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Q6.
Expert-verifiedWrite the reasons that justify the statements.
Given: $\u25b3ABC\cong \u25b3RST$. Prove: localid="1637922009427" $AK>RS$
Statement | Proof |
1. $\u25b3ABC\cong \u25b3RST$ | Given |
2. $\overline{AB}\cong \overline{RS}$, or $AB=RS$ | Corresponding parts of congruent triangles are congruent. |
3. $AK=AB+BK$ | Segment Addition Postulate. |
4. $AK>AB$ | Property of Inequality If $a=b+c$ and $c>0$ then $c>0$. |
5. $AK>RS$ | Property of Inequality If $a=b+c$ and $c>0$ then $a>c$. |
Two triangles ABC and RST as shown below are the given congruent triangles.
The corresponding parts of the congruent triangles are congruent.
Since $\u25b3ABC\cong \u25b3RST$ therefore, role="math" localid="1637922313136" $\overline{AB}\cong \overline{RS}$, or $AB=RS$
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, $AK=AB+BK$
Hence, the reason for 3rd statement is Segment Addition Postulate.
If $a=b+c$ and $c>0$ then $a>c$.
Since $AK=AB+BK$ therefore, $AK>AB$.
Hence, the answer to 4 is Property of Inequality If $a=b+c$ and $c>0$ then $a>c$.
Consider that
From 3rd statement;
$AK=AB+BK$
From 2nd statement;
$\overline{AB}\cong \overline{RS}$, or $AB=RS$
Therefore, $AK=RS+BK$
Hence from the property of inequality, If $a=b+c$ and $c>0$ then role="math" localid="1637921863353" $a>c$.
Therefore, $AK>RS$
Hence, the answer to 5 is Property of Inequality.
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