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B
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Difficulty:
65%
(hard)
Question Stats:
54% (01:54) correct
46%
(02:30)
wrong
based on 54
sessions
History
Date
Time
Result
Not Attempted Yet
Find the area of triangle with sides \(\sqrt{13}\), \(\sqrt{61}\) and \(\sqrt{80}\)?
A. 10
B. 14
C. 20
D. 22
E. 26
Try to think out of box!!!!
A. 10
B. 14
C. 20
D. 22
E. 26
Try to think out of box!!!!
14 May 2019, 05:16
Kudos
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Triangle DEF = Rectangle ABCD - Triangle ABF - Triangle ADE - triangle FCD
Triangle DEF= (8*6)- (0.5*2*3)- (0.5*4*8) - (0.5*6*5)
Triangle DEF= 48-3-16-15=14
Solution by Rohit Ner
Triangle DEF= (8*6)- (0.5*2*3)- (0.5*4*8) - (0.5*6*5)
Triangle DEF= 48-3-16-15=14
Solution by Rohit Ner
nick1816
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14 May 2019, 06:51
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in any right triangle with a,b,c sides ==> a^2 + b^2 = c^2
14 + 61 is almost 80 ==> so we can assume that this triangle is almost a right triangle
area of triangle = (1/2) * (14 * 61)^(0.5) = (1/2) * (854)^(0.5) = (almost) (1/2) * (900)^(0.5) = 30/2 = 15
==> B, which is 14, is the closest number to 15
14 + 61 is almost 80 ==> so we can assume that this triangle is almost a right triangle
area of triangle = (1/2) * (14 * 61)^(0.5) = (1/2) * (854)^(0.5) = (almost) (1/2) * (900)^(0.5) = 30/2 = 15
==> B, which is 14, is the closest number to 15
