Capgemini
Company
Numerical Ability
Geometry
2. A triangle has sides of lengths 10, 24 and n, where n is a positive n is a positive integer. The number of values of n for which this triangle has three acute angles is
A. 1
B. 2
C. 3
D. 4
Read Solution (Total 7)
-
- triangle with acute angle:a^2+b^2>c^2
i)let a=10,b=n,c=24
so 10^2+n^2>24^2
n^2>(24^2-10^2)
n^2>576-100
n^2>476
ii)let a=10,b=24,c=n
10^2+24^2>n^2
100+576>n^2
676>n^2
so from i and ii
476>n^2>676 hence root476>n>root676
ans=22,23,24,25
sol:4
- 9 years agoHelpfull: Yes(17) No(0)
- In an acute angled triangle; the sum of the squares of any two sides of a triangle have to be greater than the third side
i.e. 10^2+24^2>x^2 ; x^2+10^2>24^2 ;and x^2+>24^2>10^2 computing the minimum and maximum “x”, we get the range as 26>x>√476. the values of “x” which satisfy the above conditions are 22,23,24 and 25. Hence 4 - 9 years agoHelpfull: Yes(16) No(4)
- a^2+b^2>c^2
10^2+24^2>c^2
c^224^2
b^2>476
so 476 - 9 years agoHelpfull: Yes(4) No(11)
- I think ans:3
n should be 22 or 23 or 25 to satisfy condition
a^2+b^2>c^2 - 9 years agoHelpfull: Yes(3) No(5)
- a^2+b^2>c^2=>=>10^2+24^2>c^2
c^224^2=>=>b^2>476
so 476>>>so value is 4 - 9 years agoHelpfull: Yes(2) No(7)
- triangle with acute angle:a^2+b^2>c^2
i)let a=10,b=n,c=24
so 10^2+n^2>24^2
n^2>(24^2-10^2)
n^2>576-100
n^2>476
ii)let a=10,b=24,c=n
10^2+24^2>n^2
100+576>n^2
676>n^2
so from i and ii
476>n^2>676 hence sqrt(476)>n>sqrt(676)
ans=22,23,24,25
sol:4 - 8 years agoHelpfull: Yes(1) No(0)
- plzzz explain this properly
- 9 years agoHelpfull: Yes(0) No(3)
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