Properties Of Triangles Ques 38
The set of all real numbers a such that $a^{2}+2 a, 2 a+3$ and $a^{2}+3 a+8$ are the sides of a triangle is ……
$(1985,2 M)$
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Answer:
Correct Answer: 38.$(a>5)$
Solution:
Formula:
- Since, $a^{2}+2 a, 2 a+3$ and $a^{2}+3 a+8$ form sides of a triangle.
Now, $a^{2}+3 a+8<\left(a^{2}+2 a\right)+(2 a+3)$
$\Rightarrow \quad a^{2}+3 a+8<a^{2}+4 a+3$
$\Rightarrow \quad a>5 \quad …….(i)$
Also, $\left(a^{2}+3 a+8\right)+(2 a+3)>a^{2}+2 a$
$ \begin{aligned} \Rightarrow & 3 a >-11 \\ \Rightarrow & a >-\frac{11}{3}\quad …….(ii) \end{aligned} $
Again, $\left(a^{2}+3 a+8\right)+\left(a^{2}+2 a\right)>2 a+3$
$ \Rightarrow \quad 2 a^{2}+3 a+5>0 $
which is always true.
$\therefore$ Triangle is formed, if $a>5$
BETA
