Question

If A is the area of a triangle in $c{m}^2$, whose sides are 9 cm, 10 cm and 11 cm,then which one of the following is correct?

• A. $A<40$$c{m}^2$
• B. $40$$c{m}^2$$<A<45$$c{m}^2$
• C. $45$ $c{m}^2$$<A<50$$c{m}^2$
• D. $A>50$ $c{m}^2$

Answer 1

Answer: (B)

$40$$c{m}^2$$<A<45$$c{m}^2$

$A=\sqrt{s(s-a)(s-b)(s-c)},s=\frac{a+b+c}{2}=\frac{9+10+11}{2}=15㎝$

$\therefore A=\sqrt{15(15-9)(15-10)(15-11)}=\sqrt{15\times 6\times 5\times 4}=\sqrt{300\times 6}$

$=\sqrt{100\times 9\times 2}=30\sqrt{2}=30\times 1.41=42.30c{m}^2$

$\therefore 40c{m}^2<A<45c{m}^2$ is correct answer