What is the sum of the interior angles of a polygon that has 7 sides?
2 answers:
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Answer:
The correct option is
If two sides and one included angle are equal in triangles PQS and PRS, then their third sides are equal.
Step-by-step explanation:
Given:
RS ≅ SQ
∠PSR ≅ ∠PSQ = 90°
To Prove:
point P is equidistant from points R and Q
i.e PR ≅ PQ
Proof:
In ΔPSR and Δ PSQ
PS ≅ PS ……….{Reflexive Property}
∠PSR ≅ ∠PSQ = 90° …………..{Measure of each angle is 90° given}
RS ≅ QS ……….{Given}
ΔPSR ≅ ΔPSQ ….{By Side-Angle-Side Congruence test}
∴ PR ≅ PQ .....{Corresponding Parts of Congruent Triangles}
i.e point P is equidistant from points R and Q .......Proved
You are given the length of every side, except for two, that can be easily derived from the others:
- The long horizontal side on top is the sum of the three beneath it, so it is
- The vertical one on the right is, similarly, the sum of the opposite side to the far left, and the two small sides near it:
The sum of all sides, including these two, will give you the perimeter of the polygon.