Take two cardboard cutouts of an equilateral triangle of sidelength . Can you join them to get a quadrilateral? What type of a quadrilateral is this? Justify your answer.

Question

Homi · Accepted Answer

Understand the Question

An equilateral triangle has $3$ equal sides of length $8\text{ cm}$ and $3$ equal interior angles of $60^\circ$ .
When two congruent equilateral triangles are joined along a common edge of $8\text{ cm}$ , they form a closed four-sided polygon (a quadrilateral).
We determine the specific type of quadrilateral by examining its side lengths and interior angles.

(i) Can you join them to get a quadrilateral?

Step 1 · Join the Two Triangles

Take two equilateral triangles of side length $8\text{ cm}$ . Align one side of the first triangle with one side of the second triangle so that they share a common edge. Diagram 1

Diagram 2

The resulting shape is a closed polygon bounded by four outer line segments, which forms a quadrilateral.

Answer

(i) Yes, we can join them to get a quadrilateral.

(ii) What type of a quadrilateral is this? Justify your answer.

Step 1 · Analyze Sides and Angles

Side lengths: All four outer sides of the quadrilateral are sides of the original equilateral triangles. $\text{Side length of each side} = 8\text{ cm}$

Interior angles: Each interior angle of an equilateral triangle is $60^\circ$ .

Two opposite angles at the single vertices are $60^\circ$ each.
The other two angles are formed by combining two $60^\circ$ angles: $60^\circ + 60^\circ = 120^\circ$

The angles of the quadrilateral are $60^\circ$ , $120^\circ$ , $60^\circ$ , and $120^\circ$ .

Conclusion: A quadrilateral with all four sides equal and opposite angles equal (but not $90^\circ$ ) is a rhombus.

Answer

(ii) The quadrilateral is a rhombus because all four sides are equal ( $8\text{ cm}$ ) and opposite angles are equal ( $60^\circ$ and $120^\circ$ ).

Common Mistakes

Calling it a Square: A square has four equal sides, but its angles must each be $90^\circ$ . Here, the angles are $60^\circ$ and $120^\circ$ , so the shape is a rhombus, not a square.
Angle Addition Error: Forgetting to add the two adjacent angles ( $60^\circ + 60^\circ = 120^\circ$ ) where the two triangle bases meet.

Question 5

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