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Quadrilateral ABCD is inscribed in a circle with AC as a diameter of t
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28 Sep 2023, 01:14
28 Sep 2023, 01:14
Expert Reply
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Question Stats:
12% (02:57) correct
87% (02:57) wrong based on 8 sessions
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Quadrilateral ABCD is inscribed in a circle with AC as a diameter of the circle, such that ∠DAC is 30◦ and ∠BAC is 45◦ . The radius of the circle is 1. The area of ABCD to can be expressed as a fraction in simplest radical form as
\(\dfrac{a+\sqrt{b}}{c}\)
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
\(\dfrac{a+\sqrt{b}}{c}\)
Quantity A |
Quantity B |
a + b + c |
7 |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
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Free Materials for the GRE General Exam - Where to get it!!
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Free Materials for the GRE General Exam - Where to get it!!
GRE Geometry Formulas
GRE - Math Book
Re: Quadrilateral ABCD is inscribed in a circle with AC as a diameter of t
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29 Sep 2023, 15:33
29 Sep 2023, 15:33
I am getting 8 as answer.
How is it C?
How is it C?
Re: Quadrilateral ABCD is inscribed in a circle with AC as a diameter of t
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30 Sep 2023, 00:55
30 Sep 2023, 00:55
1
Expert Reply
We know that AC is the diameter of the circle (centered at O).
Since the radius of the circle is 1, we have AC = 2 Since the diameter always subtends a right angle at the circumference, we have
∠ABC = ∠ADC = 90◦
Also, we know that ∠DAC is 30◦ and ∠BAC is 45◦
Thus, in triangle ADC, we have ∠DCA = 180◦− (90◦+30◦) = 60◦
Also, in triangle ABC, we have ∠BCA = 180◦− (90◦+ 45◦) = 45◦
Thus, triangle ADC is a 30-90-60 triangle => AD : DC : AC =√3 : 1 : 2
Since AC = 2, we have AD = √3, and DC = 1 => Area of triangle ADC=1/2 × AD × DC=1/2×√3 × 1=√3/2 . . . (I)
Also, we have triangle ABC to be a 45-90-45 triangle => AB : BC : AC = 1 : 1 :√2
Since AC = 2, we have AB = BC = 2√/2=√2 => Area of triangle ABC =1/2 × AB × BC=1/2×√2 ×√2 = 1 . . . (ii) Thus, from (i) and (ii), we have
Area of ABCD = Area of triangle ABC + Area of triangle ADC =1 +√3/2=2 +√3/2
Thus, we have a +√b/c=2 +√3/2
=> a = 2, b = 3 & c = 2
=> a + b + c = 7
C is the answer
Since the radius of the circle is 1, we have AC = 2 Since the diameter always subtends a right angle at the circumference, we have
∠ABC = ∠ADC = 90◦
Also, we know that ∠DAC is 30◦ and ∠BAC is 45◦
Thus, in triangle ADC, we have ∠DCA = 180◦− (90◦+30◦) = 60◦
Also, in triangle ABC, we have ∠BCA = 180◦− (90◦+ 45◦) = 45◦
Thus, triangle ADC is a 30-90-60 triangle => AD : DC : AC =√3 : 1 : 2
Since AC = 2, we have AD = √3, and DC = 1 => Area of triangle ADC=1/2 × AD × DC=1/2×√3 × 1=√3/2 . . . (I)
Also, we have triangle ABC to be a 45-90-45 triangle => AB : BC : AC = 1 : 1 :√2
Since AC = 2, we have AB = BC = 2√/2=√2 => Area of triangle ABC =1/2 × AB × BC=1/2×√2 ×√2 = 1 . . . (ii) Thus, from (i) and (ii), we have
Area of ABCD = Area of triangle ABC + Area of triangle ADC =1 +√3/2=2 +√3/2
Thus, we have a +√b/c=2 +√3/2
=> a = 2, b = 3 & c = 2
=> a + b + c = 7
C is the answer
