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In trapezium ABCD,AB||CD and AC=BD.Given area and ht,find BD
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23 Sep 2019, 20:57
23 Sep 2019, 20:57
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In trapezium ABCD, AB || CD and AC=BD. If area of the trapezium is 60 sq units and height of the trapezium if 10 units, what is the length of diagonal BD?
A) 10 units
B) \(\sqrt{116}\)
C) 5\(\sqrt{5}\)
D) \(\sqrt{136}\)
E) 12 units
Source : Jamboree
A) 10 units
B) \(\sqrt{116}\)
C) 5\(\sqrt{5}\)
D) \(\sqrt{136}\)
E) 12 units
Source : Jamboree
Re: In trapezium ABCD,AB||CD and AC=BD.Given area and ht,find BD
[#permalink]
23 Sep 2019, 22:07
23 Sep 2019, 22:07
3
fdundo wrote:
In trapezium ABCD, AB || CD and AC=BD. If area of the trapezium is 60 sq units and height of the trapezium if 10 units, what is the length of diagonal BD?
A) 10 units
B) \(\sqrt{116}\)
C) 5\(\sqrt{5}\)
D) \(\sqrt{136}\)
E) 12 units
Source : Jamboree
A) 10 units
B) \(\sqrt{116}\)
C) 5\(\sqrt{5}\)
D) \(\sqrt{136}\)
E) 12 units
Source : Jamboree
Here,
Key Point:: When ever the diagonals are equal \(-\) then it is isosceles trapezoid
Meaning as per diagram below \(CN = PD\)
Now,
Given Area of the trapezoid = \(60\) and height \(BP = AN =10\)
or \(60\) = \((\frac{{AB + CD}}{2}) * Height\)
or \(AB + CD = 12\)
or \(AB + NP + x + x = 12\) (Let us take CN = PD = x , since it is a isosceles trapezoid)
or \(AB + NP + 2x = 12\)
or \(NP + NP +2x =12 ( AB =NP)\)
or \(NP + x = 6\)
i.e \(NP + x = ND = 6\)
Now in the \(\triangle\) NAD
\(BD = 6, AN = 10\)
Therefore \(AD\) = \(\sqrt{10^2 + 6^2} = \sqrt136\)
Attachments
Trapezoid.jpg [ 18.99 KiB | Viewed 1618 times ]
gmatclubot
Re: In trapezium ABCD,AB||CD and AC=BD.Given area and ht,find BD [#permalink]
23 Sep 2019, 22:07
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