Question

in the triangles, qr = de and sr = fe. which statement about the sides must be true?
○ qs = df
○ df < qs
○ qs < de
○ sr = de
images: triangle qsr with vertices s, r, q (sr has two congruence marks, qr has one; angle at r is 62°). triangle dfe with vertices f, e, d (fe has two congruence marks, de has one; angle at e is 50°).

Explanation:

Step1: Analyze triangle sides and angles

In $\triangle QSR$ and $\triangle DFE$, we know $QR = DE$ and $SR = FE$. Let's find the third angle in each triangle. In $\triangle QSR$, $\angle R = 62^\circ$, so $\angle Q = 180^\circ - 62^\circ - \angle S$. But since $SR$ is marked with two ticks (isosceles? Wait, no, $SR$ and $FE$ are equal, $QR$ and $DE$ are equal. Wait, actually, let's use the Law of Sines or compare the angles opposite the sides. In $\triangle QSR$, angle at $R$ is $62^\circ$, in $\triangle DFE$, angle at $E$ is $50^\circ$. The side opposite $\angle R$ is $QS$, and the side opposite $\angle E$ is $DF$. Also, $QR = DE$ and $SR = FE$.

Using the Law of Sines: In $\triangle QSR$, $\frac{QS}{\sin 62^\circ} = \frac{QR}{\sin \angle S}$; in $\triangle DFE$, $\frac{DF}{\sin 50^\circ} = \frac{DE}{\sin \angle F}$. But since $QR = DE$ and $SR = FE$, and let's check the angles. Wait, maybe simpler: in a triangle, the larger angle is opposite the longer side. In $\triangle QSR$, angle at $R$ is $62^\circ$, in $\triangle DFE$, angle at $E$ is $50^\circ$. The sides $QR = DE$ and $SR = FE$. Now, the side $QS$ is opposite $\angle R = 62^\circ$, and $DF$ is opposite $\angle E = 50^\circ$. Wait, no, wait: in $\triangle QSR$, sides: $SR$ (base, equal to $FE$), $QR$ (equal to $DE$), and $QS$ (unknown). In $\triangle DFE$, sides: $FE$ (equal to $SR$), $DE$ (equal to $QR$), and $DF$ (unknown). Now, let's find the angles at $S$ and $F$. In $\triangle QSR$, since $SR$ is marked with two ticks? Wait, no, the marks: $SR$ has two ticks, $FE$ has two ticks, so $SR = FE$. $QR$ has one tick, $DE$ has one tick, so $QR = DE$. So $\triangle QSR$: $SR = FE$, $QR = DE$, angle at $R$ is $62^\circ$, angle at $E$ is $50^\circ$. Now, using the Law of Cosines: For $\triangle QSR$, $QS^2 = QR^2 + SR^2 - 2 \cdot QR \cdot SR \cdot \cos 62^\circ$. For $\triangle DFE$, $DF^2 = DE^2 + FE^2 - 2 \cdot DE \cdot FE \cdot \cos 50^\circ$. Since $QR = DE$ and $SR = FE$, let's denote $QR = DE = a$, $SR = FE = b$. Then $QS^2 = a^2 + b^2 - 2ab \cos 62^\circ$, $DF^2 = a^2 + b^2 - 2ab \cos 50^\circ$. Now, $\cos 62^\circ \approx 0.4695$, $\cos 50^\circ \approx 0.6428$. So $-2ab \cos 62^\circ > -2ab \cos 50^\circ$ (because cosine is decreasing in $[0, 90^\circ]$, so smaller angle has larger cosine, so negative of that is smaller). Wait, no: $\cos 62^\circ < \cos 50^\circ$, so $-2ab \cos 62^\circ > -2ab \cos 50^\circ$ (since multiplying by negative reverses inequality). Therefore, $QS^2 = a^2 + b^2 + (\text{larger term})$, $DF^2 = a^2 + b^2 + (\text{smaller term})$. Wait, no: $QS^2 = a^2 + b^2 - 2ab \cos 62^\circ$, $DF^2 = a^2 + b^2 - 2ab \cos 50^\circ$. Since $\cos 62^\circ < \cos 50^\circ$, then $-2ab \cos 62^\circ > -2ab \cos 50^\circ$ (because if $c < d$, then $-c > -d$ when $c,d$ positive). So $QS^2 > DF^2$? Wait, no, wait: $a^2 + b^2$ is same, then subtract a smaller number (since $2ab \cos 62^\circ < 2ab \cos 50^\circ$) gives a larger result. Wait, $QS^2 = (a^2 + b^2) - (smaller value)$, $DF^2 = (a^2 + b^2) - (larger value)$. So $QS^2 > DF^2$, so $QS > DF$, which means $DF < QS$. Wait, but let's check the angles again. In $\triangle QSR$, angle at $R$ is $62^\circ$, in $\triangle DFE$, angle at $E$ is $50^\circ$. The sides opposite these angles: in $\triangle QSR$, side opposite $\angle R$ is $QS$; in $\triangle DFE$, side opposite $\angle E$ is $DF$. Wait, no, $\angle R$ is at vertex $R$, so the side opposite is $QS$. $\angle E$ is at vertex $E$, side opposite is $DF$. Now, in $\triangle QSR$, the other angles: let's find $\angle Q$. Since $SR$ and $QR$: wait, $SR$ has two ticks, $QR$ has one tick, so maybe $\triangle QSR$ is isoceles? No, $SR$ has two ticks, $QR$ has one, so $SR
eq QR$. Wait, the marks: $SR$ has two equal ticks, $FE$ has two equal ticks, so $SR = FE$. $QR$ has one tick, $DE$ has one tick, so $QR = DE$. So in $\triangle QSR$, sides: $SR$ (two ticks), $QR$ (one tick), $QS$ (unknown). In $\triangle DFE$, sides: $FE$ (two ticks, equal to $SR$), $DE$ (one tick, equal to $QR$), $DF$ (unknown). Now, the angle between $SR$ and $QR$ is $\angle R = 62^\circ$, and the angle between $FE$ and $DE$ is $\angle E = 50^\circ$. So using the Law of Cosines as above, $QS^2 = QR^2 + SR^2 - 2 \cdot QR \cdot SR \cdot \cos 62^\circ$, $DF^2 = DE^2 + FE^2 - 2 \cdot DE \cdot FE \cdot \cos 50^\circ$. Since $QR = DE$ and $SR = FE$, let's call them $x$ and $y$ respectively. Then $QS^2 = x^2 + y^2 - 2xy \cos 62^\circ$, $DF^2 = x^2 + y^2 - 2xy \cos 50^\circ$. Now, $\cos 62^\circ \approx 0.4695$, $\cos 50^\circ \approx 0.6428$. So $2xy \cos 62^\circ \approx 0.939xy$, $2xy \cos 50^\circ \approx 1.2856xy$. Therefore, $QS^2 = x^2 + y^2 - 0.939xy$, $DF^2 = x^2 + y^2 - 1.2856xy$. Subtracting the two equations: $QS^2 - DF^2 = (x^2 + y^2 - 0.939xy) - (x^2 + y^2 - 1.2856xy) = 0.3466xy$. Since $x$ and $y$ are lengths (positive), $QS^2 - DF^2 > 0$, so $QS^2 > DF^2$, so $QS > DF$, which means $DF < QS$. Let's check the other options:

- $QS = DF$: From above, $QS > DF$, so false.
- $DF < QS$: True, as shown.
- $QS < DE$: $DE = QR$, and in $\triangle QSR$, $QS$ is a side, $QR$ is another. Let's see, angle at $R$ is $62^\circ$, angle at $S$: since $SR = FE$, and $FE$ is a side in $\triangle DFE$ with angle $50^\circ$ at $E$. Wait, no, in $\triangle QSR$, angles: sum to $180^\circ$. $\angle R = 62^\circ$, let's say $\angle S = \angle F$? Wait, no, $SR = FE$, but angles at $S$ and $F$: not necessarily equal. Wait, but $QS$ and $DE$: $DE = QR$, so is $QS < QR$? In $\triangle QSR$, side opposite $62^\circ$ (angle at $R$) is $QS$, side opposite angle at $S$ is $QR$. If angle at $S$ is greater than $62^\circ$, then $QR > QS$, but we don't know angle at $S$. Wait, but from the Law of Cosines, we saw $QS^2 = x^2 + y^2 - 2xy \cos 62^\circ$, and $QR = x$, so $x^2 = x^2$, so $QS^2 = x^2 + y^2 - 2xy \cos 62^\circ$. Is $QS < x$? Let's see: $QS^2 - x^2 = y^2 - 2xy \cos 62^\circ = y(y - 2x \cos 62^\circ)$. We don't know $y$ and $x$ relationship, so can't say $QS < DE$ (since $DE = x$). So this is false.
- $SR = DE$: $SR$ is equal to $FE$, $DE$ is equal to $QR$, and $SR$ and $DE$: no, $SR$ and $DE$ are not necessarily equal. Given $QR = DE$ and $SR = FE$, but $QR$ and $SR$ are different (different tick marks), so $DE$ and $SR$ are different. So false.

Step2: Confirm the correct option

From the analysis, $DF < QS$ is the true statement.

Answer:

DF < QS (the option with "DF < QS")