Question

kdsx is an isosceles trapezoid, as shown, where ( mangle k = (35z - 20)^circ ), ( mangle x = (15z + 8)^circ ), ( mangle s = (10t + 25)^circ ).

1. what is the value of ( z ) and ( t )? round your answer to the nearest hundredth.
2. what is ( mangle s ) and ( mangle d )? round your answer to the nearest tenth.

Explanation:

Question 9:

Step1: Recall properties of isosceles trapezoid

In an isosceles trapezoid, base angles are equal, and consecutive angles between the bases are supplementary. So, $\angle K$ and $\angle X$ are supplementary? Wait, no. Wait, in trapezoid $KDSX$, $KD \parallel XS$ (since it's a trapezoid). So, $\angle K$ and $\angle X$ are same - side interior angles? Wait, no, let's look at the vertices: $K - D - S - X - K$. So, $KD$ and $XS$ are the two bases (parallel sides). Then, $\angle K$ and $\angle X$: wait, actually, in an isosceles trapezoid, base angles are equal. Wait, maybe I made a mistake. Wait, the legs are $KX$ and $DS$. So, the base angles: $\angle X$ and $\angle S$ are base angles (on base $XS$), so they should be equal? Wait, no, in an isosceles trapezoid, base angles are equal. Wait, no, actually, in an isosceles trapezoid, each pair of base angles is equal, and consecutive angles between the bases are supplementary. Wait, let's re - establish:

Since $KD \parallel XS$, then $\angle K + \angle X = 180^{\circ}$ (same - side interior angles are supplementary) and $\angle D+\angle S = 180^{\circ}$. Also, in an isosceles trapezoid, $\angle K=\angle D$ and $\angle X = \angle S$.

Given $m\angle K=(35z - 20)^{\circ}$ and $m\angle X=(15z + 8)^{\circ}$. Since $KD\parallel XS$, $\angle K$ and $\angle X$ are supplementary, so:

Step2: Solve for z

$(35z-20)+(15z + 8)=180$

Combine like terms:

$35z+15z-20 + 8=180$

$50z-12 = 180$

Add 12 to both sides:

$50z=180 + 12=192$

Divide both sides by 50:

$z=\frac{192}{50}=3.84$

Now, since $\angle X=\angle S$ (base angles of isosceles trapezoid), $m\angle X=(15z + 8)^{\circ}$ and $m\angle S=(10t + 25)^{\circ}$. We know $z = 3.84$, so first find $m\angle X$:

$m\angle X=15\times3.84+8=57.6 + 8=65.6^{\circ}$

Since $m\angle X=m\angle S$, then:

Step3: Solve for t

$10t+25=65.6$

Subtract 25 from both sides:

$10t=65.6 - 25=40.6$

Divide both sides by 10:

$t=\frac{40.6}{10}=4.06$

Question 10:

Step1: Find $m\angle S$

We know that $\angle X=\angle S$ and we found $m\angle X = 65.6^{\circ}$, so $m\angle S=65.6^{\circ}$ (wait, but let's use the formula for $m\angle S=(10t + 25)^{\circ}$ with $t = 4.06$:

$m\angle S=10\times4.06+25=40.6+25 = 65.6^{\circ}$

Step2: Find $m\angle D$

Since $\angle K+\angle D = 180^{\circ}$ (because $KD\parallel XS$ and they are same - side interior angles) and $m\angle K=(35z-20)^{\circ}$ with $z = 3.84$:

$m\angle K=35\times3.84-20=134.4 - 20=114.4^{\circ}$

Then $m\angle D=180 - m\angle K=180 - 114.4 = 65.6$? Wait, no, wait. Wait, earlier we thought $\angle K=\angle D$? Wait, no, I made a mistake. Wait, in an isosceles trapezoid, $\angle K=\angle D$ and $\angle X=\angle S$. And $\angle K+\angle X = 180^{\circ}$. So, $m\angle D=m\angle K=35z-20$. Let's recalculate $m\angle K$:

$35\times3.84-20=134.4 - 20 = 114.4^{\circ}$. Then $m\angle D = 114.4^{\circ}$

Answer:

9:
$z = 3.84$, $t = 4.06$