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02 Jun 2020, 04:30
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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AB and CD are two parallel chords in a circle with centre O and radius 5cm. If the length of AB = 6cm, CD = 8cm. Find the perpendicular distance between the two chords.
A. 4 cm
B. 5 cm
C. 6 cm
D. 7 cm
E. 8 cm
Attachment:
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02 Jun 2020, 04:51
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is it 7??
there are 2 isoceles triangles with 2 equal sides of 5.
The 1st triangle has a base of 8 (Chord AB) and the 2nd of 6 (Chord CD).
you take a perpendicular line from the Radius to both CD and AB (one base (Chord) is divided to 3 and 3, the other to 4 and 4 (SINCE THEY ARE PARALLEL) and you get the ratio of right angle triangle as: 3:4:5 and 4:3:5.
so the 1st perpendicular is 3 and the other is 4. so i go with 7 D.
there are 2 isoceles triangles with 2 equal sides of 5.
The 1st triangle has a base of 8 (Chord AB) and the 2nd of 6 (Chord CD).
you take a perpendicular line from the Radius to both CD and AB (one base (Chord) is divided to 3 and 3, the other to 4 and 4 (SINCE THEY ARE PARALLEL) and you get the ratio of right angle triangle as: 3:4:5 and 4:3:5.
so the 1st perpendicular is 3 and the other is 4. so i go with 7 D.
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02 Jun 2020, 06:40
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Connect OC OD OA and OB
Area of triangle OAB is
=> √S(S-A)(S-B)(S-C)
(Half perimeter) S = AO+OB+AB/2 => 5+5+6/2
=> S = 8
=> √8(8-5)(8-5)(8-6)
=> √8(3)(3)(2)
=> √2*2*2(3)(3)(2)
=> 2*2*3 = 12 = Area of ABO
Let the perpendicular on AB from centre be OM
Also area of ABO = 1/2*AB*OM
=> 12 = 1/2*6*OM
=> 4 = OM
Area of triangle OCD is
=> √S(S-A)(S-B)(S-C)
(Half perimeter) S = CO+OD+CD/2 => 5+5+8/2
=> S = 9
=> √9(9-5)(9-5)(9-8)
=> √9(4)(4)(1)
=> √3*3(2*2)(2*2)(1)
=> 2*2*3 = 12 = Area of CDO
Let the perpendicular on CD from centre be ON
Also area of CDO = 1/2*CD*ON
=> 12 = 1/2*8*ON
=> 3 = ON
The perpendicular distance between two chords = ON+OM = 4+3 = 7
Answer is D
Posted from my mobile device
Area of triangle OAB is
=> √S(S-A)(S-B)(S-C)
(Half perimeter) S = AO+OB+AB/2 => 5+5+6/2
=> S = 8
=> √8(8-5)(8-5)(8-6)
=> √8(3)(3)(2)
=> √2*2*2(3)(3)(2)
=> 2*2*3 = 12 = Area of ABO
Let the perpendicular on AB from centre be OM
Also area of ABO = 1/2*AB*OM
=> 12 = 1/2*6*OM
=> 4 = OM
Area of triangle OCD is
=> √S(S-A)(S-B)(S-C)
(Half perimeter) S = CO+OD+CD/2 => 5+5+8/2
=> S = 9
=> √9(9-5)(9-5)(9-8)
=> √9(4)(4)(1)
=> √3*3(2*2)(2*2)(1)
=> 2*2*3 = 12 = Area of CDO
Let the perpendicular on CD from centre be ON
Also area of CDO = 1/2*CD*ON
=> 12 = 1/2*8*ON
=> 3 = ON
The perpendicular distance between two chords = ON+OM = 4+3 = 7
Answer is D
Posted from my mobile device
