Question

∠eqh≅∠pqf
m∠eqh = 6x + 18, m∠eqf = x² + x, m∠hqf = 2x + 12
find m∠hqp.

Explanation:

Step1: Use angle - addition property

Since $\angle EQH\cong\angle PQF$, and $\angle EQF=\angle EQH+\angle HQF$, we have $x^{2}+x=(6x + 18)+(2x + 12)$.

Step2: Rearrange the equation

Rearrange the equation $x^{2}+x=6x + 18+2x + 12$ to get a quadratic equation: $x^{2}+x-(6x + 2x)-(18 + 12)=0$, which simplifies to $x^{2}+x - 8x-30 = 0$, or $x^{2}-7x - 30=0$.

Step3: Factor the quadratic equation

Factor the quadratic equation $x^{2}-7x - 30 = 0$. We need to find two numbers that multiply to $-30$ and add up to $-7$. The numbers are $-10$ and $3$. So, $(x - 10)(x+3)=0$.

Step4: Solve for x

Set each factor equal to zero: $x - 10=0$ gives $x = 10$; $x+3=0$ gives $x=-3$. But if $x=-3$, then $m\angle EQH=6x + 18=6\times(-3)+18=0$, which is not possible for an angle measure. So, $x = 10$.

Step5: Find $m\angle EQH$

Substitute $x = 10$ into the expression for $m\angle EQH$. $m\angle EQH=6x + 18=6\times10+18=60 + 18=78^{\circ}$.

Step6: Find $m\angle HQF$

Substitute $x = 10$ into the expression for $m\angle HQF$. $m\angle HQF=2x + 12=2\times10+12=20 + 12=32^{\circ}$.

Step7: Find $m\angle EQF$

$m\angle EQF=x^{2}+x=10^{2}+10=100 + 10=110^{\circ}$.

Step8: Since $\angle EQH\cong\angle PQF$

We know that $\angle HQP=\angle EQF$. So, $m\angle HQP = 110^{\circ}$.

Answer:

$110$