Quadrilaterals Class 9 Solutions Maths Ganita Manjari Chapter 12

Solving questions with the help of Ganita Manjari Class 9 Solutions and Part 2 Class 9 Maths Chapter 12 Quadrilaterals NCERT Solutions improves confidence.

Ganita Manjari Class 9 Chapter 12 Solutions

Class 9 Ganita Manjari Chapter 12 Solutions

Class 9 Maths Ganita Manjari Chapter 12 Solutions Quadrilaterals

Think and Reflect (NCERT Textbook Page No. 59)

Question 1.
The angles of a quadrilateral add up to 360°. Therefore, if the opposite angles are equal, what can we say about adjacent angles? Is the converse of your answer true? Conclude that Theorem 3 can also be stated as follows. “If each pair of adjacent angles in a quadrilateral ABCD… then ABCD is a parallelogram.” Fill in the blank.
Solution:
Let ABCD be a quadrilateral in which opposite angles are equal.

∴ ∠A = ∠C and ∠B = ∠D.
Also, ∠A + ∠B + ∠C + ∠D = 360°.
∴ ∠A + ∠B + ∠A + ∠B = 360° (∵ ∠A = ∠C; ∠B = ∠D)
∴ 2(∠A + ∠B) = 360° or ∠A + ∠B = 180°
Similarly, ∠B + ∠C = 180°, ∠C + ∠D = 180° and ∠D + ∠A = 180°.
∴ The sum of each pair of adjacent angles is 180°.
Conversely, let each pair of adjacent angles in ABCD sum to 180°.
∴ ∠A + ∠B = 180° and ∠B + ∠C = 180°
⇒ ∠A + ∠B = ∠B + ∠C
⇒ ∠A = ∠C
Similarly, ∠B = ∠D
∴ Opposite angles are equal.
∴ In a quadrilateral, opposite angles are equal if and only if the sum of any pair of adjacent angles is 180°.
Let in a quadrilateral, the opposite angles are equal.
∴ By Theorem 3, the quadrilateral is a parallelogram.
∴ Theorem 3 can also be stated as:
If each pair of adjacent angles in a quadrilateral sum to 180°, then it is a parallelogram.

Think and Reflect (NCERT Textbook Page No. 66)

Question 1.
Give another proof of the Centroid Theorem using a clever construction that we will describe only partly and ask you to complete. Once again, let M be the intersection point of the two medians CP and BQ. We want to show that the intersection point of line AM with side BC is the midpoint of BC. For this, extend AM up to a carefully chosen point S such that BSCM becomes a parallelogram. (Hint: To arrange MC || BS, first note that PM and MC are along the same line. Now where should M be on segment AS? Does your placement of S also prove MB || CS?)

Answer:
Let the chosen point S on the extended AM be such that MS = AM.
∴ M is the midpoint of AS.
In ∆ABS, P is the midpoint of AB and M is the midpoint of AS.
∴ By Midpoint Theorem 1
PM || BS and PM = \(\frac {1}{2}\)BS
∴ MC || BS (∵ M is on PC)
In ∆ACS, Q is the midpoint of AC and M is the midpoint of AS.
∴ By the Midpoint Theorem
MQ || SC and MQ = \(\frac {1}{2}\)SC
∴ BM || SC (∵ M is on BQ)
In quadrilateral BSCM, we have MC || BS and BM || SC, i.e., opposite sides are parallel.
∴ By definition, BSCM is a parallelogram.
∴ The diagonals of BSCM bisect each other.
∴ BX = XC
∴ X is the midpoint of BC.
∴ AX is a median of ∆ABC.
∴ M lies on all medians.
∴ M is the centroid of ∆ABC.
Also, MX = XS
∴ MX = \(\frac {1}{2}\)MS
∴ MX = \(\frac {1}{2}\)AM (∵ MS = AM)
∴ M divides AX in the ratio 2 : 1.
Similarly, by extending medians BQ and CP, we can prove that M divides medians BQ and CP in the ratio 2 : 1.
∴ The Centroid Theorem is proved.

Think and Reflect (NCERT Textbook Page No. 70)

Question 1.
Suppose we have a tiling of the plane. Consider any vertex. As we go around this vertex and consider the angles made by consecutive lines, the total of these angles must be 360°. This suggests an idea. What if we take 4 copies of a quadrilateral and fit them together around a common point so that each angle is used once as we go around?
Solution:
In the figure, we have a tiling of a plane.

Let P be any vertex. We measure angles 1, 2, 3, and 4.
We have ∠1 = 106°, ∠2 = 74°, ∠3 = 106° and ∠4 = 74°.
Here, ∠1 + ∠2 + ∠3 + ∠4 = 106° + 74° + 106° + 74° = 360°.
Similarly, at any other vertex, we can verify that the sum of angles made by consecutive lines is 360°.
We know that the sum of internal angles of any quadrilateral is 360°.
This gives us an idea that any quadrilateral shape can be used for tiling a surface.

Ex 12.1 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 12.1 Solutions

Exercise 12.1 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 12.1 Solutions

Question 1.
Let ABCD be a quadrilateral.
(i) List all sides of ABCD adjacent to side AB. List all sides opposite to AB.
(ii) List all angles of ABCD adjacent to ∠A. List all angles opposite to ∠A.
(iii) Define a pair of opposite sides and a pair of opposite angles without using the names of the vertices.
Solution:
Let the given quadrilateral be as shown in the figure below:

(i) Sides adjacent to the side AB are BC and AD.
The side opposite to the side AB is DC.
(ii) Angles adjacent to ∠A are ∠B and ∠D.
The angle opposite to ∠A is ∠C.
(iii) Two sides of a quadrilateral are called a pair of opposite sides if they have no point in common.
A pair of angles is called a pair of opposite angles if they are not a pair of adjacent angles.

Question 2.
You have used internal angles of quadrilaterals, but they too require an exact definition, just like how we gave one for a quadrilateral. Precisely define the internal angle of a quadrilateral at a given vertex. Your answer should work for a non-convex quadrilateral too.
Solution:
The angle formed at a vertex of a quadrilateral by a pair of adjacent sides and enclosing the opposite vertex is called an internal angle of a quadrilateral.
In all, there are four internal angles.

In the quadrilaterals given in the figure, the internal angle ∠CDA at vertex D is marked separately for a convex quadrilateral and a non-convex quadrilateral.

Question 3.
In a quadrilateral ABCD, suppose AB || DC. Can ABCD be non-convex? What if we instead assume AB = CD? What if we instead assume ∠A = ∠C?
Solution:
Let ABCD be a quadrilateral in which AB || DC.

Since AB || DC, we have ∠A + ∠D = 180°.
∴ Neither ∠A nor ∠D can be greater than 180°.
Similarly, neither ∠B nor ∠C can be greater than 180°.
∴ No angle of ABCD is greater than 180°.
∴ Quadrilateral ABCD cannot be non-convex.
Let ABCD be a quadrilateral in which AB = CD.
In this quadrilateral, the internal angle ∠BAD at vertex A is greater than 180°.

∴ Quadrilateral ABCD is non-convex.
Let ABCD be a quadrilateral in which ∠A = ∠C.
In this quadrilateral, we have ∠A = ∠C and the internal angle ∠B is greater than 180°.

∴ Quadrilateral ABCD is non-convex.

Question 4.
Consider three non-collinear points A, B, C and draw the lines AB, BC, CA. For every possible location of point D in the plane outside these lines, decide if ABCD is self-intersecting, non-convex, or convex. (Hint: the three lines divide the plane into 7 regions.)
Solution:
Let A, B, and C be three non-collinear points.
We extend lines AB, BC, and CA on both sides.
The lines divide the plane of the points into 7 regions 1, 2, …, 7 as shown in the figure.

Since the point D is not in the plane of A, B, and C, we always get a non-planar quadrilateral.
Also, the quadrilateral is always non-self-intersecting, i.e., a simple quadrilateral.
If the point D is above area 2, the quadrilateral ABCD is non-convex, because ∠ABC > 180°,
and if the point D is above area 1 or 3 or 4 or 5 or 6 or 7, the quadrilateral ABCD is convex, because no angle of the quadrilateral would be > 180°.

Question 5.
Can a quadrilateral be both self-intersecting and non-planar?
Solution:
Let ABCD be a non-planar quadrilateral.

Here, vertex D is not in the plane of A, B, and C.
Diagonal AC lies in the plane of A, B, and C.
Diagonal BD is not in the plane of A, B and C.
∴ Diagonals AC and BD cannot intersect.
∴ Quadrilateral ABCD is non-self-intersecting.
∴ There cannot be a quadrilateral which is both self-intersecting and non-planar.

Ex 12.2 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 12.2 Solutions

Exercise 12.2 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 12.2 Solutions

Question 1.
True or false?
(i) A parallelogram with a right angle is a rectangle.
(ii) A rhombus with perpendicular diagonals is a square.
(iii) If the diagonals of a parallelogram are equal, then it is a rectangle.
Solution:
(i) Let ABCD be a parallelogram with ∠A = 90°.

AD || BC and AB is a transversal.
∴ ∠A + ∠B = 180°
or ∠B = 180° – ∠A = 180° – 90° = 90°.
Similarly, ∠C = 90° and ∠D = 90°.
∴ ABCD is a rectangle.
∴ The given statement is true.

(ii) Let ABCD be a rhombus.
The diagonals of a rhombus are always perpendicular to each other. (Property of a rhombus)
Also, the internal angles of a rhombus may not be 90° each.

∴ ABCD may not be a square.
∴ The given statement is false.

(iii) Let ABCD be a parallelogram with equal diagonals.
Let the diagonals intersect at P.
We know that diagonals of a parallelogram bisect each other.

Since AC = BD (Given)
We have AP = BP = CP = DP.
In ∆APD and ∆CPB, we have
AP = CP, ∠APD = ∠CPB, and DP = BP.
∴ By SAS rule, ∆APD ≅ ∆CPB.
∴ AD = BC.
In ∆ABC and ∆BAD, we have
BC = AD, AC = BD, and side AB is common.
∴ By SSS rule, ∆ABC ≅ ∆BAD.
∴ ∠ABC = ∠BAD
Also, BC || AD and AB is a transversal.
∴ ∠ABC + ∠BAD = 180°
∴ ∠ABC + ∠ABC = 180°
∴ 2∠ABC = 180° or ∠ABC = 90°.
∴ ∠ABC = ∠BAD = 90°.
∴ Each angle of ABCD is equal to 90°, because opposite angles of a parallelogram are equal.
∴ ABCD is a rectangle.
The given statement is true.

Question 2.
The diagonal AC of a parallelogram ABCD bisects ∠A. Show that it also bisects ∠C and that ABCD is a rhombus.
Solution:
ABCD is a parallelogram such that AC bisects ∠A.
Let ∠BAC = x°.
∴ ∠DAC = x° (As AC bisects ∠A)

AB || DC and AC is a transversal.
∴ ∠BAC = ∠ACD
∴ ∠ACD = x°
BC || AD and AC is a transversal.
∴ ∠BCA = ∠DAC
∴ ∠BCA = x°
∴ ∠ACD = ∠BCA = x°
∴ Diagonal AC bisects angle C.
Also, ∠BAC = ∠BCA = x°
∴ ∆ABC is isosceles.
∴ AB = BC.
Since opposite sides of a parallelogram are equal, we have
AB = BC = CD = AD.
In a rhombus, all four sides are equal in length.
∴ ABCD is a rhombus.

Question 3.
The following questions examine converses of true properties. Answer them with Yes or No. If your answer is No, what extra condition can you add so that the answer becomes Yes?
(i) If the diagonals of a quadrilateral ABCD bisect its angles, must ABCD be a rhombus?
(ii) If the diagonals of a quadrilateral ABCD bisect each other at right angles, must ABCD be a rhombus?
(iii) If the diagonals of a quadrilateral ABCD are of equal length, must ABCD be a rectangle? (In the ancient Indian study of quadrilaterals, equality of diagonals was considered significant. Quadrilaterals were first classified according to whether their diagonals were equal or not, before considering equality of sides.)
Solution:
(i) Let ABCD be a quadrilateral such that its diagonals bisect its angles.
In ∆ABC and ∆ADC, we have
∠BAC = ∠DAC, ∠BCA = ∠DCA, and side AC is common.
∴ By ASA rule, ∆ABC ≅ ∆ADC.
∴ AB =AD and BC = CD.
Similarly, we can show that AB = BC and AD = CD.
∴ AB = BC = CD = AD

∴ All sides of quadrilateral ABCD are equal.
∴ ABCD is a rhombus.
∴ The answer is Yes.

(ii) Let ABCD be a quadrilateral such that its diagonals bisect each other at right angles.
Let the diagonals intersect at P.
In ∆APB and ∆CPD, we have
AP = CP, ∠APB = ∠CPD = 90°, and PB = PD.
∴ By SAS rule, ∆APB ≅ ∆CPD.
∴ AB = CD

In ∆APB and ∆CPB, we have
AP = CP, ∠APB = ∠CPB = 90°, and side PB is common.
∴ By SAS rule, ∆APB ≅ ∆CPB.
∴ AB = BC
Similarly, we can show that BC = AD.
∴ AB = BC = CD = AD
∴ All sides of quadrilateral ABCD are equal.
∴ ABCD is a rhombus.
∴ The answer is Yes.

(iii) Let ABCD be a quadrilateral such that its diagonals are equal.
In the figure, AC = BD.
Here, the quadrilateral ABCD is not a rectangle because opposite sides are not equal.

∴ The answer is No.
The extra condition to be added is that the diagonals should bisect each other.
Let ABCD be a quadrilateral such that its diagonals are equal and bisect each other.
Let the diagonals intersect at P.

We have PA = PB = PC = PD.
∆APB and ∆CPD are congruent.
∴ AB = DC and ∠PAB = ∠PCD.
Also, ∆APD and ∆CPB are congruent.
∴ AD = BC and ∠PAD = ∠PCB
∆ABD and ∆BAC are congruent.
∴ ∠BAD = ∠ABC.
∆ABD and ∆BAC are congruent.
∴ ∠BAD = ∠ABC
∴ All interior angles are equal.
Since the sum of all interior angles is 360°, each interior angle must be 90°.
∴ ABCD is a rectangle.

Question 4.
Let ABCD be a parallelogram with AB ≠ BC. Show that the pairwise intersection points of the four angle bisectors form the vertices of a rectangle (See Figure). Why did we assume AB ≠ BC?

Solution:
Since ABCD is a parallelogram, we have
∠A + ∠D = 180°.
∴ \(\frac {1}{2}\)∠A + \(\frac {1}{2}\)∠D = 90°
∴ ∠SAD + ∠SDA = 90°
∴ ∠ASD = 180° – (∠SAD + ∠SDA)
= 180° – 90°
= 90°
∴ ∠PSR = 90° (Vertically opposite angles)
Similarly, ∠SRQ = 90°, ∠RQP = 90° and ∠QPS = 90°.
∴ ABCD is a rectangle.
We assume that AB ≠ BC, because in case AB = BC, the points P, Q, R and S would be coincident and we shall not get rectangle PQRS.

Ex 12.3 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 12.3 Solutions

Exercise 12.3 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 12.3 Solutions

Question 1.
(i) If P, Q, R are the midpoints of sides AB, AC, BC respectively of ∆ABC, show that ∆PQR is congruent to ∆QPA and to two other triangles which you should identify.
(ii) Suppose someone erases ∆ABC, leaving only ∆PQR on the paper. Can you reconstruct ∆ABC from ∆PQR?

Solution:
(i) P, Q and R are midpoints of AB, AC and BC respectively.
∴ By the Midpoint Theorem in ∆ABC, we have
PQ || BC and PQ = \(\frac {1}{2}\)BC
∴ PQ = BR = RC
By the Midpoint Theorem in ∆BCA, we have
PR || AC and PR = \(\frac {1}{2}\)AC
∴ PR = CQ = QA
By the Midpoint Theorem in ∆CAB, we have
QR || AB and QR = \(\frac {1}{2}\)AB
∴ QR = AP = PB
∴ The given figure can also be drawn as follows.

In ∆PQR and ∆QPA, we have
QR = PA, PR = QA, and PQ is common.
∴ By SSS rule, ∆PQR and ∆QPA are congruent.
Similarly, ∆PQR is congruent to ∆RBP and ∆CRQ.

(ii) Suppose we have only ∆PQR.
From P, draw a line parallel to QR and take points A and B on it such that PA = QR and PB = QR.
Join AQ and BR. Produce these lines to meet.
Mark this point as C.
ABC is the required triangle.

Question 2.
In ∆ABC, let M and N be midpoints of AB and AC respectively. Let D be any point on BC. Show that MN bisects AD.

Solution:
M and N are midpoints of the sides AB and AC, respectively.
∴ By the Midpoint Theorem in ∆ABC, we have
MN || BC and MN = \(\frac {1}{2}\)BC
Let MN and AD intersect at P.
In ∆ABD, M is the midpoint of AB and MP || BD,
so by the Converse of the Midpoint Theorem, the point P is the midpoint of the side AD.
∴ MN bisects AD.

Question 3.
In a quadrilateral ABCD, suppose AB || DC and AB ≠ CD. Suppose G and H are the midpoints of AC and BD, respectively. Prove that GH || AB. (Why did we assume AB ≠ CD?)

Solution:
Let P be the midpoint of the side AD. Join PH.
In ∆ADB, segment PH joining the midpoints of AD and BD is parallel to AB by the Midpoint Theorem.
In ∆ADC, segment PG joining the midpoints of AD and AC is parallel to DC by the Midpoint Theorem.

∴ PH || AB and PG || DC
∴ PH || AB and PG || AB (∵ AB || DC)
Since from a point on a line there exists a unique line parallel to that line, the lines PH and PG are the same.
Points P, H, G are on the same line.
∴ Line HG is parallel to AB.
In case AB = CD, the quadrilateral ABCD would be a parallelogram and the points H and G would be coincident.

Question 4.
Suppose the midpoints of sides AB, BC, CD and DA of a quadrilateral ABCD are P, Q, R and S respectively.
(i) Show that PR and QS bisect each other.
(ii) Show that if AC = BD, then PR and QS are perpendicular. Is the converse true?
Solution:
(i) Let ABCD be a quadrilateral.
Join PQ, QR, RS and SP.
Join diagonals AC and BD.

By the Midpoint Theorem in ∆ABC, we get PQ || AC.
By the Midpoint Theorem in ∆ADC, we get SR || AC.
∴ PQ || SR
By the Midpoint Theorem in ∆BAD, we get PS || BD.
By the Midpoint Theorem in ∆BCD, we get QR || BD.
∴ PS || QR
∴ In quadrilateral PQRS, we have PQ || SR and PS || QR, i.e., opposite sides are parallel.
∴ By definition, PQRS is a parallelogram.
∴ The diagonals of PQRS bisect each other. (By Theorem 1(c)).

(ii) As a consequence of the Midpoint Theorem, we also have
PQ = \(\frac {1}{2}\)AC, SR = \(\frac {1}{2}\)AC, PS = \(\frac {1}{2}\)BD and QR = \(\frac {1}{2}\)BD.
Let AC = BD.
∴ PQ = SR = PS = QR
∴ All sides of PQRS are equal.
∴ PQRS is a rhombus.
∴ Diagonals PR and QS are perpendicular to each other.
Conversely, let the diagonals PR and QS of PQRS be perpendicular to each other.
∴ We can prove that PQRS is a rhombus.
∴ PQ = QR = RS = SP
∴ \(\frac {1}{2}\)AC = \(\frac {1}{2}\)BD
∴ AC = BD
∴ The converse is also true.

Question 5.
Suppose PQRS is the Varignon parallelogram of ABCD.

(i) Copy only PQRS on another sheet of paper. Show how you will recreate a congruent copy A’B’C’D’ of ABCD from PQRS. This will be relevant when we return to study tilings at the end of this chapter. (Hint: How will you place vertex A’?) How will you place B’, C’, and D’?
(ii) There are multiple ways to construct A’B’C’D’ in (i). Justify why the quadrilateral A’B’C’D’ you constructed is congruent to ABCD. You may need to show why S is collinear with the points A’ and D’ that you constructed and similarly for P, Q and R. (Hint: Use congruence of triangles, for example ∆SDR ≅ ∆SD’R.)
(iii) Show that if PQRS is a square, then AC and BD are perpendicular and equal. Prove the converse.
Solution:
(i) A copy of PQRS is drawn below using a tracing paper.
We choose point A’ such that A’P = AP and A’S = AS.
Draw A’P and produce it to B’ such that PB’ = PA’.
Draw B’Q and produce it to C’ such that QC’ = QB’.
Draw C’R and produce it to D’ such that RD’ = RC’.
Join D’ and A’. Here P, Q and R are midpoints of A’B’, B’C’ and C’D’ respectively.
In ∆A’B’C’, by the Midpoint Theorem, we have
PQ || A’C’ and PQ = \(\frac {1}{2}\)A’C’.
Let S’ be the midpoint of D’A’.

In ∆C’D’A’, by the Midpoint Theorem, we have
S’R || A’C’ and S’R = \(\frac {1}{2}\)A’C’.
∴ PQ || S’R
Also, PQ || SR.
∴ S’ must coincide with S.
∴ Point S is collinear with points A’ and D’.
PQRS is the Varignon parallelogram of the quadrilateral A’B’C’D’.
Also ∆SA’P ≅ ∆SAP, ∆PB’Q ≅ ∆PBQ, ∆QC’R ≅ ∆QCR and ∆RD’S ≅ ∆RDS.
∴ Quadrilateral A’B’C’D’ is a congruent copy of quadrilateral ABCD.

(ii) In general, there are unlimited choices of A such that its distances from P and S are equal.
∴ There are multiple ways to construct A’B’C’D’ such that PQRS is its Varignon parallelogram.

(iii) Let ABCD be a quadrilateral whose Varignon parallelogram PQRS is a square.

In ∆ABC, by the Midpoint Theorem, we have
PQ || AC and PQ = \(\frac {1}{2}\)AC …….(1)
In ∆BCD, by the Midpoint Theorem, we have
QR || BD and QR = \(\frac {1}{2}\)BD ……(2)
Since PQRS is a square, we have
PQ ⊥ QR and PQ = QR.
∴ AC ⊥ BD and AC = BD
∴ Diagonals AC and BD of ABCD are perpendicular and equal.
Conversely, let AC and BD be perpendicular and equal.
∴ AC ⊥ BD and AC = BD.
∴ (1) and (2) imply
PQ ⊥ QR and PQ = QR.
Adjacent sides PQ and QR of the parallelogram.
PQRS are perpendicular and equal.
∴ PQRS is a square.

Ex 12.4 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 12.4 Solutions

Exercise 12.4 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 12.4 Solutions

Question 1.
Justify why the plane cannot be tiled with copies of a regular pentagon (Hint: Read the first 3 sentences of ‘Think and Reflect’ in the section on tiling.) (There are many ways to tile the plane using a suitable irregular pentagon. The most recent method was found in 2015).

Solution:
We know that each exterior angle of a regular pentagon is \(\frac{360^{\circ}}{5}\) = 72°.
∴ Each interior angle of a regular pentagon is 180° – 72° = 108°.

Now 3(108°) = 324° < 360° and 4(108°) = 432° > 360°.
∴ If we try to fit 3 regular pentagons around a point, they will leave the same gap, and if we try to fit 4 regular pentagons around a point, two pentagons would overlap.
∴ We cannot fit any number of pentagons around a point so that there are no gaps and no overlaps.
∴ A plane cannot be tiled with copies of a regular pentagon.

Question 2.
Draw a non-convex 4-gon DART. Show how we can tile the plane with copies of DART. Both methods that we discussed earlier will work. Which do you prefer?
Solution:
Let DART be a non-convex 4-gon.
We shall tile the plane with copies of DART.

We shall proceed with the method that does not use the Varignon parallelogram of the 4-gon DART.
Cut this 4-gon along its sides and also form some copies, say 4, of this 4-gon. (Figure (I))

Paste 4-gon I on a sheet of paper. Place copy II exactly above copy I.
Rotate copy II by 180° around the midpoint of DA of copy I.
The rotated copy II aligns with copy I perfectly along the shared edge DA, as shown in the figure.

Paste copy II at its position.
Similarly, paste copy III and copy IV as shown in Figure.

Similarly, we can add any number of copies of the non-convex 4-gon DART in the figure.

Ganita Manjari Class 9 Maths Chapter 12 End of Chapter Exercise Solutions

Quadrilaterals End of Chapter Exercise Solutions

Question 1.
Using a fact about parallelograms, show how to tile the plane using any given triangle.
(Hint: Can you use the parallelogram tiling in the introduction?)
Solution:
Let ABC be any triangle.
We shall draw a tiling pattern using this triangle (Figure).

Complete the parallelogram by drawing lines CP || AB and BP || AC (Figure)
∆ABC and ∆PCB are congruent.

∴ ∆PCB is a copy of ∆ABC.
Make sufficient copies of ∆ABC using tracing paper.
Arrange these triangles as shown in Figure.

This shows a tiling pattern using the given triangle ABC.

Question 2.
Mark the midpoint of the line drawn on the paper (see Figure), given that the horizontal lines are equally spaced. Justify your answer.

Solution:
Mark points A, B, C on the given line.
From B, draw PQ perpendicular to the given parallel lines.
In ∆ABP and ∆CBQ, we have
∠ABP = ∠CBQ (Vertically opposite angles)
BP = BQ (Each is twice the distance between two parallel lines)

∠APB = ∠CQP (Each = 90°)
∴ By ASA rule, ∆ABP ≅ ∆CBQ.
∴ AB = CB
∴ B is the midpoint of AC.

Question 3.
You know that the sum of angles of a quadrilateral is 360°, even for a non-convex quadrilateral. (Recall the proof) Now consider a self-intersecting quadrilateral ABCD, where AB and CD intersect at point E. Show that ∠A + ∠B + ∠C + ∠D < 360°. Can you construct ABCD such that ∠A + ∠B + ∠C + ∠D = 2°?
Solution:
Let ABCD be a non-convex quadrilateral.
Join BD.
In ∆ABD, we have
∠1 + ∠2 + ∠3 = 180°.

In ∆BCD, we have
∠4 + ∠6 + ∠5 = 180°.
∴ (∠1 + ∠2 + ∠3) + (∠4 + ∠6 + ∠5) = 360°
∴ ∠1 + (∠2 + ∠4) + ∠6 + (∠3 + ∠5) = 360°
∴ ∠A + ∠B + ∠C + ∠D = 360°.
∴ The sum of all angles of the non-convex quadrilateral ABCD is 360°.
Let ABCD be a self-intersecting quadrilateral.

Join BD and AC.
ADBC is a non-self-intersecting quadrilateral, and so
(∠1 + ∠8) + (∠2 + ∠3) + (∠4 + ∠5) + (∠6 + ∠7) = 360°.
∴ (∠1 + ∠2 + ∠5 + ∠6) + (∠3 + ∠4 + ∠7 + ∠8) = 360°.
∴ ∠1 + ∠2 + ∠5 + ∠6 < 360°.
∴ The sum of angles of the self-intersecting quadrilateral ABCD is less than 360°.
Theoretically, it is possible to draw a self- intersecting quadrilateral such that the sum of angles is 2°, but it is practically impossible to draw such a self-intersecting quadrilateral.

Question 4.
In a parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Figure). Show that APCQ is a parallelogram.

Solution:
We have DP = BQ.

In ∆ABQ and ∆CDP, we have
BQ = DP, ∠ABQ = ∠CDP (Alternate angles)
and AB = CD.
∴ By SAS rule, ∆ABQ ≅ ∆CDP.
∴ AQ = CP
Similarly, AP = CQ.
∴ By Theorem 2, APCQ is a parallelogram.

Question 5.
A right triangle-shaped cutout of a paper is folded such that point A touches point B (Figure). Show that the crease line can be used to find the midpoint of not only AB but also that of AC.

Solution:
Let DE be the crease line.
Since point A touches point B, the line DE is the perpendicular bisector of AB.
∴ AD = DB and DE ⊥ AB.

∴ DE || BC, because corresponding angles ∠ADE and ∠ABC are equal.
∴ By the converse of the Midpoint Theorem, E is the midpoint of AC.

Question 6.
You saw how to use the Midpoint Theorem to divide a given triangle into 4 congruent triangles. Can we divide a triangle into 3 congruent triangles? This exercise shows us how to do that and more, provided we are allowed to cut and reassemble.
(i) Draw medians AP, BQ and CR of ∆ABC, meeting in a common point M. Cut along each median to get 6 triangles. Show with justification how to assemble the 6pieces into 3 congruent triangles! (Hint: Align ∆MPB and ∆MPC along equal sides PB and PC, matching P with itself and B with C.) Prove that you get a triangle. What are its side lengths?
(ii) Repeat the procedure with each of the 3 assembled triangles you got in (i). Show that each of the resulting 9 triangles has sides \(\frac {AB}{3}\), \(\frac {BC}{3}\) and \(\frac {AC}{3}\). The procedure and the result above were unnoticed till 2014, when they were discovered by Lee Sallows, an amateur mathematician!
(iii) Naturally, we can next ask about the possibility of cutting a triangle and reassembling it into 2 congruent triangles. You were asked this question in the chapter Measuring Space: Perimeter and Area. Can you solve this now? (Hint: Use median AP in ∆ABC and then a median of ∆APB.)
For any two given figures with straight sides, we can ask whether one figure can be cut (using only straight cuts) into pieces and reassembled to make the other. This notion is called scissors congruence. The figures need not be in the plane. Research on such questions at a higher level is ongoing.
Solution:
(i) Let ABC be any triangle. We will divide this triangle into 3 congruent triangles by cutting it and reassembling its parts.

Draw medians AP, BQ, and CR of ∆ABC, meeting in a common point M.
Cut along each median to get 6 triangle pieces.

We align ∆MPB and ∆MPC along equal sides PB and PC, matching P with itself and B with C.
Similarly, we align triangles 2, 4 and 1, 3.
Thus, we see that  ∆ABC is divided into 3 coloured small triangles as shown in the figure.

In ∆MEF, ∠MPE + ∠MPF = 180°.
∴ ∠MPE + ∠EPY = 180° (∵ ∠EPY = ∠MPF)
∴ MPY is a straight line.
Also, MY = MP + PY
= 2MP (∵ PY = MP)
= 2(\(\frac {1}{2}\)DM)
= DM
= \(\frac {2}{3}\)AP
∴ MY = \(\frac {2}{3}\)(Median DP)
Similarly, MZ = \(\frac {2}{3}\)(Median EQ)
and MX = \(\frac {2}{3}\)(Median FR)
Thus, we see that all sides in ∆MEY, ∆MFZ and ∆MDX are two-thirds of a median of ∆ABC.
∴ By the SSS rule, these three triangles are congruent to each other, and the given ∆ABC is divided into these three congruent triangles.

(ii) We consider ∆DXM. This is one of the 3 assembled triangles in Figure.
Draw medians DG, XH and MI of ∆DXM.

We have DG = \(\frac {1}{2}\)DE = \(\frac {1}{2}\)AB
XH = EP = \(\frac {1}{2}\)EF = \(\frac {1}{2}\)BC
MI = FQ = \(\frac {1}{2}\)FD = \(\frac {1}{2}\)CA
Dividing ∆DXM into 6 parts and assembling these into 3 congruent triangles as we did earlier, the sides of each of the three congruent triangles would be
\(\frac{2}{3}(\text { Median } \mathrm{DG})=\frac{2}{3}\left(\frac{1}{2} \mathrm{AB}\right)=\frac{\mathrm{AB}}{3}\)
\(\frac{2}{3}(\operatorname{Median} \mathrm{XH})=\frac{2}{3}\left(\frac{1}{2} \mathrm{BC}\right)=\frac{\mathrm{BC}}{3}\)
and \(\frac{2}{3}(\text { Median } \mathrm{MI})=\frac{2}{3}\left(\frac{1}{2} \mathrm{CA}\right)=\frac{\mathrm{CA}}{3}\)

(iii) Let ABC be any triangle. We shall divide this triangle into 2 congruent triangles by cutting and reassembling its parts.
Draw the median AP of ∆ABC.

Let PQ be a median of ∆ABP.
Make the figure on a paper using a tracing paper.
Cut this figure along PQ. We get two pieces of ∆ABC.

We align ∆BPQ and quadrilateral AQPC, matching Q with itself and B with A.

We shall show that ∆DEF and ∆FGD are congruent.
∠BFQ = ∠DEQ (By alignment)
∴ Lines FG and ED are parallel.
Also FG = ED (∵ ED = BF)
∴ EFGD is a parallelogram.
DF is a diagonal of EFGD.
∴ ∆DEF and ∆FGD are congruent.
∴ Given ∆ABC is divided into 2 congruent triangles.

Question 7.
A more general midpoint theorem and its converse:
In a quadrilateral ABCD, suppose AB || DC. Recall that such a quadrilateral ABCD is called a trapezium. Let E be the midpoint of AD. A line drawn through E intersects side BC at F.

(i) If EF || AB, then show that F is the midpoint of BC. Conclude that EF = \(\frac{(\mathrm{AB}+\mathrm{CD})}{2}\)
(ii) If F is the midpoint of BC, then show that EF || AB.
There are at least two ways to solve (ii). Do it directly by using the midpoint M of BD and showing that EM and FM are the same line. Or use (i) along with the same idea used in the second proof of the converse of the Midpoint Theorem. Which way do you think is simpler?
Solution:
(i) In the given trapezium ABCD, join BD.
In ∆ABD, E is the midpoint of AD and EG || AB.

∴ By the converse of the Midpoint Theorem,
EG = \(\frac {1}{2}\)AB and G is the midpoint of BD.
In ∆BCD, G is the midpoint of BD and GF || DC.
∴ By the converse of the Midpoint Theorem,
GF = \(\frac {1}{2}\)DC and F is the midpoint of BC.
∴ EG + GF = \(\frac {1}{2}\)AB + \(\frac {1}{2}\)DC
∴ EF = \(\frac{(\mathrm{AB}+\mathrm{CD})}{2}\)

(ii) Here, we are given that F is the midpoint of BC. Join BD.
Let BD meet EF at M.
Let M’ be the midpoint of BD.

In ∆ABD, E and M’ are respectively midpoints of AD and BD.
∴ By the Midpoint Theorem, we have
EM’ || AB and EM’ = \(\frac {1}{2}\)AB.
In ∆BCD, M’ and F are respectively midpoints of BD and BC.
∴ By the Midpoint Theorem, we have
M’F || DC and M’F = \(\frac {1}{2}\)DC.
∴ EM’ || AB, M’F || AB (∵ AB || DC)
From E, there exists only one line which is parallel to AB.
∴ Point M’ is coincident with M.
∴ E, M and F are on the same line.
∴ EF || AB.

Question 8.
The diagonals AC and BD of a parallelogram ABCD intersect at O. A line through O meets AB and CD at points P and Q respectively. Show that O is the midpoint of PQ. (Multiple proofs are possible. Which is the simplest?)

Solution:
ABCD is a parallelogram.
∴ AB || DC and BC || AD.
Diagonals of a parallelogram always bisect each other.
∴ AO = OC and OB = OD.

In ∆APO and ∆CQO, we have
∠PAO = ∠QCO (Alternate angles)
AO = CO
and ∠AOP = ∠COQ (Vertically opposite angles)
∴ By ASA rule, ∆APO ≅ ∆CQO.
∴ OP = OQ
∴ O is the midpoint of PQ.

Question 9.
ABCD is a trapezium with parallel sides AD = 3 cm and BC = 5 cm. E and F are the midpoints of the non-parallel sides. Find the ratio of the areas of the 4-gons AEFD and EBCF.

Solution:
Join AC. Let AC intersect EF at M.
Let M’ be the midpoint of AC.
In ∆ABC, E and M’ are respectively midpoints of AB and AC.

∴ By the Midpoint Theorem, we have EM’ || BC
and EM’ = \(\frac {1}{2}\)BC = \(\frac {1}{2}\)(5) = 2.5 cm.
In ∆ACD, M’ and F are respectively midpoints of AC and CD.
∴ By the Midpoint Theorem, we have M’F || AD
and M’F = \(\frac {1}{2}\)AD = \(\frac {1}{2}\)(3) = 1.5 cm
∴ EM’ || AD and M’F || AD (∵ AD || BC)
∴ Points M’ and M are coincident.
∴ EF = EM + MF
= EM’ + M’F
= 2.5 + 1.5
= 4 cm
∴ M lies on EF and EF is parallel to AD.
Draw AQ ⊥ BC.
In ∆ABQ, by the converse of the Midpoint Theorem,
P is the midpoint of AQ.
Let AP = PQ = h
Area AEFD = \(\frac {1}{2}\)(AD + EF) × AP
= \(\frac {1}{2}\)(3 + 4 )h
= \(\frac {7h}{2}\)
Area EBCF = \(\frac {1}{2}\)(EF + BC ) × PQ
= \(\frac {1}{2}\)(4 + 5 )h
= \(\frac {9h}{2}\)
∴ Required ratio = \(\frac{7 h}{2}: \frac{9 h}{2}\) or 7 : 9.

Question 10.
Consider 4 points A, B, C, D in the plane with no three collinear. Answer the following questions. Some answers may require you to consider different cases, depending on how the points are positioned in the plane.
(i) How many different quadrilaterals do they form if the quadrilateral is allowed to be self-intersecting or non-convex?
(ii) How many of these quadrilaterals are self-intersecting? How many are convex?
Solution:
We know that using four given distinct points A, B, C, and D, we can form 3 distinct quadrilaterals: ABCD, ABDC, and ACBD.
(i) Case I: Points A, B, C and D are in convex configuration

Here, ABCD is a convex quadrilateral.
ABDC is a self-intersecting quadrilateral.
ACBD is a self-intersecting quadrilateral.
∴ Total number of self-intersecting or non-convex quadrilaterals is 2.
Case II: Points A, B, C and D are in a non-convex configuration

Here, ABCD is a non-convex quadrilateral.
Similarly, ABDC and ACBD are non-convex quadrilaterals.
∴ Total number of self-intersecting or non-convex quadrilaterals is 3.

(ii) Case I: Points A, B, C and D are in convex configuration
In this case, quadrilaterals ABDC and ACBD are self-intersecting quadrilaterals, and ABCD is a convex quadrilateral.

∴ We have 2 self-intersecting quadrilaterals and 1 convex quadrilateral.
Case II: Points A, B, C and D are in a non-convex configuration.
In this case, all quadrilaterals ABCD, ABDC and ACBD are non-convex.

∴ There is no self-intersecting or convex quadrilateral.

Question 11.
Suppose P is a point on side AB of ΔABC, and the line through P parallel to BC meets AC in point Q. For parts (i) to (iii), assume AP = 1, AQ = √5, and see Figure.

(i) If PB = 2, find QC. (Hint: See the dotted line)
(ii) If PB = \(\frac {1}{3}\), find QC. (Hint: See the dotted lines)
(iii) If PB = \(\frac {2}{3}\), find QC. (Hint: Divide both PB and AP suitably.)
(iv) Show that if \(\frac {AP}{PB}\) is a rational number, then \(\frac{A P}{P B}=\frac{A Q}{Q C}\). In Grade 10, you will prove this equality without assuming \(\frac {AP}{PB}\) is rational. That will require a new idea.
Solution:
(i) We consider Figure (I).
Let R be the midpoint of BR.
∴ P is the midpoint of AR.
By the converse of the Midpoint Theorem in ∆ARS, we have AQ = QS.
∴ QS = AQ = √5
Join BQ.
By the converse of the Midpoint Theorem in ∆PBQ,
we have that T is the midpoint of BQ.

By the converse of the Midpoint Theorem in ∆BQC,
we have that S is the midpoint of QC.
∴ SC = QS = √5
∴ QC = QS + SC = √5 + √5 = 2√5.

(ii) We consider Figure (II).
We have AS = \(\frac {2}{3}\) and SB = \(\frac{1}{3}+\frac{1}{3}=\frac{2}{3}\)
∴ S is the midpoint of AB.
∴ By the converse of the Midpoint Theorem in ∆ABC, we have AU = UC.
Join BU. The midpoint of BS is P.

∴ By the converse of the Midpoint Theorem in ∆BUS, we have BV = VU.
By the converse of the Midpoint Theorem in ∆BCU, we have UQ = QC.
Let QC = x
∴ UQ = x and UC = x + x = 2x
∴ AU = UC = 2x
∴ AQ = AU + UQ = 2x + x = 3x
∴ √5 = 3x or x = \(\frac{\sqrt{5}}{3}\)
∴ QC = x = \(\frac{\sqrt{5}}{3}\)

(iii) In ∆ABC, we have
AP = 1, AQ = √5 and PB = \(\frac {2}{3}\)
∆APQ and ∆ABC are similar.

(iv) In part (iii), we have

Question 12.
(i) Suppose ABCD is a parallelogram and M and N are midpoints of AB and CD respectively. Show that segments DM and BN trisect segment AC.

(ii) Use part (i) to find a procedure to trisect any given segment PQ. Find two other ways to trisect PQ, one using Exercise 11 above and a third using the Centroid Theorem.
Solution:
(i) ABCD is a parallelogram.
∴ AB = CD
⇒ \(\frac{A B}{2}=\frac{C D}{2}\)
⇒ MB = DN
Also, MB || DN

∴ MBND is a parallelogram.
∴ BN || MD
In ∆ABQ, M is the midpoint of AB and MP || BQ,
so by converse of Midpoint Theorem,
P is the midpoint of AQ, i.e., AP = PQ.
In ∆CDP, N is the midpoint of CD and QN || PD,
so by the converse of the Midpoint Theorem,
Q is the midpoint of CP, i.e., CQ = PQ.
∴ AP = PQ = CQ
∴ DM and BN trisect segment AC.

(ii) Let PQ be any given segment.
At P, draw an angle ∠APQ.
At Q, draw ∠BQP such that
∠BQP = ∠APQ and AP = BQ.

∴ AP || BQ, because ∠APQ = ∠BQP.
∴ APBQ is a parallelogram, because AP || BQ and AP = BQ.
Let X and Y be the midpoints of AP and BQ, respectively.
Draw BX and AY. Let these lines intersect PQ at points M and N.
By part (i), segments BX and AY trisect segment PQ.
∴ PM = MN = NQ.

Question 13.
Is the Midpoint Theorem for Quadrilaterals (Theorem 9) true when the quadrilateral is non-convex? How about when it is self-intersecting? Experiment and check. The pictures may look strange, but the theorem appears to be true.
(i) Can you explain why this is so? See the geometric reasoning in the text.
(ii) There is one exception: In a very special case, the Varignon parallelogram becomes a single segment. When will this happen?
(iii) Does your reasoning apply even when ABCD is non-planar?
Solution:
(i) Let ABCD be a quadrilateral.
Consider the cases where ABCD is a non-convex and where ABCD is a self-intersecting.

Let P, Q, R, and S be the midpoints of the sides AB, BC, CD, and DA, respectively.
Join PQ, QR, RS and SP.
Also, join diagonals AC and BD.
By the Midpoint Theorem in ∆ABC, we get PQ || AC.
By the Midpoint Theorem in ∆ADC, we get SR || AC.
∴ PQ || SR
By the Midpoint Theorem in ∆BAD, we get PS || BD.
By the Midpoint Theorem in ∆BCD, we get QR || BD.
∴ PS || QR
∴ In quadrilateral PQRS, we have PQ || SR and PS || QR,
i.e., opposite sides are parallel.
∴ PQRS is a parallelogram in both types of quadrilaterals.

(ii) Consider the self-intersecting quadrilateral ABCD formed in a manner that ABDC is a rectangle.
In this case, the quadrilateral PQRS reduces to a single segment PR.

(iii) Let ABCD be a non-planar quadrilateral.
Here, vertex D is not in the plane of A, B, and C.
Let P, Q, R, S be the midpoints of AB, BC, CD, DA respectively.
Join PQ, QR, RS, and PS.

PQRS is the Varignon parallelogram of the non-planar quadrilateral ABCD.
It should be noted here that PQRS is a planar parallelogram.

Question 14.
Let P, Q, R, S be four points on sides AB, BC, CD and DA respectively of a quadrilateral ABCD. Suppose PQRS is a parallelogram. Must P, Q, R and S be midpoints of the respective sides? This can be considered a possible converse question to Theorem 9.
Solution:
If PQRS is a parallelogram, then P, Q, R and S may not be the midpoints of the corresponding sides AB, BC, CD and DA of the quadrilateral ABCD.
Let us see one counterexample.
Let ABCD be a quadrilateral, which is also a parallelogram.
Let P, Q, R and S be respectively the points which divide AB, BC, CD, and DA in the ratio 1 : 2.

Since ABCD is a parallelogram, we have
AB = CD and BC = DA.
∴ AP = BQ = CR = DS and PB = QC = RD = SA.
∴ ∆APS ≅ ∆CRQ and ∆PBQ ≅ ∆RDS.
∴ PS = RQ and PQ = RS
∴ PQRS is a parallelogram.
Here P, Q, R and S are not midpoints of AB, BC, CD and DA respectively.
∴ The converse of the Midpoint Theorem for quadrilaterals is not true.

Question 15.
(i) Complete the following proof of the Midpoint Theorem. Extend segment PQ beyond Q until point S. By how much should we extend PQ?

It would be good to be able to prove that ∆APQ ≅ ∆CSQ. (Why) Use this as a guide to specify the location of S and then complete this proof.
(ii) Give a similar proof of the converse of the Midpoint Theorem (Theorem 7). Start by extending PQ up to suitable point S.
Solution:
(i) We extend segment PQ beyond Q until point S so that PQ = QS.
Join C and S.

Consider ∆APQ and ∆CSQ.
AQ = CQ, ∠AQP = ∠CQS (Vertically opposite angles)
and PQ = QS.
∴ By SAS rule, ∆APQ ≅ ∆CSQ.
∴ AP = CS and ∠APQ = ∠CSQ.
∴ BP = CS (∵ AP = BP)
and AB || CS (∵ ∠APQ = ∠CSQ)
BCSP is a parallelogram.
∴ PS || BC and PS = BC
∴ PQ || BC and PQ = \(\frac {1}{2}\)PS = \(\frac {1}{2}\)BC
∴ Midpoint Theorem is proved.

(ii) Let P be the midpoint of AB and PQ parallel to AC.
Draw a line parallel to AB through C.
Let this line intersect PQ produced to S.

In BCSP, we have BC || PS and BP || CS
∴ BCSP is a parallelogram.
∴ PS = BC and BP = CS
∴ AP = CS (∵ AP = BP)
In ∆APQ and ∆CSQ, we have
∠APQ = ∠CSQ (Alternate angles)
AP = CS and ∠PAQ = ∠SCQ (Alternate angles)
∴ By ASA rule, ∆APQ ≅ ∆CSQ.
AQ = CQ and PQ = QS
∴ PQ = \(\frac {1}{2}\)PS = \(\frac {1}{2}\)BC
∴ Q is the midpoint of AC and PQ = \(\frac {1}{2}\)BC.
∴ The converse of Midpoint Theorem is proved.

Question 16.
Review all the properties of a rhombus/rectangle/square that you proved in Grade 8. Formulate a converse of each. Decide if the converse is true. There are many possibilities here!
Solution:
(i) Rhombus
Property 1: If a quadrilateral is a rhombus, then all its sides are equal.

Converse: If all sides of a quadrilateral are equal, then it is a rhombus.
The converse is True.
Property 2: If a quadrilateral is a rhombus, then its diagonals are perpendicular to each other.

Converse: If the diagonals of a quadrilateral are perpendicular to each other, then it is a rhombus.
The converse is not true, because in the figure, diagonals AC and BD are perpendicular to each other, but ABCD is not a rhombus.

(ii) Rectangle
Property 1: If a quadrilateral is a rectangle, then all its angles are 90°.

Converse: If all angles of a quadrilateral are 90° then it is a rectangle.
The converse is True.
Property 2: If a quadrilateral is a rectangle, then its diagonals are equal in length.

Converse: If the diagonals of a quadrilateral are equal, then it is a rectangle.
The converse is not true, because in the figure, diagonals AC and BD are equal, but ABCD is not a rectangle.

(iii) Square:
Property 1: If a quadrilateral is a square, then all its sides are equal, and all its angles are 90°.

Converse: If a quadrilateral has all sides equal and all angles equal to 90°, then it is a square.
The converse is True.
Property 2: If a quadrilateral is a square then its diagonals are equal and perpendicular bisectors of each other.

Converse: If the diagonals of a quadrilateral are equal and perpendicular bisectors of each other, then it is a square.
The converse is True.

Question 17.
Show that the sum of angles of a non-planar quadrilateral is always less than 360°. Can you find a non-planar quadrilateral ABCD for which ∠A + ∠B + ∠C + ∠D = 2°?
(Hint: Think of a diagonal, say AC, as a hinge around which triangles ABC and ADC can rotate.) What happens to each angle of ABCD as you do this rotation?
Solution:
Let ABCD be a quadrilateral, where vertices A, B, and C are in plane P and vertex D is not in plane P.

∴ ABCD is a non-planar quadrilateral.
We have ∠CAB + ∠ABC + ∠BCA = 180°
∴ ∠ABC = 180° – (∠CAB + ∠BCA) …..(1)
Also, ∠CAD + ∠ACD + ∠CDA = 180°
∴ ∠CDA = 180° – (∠CAD + ∠ACD) …..(2)
Non-planar lines AB, AC, AD meet at A, and non-planar lines CB, CA, CD meet at C.
∴ ∠DAB < ∠BAC + ∠CAD and ∠BCD < ∠BCA + ∠ACD
∴ Sum of angles of ABCD = ∠BAD + ∠ABC + ∠BCD + ∠CDA < (∠BAC + ∠CAD) + 180° – (∠CAB + ∠BCA) + (∠BCA + ∠ACD) + 180° – (∠CAD + ∠ACD)
= 180° + 180°
= 360°
∴ Sum of angles of ABCD < 360°.
∴ Sum of angles of a non-planar quadrilateral is always less than 360°.
Yes, we can find a non-planar quadrilateral ABCD for which ∠A + ∠B + ∠C + ∠D = 2°.

Question 18.
Let us see a third method to tile the plane using a 4-gon.
Focus on only two coloured copies of SOME sharing a vertex. What do you see? It appears that each 4-gon is just a shifted copy of the other. Let us see exactly how. Draw two copies of SOME as shown in Figure, so that the diagonals EO and E’O’are collinear with O = E’. Now place a cutout of one copy on top of SOME with diagonal EO drawn on it. Slide this cutout so that segment EO moves along EO’ until EO matches E’O’. Verify that your cutout exactly matches S’O’M’E’.

(i) Prove the exact match of SOME with S’O’M’E’ using four parallelograms.
(ii) Follow the described procedure along each diagonal of the starting coloured 4-gon in both directions to get 4 new 4-gons. Using these 4 copies, repeat the procedure 4 more times to place 4 new copies, resulting in a 3 by 3 grid of 9 copies exactly like the 9 coloured copies in Figure. These 9 copies meet each other only at vertices, and once again the blank spaces among them also form 4-gons congruent to SOME!
Solution:
(i) After sliding EO to E’O’ with O = E’,
Let the new positions of S, E and M be respectively S’, E’ and M’.
∴ EE’ || MM’ and EE’ = MM’
∴ EMM’E’ is a parallelogram.
∴ E’M’ || EM and E’M’ = EM
Similarly, by considering 3 other parallelograms, we have
M’O’ || MO and M’O’ = OM,
O’S’ || OS and O’S’ = OS,
S’E’ || SE and S’E’ = SE.
∴ The sides E’M’, M’O’, O’S’ and S’E’ are respectively equal to the sides EM, MO, OS and SE.
Also, ∠SEM = ∠SEO + ∠OEM
= ∠S’E’O’ + ∠O’E’M’
= ∠S’E’M’
∴ ∠SEM = ∠S’E’M’.
Similarly, other angles of the 4-gon S’O’M’E’ are equal to the corresponding angles of the 4-gon SOME.
∴ The 4-gon S’O’M’E’ is the exact match of the 4-gon SOME.

(ii) ∆OEM is congruent to ∆MM’E’.
∴ A copy of 4-gon SOME can be arranged so that ∆OEM of the copy coincides with ∆MM’E’.
∆OES is congruent to ∆E’S’S.
∴ A copy of 4-gon SOME can be arranged so that ∆OES of the copy coincides with ∆E’S’S.
Similarly, we can arrange any number of copies of the 4-gon SOME to form a plane tiling.

Question 19.
Is there a 4-gon with given side lengths?
(i) Recall the following fact about triangles and check it by construction. For given positive numbers a, b, c, is there a triangle whose sides have these lengths? The answer is yes exactly when the sum of any two numbers is greater than the third. If we arrange the numbers in increasing order (suppose a ≤ b ≤ c), then this amounts to requiring a + b > c.
(Hint: Start by drawing a segment of length c.)
(ii) Suppose a 4-gon has 2, 5, 11 as three side lengths. Can the length of the fourth side be 100? Can it be 10? Can it be 1? What are the possible lengths of the fourth side?
(iii) For given positive numbers a, b, c, d, how will you decide if there is a 4-gon whose sides have these lengths?
Solution:
(i) We know that a triangle with sides a, b, c can be drawn only when the sum of any two sides is greater than the third side
i.e., if a + b > c, b + c > a and a + c > b.
In case a ≤ b ≤ c, then the requirement is reduced only to a + b > c.

(ii) and (iii) Let ABCD be a quadrilateral with sides a, b, c and d.
Let AC be p.
In ∆ABC, a + b > p
In ∆ADC, c + p > d

∴ d < c + p < c + (a + b)
∴ d < a + b + c i.e., a + b + c > d.
Similarly, we can show that: b + c + d > a, c + d + a > b and d + a + b > c.
This condition is also sufficient.
∴ A quadrilateral with sides a, b, c, d can be drawn only when the sum of any three sides is greater than the fourth side.
In case a ≤ b ≤ c ≤ d, then the condition reduces only to a + b + c > d.
Given sides of a 4-gon are 2, 5, 11.
Let the fourth side be 100.
∴ Sides are 2, 5, 11, and 100.
Here, 2 ≤ 5 ≤ 11 ≤ 100 and 2 + 5 + 11 > 100
i.e., 18 > 100 is not true.
∴ 100 cannot be the fourth side.
Let the fourth side be 10.
∴ Sides are 2, 5, 11, and 10.
Here, 2 ≤ 5 ≤ 10 ≤ 11 and 2 + 5 + 10 > 11
i.e., 17 > 11 is true,
∴ 10 can be the fourth side.
Let the fourth side be 1.
∴ Sides are 2, 5, 11, and 1.
Here, 1 ≤ 2 ≤ 5 ≤ 11 and 1 + 2 + 5 > 11
i.e., 8 > 11 is not true.
∴ 1 cannot be the fourth side.
Now we find the possible lengths of the fourth side.
Let the fourth side be x.
∴ The sides are 2, 5, 11, and x.
∴ We must have 2 + 5 + 11 > x, 5 + 11 + x > 2, 11 + x + 2 > 5 and x + 2 + 5 > 11.
∴ x < 18, x > -14, x > -8 and x > 4.
∴ 4 < x and x < 18
∴ The length of the fourth side must be greater than 4 but less than 18.

Question 20.
Counting diagonals of a polygon.
(i) How should we define a diagonal of an n-gon? How many diagonals does an n-gon have? Make a table for small values of n. A 3-gon has no diagonals. A 4-gon has 2. How many diagonals does a 5-gon have? A 6-gon? Try to find a pattern and guess the answers for n = 7 and n = 8. Check your guesses by systematic counting.
(ii) Can you guess a formula for the number of diagonals? How many diagonals get added when we increase the number of sides by 1? Can you now justify why the formula you guessed is true for all n?
Solution:
(i) A diagonal of a n-gon is defined as a line segment connecting two non-adjacent vertices.
Let D(n) be the number of diagonals of a n-gon, where n > 3.
If n = 3, then there are no two non-adjacent vertices and hence no diagonal.
Let P be a vertex of the n-gon. There are n-3 vertices excluding P and the adjacent vertices of P.
∴ The number of lines through P is n – 3.
Since there are n vertices, the total number of lines is n(n – 3).
A diagonal of an n-gon connects two vertices, so each line is counted twice (once for each end).
∴ Total number of diagonals = \(\frac{n(n-3)}{2}\)

No. of sides of n-gon	No. of diagonals
3	0
4	2
5	5
6	9
7	14
8	20

(ii) The number of diagonals in a n-gon is given by \(\frac{n(n-3)}{2}\).
Let the number of sides of a n-gon be increased by 1.
∴ New number of sides = n + 1
∴ Number of diagonals = \(\frac{(n+1)(n+1-3)}{2}\) = \(\frac{(n+1)(n-2)}{2}\)
∴ Increase in diagonals = \(\frac{(n+1)(n-2)}{2}-\frac{n(n-3)}{2}\)
= \(\frac{n^2-n-2}{2}-\frac{n^2-3 n}{2}\)
= \(\frac{n^2-n-2-n^2+3 n}{2}\)
= n – 1

Question 21.
Sum of angles of a polygon. What is the sum of angles of a (planar non-self-intersecting) n-gon? We know that the answer is 180° for n = 3 and 360° for n = 4. Find the next few values. Then find a formula in terms of n and prove it.
Solution:
We consider a planar non-self-intersecting n-gon.
We shall prove that the sum of all angles of the n-gon is (n – 2)180°.
Let P be any vertex of the n-gon.
There are n-3 vertices which are other than P and not adjacent to P.
∴ The number of diagonals drawn from P is n – 3.
These n – 3 diagonals will divide the whole n-gon into n – 2 triangles.
Also, the sum of angles in any triangle is 180°.
∴ Sum of all angle of n-gon = Sum of angles of all n – 2 triangles = (n – 2)180°
∴ Sum of all angles of a n-gon = (n – 2)180°.

Question 22.
Multiple converses to a theorem: Let us see how multiple statements can be considered converses to the Midpoint Theorem and how Theorem 7 is one of them.
To formulate a converse, we should express the original statement in “If … then …” form. For a complex statement, there may be multiple ways to do that. The Midpoint Theorem starts with ∆ABC and points P and Q on sides AB and AC, respectively. The theorem has two assumptions and two conclusions, which we have named for further discussion.

Let us use these four named conditions to discuss various possible statements.
(i) A natural converse of the Midpoint Theorem would be: “If (PRLL) and (HALF) are true, then (P MID) and (Q MID) are true.” Write out this statement fully. Experiment and see that it seems to be true! Prove the statement. (Hint: Try to run a proof of the Midpoint Theorem backwards.)
(ii) To see Theorem 7 as a converse of the Midpoint Theorem, we first write the Midpoint Theorem as follows: “Suppose P is the midpoint of side AB of ∆ABC and Q is a point on side AC. If Q is the midpoint of AC, then PQ || BC.” (Also PQ = \(\frac {BC}{2}\), but let us set that aside for now.) Write the converse of the second sentence in quotes by keeping the first sentence the same. Verify that Theorem 7 is what you get! (And PQ = \(\frac {BC}{2}\), also follows.) Now write Theorem 7 in terms of the four named conditions.
(iii) The discussion so far suggests that it is reasonable to take two of the four listed statements as assumptions and ask if the other two are true. Verify that only one such combination remains to be examined. “If (P MID) and (HALF) are true, then can we conclude (Q MID) and/or (PRLL)?”
Write this out in words. This is a precise version of question (1) stated before Theorem 7. Can you answer it?
In the text, why did we focus only on Theorem 7 among the multiple possibilities? Because Theorem 7 turns out to be particularly useful. It is still important to pursue various avenues while exploring a subject: often it is only after examining multiple possibilities that we can judge what is most valuable.
Solution:
(i) Converse of Midpoint Theorem
Let ABC be a triangle, and P and Q be points on AB and AC, respectively.
If (PRLL) PQ || BC and (HALF) PQ = \(\frac {BC}{2}\), then prove that (P MID) P is the midpoint of AB and (Q MID) Q is the midpoint of AC.
Proof: We have PQ || BC and PQ = \(\frac {1}{2}\)BC.
From C, draw a line l parallel to BA.
Let l intersect PQ produced at R.

In BCRP, opposite sides are parallel.
∴ BCRP is a parallelogram,
∴ PB = RC and PR = BC
∴ PQ = \(\frac {1}{2}\)PR (∵ PR = BC)
∴ PQ = QR
In ∆APQ and ∆CRQ, we have
∠APQ = ∠CRQ (Alternate angles)
PQ = QR
∠AQP = ∠CQR (Vertically opposite angles)
∴ By ASA rule, ∆APQ ≅ ∆CRQ.
∴ AP = CR, AQ = QC
∴ AP = PB (∵ PB = CR)
∴ P is the midpoint of AB, and Q is the midpoint of AC.

(ii) Converse of Midpoint Theorem
Suppose P is the midpoint of side AB of ∆ABC and Q is a point on side AC.
If PQ || BC, then Q is the midpoint of AC and PQ = \(\frac {1}{2}\)BC.
This is the required statement.

(iii) We have proved that P is the midpoint of AB and PQ = \(\frac {1}{2}\)BC.
Let Q’ be the midpoint of AC.

∴ By the Midpoint Theorem,
PQ’ || BC and PQ’ = \(\frac {1}{2}\)BC.
∴ PQ = PQ’
∴ Points Q and Q’ are coincident, because Q and Q’ are on the same side of P.
∴ Q is the midpoint of AC and PQ || BC.

Question 23.
Validity of tiling methods: Show using reasoning that each of the three tiling methods we saw produces a tiling of the plane using the given 4-gon. You have to prove that when we place new copies using any of the three procedures, the entire plane is covered with no gaps and no overlaps. First, think carefully about what it means to prove this. Then find a proof for each procedure.
Solution:
To show that a method of tiling produces a valid tiling of the plane using a given 4-gon, we must ensure that:
(i) Every point in the plane is enclosed by atleast one copy of the 4-gon.
(ii) The interiors of the copies should not intersect, i.e., there should not be any overlapping.
All the methods which we have already discussed satisfy these requirements.

Question 24.
What fraction of the square is shaded?

Solution:
We have AB = BC = CD = DA.
∴ \(\frac {1}{2}\)AB = \(\frac {1}{2}\)CD
∴ EB = DG

Also EB || DG
∴ EBGD is a parallelogram.
∴ DE || GB
Similarly, AF || HC.
In ∆ABQ, E is the midpoint of AB and EP || BQ. (∵ DE || GB)
∴ By the converse of the midpoint theorem,
P is the midpoint of AQ and EP = \(\frac {1}{2}\)BQ.
Similarly,
(i) Q is the midpoint of BR and FQ = \(\frac {1}{2}\)CR.
(ii) R is the midpoint of CS and GR = \(\frac {1}{2}\)DS.
(iii) S is the midpoint of DP and HS = \(\frac {1}{2}\)AP.
By the symmetry properties of a square, PQRS is a square.
Let PQ = x.
∴ PQ = QR = RS = SP = x
Using the above findings:
AP = BQ = CR = DS = x
∴ HS = EP = FQ = GR = \(\frac {x}{2}\)
Area ∆ABQ = \(\frac {1}{2}\)(BQ)(AQ)
= \(\frac {1}{2}\)(x)(2x)
= x²
Similarly, area ∆BCR = x², area ∆CDS = x² and area ∆DAP = x².
∴ Area of 4 triangles = 4(x²) = 4x²
Also, area of square PQRS = x²
∴ Area of square ABCD = 4x² + x² = 5x²
∴ \(\frac{\text { Area of square PQRS }}{\text { Area of square ABCD }}=\frac{x^2}{5 x^2}=\frac{1}{5}\)
∴ The shaded square is one-fifth of the outer square.

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