QUESTION IMAGE
Question
are these shapes similar?
(left parallelogram: vertices a, b, c, d; sides ab=42 mm, bc=42 mm, cd=42 mm, da=42 mm; angles at a, c marked similarly, angles at b, d marked similarly. right parallelogram: vertices g, h, i, f; sides gh=39 mm, hi=39 mm, if=39 mm, fg=39 mm; angles at g, i marked similarly, angles at h, f marked similarly. below left parallelogram: two buttons labeled yes and no)
Step1: Analyze the first shape
The first quadrilateral (ABCD) has all sides equal to 42 mm, so it's a rhombus (or a square - type rhombus with equal sides). The marked angles: angle A and angle C are equal, angle B and angle D are equal (from the red arc markings).
Step2: Analyze the second shape
The second quadrilateral (GFHI) has all sides equal to 39 mm, so it's also a rhombus. The marked angles: angle G and angle I are equal, angle H and angle F are equal (from the red arc markings).
Step3: Check similarity conditions
For two polygons to be similar, their corresponding angles must be equal, and their corresponding sides must be in proportion.
First, check the side ratios: The ratio of the side of the first rhombus to the second is $\frac{42}{39}=\frac{14}{13}$.
Now, check the angles: In a rhombus, opposite angles are equal. Looking at the angle markings: In the first rhombus (ABCD), angle A and angle C are marked with one arc, angle B and angle D with two arcs. In the second rhombus (GFHI), angle G and angle I are marked with one arc, angle H and angle F with two arcs. So the corresponding angles (angle A corresponds to angle G, angle B corresponds to angle H, etc.) have the same marking pattern, meaning they are equal (since in a rhombus, the angle types (acute/obtuse) are consistent for corresponding marked angles).
Since all corresponding angles are equal (because in rhombuses, opposite angles are equal and the marking shows corresponding angles have the same measure relationship) and the sides are in proportion ($\frac{42}{39}$ is constant for all sides), the two rhombuses are similar.
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yes