NCERT Solutions Class 9 Maths Ch 12: Quadrilaterals

Yes. Every quadrilateral, convex or non-convex, can tile the plane with congruent copies.

1. The ideaTake a quadrilateral ABCD and a copy of it turned through 180° (a half-turn) about the midpoint of one of its sides. The two fit together along that side. Repeating half-turns about the midpoints of the sides and sliding the pair gives a pattern with no gaps and no overlaps.
2. Why there are no gapsThe four angles of the quadrilateral add up to 360°. At every corner of the pattern all four angles A, B, C and D meet, and together they fill the whole angle of 360° around the point.
3. Which ones cannot?A figure that is not a quadrilateral in the plane (a crossed figure, or one that is not flat) is not an ordinary tile. For ordinary quadrilaterals there is no exception.

Yes, but we need more conditions than just “four non-collinear points”.

1. The definitionTake four distinct points A, B, C, D in one plane, with no three of them collinear. Join AB, BC, CD and DA. The figure is a quadrilateral ABCD if the sides meet only at their shared endpoints: that is, the opposite sides AB and CD do not meet, and the opposite sides BC and DA do not meet.
2. What the conditions excludeThey exclude a crossed figure (where two sides cross), overlapping sides, and a figure that is not flat. They allow both convex and non-convex quadrilaterals.

Adjacent angles are supplementary (add up to 180°). The converse is also true. Blank: “is supplementary”.

1. Equal opposite anglesLet ∠A = ∠C = x and ∠B = ∠D = y. Then 2x + 2y = 360°, so x + y = 180°. Each pair of adjacent angles is supplementary.
2. The converseIf each pair of adjacent angles adds up to 180°, then ∠A + ∠B = ∠B + ∠C gives ∠A = ∠C, and ∠B + ∠C = ∠C + ∠D gives ∠B = ∠D. So the opposite angles are equal.
3. Theorem 3 restated“If each pair of adjacent angles in a quadrilateral ABCD is supplementary, then ABCD is a parallelogram.”

Once the Midpoint Theorem gives PQ = BC ÷ 2, QR = AB ÷ 2 and PR = AC ÷ 2, the four triangles are congruent by SSS.

1. Why a direct test failsP, Q, R are the midpoints of AB, AC, BC. This gives AP = PB, AQ = QC, BR = RC, but we do not know the lengths or directions of PQ, PR, QR. So no congruence test (SSS, SAS, ASA) can yet be used.
2. First prove a new factBy the Midpoint Theorem, PQ = BC ÷ 2, PR = AC ÷ 2 and QR = AB ÷ 2 (and PQ ∥ BC, and so on).
3. Now use SSSEach of △APQ, △PBR, △QRC and △RQP has sides AB ÷ 2, AC ÷ 2 and BC ÷ 2. So the four triangles are congruent by SSS.
4. The ideaProve one new fact first (here the Midpoint Theorem), and use it to solve the larger problem.

Choose S so that AM = MS. Then BSCM is a parallelogram, so AM passes through the midpoint X of BC, and AM : MX = 2 : 1.

1. The constructionExtend AM beyond M to S so that AM = MS (M is the midpoint of AS).
2. BSCM is a parallelogramIn △ABS, P is the midpoint of AB and M is the midpoint of AS, so PM ∥ BS. P, M, C are on one line, so MC ∥ BS. In △ACS, Q and M are midpoints of AC and AS, so QM ∥ CS. Q, M, B are on one line, so MB ∥ CS. So BSCM has both pairs of opposite sides parallel: it is a parallelogram.
3. AM passes through the midpoint of BCLet X be the point where BC meets MS. The diagonals of a parallelogram bisect each other, so BX = XC and MX = XS. So the line AM meets BC at its midpoint X: AX is the third median.
4. The ratioAM = MS = 2 MX, so AM : MX = 2 : 1. Similarly CM : MP = 2 : 1 and BM : MQ = 2 : 1. So all three medians pass through M, and M divides each in the ratio 2 : 1 from the vertex.

(i) Adjacent: AD and BC. Opposite: CD. (ii) Adjacent: ∠B and ∠D. Opposite: ∠C.

1. (i) SidesThe sides adjacent to AB (sharing an endpoint with it) are AD and BC. The side opposite to AB is CD.
2. (ii) AnglesThe angles adjacent to ∠A are ∠B and ∠D. The angle opposite to ∠A is ∠C.
3. (iii) DefinitionsA pair of opposite sides is two sides of the quadrilateral that have no common endpoint. A pair of opposite angles is two angles whose vertices are not joined by a side of the quadrilateral.

The internal angle at a vertex is the angle between the two sides at that vertex, measured through the inside of the quadrilateral. At a dent it is greater than 180°.

1. DefinitionThe internal angle at a vertex is the angle between the two sides that meet there, measured on the side where the inside of the quadrilateral lies near that vertex.
2. A rule that works for every quadrilateralLet ABCD be a quadrilateral and look at the vertex A, with the triangle BCD formed by the other three vertices. Let θ be the smaller angle between the rays AB and AD. If A lies inside triangle BCD, then A is a dent (an inward-pointing vertex) and the internal angle at A is 360° − θ (more than 180°). Otherwise the internal angle at A is θ.
3. WarningDo not simply choose the angle that contains the direction of the opposite vertex: for some non-convex quadrilaterals that rule gives the wrong angle.

If AB ∥ DC: no, it must be convex. If AB = CD: yes, it can be non-convex. If ∠A = ∠C: yes, it can be non-convex.

1. If AB ∥ DCNo. Draw the two parallel lines. The sides AD and BC join their ends without crossing, so the four vertices go round the boundary of a trapezium, and none can be a dent. So ABCD is convex.
2. If AB = CDYes. Take A(0, 0), B(2, 0), C(1, 1), D(1, 3) in this order. Then AB = CD = 2, but C lies inside triangle ABD, so C is a dent and ABCD is non-convex.
3. If ∠A = ∠CYes. Take A(−2, 0), B(0, 1), C(2, 0), D(0, 3). The figure is symmetric about the vertical axis, which swaps A and C, so ∠A = ∠C. But B lies inside triangle ACD, so B is a dent: the quadrilateral is non-convex.

The three lines cut the plane into 7 regions: D in 1 region gives a convex quadrilateral, in 2 regions a self-intersecting one, and in 4 regions a non-convex one.

1. The seven regionsThe lines AB, BC and CA divide the plane into the inside of the triangle, three regions next to the sides, and three regions at the corners. The order of the vertices is always A-B-C-D.
2. The result

   Region containing D	Type of ABCD
   Inside triangle ABC	Non-convex (D is the dent)
   Beyond side AC (opposite B)	Convex
   Beyond side AB (opposite C)	Self-intersecting (CD crosses AB)
   Beyond side BC (opposite A)	Self-intersecting (AD crosses BC)
   Corner region at A	Non-convex (A is the dent)
   Corner region at B	Non-convex (B is the dent)
   Corner region at C	Non-convex (C is the dent)

3. Why the type does not change inside a regionTo change the type, D would have to cross one of the three lines. So all the points of one region give the same type.

No (if “self-intersecting” means that two sides really cross).

1. ArgumentSuppose two opposite sides, say AB and CD, meet at a point. Two lines that meet lie in one plane. A and B lie on the first line and C and D lie on the second, so all four vertices lie in that plane. The same is true if AD and BC meet.
2. ConclusionA quadrilateral whose sides really cross is flat (planar). A drawing of a non-planar quadrilateral can look crossed, but its sides in space do not meet.

True.

1. ReasonIn a parallelogram, adjacent angles add up to 180° and opposite angles are equal. If one angle is 90°, then its adjacent angles are 90° and its opposite angle is 90°. All four angles are 90°, so it is a rectangle.

False.

1. ReasonEvery rhombus has perpendicular diagonals. A rhombus with angles 60° and 120° has perpendicular diagonals but is not a square. To be a square it also needs a right angle (or equal diagonals).

True.

1. ProofIn parallelogram ABCD, compare △ABC and △BAD. AB is common, BC = AD (opposite sides), and AC = BD (given). So the triangles are congruent by SSS, and ∠ABC = ∠BAD.
2. ConcludeThese are adjacent angles of a parallelogram, so they add up to 180°. Equal angles that add up to 180° are each 90°. So the parallelogram has a right angle, and it is a rectangle.

AC bisects ∠C too, and AB = BC = CD = DA, so ABCD is a rhombus.

1. AC bisects ∠CGiven ∠BAC = ∠CAD. Since AB ∥ CD, ∠BAC = ∠ACD (alternate angles). Since AD ∥ BC, ∠CAD = ∠BCA (alternate angles). So ∠ACD = ∠BCA: AC bisects ∠C.
2. A rhombusNow ∠BAC = ∠BCA in △ABC, so AB = BC (sides opposite equal angles). Opposite sides of a parallelogram are equal: AB = CD and BC = AD. So AB = BC = CD = DA: ABCD is a rhombus.

Yes.

1. Using ACAC bisects the angles at A and C. In △ABC and △ADC, ∠BAC = ∠DAC, ∠BCA = ∠DCA and AC is common. So the triangles are congruent by ASA, and AB = AD and BC = CD.
2. Using BDIn the same way, BD bisects the angles at B and D. Comparing △ABD and △CBD by ASA gives AB = CB and AD = CD.
3. ConclusionSo AB = BC = CD = DA, and ABCD is a rhombus.

Yes.

1. A parallelogramDiagonals that bisect each other make ABCD a parallelogram. Let them meet at O.
2. Equal adjacent sidesThe right triangles AOB and COB have AO = CO, the common side OB, and ∠AOB = ∠COB = 90°. They are congruent by SAS, so AB = CB.
3. ConclusionA parallelogram with two equal adjacent sides is a rhombus.

No. An extra condition is needed: ABCD must also be a parallelogram (or its diagonals must bisect each other).

1. A counter-exampleAn isosceles trapezium that is not a rectangle has equal diagonals. For example A(0, 0), B(4, 0), C(3, 2), D(1, 2): AC = BD = √13, but the angles are not all right angles.
2. The extra conditionIf ABCD is also a parallelogram (diagonals bisect each other), then equal diagonals make it a rectangle, as we proved in Question 13.

The four meeting points form a quadrilateral with four right angles, so it is a rectangle. If AB = BC the parallelogram is a rhombus and all the points coincide.

1. One angle of the new quadrilateralThe bisectors of ∠A and ∠B meet at a point. In the triangle formed by AB and these two bisectors, the angles at A and B are ½∠A and ½∠B. In a parallelogram ∠A + ∠B = 180°, so ½∠A + ½∠B = 90°. The angle at their meeting point is 180° − 90° = 90°.
2. The other three anglesThe same argument works for each pair of bisectors at adjacent vertices (B and C, C and D, D and A). So all four angles of the quadrilateral formed by the meeting points are 90°.
3. A rectangleA quadrilateral with four right angles is a rectangle.
4. Why AB ≠ BCIf AB = BC, then ABCD is a rhombus, and its angle bisectors lie along the two diagonals. The four meeting points then all coincide at the centre, and there is no rectangle with four distinct corners.

(i) △PQR ≅ △QPA ≅ △RBP ≅ △CRQ. (ii) Yes, by drawing lines through P, Q, R parallel to the opposite sides of △PQR.

1. (i) The sidesBy the Midpoint Theorem, PQ = BC ÷ 2 = BR = RC, QR = AB ÷ 2 = AP = PB, and PR = AC ÷ 2 = AQ = QC.
2. Congruent by SSS△PQR and △QPA have PQ = QP, QR = PA and PR = QA, so △PQR ≅ △QPA. In the same way △PQR ≅ △RBP and △PQR ≅ △CRQ. The four triangles are congruent.
3. (ii) Reconstruct △ABCThrough P draw a line parallel to QR, through Q a line parallel to PR, and through R a line parallel to PQ. Let the first two lines meet at A, the first and third at B, and the second and third at C. Then APQR is a parallelogram, so AP = RQ, and PBRQ is a parallelogram, so PB = RQ. So AP = PB. Similarly AQ = QC and BR = RC. So P, Q, R are the midpoints of the sides of △ABC, and it is the required triangle.

MN meets AD at its midpoint.

1. MN is parallel to BCBy the Midpoint Theorem, MN ∥ BC.
2. In △ABDLet MN meet AD at X. Since D is on BC, MX ∥ BD. M is the midpoint of AB and MX ∥ BD, so by the converse of the Midpoint Theorem, X is the midpoint of AD. So AX = XD.
3. Special casesIf D is B or C, the result follows at once because M or N is a midpoint.

GH ∥ AB, and GH = |CD − AB| ÷ 2. If AB = CD, then G = H and there is no segment GH.

1. Use the midpoint E of ADLet E be the midpoint of AD. In △ADC, E and G are midpoints, so EG ∥ DC and EG = DC ÷ 2. In △ADB, E and H are midpoints, so EH ∥ AB and EH = AB ÷ 2.
2. CollinearAB ∥ DC, so EG and EH are both parallel to AB through the same point E. They lie on the same line: E, G, H are collinear, and GH ∥ AB.
3. The lengthG and H lie on the same side of E, so GH = |EG − EH| = |CD − AB| ÷ 2.
4. Why AB ≠ CDIf AB = CD, then ABCD is a parallelogram and its diagonals have the same midpoint. So G and H coincide, and there is no line GH.

PQRS is a parallelogram, so PR and QS bisect each other. If AC = BD, PQRS is a rhombus and PR ⊥ QS. The converse is true.

1. PQRS is a parallelogramBy the Midpoint Theorem in △ABC and △ADC, PQ ∥ AC, PQ = AC ÷ 2, SR ∥ AC and SR = AC ÷ 2. So PQ ∥ SR and PQ = SR: PQRS is a parallelogram (the Varignon parallelogram).
2. (i)The diagonals of a parallelogram bisect each other, so PR and QS bisect each other.
3. (ii) If AC = BDIn the same way QR = BD ÷ 2 (from △BCD). If AC = BD, then PQ = QR, so the parallelogram PQRS has equal adjacent sides: it is a rhombus. The diagonals of a rhombus are perpendicular, so PR ⊥ QS.
4. The converseIf PR ⊥ QS, then PQRS is a parallelogram with perpendicular diagonals, so it is a rhombus and PQ = QR. So AC ÷ 2 = BD ÷ 2, and AC = BD. The converse is true.

(i) Place A′, then turn A′ half a turn about P to get B′, B′ about Q to get C′, C′ about R to get D′. (ii) S is the midpoint of D′A′, and A′B′C′D′ is congruent to ABCD. (iii) PQRS is a square exactly when AC ⊥ BD and AC = BD.

1. (i) The constructionPlace A′ so that △SPA′ is congruent to the original △SPA (copy the lengths SP, PA, SA). Then:
   • extend A′P beyond P to B′ with PB′ = PA′;
   • extend B′Q beyond Q to C′ with QC′ = QB′;
   • extend C′R beyond R to D′ with RD′ = RC′;
   • join D′ to A′.
   In other words, B′, C′, D′ are got by half-turns about P, Q and R.
2. (ii) S is the midpoint of A′D′Use coordinates: a half-turn about a point P takes a point X to 2P − X. So B′ = 2P − A′, C′ = 2Q − B′ = 2Q − 2P + A′, and D′ = 2R − C′ = 2R − 2Q + 2P − A′. The midpoint of A′D′ is A′ + D′2 = R − Q + P. In the parallelogram PQRS the diagonals PR and QS have the same midpoint, so P + R = Q + S, that is, R − Q + P = S. So S is the midpoint of A′D′: A′, S, D′ are collinear and SA′ = SD′. P, Q, R are midpoints of A′B′, B′C′, C′D′ by construction.
3. A′B′C′D′ is congruent to ABCDWe placed A′ in the same position relative to PQRS as A. In the original, B is the half-turn image of A about P (P is the midpoint of AB), C of B about Q, and D of C about R. Our construction makes B′, C′, D′ in exactly the same way, so A′B′C′D′ has the same vertices as ABCD and is congruent to it.
4. (iii) A squarePQ ∥ AC, QR ∥ BD, PQ = AC ÷ 2 and QR = BD ÷ 2. If PQRS is a square, then PQ ⊥ QR and PQ = QR, so AC ⊥ BD and AC = BD. Conversely, if AC ⊥ BD and AC = BD, then the adjacent sides PQ and QR of the parallelogram PQRS are perpendicular and equal, so PQRS is a square.

The angle of a regular pentagon is 108°, and 108° does not fit a whole number of times into 360°.

1. The angleEach interior angle of a regular pentagon is (5 − 2) × 180° ÷ 5 = 108°.
2. At a cornerWhere the corners of tiles meet, their angles must add up to 360°. But 360 ÷ 108 = 10/3, which is not a whole number: three corners give 324° (a gap of 36°) and four give 432° (an overlap).
3. A corner meeting the side of another tileThen that side contributes 180°, leaving 180° for corners, and 180 ÷ 108 is not a whole number either. In general we would need 108k + 180l = 360 with k ≥ 1 and l ≥ 0. Dividing by 36 gives 3k + 5l = 10, which has no such solution (l = 0, 1, 2 give 3k = 10, 5, 0).
4. ConclusionSo congruent regular pentagons cannot tile the plane.

A DART can tile the plane by half-turns about the midpoints of its sides. Both methods work; the half-turn method is simpler to describe.

1. A DARTTake D(1, 1), A(0, 3), R(0, 0), T(3, 0), joined in the order D-A-R-T. D lies inside triangle ART, so D is the dent (a reflex vertex) and DART is non-convex.
2. Method 1: half-turnsTurn a copy of DART through 180° about the midpoint of one side. The two pieces fit together along that side. Then slide this pair along the two diagonals of DART (the arrows from D to R and from A to T) again and again to fill the plane. At every corner the four angles of DART meet and add up to 360°.
3. Method 2: the Varignon parallelogram gridCut the plane into the parallelogram grid given by the Varignon parallelogram, put equally turned copies at the chosen grid positions, and fill the gaps with half-turned copies.
4. My preferenceThe half-turn method: each new tile is placed by one half-turn about a known midpoint, so it is easy to describe and to check.

Two copies of a triangle make a parallelogram, and parallelograms tile the plane.

1. A parallelogram from two trianglesTake any triangle ABC and a congruent copy turned through 180° about the midpoint of BC. The two triangles together form a parallelogram, because the half-turn makes the opposite sides equal and parallel.
2. Tile with the parallelogramParallelograms tile the plane: slide copies of this parallelogram along its two sides. Keep the diagonal that separates the two triangles in every copy. Then the plane is tiled by copies of the original triangle.

The midpoint is the point where the segment crosses the ruled line that is two spaces from each end (the middle line between the end lines).

1. Where the ends areThe two ends of the segment are on ruled lines that are 4 equal spaces apart (the 2nd and 6th lines in the figure). The middle line between them is 2 spaces from each end.
2. The midpointMark the point where the segment crosses this middle ruled line. This is the midpoint.
3. JustificationDraw a vertical line through one end and a horizontal line through the other end, forming a right triangle with the segment as its sloping side. The vertical side crosses the ruled lines at equal gaps, so the middle ruled line passes through the midpoint of the vertical side and is parallel to the horizontal side. By the converse of the Midpoint Theorem, it also bisects the sloping side (the segment).