Introduction
?In-text Question — Observe the following figures.(i)(ii)(iii)(iv)(v)Figs. (i), (ii), and (iii) are quadrilaterals, and the others are not. Why?
A quadrilateral is a closed figure bounded by four straight line segments (its four sides).
Figs. (i), (ii) and (iii) are each bounded by exactly four straight sides, so they are quadrilaterals.
Fig. (iv) has a curved side.
A curved side is not a line segment, so Fig. (iv) is not a quadrilateral.
Fig. (v) also has curved sides, so it is not a quadrilateral either.
Rectangles and Squares
AActivity — Verify this property by constructing some rectangles and measuring their diagonals and the points of intersection.Rectangle ABCD with its diagonals AC and BD meeting at O.
Step 1: Draw AB = 6 cm.
Step 2: At A and B draw perpendiculars (use a set-square) and mark AD = BC = 4 cm.
Step 3: Join DC to complete the rectangle ABCD.
Step 4: Draw the diagonals AC and BD and mark their point of intersection O.
Measure: \(AC\approx7.2\) cm and \(BD\approx7.2\) cm.
Measure: \(AO\approx CO\approx3.6\) cm and \(BO\approx DO\approx3.6\) cm.
Check by Pythagoras: \(AC=\sqrt{6^2+4^2}=\sqrt{52}\approx7.2\) cm.
Rectangle 2: 8 cm by 3 cmRepeat the same steps.
Measure: \(AC\approx BD\approx8.5\) cm, and \(AO\approx BO\approx CO\approx DO\approx4.3\) cm.
Check: \(\sqrt{8^2+3^2}=\sqrt{73}\approx8.5\) cm.
MTMath Talk — Can the following equalities be used to establish \(\triangle AOD \cong \triangle COB\)? \(AO = CO\) (proved above), \(\angle AOB = \angle COD\) (vertically opposite angles), \(AD = CB\).Rectangle \(ABCD\) with diagonals meeting at \(O\)
In \(\triangle AOD\) and \(\triangle COB\), the sides \(AO, AD\) and \(CO, CB\) are given equal.
But \(\angle AOB\) and \(\angle COD\) are angles of \(\triangle AOB\) and \(\triangle COD\).
They are not angles of \(\triangle AOD\) or \(\triangle COB\).
Also \(\angle AOB\) is not the angle included between \(AO\) and \(AD\).
Two sides and a non-included angle do not prove congruence.
So these three equalities cannot be used.
A correct way: in \(\triangle AOD\) and \(\triangle COB\),
\(AO = CO\) (proved above)
\(OD = OB\) (proved above, diagonals bisect each other)
\(\angle AOD = \angle COB\) (vertically opposite angles)
So \(\triangle AOD \cong \triangle COB\) (SAS).
?In-text Question — Can you similarly show that \(AB\) is parallel to \(DC\) \((AB \parallel DC)\)?Rectangle \(ABCD\); \(AD \parallel BC\) was shown using \(AB\) as the transversal
Take \(AD\) as the transversal to the lines \(AB\) and \(DC\).
\(\angle A + \angle D = 90^\circ + 90^\circ = 180^\circ\)
These are internal angles on the same side of the transversal \(AD\).
When the sum of the internal angles on the same side of a transversal is \(180^\circ\), the lines are parallel.
\(\therefore AB \parallel DC\)
?In-text Question — Is it wrong to write \(\triangle BAD \cong \triangle CDB\)? Why?Quadrilateral \(ABCD\) with all angles \(90^\circ\), and the diagonal \(BD\) joined
The order of the letters in a congruence statement shows which vertices correspond.
\(\triangle BAD \cong \triangle DCB\) means \(B \leftrightarrow D,\ A \leftrightarrow C,\ D \leftrightarrow B\).
This is what we proved.
\(\triangle BAD \cong \triangle CDB\) would mean \(B \leftrightarrow C,\ A \leftrightarrow D,\ D \leftrightarrow B\).
Then \(\angle A = 90^\circ\) would correspond to \(\angle CDB\), which is less than \(90^\circ\).
Also side \(AD\) would correspond to side \(DB\), the diagonal, and these are not equal.
So the corresponding parts do not match.
AActivity — Using this fact, construct a square with a diagonal of length 8 cm.(The fact: the diagonals of a square are of equal length and bisect each other at right angles.)
Steps:
Step 1: Draw a line segment \(PR = 8\) cm.
Step 2: With the compass, draw the perpendicular bisector of \(PR\).
It meets \(PR\) at its midpoint \(O\), so \(PO = OR = 4\) cm.
Step 3: On the perpendicular bisector, mark \(Q\) and \(S\) on opposite sides of \(O\) with \(OQ = OS = 4\) cm.
Step 4: Join \(PQ,\ QR,\ RS\) and \(SP\).
Why this is a square: the diagonals \(PR\) and \(QS\) are equal (8 cm each).
They bisect each other at \(O\) and cross at \(90^\circ\).
Equal diagonals that bisect each other give a rectangle.
Diagonals at right angles make the four triangles at \(O\) congruent, so all sides are equal.
?In-text Question — Verify if this is true by going through geometric reasoning in Deduction 1 and Deduction 2, and see if they apply to a square as well.Square \(ABCD\) with diagonals meeting at \(O\)
Let \(ABCD\) be a square. A square has all angles \(90^\circ\) and all sides equal.
Deduction 1 (equal diagonals): in \(\triangle ADC\) and \(\triangle DAB\):
\(AB = CD\) (sides of a square)
\(\angle BAD = \angle CDA = 90^\circ\)
\(AD\) is common to both triangles
\(\triangle ADC \cong \triangle DAB\) (SAS), so \(AC = BD\)
Deduction 2 (diagonals bisect each other): let the diagonals meet at \(O\).
\(\angle 3 + \angle 1 = 90^\circ\) (\(\angle B = 90^\circ\)) and \(\angle 3 + \angle 2 = 90^\circ\) (in \(\triangle BCD\)), so \(\angle 1 = \angle 2\)
\(\angle AOB = \angle COD\) (vertically opposite angles) and \(AB = CD\)
\(\triangle AOB \cong \triangle COD\) (AAS), so \(OA = OC\) and \(OB = OD\)
These arguments use only the right angles and the equal opposite sides, and a square has both.
?In-text Question — What are the measures of \(\angle 1\), \(\angle 2\), \(\angle 3\), and \(\angle 4\)? See if you can reason and/or experiment to figure this out!
Let \(ABCD\) be the square, with the diagonal \(AC\).
\(\angle 1\) and \(\angle 3\): in \(\triangle ADC\), \(\angle 1 + \angle 3 + 90^\circ = 180^\circ\)
\(AD = DC\) (sides of a square), so \(\angle 1 = \angle 3\) (angles opposite equal sides)
\(2\angle 1 = 90^\circ\), so \(\angle 1 = \angle 3 = 45^\circ\)
\(\angle 2\) and \(\angle 4\): in \(\triangle ABC\), \(\angle 2 + \angle 4 + 90^\circ = 180^\circ\)
\(AB = BC\) (sides of a square), so \(\angle 2 = \angle 4\) (angles opposite equal sides)
\(2\angle 2 = 90^\circ\), so \(\angle 2 = \angle 4 = 45^\circ\)
Figure it Out — Set 1 (Page 94)
Five questions from page 94 of the textbook.
1Find all the other angles inside the following rectangles.(i)(ii)
(i) Let the diagonals \(AC\) and \(BD\) meet at \(O\).
Given \(\angle CAB = 30^\circ\).
\(\angle DAC = 90^\circ - 30^\circ = 60^\circ\) (\(\angle DAB = 90^\circ\))
In \(\triangle ABC\): \(\angle ACB = 180^\circ - 90^\circ - 30^\circ = 60^\circ\)
\(\angle ACD = \angle CAB = 30^\circ\) (alternate angles, \(AB \parallel DC\))
\(OA = OB\) (diagonals of a rectangle are equal and bisect each other)
\(\angle ABD = \angle OAB = 30^\circ\) (angles opposite equal sides)
\(\angle DBC = 90^\circ - 30^\circ = 60^\circ\)
\(\angle BDC = \angle ABD = 30^\circ\) (alternate angles, \(AB \parallel DC\))
\(\angle ADB = \angle DBC = 60^\circ\) (alternate angles, \(AD \parallel BC\))
In \(\triangle AOB\): \(\angle AOB = 180^\circ - 30^\circ - 30^\circ = 120^\circ\)
\(\angle COD = 120^\circ\) (vertically opposite angles)
\(\angle AOD = \angle BOC = 180^\circ - 120^\circ = 60^\circ\) (linear pair)
(ii) Let the diagonals meet at \(O\).
Given \(\angle QOR = 110^\circ\).
\(\angle POS = 110^\circ\) (vertically opposite angles)
\(\angle POQ = \angle ROS = 180^\circ - 110^\circ = 70^\circ\) (linear pair)
\(OQ = OR\) (diagonals are equal and bisect each other)
\(\angle OQR = \angle ORQ = \dfrac{180^\circ - 110^\circ}{2} = 35^\circ\)
\(\angle OPS = \angle OSP = 35^\circ\) (\(\triangle POS\) is isosceles with apex angle \(110^\circ\))
\(\angle OQP = \angle OPQ = \dfrac{180^\circ - 70^\circ}{2} = 55^\circ\)
\(\angle ORS = \angle OSR = 55^\circ\)
Check: \(\angle PQR = 35^\circ + 55^\circ = 90^\circ\)
(ii) \(\angle POS = 110^\circ\); \(\angle POQ = \angle ROS = 70^\circ\); \(\angle OQR = \angle ORQ = \angle OPS = \angle OSP = 35^\circ\); \(\angle OQP = \angle OPQ = \angle ORS = \angle OSR = 55^\circ\)
2Draw 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°
Equal diagonals that bisect each other always give a rectangle.
The angle between them only changes its shape.
Steps (the same for each angle \(\theta\)):
Step 1: Draw \(AB = 8\) cm and mark its midpoint \(O\) (4 cm from \(A\)).
Step 2: At \(O\), draw a line making an angle \(\theta\) with \(OB\), using the protractor.
Step 3: On this line mark \(C\) and \(D\) on opposite sides of \(O\) with \(OC = OD = 4\) cm.
Step 4: Join \(AC,\ CB,\ BD\) and \(DA\).
Then \(ACBD\) is the required quadrilateral.
Why it works: \(AB = CD = 8\) cm and they bisect each other at \(O\).
(i) \(\theta = 30^\circ\): a long thin rectangle.
(ii) \(\theta = 40^\circ\): a rectangle.
(iii) \(\theta = 90^\circ\): the diagonals are also perpendicular, so it is a square.
(iv) \(\theta = 140^\circ\): the other angle between the diagonals is \(180^\circ - 140^\circ = 40^\circ\), so this is the same rectangle as in (ii).
(ii) rectangle
(iii) square
(iv) rectangle (the same shape as in part ii)
3Consider 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.Circle with centre \(O\) and perpendicular diameters \(PL\) and \(AM\)
\(PL\) and \(AM\) are diameters, so each has length \(2r\).
\(PL = AM\) (diagonals of \(APML\) are equal)
\(O\) is the centre, so it is the midpoint of both diameters.
\(OP = OL\) and \(OA = OM\) (diagonals bisect each other)
Equal diagonals that bisect each other make \(APML\) a rectangle.
The diameters are perpendicular, so the diagonals meet at \(90^\circ\).
In \(\triangle AOP\) and \(\triangle MOP\): \(OA = OM = r\), \(\angle AOP = \angle MOP = 90^\circ\), \(OP\) is common.
\(\triangle AOP \cong \triangle MOP\) (SAS), so \(AP = PM\).
A rectangle with two adjacent sides equal has all four sides equal.
4Math Talk — 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?
Let \(AB\) and \(CD\) be the two equal sticks.
Place them one over the other so that their midpoints coincide at \(O\).
The sticks may cross at any angle.
Fix one end of the thread at \(A\), pass it round \(C\), and take it to \(B\).
Pull it tight.
The thread forms \(AC\) and \(CB\).
\(AB = CD\) (the sticks are equal)
\(OA = OB\) and \(OC = OD\) (\(O\) is the midpoint of both sticks)
Equal diagonals that bisect each other form a rectangle, so \(ACBD\) is a rectangle.
\(\therefore \angle ACB = 90^\circ\)
Another way (isosceles triangles): \(OA = OB = OC\), since each is half of a stick.
In \(\triangle OAC\): \(OA = OC\), so \(\angle OAC = \angle OCA = a\).
In \(\triangle OBC\): \(OB = OC\), so \(\angle OBC = \angle OCB = b\).
In \(\triangle ABC\): \(a + b + (a + b) = 180^\circ\), so \(a + b = 90^\circ\).
\(\angle ACB = a + b = 90^\circ\)
5Try This — 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?
No.
A rectangle needs right angles, and parallel and equal opposite sides do not give them.
Counter-example: draw a parallelogram with sides \(4\) cm and \(5\) cm and an angle of \(30^\circ\).
Its opposite sides are parallel and equal.
But its angles are \(30^\circ,\ 150^\circ,\ 30^\circ,\ 150^\circ\).
They are not \(90^\circ\), so it is not a rectangle.
Every rectangle has parallel and equal opposite sides, but the converse is false.
Parallelograms
?In-text Question — Construct such a figure by recalling how parallel lines can be constructed using a ruler and a set-square, or a compass and a ruler.
Step 1: Draw a line AB.
Step 2: Place a set-square with one edge along AB and a ruler against another edge of the set-square.
Step 3: Hold the ruler still and slide the set-square along it. Draw the line CD along the same edge through a point away from AB. So \(CD\parallel AB\).
Step 4: Repeat with a different angle of the set-square to draw AD and BC, so that \(AD\parallel BC\) and the angle at A is not \(90^\circ\).
Step 5: Mark the corners A, B, C and D.
ABCD has both pairs of opposite sides parallel, but its angles are not \(90^\circ\).
?In-text Question — Draw a parallelogram with adjacent sides of lengths 4 cm and 5 cm, and an angle of 30° between them. What are the remaining angles of the parallelogram? What are the lengths of the remaining sides? See if you can reason out and/or experiment to figure these out.Step 1 (left) and Step 2 (right): the parallelogram ABCD with AB = 4 cm, AD = 5 cm and ∠A = 30°.
Step 1: Draw AB = 4 cm. At A, draw AD = 5 cm making \(\angle A=30^\circ\).
Step 2: Through D draw a line parallel to AB.
Step 3: Through B draw a line parallel to AD.
Step 4: Mark the point where the two lines meet as C. ABCD is the parallelogram.
Remaining angles\(AB\parallel DC\) and AD is a transversal.
So \(\angle A+\angle D=180^\circ\).
\(\angle D=180^\circ-30^\circ=150^\circ\)
\(AD\parallel BC\) and AB is a transversal.
So \(\angle A+\angle B=180^\circ\).
\(\angle B=180^\circ-30^\circ=150^\circ\)
\(\angle C+\angle D=180^\circ\), so \(\angle C=180^\circ-150^\circ=30^\circ\).
Remaining sidesMeasuring: \(DC\approx4\) cm and \(BC\approx5\) cm.
So the opposite sides are equal.
\(BC=5\) cm and \(CD=4\) cm
?In-text Question — Is it wrong to write \(\triangle ABD \cong \triangle CBD\)? Why?Parallelogram \(ABCD\) with the diagonal \(BD\)
\(\triangle ABD \cong \triangle CDB\) means \(A \leftrightarrow C,\ B \leftrightarrow D,\ D \leftrightarrow B\).
This is what we proved.
\(\triangle ABD \cong \triangle CBD\) would mean \(A \leftrightarrow C,\ B \leftrightarrow B,\ D \leftrightarrow D\).
Then side \(AB\) would correspond to \(CB\), and side \(AD\) to \(CD\).
But \(AB = CD\) and \(AD = CB\), and in general \(AB \neq CB\).
Also \(\angle ABD\) would correspond to \(\angle CBD\), and these are not equal in general.
?In-text Question — Are the diagonals of a parallelogram always equal? Check with the parallelogram that you have constructed.AC and BD are not equal in length.
Take the parallelogram ABCD with AB = 4 cm, AD = 5 cm and \(\angle A=30^\circ\).
Join AC and BD.
Measure: \(AC\approx8.7\) cm.
Measure: \(BD\approx2.5\) cm.
\(AC\neq BD\)
Check by calculation: \(AC^2=4^2+5^2+2\times4\times5\cos30^\circ\approx75.6\), so \(AC\approx8.7\) cm.
\(BD^2=4^2+5^2-2\times4\times5\cos30^\circ\approx6.4\), so \(BD\approx2.5\) cm.
?In-text Question — Is it wrong to write \(\triangle AOE \cong \triangle SOY\)? Why?Parallelogram \(EASY\) with diagonals \(AY\) and \(ES\) meeting at \(O\)
\(\triangle AOE \cong \triangle YOS\) means \(A \leftrightarrow Y,\ O \leftrightarrow O,\ E \leftrightarrow S\).
This is what we proved.
\(\triangle AOE \cong \triangle SOY\) would mean \(A \leftrightarrow S,\ E \leftrightarrow Y\).
Then \(OA\) would equal \(OS\), and \(OE\) would equal \(OY\).
But \(OA = OY\) and \(OE = OS\), and the diagonals of a parallelogram are not equal in general.
So \(OA \neq OS\) in general.
?In-text Question — Do the diagonals of a parallelogram intersect at a particular angle?
No.
The angle between the diagonals of a parallelogram changes from one parallelogram to another.
In the rectangle (Deduction 3), the diagonals can meet at \(60^\circ\), or at any other angle.
In Figure it Out Set 2, Question 2, the diagonals of a parallelogram meet at \(140^\circ\).
Only in special parallelograms, the rhombus and the square, are the diagonals always at \(90^\circ\).
Rhombus
?In-text Question — It can be seen that \(\triangle GAE \cong \triangle MAE\). (How?)Rhombus \(GAME\) with the diagonal \(AE\) and the angles \(a, b, c, d\)
In \(\triangle GAE\) and \(\triangle MAE\):
\(GA = MA\) (sides of a rhombus)
\(GE = ME\) (sides of a rhombus)
\(AE = AE\) (common side)
\(\therefore \triangle GAE \cong \triangle MAE\) (SSS)
?In-text Question — Are the diagonals of a rhombus equal?
No, not in general.
In a rhombus with a \(50^\circ\) angle, the diagonals cut each other at right angles into four triangles with angles \(25^\circ, 65^\circ\).
The two diagonals are the long one and the short one, so they differ.
Example: in the rhombus constructed in Figure it Out Set 2, Question 3, the diagonals are 5 cm and 4 cm.
They are equal only when the rhombus is a square.
?In-text Question — In the rhombus \(GAME\), we have \(\triangle GEO \cong \triangle MEO\). (Why?)Rhombus \(GAME\) with diagonals \(GM\) and \(AE\) meeting at \(O\)
In \(\triangle GEO\) and \(\triangle MEO\):
\(GE = ME\) (sides of a rhombus)
\(GO = MO\) (the diagonals of a rhombus bisect each other)
\(EO = EO\) (common side)
\(\therefore \triangle GEO \cong \triangle MEO\) (SSS)
?In-text Question — Verify that the arguments in Deduction 6 can be applied to a rhombus as well!Rhombus ABCD with ∠A = 50°.
A rhombus ABCD is a parallelogram, so \(AB\parallel DC\) and \(AD\parallel BC\).
Take \(\angle A=50^\circ\).
\(AB\parallel DC\) and AD is a transversal, so \(\angle A+\angle D=180^\circ\).
\(\angle D=180^\circ-50^\circ=130^\circ\)
\(AD\parallel BC\) and AB is a transversal, so \(\angle A+\angle B=180^\circ\).
\(\angle B=180^\circ-50^\circ=130^\circ\)
\(\angle C+\angle D=180^\circ\), so \(\angle C=180^\circ-130^\circ=50^\circ\).
So the adjacent angles add up to \(180^\circ\) and the opposite angles are equal: \(\angle A=\angle C=50^\circ\) and \(\angle B=\angle D=130^\circ\).
Figure it Out — Set 2 (Page 102)
Three questions from page 102 of the textbook.
1Find the remaining angles in the following quadrilaterals.(i)(ii)(iii)(iv)
(i) \(PEAR\) is a parallelogram with \(\angle P = 40^\circ\).
\(\angle E = 180^\circ - 40^\circ = 140^\circ\) (adjacent angles of a parallelogram add up to \(180^\circ\))
\(\angle A = \angle P = 40^\circ\) (opposite angles are equal)
\(\angle R = \angle E = 140^\circ\) (opposite angles are equal)
(ii) \(PQRS\) is a parallelogram with \(\angle P = 110^\circ\).
\(\angle Q = 180^\circ - 110^\circ = 70^\circ\)
\(\angle R = \angle P = 110^\circ\) (opposite angles are equal)
\(\angle S = \angle Q = 70^\circ\) (opposite angles are equal)
(iii) All four sides are equal, so \(UVWX\) is a rhombus.
Given \(\angle XVU = 30^\circ\).
The diagonal \(XV\) bisects \(\angle V\), so \(\angle XVW = 30^\circ\).
\(\angle UVW = 30^\circ + 30^\circ = 60^\circ\)
\(\angle WXU = \angle UVW = 60^\circ\) (opposite angles are equal)
The diagonal \(XV\) bisects \(\angle X\), so \(\angle UXV = \angle WXV = 30^\circ\).
\(\angle U = 180^\circ - 60^\circ = 120^\circ\) (adjacent angles add up to \(180^\circ\))
\(\angle W = \angle U = 120^\circ\)
(iv) All four sides are equal, so \(OIEA\) is a rhombus.
Given \(\angle OEA = 20^\circ\).
The diagonal \(OE\) bisects \(\angle E\), so \(\angle OEI = 20^\circ\).
\(\angle AEI = 20^\circ + 20^\circ = 40^\circ\), so \(\angle AOI = 40^\circ\) (opposite angles are equal)
The diagonal \(OE\) bisects \(\angle O\), so \(\angle AOE = \angle EOI = 20^\circ\).
\(\angle A = 180^\circ - 40^\circ = 140^\circ\) (adjacent angles add up to \(180^\circ\))
\(\angle I = \angle A = 140^\circ\)
(ii) \(\angle Q = \angle S = 70^\circ\), \(\angle R = 110^\circ\)
(iii) \(\angle XVW = \angle UXV = \angle WXV = 30^\circ\), \(\angle UVW = \angle WXU = 60^\circ\), \(\angle U = \angle W = 120^\circ\)
(iv) \(\angle OEI = \angle AOE = \angle EOI = 20^\circ\), \(\angle AOI = 40^\circ\), \(\angle A = \angle I = 140^\circ\)
2Using the diagonal properties, construct a parallelogram whose diagonals are of lengths 7 cm and 5 cm, and intersect at an angle of 140°.
The diagonals of a parallelogram bisect each other.
So we draw two segments of 7 cm and 5 cm with a common midpoint.
Steps:
Step 1: Draw \(AB = 7\) cm and mark its midpoint \(O\) (3.5 cm from \(A\)).
Step 2: At \(O\), draw a line \(l\) making an angle of \(140^\circ\) with \(OB\).
Step 3: On \(l\), mark \(C\) and \(D\) on opposite sides of \(O\) with \(OC = OD = 2.5\) cm.
Step 4: Join \(AC,\ CB,\ BD\) and \(DA\).
The diagonals bisect each other at \(O\), so \(ACBD\) is a parallelogram.
3Using the diagonal properties, construct a rhombus whose diagonals are of lengths 4 cm and 5 cm.
The diagonals of a rhombus bisect each other at right angles.
Steps:
Step 1: Draw \(AB = 5\) cm and mark its midpoint \(O\) (2.5 cm from \(A\)).
Step 2: At \(O\), draw the perpendicular to \(AB\) (an angle of \(90^\circ\)).
Step 3: On it, mark \(C\) and \(D\) on opposite sides of \(O\) with \(OC = OD = 2\) cm.
Step 4: Join \(AC,\ CB,\ BD\) and \(DA\).
The diagonals bisect each other at right angles, so the four triangles at \(O\) are congruent (SAS).
Hence all four sides are equal.
Each side is \(\sqrt{2.5^2 + 2^2} \approx 3.2\) cm.
Playing with Quadrilaterals
AActivity — Place two rubber bands perpendicular to each other, forming diagonals of equal length. Join the ends. What is the quadrilateral that you get? Justify your answer.Two equal rubber bands, perpendicular to each other, on a geoboard
The two bands cross at their midpoints, as on the geoboard.
So the diagonals are equal and bisect each other, which makes the figure a rectangle.
The diagonals are also perpendicular.
So the four triangles at the centre \(O\) are congruent (SAS), with two equal legs each.
Hence all four sides are equal.
A rectangle with four equal sides is a square.
AActivity — Extend one of the diagonals on both sides by 2 cm. What quadrilateral will you get now? Justify your answer.The vertical band extended by \(2\) cm on both sides
The extended diagonal still has the same midpoint \(O\), since we added 2 cm on both sides.
So the diagonals still bisect each other, and they are still perpendicular.
Diagonals that bisect each other give a parallelogram.
Let the half-diagonals be \(a\) and \(b\).
The four right triangles at \(O\) have legs \(a\) and \(b\), so they are congruent (SAS).
Hence all four sides are equal, each \(\sqrt{a^2 + b^2}\).
The diagonals are now unequal, so it is not a square.
?In-text Question — Take two cardboard cutouts of an equilateral triangle of sidelength 8 cm. Can you join them to get a quadrilateral? What type of a quadrilateral is this? Justify your answer.Two equilateral triangles of side \(8\) cm
Yes.
Join the two triangles along a full side (8 cm).
All sides of each triangle are 8 cm, so the four outer sides of the quadrilateral are 8 cm each.
A quadrilateral with four equal sides is a rhombus.
Its angles are \(60^\circ,\ 60^\circ + 60^\circ = 120^\circ,\ 60^\circ,\ 120^\circ\).
It is not a square, since no angle is \(90^\circ\).
?In-text Question — Take two cardboard cutouts of an isosceles triangle with sidelengths 8 cm, 8 cm, and 6 cm. What are the different ways they can be joined to get a quadrilateral? What quadrilaterals are these? Justify your answers.Two isosceles triangles with sides \(8\) cm, \(8\) cm and \(6\) cm
The triangles can be joined along the 6 cm side or along an 8 cm side.
Along an 8 cm side they can be joined in two ways, flipped over or turned round.
Joined along the 6 cm side: the four outer sides are all 8 cm, so it is a rhombus.
Joined along an 8 cm side, flipped over (mirror image): the sides are 8, 8, 6, 6 with the equal sides adjacent.
So \(AB = AD\) and \(CB = CD\) (two pairs of equal adjacent sides).
It is a kite.
Joined along an 8 cm side, turned round (rotated by \(180^\circ\)): the sides are 8, 6, 8, 6 with the equal sides opposite.
A quadrilateral with equal opposite sides is a parallelogram (see Figure it Out Set 3, Question 9).
?In-text Question — Take two cardboard cutouts of a scalene triangle with sides 6 cm, 9 cm, and 12 cm. What are the different ways they can be joined to get a quadrilateral? Are you able to identify the different quadrilaterals that are obtained by joining the triangles? Justify your answer whenever you identify a quadrilateral.Two scalene triangles with sides \(6\) cm, \(9\) cm and \(12\) cm
The triangles can be joined along any of the three sides (6 cm, 9 cm or 12 cm).
Along each side they can be joined in two ways, flipped over (mirror image) or turned round (rotated by \(180^\circ\)).
This gives 6 quadrilaterals.
Flipped over: the two copies of each of the other two sides are equal and meet at a corner.
So we get two pairs of equal adjacent sides, which is a kite.
Along 12 cm the sides are 6, 6, 9, 9.
Along 9 cm they are 6, 6, 12, 12.
Along 6 cm they are 9, 9, 12, 12.
The kite along 12 cm is the one in the textbook.
The kites along 9 cm and 6 cm have a dent (the angle \(2 \times 104.5^\circ\) at the joined corner is more than \(180^\circ\)), but they still satisfy the definition of a kite.
Turned round: the opposite sides are equal, which is a parallelogram.
Along 12 cm the sides are 6, 9, 6, 9.
Along 9 cm they are 6, 12, 6, 12.
Along 6 cm they are 9, 12, 9, 12.
No rhombus, rectangle or square can be formed, since the triangle has no equal sides and no right angle.
Kite and Trapezium
?In-text Question — Property 1: In the kite, show that the diagonal \(BD\)
(i) bisects \(\angle ABC\) and \(\angle ADC\),
(ii) bisects the diagonal \(AC\), that is, \(AO = OC\), and is perpendicular to it.
Hint: Is \(\triangle AOB \cong \triangle COB\)?Kite \(ABCD\) with \(AB = BC\) and \(CD = DA\); the diagonals meet at \(O\)
(i) bisects \(\angle ABC\) and \(\angle ADC\),
(ii) bisects the diagonal \(AC\), that is, \(AO = OC\), and is perpendicular to it.
Hint: Is \(\triangle AOB \cong \triangle COB\)?
(i) In \(\triangle ABD\) and \(\triangle CBD\):
\(AB = CB\) (given)
\(AD = CD\) (given)
\(BD = BD\) (common side)
\(\therefore \triangle ABD \cong \triangle CBD\) (SSS)
\(\angle ABD = \angle CBD\) and \(\angle ADB = \angle CDB\) (corresponding parts of congruent triangles)
So \(BD\) bisects \(\angle ABC\) and \(\angle ADC\).
(ii) In \(\triangle AOB\) and \(\triangle COB\):
\(AB = CB\) (given)
\(\angle ABO = \angle CBO\) (shown in (i))
\(BO = BO\) (common side)
\(\therefore \triangle AOB \cong \triangle COB\) (SAS)
\(AO = OC\) (corresponding parts of congruent triangles)
\(\angle AOB = \angle COB\) (corresponding parts of congruent triangles)
\(\angle AOB + \angle COB = 180^\circ\) (linear pair)
\(2\angle AOB = 180^\circ\), so \(\angle AOB = \angle COB = 90^\circ\).
(ii) \(AO = OC\) and \(BD \perp AC\).
?In-text Question — Construct a trapezium. Measure the base angles (marked in the figure). Can you find the remaining angles without measuring them?
Construction (one example):
Step 1: Draw \(SR = 6\) cm.
Step 2: At \(S\), draw a ray making an angle of \(70^\circ\) with \(SR\), and mark \(P\) on it with \(SP = 3.5\) cm.
Step 3: Through \(P\), draw a line parallel to \(SR\) (using a ruler and a set-square).
Step 4: At \(R\), draw a ray making an angle of \(60^\circ\) with \(RS\). It meets the parallel line at \(Q\).
Then \(PQ \parallel SR\), so \(PQRS\) is a trapezium. Measuring gives the base angles \(\angle S = 70^\circ\) and \(\angle R = 60^\circ\).
Remaining angles: \(PQ \parallel SR\), so the angles on the same side of a transversal add up to \(180^\circ\).
\(\angle S + \angle P = 180^\circ\), so \(\angle P = 180^\circ - 70^\circ = 110^\circ\)
\(\angle R + \angle Q = 180^\circ\), so \(\angle Q = 180^\circ - 60^\circ = 120^\circ\)
Check: \(70^\circ + 60^\circ + 110^\circ + 120^\circ = 360^\circ\)
?In-text Question — Construct an isosceles trapezium \(UVWX\), with \(UV \parallel XW\). Measure \(\angle U\). Can you find the remaining angles without measuring them?
Construction (one example):
Step 1: Draw \(UV = 6\) cm.
Step 2: At \(U\), draw a ray making \(70^\circ\) with \(UV\), and mark \(X\) on it with \(UX = 3\) cm.
Step 3: Through \(X\), draw a line parallel to \(UV\).
Step 4: Mark \(W\) on this line so that \(VW = UX = 3\) cm, on the same side of \(UV\) as \(X\) and with \(W\) nearer to \(V\) than \(X\) is.
Measuring gives \(\angle U = 70^\circ\).
Remaining angles: in an isosceles trapezium the angles opposite to the equal sides are equal, so \(\angle V = \angle U = 70^\circ\).
\(XW \parallel UV\), so \(\angle U + \angle X = 180^\circ\)
\(\angle X = 180^\circ - 70^\circ = 110^\circ\)
\(\angle W = 180^\circ - \angle V = 110^\circ\)
Check: \(70^\circ + 70^\circ + 110^\circ + 110^\circ = 360^\circ\)
?In-text Question — Now, it can be shown that \(\triangle UXY \cong \triangle VWZ\). (How?)Isosceles trapezium \(UVWX\) with \(UV \parallel XW\) and \(UX = VW\); \(XY\) and \(WZ\) are perpendicular to \(UV\)
\(XWZY\) is a rectangle (shown in the text), so \(XY = WZ\) (opposite sides of a rectangle).
In \(\triangle UXY\) and \(\triangle VWZ\):
\(\angle UYX = \angle VZW = 90^\circ\)
\(UX = VW\) (given, the equal sides)
\(XY = WZ\) (shown above)
\(\therefore \triangle UXY \cong \triangle VWZ\) (RHS: right angle, hypotenuse and one side)
\(\angle U = \angle V\) (corresponding parts of congruent triangles)
The remaining angles are then \(\angle X = \angle W = 180^\circ - \angle U\), since \(UV \parallel XW\).
Figure it Out — Set 3 (Page 107–109)
Eleven questions from pages 107 to 109 of the textbook.
1Find all the sides and the angles of the quadrilateral obtained by joining two equilateral triangles with sides 4 cm.
Join the two triangles \(ABD\) and \(CBD\) along the common side \(BD\).
Sides: \(AB = BC = CD = DA = 4\) cm.
Angles: each angle of an equilateral triangle is \(60^\circ\).
\(\angle A = 60^\circ\)
\(\angle C = 60^\circ\)
\(\angle ABC = 60^\circ + 60^\circ = 120^\circ\)
\(\angle ADC = 60^\circ + 60^\circ = 120^\circ\)
Check: \(60^\circ + 120^\circ + 60^\circ + 120^\circ = 360^\circ\)
2Construct a kite whose diagonals are of lengths 6 cm and 8 cm.
In a kite, one diagonal bisects the other at right angles (Property 1).
Steps:
Step 1: Draw \(PQ = 6\) cm.
Step 2: Draw the perpendicular bisector of \(PQ\).
It meets \(PQ\) at \(T\), where \(PT = TQ = 3\) cm.
Step 3: On the perpendicular bisector, mark \(S\) and \(R\) on opposite sides of \(T\) such that \(RS = 8\) cm (for example \(TS = 5\) cm and \(TR = 3\) cm).
Step 4: Join \(PR,\ RQ,\ QS\) and \(SP\).
\(SP = SQ\) and \(RP = RQ\), since every point on the perpendicular bisector of \(PQ\) is equally far from \(P\) and \(Q\).
So \(PRQS\) has two pairs of equal adjacent sides, which is a kite.
Note: any split of the 8 cm as \(TS + TR\) works.
If \(T\) is the midpoint of \(RS\) (4 cm each), the kite becomes a rhombus.
3Find the remaining angles in the following trapeziums —(i)(ii)
(i) In trapezium \(PQRS\), \(PQ \parallel SR\).
Given \(\angle P = 135^\circ\) and \(\angle Q = 105^\circ\).
\(\angle S + \angle P = 180^\circ\) (internal angles on the same side of the transversal \(PS\))
\(\angle S = 180^\circ - 135^\circ = 45^\circ\)
\(\angle R + \angle Q = 180^\circ\) (internal angles on the same side of the transversal \(QR\))
\(\angle R = 180^\circ - 105^\circ = 75^\circ\)
(ii) In trapezium \(ABCD\), \(AB \parallel DC\) and \(AD = BC\).
Given \(\angle D = 100^\circ\).
\(\angle A + \angle D = 180^\circ\) (internal angles on the same side of the transversal \(AD\))
\(\angle A = 180^\circ - 100^\circ = 80^\circ\)
\(AD = BC\), so it is an isosceles trapezium and the angles opposite the equal sides are equal.
\(\angle B = \angle A = 80^\circ\)
\(\angle C = 180^\circ - 80^\circ = 100^\circ\)
(ii) \(\angle A = \angle B = 80^\circ,\ \angle C = 100^\circ\)
4Draw 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?
Facts used:
Every rectangle is a parallelogram.
Every rhombus is a parallelogram.
Every rhombus is a kite (its four equal sides satisfy \(AB = BC\) and \(CD = DA\)).
A square is both a rectangle and a rhombus.
(i) The overlap of the kite region and the parallelogram region is the rhombus (this includes the square).
A kite that is a parallelogram must have all four sides equal.
(ii) Yes.
A square is a rectangle, and it is also a rhombus and hence a kite.
A rectangle that is not a square has unequal adjacent sides, so it is not a kite.
(iii) No.
A kite such as the one with sides 6, 6, 9, 9 has unequal sides, so it is not a rhombus.
Every rhombus is a kite, but not every kite is a rhombus.
(ii) Yes, a square
(iii) No. Every rhombus is a kite, but a kite need not be a rhombus.
5If PAIR and RODS are two rectangles, find ∠IOD.
Given: \(\angle ORI = 30^\circ\).
In \(\triangle ORI\): \(\angle RIO = 90^\circ\) (angle of rectangle \(PAIR\)).
\(\angle ROI = 180^\circ - 90^\circ - 30^\circ = 60^\circ\)
\(\angle ROD = 90^\circ\) (angle of rectangle \(RODS\))
\(I\) lies inside \(\angle ROD\), so \(\angle IOD = \angle ROD - \angle ROI\).
\(\angle IOD = 90^\circ - 60^\circ = 30^\circ\)
6Construct a square with diagonal 6 cm without using a protractor.
Steps:
Step 1: Draw \(PR = 6\) cm.
Step 2: With the compass, draw the perpendicular bisector of \(PR\): with centres \(P\) and \(R\) and the same radius (more than 3 cm), draw arcs above and below \(PR\).
Join the points where they meet.
Step 3: This line meets \(PR\) at its midpoint \(O\), and is perpendicular to \(PR\).
Step 4: With centre \(O\) and radius 3 cm, cut the perpendicular bisector at \(Q\) and \(S\), so \(OQ = OS = 3\) cm.
Step 5: Join \(PQ,\ QR,\ RS\) and \(SP\).
The diagonals \(PR\) and \(QS\) are equal (6 cm each), bisect each other and are perpendicular.
So \(PQRS\) is a square.
7CASE 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).(a)(b)
Geometric reasoning: let the side of \(CASE\) be \(2a\).
Then \(CU = CV = a\), and similarly for the other corners.
The four corner triangles \(CUV,\ AUX,\ SXW,\ EWV\) are right-angled with both legs \(a\), so they are congruent (SAS).
\(UV = UX = XW = WV\) (corresponding sides), so \(UVWX\) is a rhombus.
In \(\triangle CUV\): \(CU = CV\) and \(\angle C = 90^\circ\), so \(\angle CUV = 45^\circ\).
In the same way \(\angle AUX = 45^\circ\).
\(\angle VUX = 180^\circ - 45^\circ - 45^\circ = 90^\circ\) (straight angle at \(U\))
A rhombus with one angle \(90^\circ\) is a square (Question 8).
So \(UVWX\) is a square.
Construction and measurement: draw a square of side 8 cm and mark the midpoints.
Join them.
Measure: \(UV = UX = XW = WV \approx 5.7\) cm, each angle \(= 90^\circ\), and the diagonals \(UW = VX = 8\) cm.
(Check: \(UV = \sqrt{4^2 + 4^2} = 4\sqrt{2} \approx 5.66\) cm.)
Other squares inside a square: on each side, mark a point at the same distance \(d\) from a corner, going round in the same direction.
Take \(CU = AX = SW = EV = d\) (any \(d\) between 0 and the side \(s\)).
Join \(U, X, W, V\).
The four corner triangles have a right angle and legs \(d\) and \(s - d\), so they are congruent (SAS).
So \(UX = XW = WV = VU\).
\(\angle CUV = \angle AXU\) (corresponding angles), and \(\angle CUV + \angle CVU = 90^\circ\).
So \(\angle VUX = 180^\circ - (\angle CUV + \angle AUX) = 90^\circ\), and the same holds at every corner.
Figure (b) uses \(d = 2\) cm on a square of side 8 cm.
Choosing different values of \(d\) gives different squares.
8If 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.
Geometric reasoning: a quadrilateral with four equal sides is a rhombus.
A rhombus is a parallelogram, so adjacent angles add up to \(180^\circ\) and opposite angles are equal.
Let \(\angle A = 90^\circ\).
\(\angle B = 180^\circ - 90^\circ = 90^\circ\) and \(\angle D = 180^\circ - 90^\circ = 90^\circ\)
\(\angle C = \angle A = 90^\circ\) (opposite angles)
All four angles are \(90^\circ\) and all sides are equal, so it is a square.
Construction and measurement: draw \(AB = 4\) cm and \(AD = 4\) cm with \(\angle A = 90^\circ\).
With the compass, draw arcs of radius 4 cm from \(B\) and from \(D\).
They meet at \(C\).
Join \(BC\) and \(DC\).
Measure: \(\angle B = \angle C = \angle D = 90^\circ\) and \(AC = BD \approx 5.7\) cm.
9What 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.
Let \(ABCD\) have \(AB = CD\) and \(BC = DA\).
Draw the diagonal \(AC\).
In \(\triangle ABC\) and \(\triangle CDA\):
\(AB = CD\) (given)
\(BC = DA\) (given)
\(AC = CA\) (common side)
\(\therefore \triangle ABC \cong \triangle CDA\) (SSS)
\(\angle 1 = \angle 4\), that is, \(\angle BAC = \angle DCA\) (corresponding parts)
These are alternate angles for the lines \(AB\) and \(CD\) with the transversal \(AC\), so \(AB \parallel CD\).
\(\angle 3 = \angle 2\), that is, \(\angle BCA = \angle DAC\) (corresponding parts)
These are alternate angles for the lines \(BC\) and \(AD\) with the transversal \(AC\), so \(BC \parallel AD\).
Both pairs of opposite sides are parallel.
10Try This — 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.Quadrilateral \(ABCD\); the angle at \(D\) is more than \(180^\circ\)
Here the angle at \(D\) is a reflex angle (more than \(180^\circ\)), measured inside the quadrilateral \(ABCD\).
Geometric reasoning: join \(BD\).
It lies inside the quadrilateral.
In \(\triangle ABD\): \(\angle A + \angle ABD + \angle ADB = 180^\circ\)
In \(\triangle CBD\): \(\angle C + \angle CBD + \angle CDB = 180^\circ\)
Adding: \(\angle A + \angle C + (\angle ABD + \angle CBD) + (\angle ADB + \angle CDB) = 360^\circ\)
\(\angle ABD + \angle CBD = \angle B\), and \(\angle ADB + \angle CDB = \angle D\) (the reflex angle).
\(\therefore \angle A + \angle B + \angle C + \angle D = 360^\circ\)
Construction and measurement: draw \(AC = 6\) cm, \(B\) at height 5 cm above its midpoint, and \(D\) at height 1.5 cm above its midpoint.
Measure: \(\angle A \approx 32.5^\circ\), \(\angle C \approx 32.5^\circ\), \(\angle B \approx 61.9^\circ\), reflex \(\angle D \approx 233.1^\circ\).
Sum \(= 32.5^\circ + 61.9^\circ + 32.5^\circ + 233.1^\circ = 360^\circ\)
11State 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.
(i) False. A rectangle with sides 6 cm and 8 cm has equal diagonals (10 cm each) that bisect each other, but it is not a square.
(ii) True. The sum of the angles is \(360^\circ\).
The fourth angle is \(360^\circ - 3 \times 90^\circ = 90^\circ\).
So all four angles are \(90^\circ\), and the quadrilateral is a rectangle.
(iii) True. Let the diagonals \(AC\) and \(BD\) meet at \(O\), with \(OA = OC\) and \(OB = OD\).
\(\triangle AOB \cong \triangle COD\) (SAS, vertically opposite angles), so \(\angle OAB = \angle OCD\).
These are alternate angles, so \(AB \parallel CD\).
In the same way \(\triangle AOD \cong \triangle COB\), so \(AD \parallel BC\).
Both pairs of opposite sides are parallel, so it is a parallelogram.
(iv) False. A kite has perpendicular diagonals but need not be a rhombus (for example, the kite in Question 2 with diagonals 6 cm and 8 cm).
(v) True. Let \(\angle A = \angle C = x\) and \(\angle B = \angle D = y\).
\(2x + 2y = 360^\circ\), so \(x + y = 180^\circ\).
So \(\angle A + \angle B = 180^\circ\) and \(\angle A + \angle D = 180^\circ\).
These are internal angles on the same side of a transversal, so \(AD \parallel BC\) and \(AB \parallel DC\).
Hence it is a parallelogram.
(vi) True. The four equal angles add up to \(360^\circ\), so each is \(360^\circ \div 4 = 90^\circ\).
So it is a rectangle.
(vii) False. An isosceles trapezium has only one pair of parallel sides (the other two sides are equal but not parallel).
For example, take \(ABCD\) with \(AB \parallel DC\), \(AD = BC\) and \(\angle A = \angle B = 80^\circ\).
With the transversal \(AB\), \(\angle A + \angle B = 160^\circ \neq 180^\circ\), so \(AD\) is not parallel to \(BC\).
Only one pair of opposite sides is parallel, so it is not a parallelogram.
(ii) True
(iii) True
(iv) False
(v) True
(vi) True
(vii) False
Puzzle Time — Which Quad?
Which Quad?
Gameplay
- Fold a sheet into half.
- Now, fold it once more into a quarter.
- Make a triangular crease at the corner that is at the middle of the paper.
- Open the sheet. What is the shape formed by the creases?
- How would you fold the quarter paper to get the kinds of creases shown in the following image.
- How would you fold the quarter paper such that a square is formed?
AActivity — Fold a sheet into half. Now, fold it once more into a quarter. Make a triangular crease at the corner that is at the middle of the paper. Open the sheet. What is the shape formed by the creases?
Let the two fold lines meet at the middle point \(O\) of the sheet.
The crease cuts the two folded edges at \(P\) and \(Q\).
When the sheet is opened, the crease is repeated in the four quarters.
It is reflected about both fold lines.
So the creases join \(P, Q, P', Q'\), with \(OP = OP'\) and \(OQ = OQ'\).
The diagonals \(PP'\) and \(QQ'\) lie along the two fold lines, so they are perpendicular.
They bisect each other at \(O\).
Diagonals that bisect each other at right angles give a rhombus.
AActivity — How would you fold the quarter paper to get the kinds of creases shown in the following image?The creases to be obtained: nested diamonds
Each diamond is a rhombus made by one triangular crease at the corner \(O\) (as in the previous question).
To get nested diamonds of the same shape, make several creases one after another, each parallel to the previous one.
Each crease cuts off a smaller or larger corner of the quarter paper.
Parallel creases cut the two folded edges in the same ratio, so each unfolded diamond has the same shape.
When the paper is opened, we get a set of rhombuses, one inside the other, with the same centre \(O\).
AActivity — How would you fold the quarter paper such that a square is formed?
The creases form a square when \(OP = OQ\).
That is, the crease must cut both folded edges at the same distance from the corner \(O\).
Method: from the corner \(O\), measure the same length (say 5 cm) along both folded edges and mark \(P\) and \(Q\).
Crease along \(PQ\), as in the earlier question.
On opening, the diagonals \(PP'\) and \(QQ'\) are equal, since \(OP = OQ\) and each is twice that length.
They bisect each other at right angles at \(O\).
Equal diagonals that bisect each other at right angles give a square.
Continue with Chapter 5
Move on to Number Play next.