Quadrilaterals | FIO

Question 4

We have seen how to get 90° 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° using these?

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Solution

We can use the properties of congruent triangles to make a 90-degree angle.

Step 1 — Setting up the sticks

Let us take two sticks. Let their length be LL. We will call them Stick AB and Stick CD. We find the midpoint of each stick. Let M be the midpoint of Stick AB. So, AM = MB = L/2L/2. Let M also be the midpoint of Stick CD. So, CM = MD = L/2L/2. We place the sticks so their midpoints meet. We fix them at this common point M. This allows the sticks to rotate around M.

Diagram 1

Step 2 — Using the thread

We use the thread to measure distances. We measure the distance from point A to point D. Let us call this distance d1d_1. We then measure the distance from point B to point D. Let us call this distance d2d_2. We slowly move the sticks around M. We adjust their angle until d1d_1 and d2d_2 are equal. So, we make sure that AD = BD. Once AD = BD, we tighten the screw at M. This fixes the sticks in their position.

Diagram 2

Step 3 — Proving congruence

Now, let us look at two triangles. Consider triangle AMD (AMD\triangle AMD). Consider triangle BMD (BMD\triangle BMD). We know that AM = BM. This is because M is the midpoint of AB. We also know that AD = BD. We made these distances equal using the thread. The side MD is common to both triangles. So, MD = MD. We have three pairs of equal sides. This is the SSS (Side-Side-Side) congruence condition. Therefore, AMD\triangle AMD and BMD\triangle BMD are congruent.

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

AMDBMD(by SSS rule)\boxed{\triangle AMD \cong \triangle BMD \quad (\text{by SSS rule})}

Step 4 — Finding the angle

Since the triangles are congruent, their corresponding angles are equal. So, angle AMD (AMD\angle AMD) equals angle BMD (BMD\angle BMD). Points A, M, and B lie on a straight line. So, angles AMD\angle AMD and BMD\angle BMD form a linear pair. The sum of angles in a linear pair is 180 degrees. So, AMD+BMD=180\angle AMD + \angle BMD = 180^\circ. Since AMD=BMD\angle AMD = \angle BMD, we can substitute. AMD+AMD=180\angle AMD + \angle AMD = 180^\circ 2×AMD=1802 \times \angle AMD = 180^\circ AMD=1802\angle AMD = \frac{180^\circ}{2}

AMD=90\boxed{\angle AMD = 90^\circ} Therefore, the angle between the sticks is 90 degrees.

Answer

The angle between the sticks is 90 degrees.

More questions in FIO

Q1

Find all the other angles inside the following rectangles.

Q2

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

(i) 30° (ii) 40° (iii) 90° (iv) 140°

Q3

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.

Q4

We have seen how to get 90° 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° using these?

Q5

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?

Q6

Find the remaining angles in the following quadrilaterals.

Q7

Using the diagonal properties, construct a parallelogram whose diagonals are of lengths 7 cm and 5 cm, and intersect at an angle of 140°.

Q8

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

Q9

Find all the sides and the angles of the quadrilateral obtained by joining two equilateral triangles with sides 4 cm.

Q10

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

Q14

Construct a square with diagonal 6 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°, 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°? 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.

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