Quadrilaterals | FIO

Question 4

We have seen how to get $90^\circ$ using paper folding. Now, suppose we do not have any paper but two sticks of equal length, and a thread. How do we make an exact $90^\circ$ using these?

Solution

Understand the Question

Two sticks $\text{AB}$ and $\text{CD}$ of equal length are pivoted together at their midpoints $M$ .
Using a thread, we adjust the angle between the sticks until the distance from one end of a stick to both ends of the other stick is equal ( $\text{AD} = \text{BD}$ ).
By the SSS congruence criterion, $\triangle \text{AMD} \cong \triangle \text{BMD}$ , which makes the adjacent supplementary angles equal ( $\angle \text{AMD} = \angle \text{BMD} = 90^\circ$ ).

Step 1 · Set Up the Sticks at Their Midpoints

Let the two sticks be $\text{AB}$ and $\text{CD}$ of equal length $L$ . Mark their midpoints at $M$ and fix them at $M$ so they can rotate freely: $\text{AM} = \text{MB} = \dfrac{L}{2}, \quad \text{CM} = \text{MD} = \dfrac{L}{2}$ Diagram 1

Step 2 · Adjust with Thread to Make $\text{AD} = \text{BD}$

Using the thread, measure the distance from point $D$ to point $A$ ( $d_1 = \text{AD}$ ) and from point $D$ to point $B$ ( $d_2 = \text{BD}$ ). Adjust the angle between the sticks until these two distances are equal: $\text{AD} = \text{BD}$ Diagram 2

Step 3 · Prove Congruence of Triangles

In $\triangle \text{AMD}$ and $\triangle \text{BMD}$ :

\begin{aligned} \text{AM} &= \text{BM} \quad (M \text{ is midpoint of } \text{AB}) \\ \text{AD} &= \text{BD} \quad (\text{by construction}) \\ \text{MD} &= \text{MD} \quad (\text{common side}) \end{aligned}

Therefore, by the SSS congruence rule: $\triangle \text{AMD} \cong \triangle \text{BMD}$

Step 4 · Calculate the Angle Between the Sticks

Since corresponding parts of congruent triangles are equal: $\angle \text{AMD} = \angle \text{BMD}$

Since $\text{AB}$ is a straight line, $\angle \text{AMD}$ and $\angle \text{BMD}$ form a linear pair:

\begin{aligned} \angle \text{AMD} + \angle \text{BMD} &= 180^\circ \\[0.6em] \angle \text{AMD} + \angle \text{AMD} &= 180^\circ \\[0.6em] 2 \times \angle \text{AMD} &= 180^\circ \\[0.6em] \angle \text{AMD} &= \dfrac{180^\circ}{2} = 90^\circ \end{aligned}

Answer

The angle between the sticks is $90^\circ$ .

Common Mistakes

Not pivoting at the exact midpoint: If the sticks are not joined at their midpoints, $\text{AM} \neq \text{BM}$ , so $\triangle \text{AMD}$ and $\triangle \text{BMD}$ will not be congruent by SSS.
Linear Pair Assumption: Forgetting that $\angle \text{AMD} + \angle \text{BMD} = 180^\circ$ only holds because $A$ , $M$ , and $B$ lie on the same straight stick.

More questions in FIO

Find all the other angles inside the following rectangles.

Draw a quadrilateral whose diagonals have equal lengths of $8\text{ cm}$ that bisect each other, and intersect at an angle of

(i) $30^\circ$

(ii) $40^\circ$

(iii) $90^\circ$

(iv) $140^\circ$

Consider a circle with centre $O$ . Line segments $PL$ and $AM$ are two perpendicular diameters of the circle. What is the figure $APML$ ? Reason and/or experiment to figure this out.

We have seen how to get $90^\circ$ using paper folding. Now, suppose we do not have any paper but two sticks of equal length, and a thread. How do we make an exact $90^\circ$ using these?

We saw that one of the properties of a rectangle is that its opposite sides are parallel. Can this be chosen as a definition of a rectangle? In other words, is every quadrilateral that has opposite sides parallel and equal, a rectangle?

Find the remaining angles in the following quadrilaterals.

Using the diagonal properties, construct a parallelogram whose diagonals are of lengths $7\text{ cm}$ and $5\text{ cm}$ , and intersect at an angle of $140^\circ$ .

Using the diagonal properties, construct a rhombus whose diagonals are of lengths 4 cm and 5 cm.

Find all the sides and the angles of the quadrilateral obtained by joining two equilateral triangles with sides $4\text{ cm}$ .

Q10

Construct a kite whose diagonals are of lengths $6\text{ cm}$ and $8\text{ cm}$ .

Q11

Find the remaining angles in the following trapeziums—

Q12

Draw a Venn diagram showing the set of parallelograms, kites, rhombuses, rectangles, and squares. Then, answer the following questions—

(i) What is the quadrilateral that is both a kite and a parallelogram?

(ii) Can there be a quadrilateral that is both a kite and a rectangle?

(iii) Is every kite a rhombus? If not, what is the correct relationship between these two types of quadrilaterals?

Q13

If PAIR and RODS are two rectangles, find $\angle \text{IOD}$ .

Q14

Construct a square with diagonal $6\text{ cm}$ without using a protractor.

Q15

CASE is a square. The points U, V, W and X are the midpoints of the sides of the square. What type of quadrilateral is UVWX? Find this by using geometric reasoning, as well as by construction and measurement. Find other ways of constructing a square within a square such that the vertices of the inner square lie on the sides of the outer square, as shown in Figure (b).

Q16

If a quadrilateral has four equal sides and one angle of $90^\circ$ , will it be a square? Find the answer using geometric reasoning as well as by construction and measurement.

Q17

What type of a quadrilateral is one in which the opposite sides are equal? Justify your answer.

Hint: Draw a diagonal and check for congruent triangles.

Q18

Will the sum of the angles in a quadrilateral such as the following one also be $360^\circ$ ? Find the answer using geometric reasoning as well as by constructing this figure and measuring.

Q19

State whether the following statements are true or false. Justify your answers.

(i) A quadrilateral whose diagonals are equal and bisect each other must be a square.

(ii) A quadrilateral having three right angles must be a rectangle.

(iii) A quadrilateral whose diagonals bisect each other must be a parallelogram.

(iv) A quadrilateral whose diagonals are perpendicular to each other must be a rhombus.

(v) A quadrilateral in which the opposite angles are equal must be a parallelogram.

(vi) A quadrilateral in which all the angles are equal is a rectangle.

(vii) Isosceles trapeziums are parallelograms.