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Step-3

Title:

Grade: 10-a Lesson: S2-L3

Explanation:

From the figure we can observe that
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Consider \$\triangle PAQ\$ and \$\triangle PBQ\$.

\begin{align} PQ &= PQ \tag{Common side} \\ AP &= BP \tag{Given} \\ AQ &= BQ \tag{Given} \\ \end{align}

So, \begin{align} \triangle PAQ \cong \triangle PBQ \\ \end{align}

Henced proved that \$\trianglePAQ\$ \$\cong\$ \$\trianglePBQ\$ by the SSS congruence rule.

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Steps	Statment	Solution
1	Given	From the figure we can observe that
2	Side of triangles	PA = PB and QA = QB and PQ ⊥ AB
3	Consider triangles	\$\triangle PAQ and \triangle PBQ\$
4	Common side	PQ = PQ
5	Side of triangles	AP = BP and AQ = BQ
6	So, Congruency	\$\triangle PAQ \cong \triangle PBQ\$
7	Prove that	Henced proved that \$\trianglePAQ\$ \$\cong\$ \$\trianglePBQ\$ by the SSS congruence rule.

Steps

Statment

Solution

Given

From the figure we can observe that

Side of triangles

PA = PB and QA = QB and PQ ⊥ AB

Consider triangles

\$\triangle PAQ and \triangle PBQ\$

Common side

PQ = PQ

Side of triangles

AP = BP and AQ = BQ

So, Congruency

\$\triangle PAQ \cong \triangle PBQ\$

Prove that

Henced proved that \$\trianglePAQ\$ \$\cong\$ \$\trianglePBQ\$ by the SSS congruence rule.