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31 Aug 2020, 10:02
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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:
65%
(hard)
Question Stats:
49% (01:35) correct
51%
(01:53)
wrong
based on 98
sessions
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In an isosceles triangle ABC, AD is a median of length \(5\sqrt{2}\). What is the perimeter of the triangle ABC?
1. AD=BD=CD
2. Triangle ABC is right-angled at A.
1. AD=BD=CD
2. Triangle ABC is right-angled at A.
yashikaaggarwal
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31 Aug 2020, 11:05
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Given: In an isosceles triangle ABC, AD is a median of length 5√2.
To find: What is the perimeter of the triangle ABC?
1. AD=BD=CD
Angle DAC= Angle DCA = Angle DBA = Angle DAB = 45
Angle A = 90
Area of triangle = 1/2*Height*base
Area of triangle = 1/2*5√2*2*5√2.
Area of triangle = 50
(Sufficient)
2. Triangle ABC is right-angled at A.
AD is median
therefore Angle DBA = Angle DAB = 45
the remaining angles by triangle sum property will be Angle DAC= Angle DCA = 45
Area of triangle = 1/2*Height*base
Area of triangle = 1/2*5√2*2*5√2.
Area of triangle = 50
(Sufficient)
Answer is D
To find: What is the perimeter of the triangle ABC?
1. AD=BD=CD
Angle DAC= Angle DCA = Angle DBA = Angle DAB = 45
Angle A = 90
Area of triangle = 1/2*Height*base
Area of triangle = 1/2*5√2*2*5√2.
Area of triangle = 50
(Sufficient)
2. Triangle ABC is right-angled at A.
AD is median
therefore Angle DBA = Angle DAB = 45
the remaining angles by triangle sum property will be Angle DAC= Angle DCA = 45
Area of triangle = 1/2*Height*base
Area of triangle = 1/2*5√2*2*5√2.
Area of triangle = 50
(Sufficient)
Answer is D
Attachments
Triangle.png [ 39.37 KiB | Viewed 2200 times ]
01 Sep 2020, 07:24
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Given that ABC is an Isosceles triangle and AD is a median of length of \(5\sqrt{2}\)
To find the perimeter of triangle ABC
Statement 1: AD = BD = CD
Let \(\angle{DAC} = x\)
Then \(\angle{DAC} = \angle{DAB} = \angle{ABD} = \angle{ACD} = x\)
In \(\triangle{DAC}: \) x + x + x + x = 180 (Sum of angles of a triangle = \(180^o\))
\(4x = 180^o\), or \(x = 45^o\)
The triangle is an isosceles right angled triangle, right angled at A and hypotenuse BC = \(5\sqrt{2}+ 5\sqrt{2} = 10\sqrt{2}\)
The equal sides AC = AB = \(\frac{1}{\sqrt{2}} * Hypotenuse = 10\)
Therefore the perimeter = \(10 + 10 + 10\sqrt{2} = 20 + 10\sqrt{2}\)
Statement 1 is sufficient.. The answer Options could be A or D
Statement 2: Triangle ABC is right angled at A
Since it is right angled at A, again we have an isosceles right angled triangle, with \(\angle A = 90^o and \angleB = \angle C = 45^o\)
Therefore, the median from A is a perpendicular bisector and an angular bisector. (Property of an isosceles triangle for the median to the unequal side)
\(\triangle{ABD}\) is an isosceles right angled triangle, with AD = BD = \(5\sqrt{2}\)
\(\triangle{ADC}\) is an isosceles right angled triangle, with AD = DC = \(5\sqrt{2}\)
\(\triangle{ABC}\) is an isosceles right angled triangle, with AB = AC and hypotenuse = \(5\sqrt{2} + 5\sqrt{2} = 10\sqrt{2} \)
The equal sides AC = AB = \(\frac{1}{\sqrt{2}} * Hypotenuse = 10\)
Therefore the perimeter = \(10 + 10 + 10\sqrt{2} = 20 + 10\sqrt{2}\)
Statement 2 is sufficient..
Option D
Arun Kumar
To find the perimeter of triangle ABC
Statement 1: AD = BD = CD
Let \(\angle{DAC} = x\)
Then \(\angle{DAC} = \angle{DAB} = \angle{ABD} = \angle{ACD} = x\)
In \(\triangle{DAC}: \) x + x + x + x = 180 (Sum of angles of a triangle = \(180^o\))
\(4x = 180^o\), or \(x = 45^o\)
The triangle is an isosceles right angled triangle, right angled at A and hypotenuse BC = \(5\sqrt{2}+ 5\sqrt{2} = 10\sqrt{2}\)
The equal sides AC = AB = \(\frac{1}{\sqrt{2}} * Hypotenuse = 10\)
Therefore the perimeter = \(10 + 10 + 10\sqrt{2} = 20 + 10\sqrt{2}\)
Statement 1 is sufficient.. The answer Options could be A or D
Statement 2: Triangle ABC is right angled at A
Since it is right angled at A, again we have an isosceles right angled triangle, with \(\angle A = 90^o and \angleB = \angle C = 45^o\)
Therefore, the median from A is a perpendicular bisector and an angular bisector. (Property of an isosceles triangle for the median to the unequal side)
\(\triangle{ABD}\) is an isosceles right angled triangle, with AD = BD = \(5\sqrt{2}\)
\(\triangle{ADC}\) is an isosceles right angled triangle, with AD = DC = \(5\sqrt{2}\)
\(\triangle{ABC}\) is an isosceles right angled triangle, with AB = AC and hypotenuse = \(5\sqrt{2} + 5\sqrt{2} = 10\sqrt{2} \)
The equal sides AC = AB = \(\frac{1}{\sqrt{2}} * Hypotenuse = 10\)
Therefore the perimeter = \(10 + 10 + 10\sqrt{2} = 20 + 10\sqrt{2}\)
Statement 2 is sufficient..
Option D
Arun Kumar
Attachments
DS .jpg [ 444.01 KiB | Viewed 2060 times ]
