In the diagram, $A B C D$ and $D E F G$ are squares with equal side lengths, and $\angle D C E=70^{\circ}$. The value of $y$ is
(A) 120
(D) 110
(B) 160
(C) 130
(E) 140

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-129.jpg?height=276&width=436&top_left_y=198&top_left_x=1316)
Since squares $A B C D$ and $D E F G$ have equal side lengths, then $D C=D E$, ie. $\triangle C D E$ is isosceles. Therefore,

$$
\angle D E C=\angle D C E=70^{\circ} \text { and so }
$$

$$
\angle C D E=180^{\circ}-70^{\circ}-70^{\circ}=40^{\circ}
$$

and

$$
\begin{aligned}
y^{\circ} & =360^{\circ}-\angle A D C-\angle C D E-\angle G D E \\
y^{\circ} & =360^{\circ}-90^{\circ}-40^{\circ}-90^{\circ} \\
y^{\circ} & =140^{\circ} \\
y & =140
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-251.jpg?height=339&width=520&top_left_y=253&top_left_x=1301)

ANSWER: (E)