Sequence-And-Series Question 647
Question: The perimeter of a triangle whose sides are in A.P. is 21 cm and the product of lengths of the shortest side and the longest side exceeds the length of the other side by 6 cm. The longest side of the triangle is
Options:
A) 1 cm
B) 7 cm
C) 13 cm
D) None
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Let the sides of the triangle be $ a-d,a,a+d, $ then Perimeter $ =(a+d)+a+(a-d)=21 $
$ \therefore ,a=7 $ Again $ (a-d)(a+d)=a+6 $
$ \Rightarrow ,a^{2}-d^{2}=a+6\Rightarrow 49-d^{2}=13 $
$ \therefore d=\pm 6. $ Hence, the sides of the triangle are $ 1cm,7,cm,13,cm. $
BETA
