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Updated on: 07 Aug 2018, 21:15
Originally posted by EgmatQuantExpert on 29 Nov 2016, 08:40.
Last edited by EgmatQuantExpert on 07 Aug 2018, 21:15, edited 6 times in total.
Last edited by EgmatQuantExpert on 07 Aug 2018, 21:15, edited 6 times in total.
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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Mastering Important Concepts Tested by GMAT in Triangles - II : Exercise Question
Q. In the figure given below, AB = BC = 2√2 units and AC = 4 units. If BD bisects the side AC, find the length of BD.
Answer Choices
A. 1
B. √2
C. 2
D. 2√2
E. 3
Key concepts on Triangles are explained in detail in the following posts:
1. Mastering Important Concepts Tested By GMAT in Triangle - I
2. Mastering Important Concepts Tested By GMAT in Triangle - II
Detailed solution will be posted soon.
Thanks,
Saquib
Quant Expert
e-GMAT
Q. In the figure given below, AB = BC = 2√2 units and AC = 4 units. If BD bisects the side AC, find the length of BD.
Answer Choices
A. 1
B. √2
C. 2
D. 2√2
E. 3
Key concepts on Triangles are explained in detail in the following posts:
1. Mastering Important Concepts Tested By GMAT in Triangle - I
2. Mastering Important Concepts Tested By GMAT in Triangle - II
Detailed solution will be posted soon.
Thanks,
Saquib
Quant Expert
e-GMAT
Updated on: 07 Aug 2018, 21:18
Originally posted by EgmatQuantExpert on 29 Nov 2016, 08:41.
Last edited by EgmatQuantExpert on 07 Aug 2018, 21:18, edited 3 times in total.
Last edited by EgmatQuantExpert on 07 Aug 2018, 21:18, edited 3 times in total.
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Let's look at the detailed solution to the above question.
Since it is given in the question that AB = BC, we can infer that –
Also notice that,
\(AB^2 + BC^2 = (2√2)^2 + (2√2)^2\)
\(= 16\)
\(= 4^2\)
\(AB^2 + BC^2 = AC^2\)
Therefore, ABC is a right angled triangle. (Since it satisfies Pythagorus Theorem)
As we have mentioned in the article, in an isosceles triangle the median is also the perpendicular to side on which it is drawn.
Thus, another conclusion that we can draw is that BD is perpendicular to AC, since BD is the median as per the information given in the question.
To find the length of BD, we can use the formula of area of the triangle –
or we can write the area of triangle \(ABC = ½ * AC * BD\)……..(ii)
(where BD is the perpendicular and AC is the base)
Equating equation (i) and (ii) we get –
Hence the length of BD = 2 units …(option C)
Since it is given in the question that AB = BC, we can infer that –
- • ABC is an isosceles triangle.
Also notice that,
\(AB^2 + BC^2 = (2√2)^2 + (2√2)^2\)
\(= 16\)
\(= 4^2\)
\(AB^2 + BC^2 = AC^2\)
Therefore, ABC is a right angled triangle. (Since it satisfies Pythagorus Theorem)
As we have mentioned in the article, in an isosceles triangle the median is also the perpendicular to side on which it is drawn.
Thus, another conclusion that we can draw is that BD is perpendicular to AC, since BD is the median as per the information given in the question.
To find the length of BD, we can use the formula of area of the triangle –
- • Area of triangle \(ABC =½ * base * height\)
\(=½ *AB *BC\)……….. (i)
or we can write the area of triangle \(ABC = ½ * AC * BD\)……..(ii)
(where BD is the perpendicular and AC is the base)
Equating equation (i) and (ii) we get –
- • \(½ * AB * BC = ½ * AC * BD\)
• \(BD = (AB * BC)/AC\)
• \(BD = (2√2 * 2√2) / 4\)
• \(BD = 2\)
Hence the length of BD = 2 units …(option C)
29 Nov 2016, 10:05
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By Apollonius theorem,
\(AB^2+BC^2=2(AD^2+BD^2)\)
\(4X2+4X2=2(2^2+BD^2)\)
\(16=2(4+BD^2)\)
BD=2
Answer is C.
If u find my post useful, please press kudos!
\(AB^2+BC^2=2(AD^2+BD^2)\)
\(4X2+4X2=2(2^2+BD^2)\)
\(16=2(4+BD^2)\)
BD=2
Answer is C.
If u find my post useful, please press kudos!
