Answer
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Hint: We will use the fact that: Tangents from the same external points are equal in length. Using this fact and doing some modification, we will prove that OQBP is a square and thus find the radius using given data.
Complete step-by-step answer:
We see that AD and CD touch the circle at R and S respectively and nowhere else.
Hence, we can call AD and CD the tangents to the circle.
Because tangent is a straight line or plane that touches a curve or curved surface at a point, but if extended does not cross it at that point.
Now, we will use the fact that: Tangents from the same external points are equal in length.
Now tangents AD and CD cut at R and S respectively coming from the same point D.
Hence, DR = DS ………..(1)
Similarly using the same argument again and again, we will have:-
AR = AQ ……..(2)
CP = CS ……..(3)
BQ = BP ……..(4)
Adding (1) and (2), we will get:-
DR + AR = DS + AQ ……….(5)
We can clearly see from the figure that DR + AR = AD = 23 cm (given).
Hence, (5) becomes DS + AQ = AD = 23 cm.
So, DS + AQ = 23 cm.
But we are given DS = 5 cm.
So, 5 + AQ = 23.
So, AQ = 18 cm.
Now, we know that AQ + QB = AB = 23 cm.
So, 18 cm + QB = 23 cm
SO, QB = 5 cm ………..(6)
By using (4), we will have:-
BQ = BP = 5 cm.
And we see that OQ and OP are the radii of the same circle.
Hence, OQ = OP …..(7)
We also are already given that \[\angle B = {90^ \circ }\] ……..(8)
By using (6), (7) and (8):-
We have OQBP is a square.
Hence, all sides must be equal.
So, OQ = 5cm.
Hence, the radius is 5 cm.
Note: The students might just prove the sides to be equal to forget to refer to the one angle as the right angle which is a must for the quadrilateral to be a square. So, be aware to mention the given angle as well.
If the figure is not provided in such questions, after prefer drawing it first before starting off with the solution to see things better.
Complete step-by-step answer:
We see that AD and CD touch the circle at R and S respectively and nowhere else.
Hence, we can call AD and CD the tangents to the circle.
Because tangent is a straight line or plane that touches a curve or curved surface at a point, but if extended does not cross it at that point.
Now, we will use the fact that: Tangents from the same external points are equal in length.
Now tangents AD and CD cut at R and S respectively coming from the same point D.
Hence, DR = DS ………..(1)
Similarly using the same argument again and again, we will have:-
AR = AQ ……..(2)
CP = CS ……..(3)
BQ = BP ……..(4)
Adding (1) and (2), we will get:-
DR + AR = DS + AQ ……….(5)
We can clearly see from the figure that DR + AR = AD = 23 cm (given).
Hence, (5) becomes DS + AQ = AD = 23 cm.
So, DS + AQ = 23 cm.
But we are given DS = 5 cm.
So, 5 + AQ = 23.
So, AQ = 18 cm.
Now, we know that AQ + QB = AB = 23 cm.
So, 18 cm + QB = 23 cm
SO, QB = 5 cm ………..(6)
By using (4), we will have:-
BQ = BP = 5 cm.
And we see that OQ and OP are the radii of the same circle.
Hence, OQ = OP …..(7)
We also are already given that \[\angle B = {90^ \circ }\] ……..(8)
By using (6), (7) and (8):-
We have OQBP is a square.
Hence, all sides must be equal.
So, OQ = 5cm.
Hence, the radius is 5 cm.
Note: The students might just prove the sides to be equal to forget to refer to the one angle as the right angle which is a must for the quadrilateral to be a square. So, be aware to mention the given angle as well.
If the figure is not provided in such questions, after prefer drawing it first before starting off with the solution to see things better.
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