Three girls Radha, Seema and Medha are playing a game by standing on a circle of in a park. Radha
throws a ball to Seema, Seema to Medha, Medha to Radha. The distance between Radha and Seema and
between Seema and Medha is 6 m each. The distance between Radha and Medha is 10 m. Which condition cannot be applied to prove Δ OCB congruent to Δ OAB?
(a) SAS (b) ASA (c) RHS (d) AAS
1
(ii) Name the type of quadrilateral OABC
(a) Rhombus (b) Kite (c) Rectangle (d) Square
1
(iii) Radius of the circle is
(a) 18√11 m (b) 9√11 m (c) 18
11
√11 m (d) 18
11
m
1
(iv) Area of the circle is
(a) 3564 m2
(b) 891 m2
(c) 324
11
m2
(d) 324
121
m2
1
(v) Another girl Priya joins in and sits at position D. If ∠ ADC = 54°, then ∠ CAB is
(a) 54° (b) 27° (c) 126° (d) 108°
Answers
c is the correct answer of this question
Given :- Three girls Radha, Seema and Medha are playing a game by standing on a circle of in a park. Radha throws a ball to Seema, Seema to Medha, Medha to Radha. The distance between Radha and Seema and between Seema and Medha is 6 m each. The distance between Radha and Medha is 10 m.
To Find :-
A) Which condition cannot be applied to prove ΔOCB congruent to ΔOAC ?
(a) SAS (b) ASA (c) RHS (d) AAS
B) Name the type of quadrilateral OABC ?
(a) Rhombus (b) Kite (c) Rectangle (d) Square
C) Radius of the circle is ?
(a) 18√11 m (b) 9√11 m (c) 18/√11 m . d) 9/√11 m .
D) Area of the circle is ?
(a) 356.4 m² (b) 92.6 m² (c) 324m² (d) 121m² .
Solution :-
from image we can see that, Both ∆OAC and ∆OBC are Equaliteral ∆'s.
→ OA = OB = r (radius)
→ AC = BC (given 6 cm.)
→ OC = OC (common.)
therefore, we can conclude that, (c) RHS condition cannot be applied to prove ∆OAC congruent to ∆OBC .
Now, since Diagonals of quadrilateral OABC intersect at 90° and all sides are not same .
Therefore, quadrilateral OABC is a (b) kite.
now, Let us assume that, radius of the circle is r cm.
so,
→ OA = OC = OB = r cm.
now, in right ∆OEC ,
→ OE = √(OC² - EC²)
→ OE = √(r² - 9)
then, comparing ∆OAS area , we get,
→ (1/2) * AD * OC = (1/2) * AC * OE
→ 5 * r = 6 * √(r² - 9)
→ (5r/6) = √(r² - 9)
squaring both sides,
→ (r² - 9) = 25r²/36
→ r² - 25r²/36 = 9
→ (36r² - 25r²)/36 = 9
→ 11r² = 324
→ r² = (324/11)
→ r = (18/√11) m (c) (Ans.)
therefore,
→ Area of circle = π * r²
→ Area = (22/7) * (324/11)
→ Area = (2 * 324)/7
→ Area = 648/7
→ Area = 92.6 m² (b) (Ans.)
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