Question

The shorter sides of an acute triangle are x cm and 2x cm. The longest side of the triangle is 15 cm. What is the smallest possible whole-number value of x? 6 7 8 9

Answer 1

The longest side in the triangle is opposite to the largest angle of this triangle. If triangle is acute, then all angles are acute. Acute angle has cosine that is positive.

Use cosine theorem to determine the cosine of the largest angle:

$15^2=x^2+(2x)^2-2\cdot x\cdot 2x\cdot \cos \alpha,$ where $\alpha$ is the largest angle.

Then

$225=x^2+4x^2-4x^2\cos \alpha,\\ \\\cos \alpha=\dfrac{225-5x^2}{-4x^2}=\dfrac{5x^2-225}{4x^2}.$

Since $\cos \alpha>0,$ then

$\dfrac{5x^2-225}{4x^2}>0\Rightarrow 5x^2-225>0.$

Divide this inequality by 5:

$x^2-45>0,\\ \\(x-3\sqrt{5})(x+3\sqrt{5})>0,\\ \\x\in (-\infty,-3\sqrt{5})\cup (3\sqrt{5},\infty).$

Note that $3\sqrt{5}\approx 6.71,$ then the smallest possible whole-number value of x is 7.

Answer: correct choice is B

Answer 2

Answer:

The value of x is 7

B is correct

Step-by-step explanation:

Three sides of triangle are x cm, 2x cm and 15 cm.

In a triangle

• Sum of two smallest side is greater than third side.

• Difference of two smallest side is less than third side.

Sum of two side is greater than 3rd side

x + 2x > 15

3x > 15

x > 5

Thus, x must be greater than 5

Difference of two sides is less than 3rd side

2x - x < 15

      x < 15

Thus, x must be less than 15

Using cosine rule,

$\cos A=\dfrac{b^2+c^2-a^2}{2bc}$

Triangle is acute, Cos A>0

where, a=x , b=2x  and c=15

$\dfrac{x^2+(2x)^2-15^2}{2\cdot x\cdot 2x}>0$

$5x^2-225>0$

$x> 3\sqrt5\approx 6.7$

$x< -3\sqrt5\approx -6.7$

x and 2x is shorter sides.

2x<15 or x<7.5

Possible value of x>6.7 and x<7.5

Hence, The value of x is 7