarrow}{R T}$ and $\overset{\leftrightarrow}{S W}$ are skew.

Want to see more solutions like these?

Sign up for free to discover our expert answers

Get Started - It’s free

Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.

Sign up for free