ΔRSP ≅ ΔQSP by SAS. Therefore, PR ≅ PQ by CPCTC theorem.
<h3>What is the CPCTC Theorem?</h3>
If two triangles are congruent, the CPCTC theorem states that all corresponding parts of the two triangles would also be congruent to each other.
Since S is the midpoint of QR, therefore, RS ≅ QS (one pair of congruent sides)
∠RSP ≅ ∠QSP [right angles] (one pair of congruent included angles)
SP ≅ SP based on the reflexive property (one pair of congruent sides)
Therefore, ΔRSP ≅ ΔQSP by SAS.
Conclusively, PR ≅ PQ by CPCTC theorem.
Learn more about the CPCTC theorem on:
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<span>This characteristic of non-functions was noticed by I-don't-know-who, and was codified in "The Vertical Line Test": Given the graph of a relation, if you can draw a vertical line that crosses the graph in more than one place, then the relation is not a function.</span>
Answer:
D.
Step-by-step explanation:
It's not linear but it is decreasing.
Answer:
Area of Triangle QRP = 3
Step-by-step explanation:
According to Question , We have a circle With Centre 'O' & Area 48
.
Area Of Circle = 48
= 48
r = 
Now We Have Two Points Given On Circle Q & R , P Is Circumcentre Of Triangle QRO .
Thus A Circle Can Also Be Formed with Centre P . ( See attachment For Diagram )
Now The Diameter of Circle With Centre P = Radius Of Circle with Centre O
so Radius Of Circle With Centre P(
) = 
Now We Have To Find Area Of Equilateral Triangle .
A = 
A= 
The Area Of PQR is = 3
For Diagram , Please Find In Attachment
Answer:
He would run a half of a lap because if he ran a full lap in 2 minutes than in 1 minute it would be half of 1.
Step-by-step explanation:
