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solve the problem. a kite is flying on 149 feet of string. how high is …

Question

solve the problem. a kite is flying on 149 feet of string. how high is it above the ground if its height is 89 feet more than the horizontal distance from the person flying it? assume the string is being released at ground level. the kite is (square) feet above the ground.

Explanation:

Step1: Apply Pythagorean theorem

Let the horizontal distance be $x$ and the height of the kite above the person be $x - 89$, and the length of the string be the hypotenuse $c = 149$. According to the Pythagorean theorem $a^{2}+b^{2}=c^{2}$, so $(x - 89)^{2}+x^{2}=149^{2}$.
Expand $(x - 89)^{2}$ using the formula $(a - b)^{2}=a^{2}-2ab + b^{2}$. We get $x^{2}-178x + 89^{2}+x^{2}=149^{2}$.
Combining like - terms: $2x^{2}-178x+7921 = 22201$.
Rearrange to get a quadratic equation: $2x^{2}-178x+7921 - 22201=0$, which simplifies to $2x^{2}-178x - 14280 = 0$. Divide through by 2: $x^{2}-89x - 7140 = 0$.

Step2: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ (here $a = 1$, $b=-89$, $c = - 7140$), we can use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.
First, calculate the discriminant $\Delta=b^{2}-4ac=(-89)^{2}-4\times1\times(-7140)=7921 + 28560=36481$.
Then $x=\frac{89\pm\sqrt{36481}}{2}=\frac{89\pm191}{2}$.
We have two solutions for $x$: $x_1=\frac{89 + 191}{2}=\frac{280}{2}=140$ and $x_2=\frac{89 - 191}{2}=\frac{-102}{2}=-51$. Since distance cannot be negative, we take $x = 140$.

Step3: Find the height of the kite above the ground

The height of the kite above the ground is $x+(x - 89)$. Substitute $x = 140$ into it. The height $h=x+(x - 89)=140+(140 - 89)=191$.

Answer:

191