NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 12 Quadrilaterals

Class 9 Ganita Manjari Chapter 12 Quick Links:

• Think and Reflect
• Exercise Set 12.1
• Exercise Set 12.2
• Exercise Set 12.3
• Exercise Set 12.4
• End of Chapter Exercises

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Class 9 Maths Ganita Manjari Chapter 12 Think and Reflect Solutions

Answer:
Yes. Every ordinary quadrilateral, whether convex or non-convex, can tile the plane with congruent copies, without gaps or overlaps.
One construction is to rotate a copy through 180° about the midpoint of a side, so that it fits against the original along that side. Continue using half-turns about the midpoints of the exposed sides. This produces a repeating tiling.
The four different vertex angles can meet around a point because their sum is 360°. A non-planar four-sided figure is not a flat tile, and a crossed figure is not an ordinary quadrilateral tile.

Answer:
No. Not every figure shown is a quadrilateral in the usual sense.

• ABCD: Yes; it is a convex quadrilateral.
• NOPE: No; three named vertices lie on one straight side, so the outline is a triangle.
• SILY: No; some of the supposed sides overlap.
• DART: Yes; it is a non-convex quadrilateral, with a reflex internal angle at D.
• CUTS: It is a self-intersecting four-sided figure, excluded from the usual definition used in this chapter.
• OPENS: No; it is an open chain, not a closed four-sided figure.
• BENT: It is non-planar, so it is excluded from the usual planar definition.

A suitable definition is: A quadrilateral is a closed figure formed by joining four distinct points in one plane in order, with no three vertices collinear, such that adjacent sides meet only at their common endpoint and non-adjacent sides do not intersect.
This includes both convex quadrilaterals and non-convex quadrilaterals such as DART.

Answer:
Yes, but more conditions are needed than merely saying “four non-collinear points”.
Take four distinct points A, B, C, D in one plane, with no three collinear. Join AB, BC, CD and DA. Require adjacent sides to have only their shared endpoint in common and require opposite sides not to intersect.
These four segments then form quadrilateral ABCD. The conditions exclude overlapping sides, a crossed figure, an open chain and a non-planar figure, while allowing a non-convex quadrilateral.

Answer:
Let ∠A = ∠C = x and ∠B = ∠D = y.
Then 2x + 2y = 360°, so x + y = 180°.
Thus, each pair of adjacent angles is supplementary.

The converse is also true. If each pair of adjacent angles sums to 180°, then
∠A + ∠B = ∠B + ∠C, giving ∠A = ∠C;
∠B + ∠C = ∠C + ∠D, giving ∠B = ∠D.
Therefore the opposite angles are equal and Theorem 3 makes ABCD a parallelogram.
Completed statement: “If each pair of adjacent angles in a quadrilateral ABCD is supplementary, then ABCD is a parallelogram.”

Answer:
Let P, Q, R be the midpoints of AB, AC, BC respectively. The midpoint information gives AP = PB, AQ = QC and BR = RC, but it does not immediately give the lengths or directions of PQ, PR and QR. Thus we cannot yet apply a congruence test to all four small triangles using only these known equalities.
First prove the Midpoint Theorem. It gives PQ = BC/2, PR = AC/2 and QR = AB/2.
Now △APQ, △BPR, △CQR and △PQR each have side lengths AB/2, AC/2 and BC/2. They are congruent by SSS.
The useful idea is to establish one new segment property first, then use it to solve the larger problem.

Answer:
(1) If PQ ∥ BC and PQ = BC/2, then P and Q are the midpoints of AB and AC. A proof is given in End-of-Chapter Question 22(i).
(2) The line through the midpoint of one side, parallel to a second side, bisects the third side.
Proof: Let P be the midpoint of AB and let the parallel through P meet AC at Q. Let N be the actual midpoint of AC. The Midpoint Theorem gives PN ∥ BC. There is only one parallel to BC through P, so lines PN and PQ are the same. They meet AC at the same point; hence Q = N. Therefore AQ = QC.

(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:
Extend AM beyond M to S so that AM = MS. Thus M is the midpoint of AS.
In △ABS, P is the midpoint of AB and M is the midpoint of AS. By the Midpoint Theorem, PM ∥ BS. Since P, M, C are collinear, MC ∥ BS.
In △ACS, Q and M are midpoints of AC and AS. Therefore QM ∥ CS. Since Q, M, B are collinear, MB ∥ CS.
Hence BSCM is a parallelogram.

Let X be the intersection of BC and MS. Diagonals of a parallelogram bisect each other, so BX = XC and MX = XS.
Thus line AM meets BC at its midpoint X: AX is the third median, and it passes through M.
Also, AM = MS = 2MX, giving AM : MX = 2 : 1.
From △ABS, PM = BS/2 = MC/2, so CM : MP = 2 : 1.
From △ACS, QM = CS/2 = MB/2, so BM : MQ = 2 : 1.
Therefore the three medians are concurrent at M and each is divided in the ratio 2 : 1 from its vertex.

Answer:
The four internal angles of SOME add to 360°. Therefore, placing one copy of each angle around the common point fills one complete turn, with no angular gap or overlap near that point.
For a pattern that continues consistently, use half-turns about shared-side midpoints. These make neighbouring sides coincide exactly and place the vertex angles in a compatible order. Repeating the construction gives the tiling illustrated in the chapter.
Matching the angle sum at one point is a useful first check, but it does not alone prove that an arbitrary arrangement will extend to a whole-plane tiling.

Class 9 Maths Ganita Manjari Chapter 12 Exercise Set 12.1 Solutions

(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.
Answer:
(i) Sides adjacent to AB: AD and BC. The side opposite to AB: CD.
(ii) Angles adjacent to ∠A: ∠B and ∠D. The angle opposite to ∠A: ∠C.
(iii) Opposite sides are two sides of a quadrilateral that have no common endpoint.
Opposite angles are two internal angles whose vertices are not joined by a side of the quadrilateral.

(Hint: use the opposite vertex as well.)
Answer:
The internal angle at a vertex is the angle between the two adjacent sides measured through the region enclosed by the quadrilateral near that vertex. It is the smaller angle at a convex vertex and the reflex angle, greater than 180°, at a non-convex vertex.

Here is a precise way to choose the correct angle at A using the other vertices, including the opposite vertex C:

• Draw triangle BCD and consider the smaller angle θ between rays AB and AD.
• If A lies inside triangle BCD, A is the inward-pointing vertex; define the internal angle at A to be 360° − θ.
• Otherwise, define the internal angle at A to be θ.
• Apply the same rule at each of the other vertices.

Do not simply choose the angle containing the ray towards the opposite vertex: in some non-convex quadrilaterals that rule gives the wrong angle.

Answer:
If AB ∥ DC: No. A simple quadrilateral with a pair of parallel opposite sides is convex. To see this, draw the two parallel lines. The connecting sides AD and BC must join their endpoints without crossing. The four vertices then occur around the boundary of a convex trapezium; none can be an inward-pointing vertex.

• If AB = CD: Yes. Equal opposite side lengths alone do not prevent a dent.
  Example: Plot A(0, 0), B(2, 0), C(1, 1), D(1, 3) and join them in that order. AB = CD = 2 units, but C lies inside triangle ABD, so ABCD is non-convex.
• If ∠A = ∠C: Yes. A symmetric dart can have equal convex angles at A and C and a reflex angle at B.
  Example: A(−2, 0), B(0, 1), C(2, 0), D(0, 3). Reflection in the vertical axis exchanges A and C and preserves the figure, so ∠A = ∠C; B is the inward-pointing vertex.

(Hint: the three lines divide the plane into 7 regions.)
Answer:
There are seven regions. The order of the vertices is fixed as A-B-C-D.

Thus, four regions give non-convex quadrilaterals, two give crossed quadrilaterals and one gives convex quadrilaterals.
To check each case, place D in the region, join A-B-C-D-A and examine whether opposite sides cross or one vertex lies inside the triangle of the other three. The type cannot change while D stays in the same region, because a change would require D to pass through one of the three boundary lines.

Answer:
No, if “self-intersecting” means an actual crossing of its straight sides.
Suppose opposite sides AB and CD meet. Two intersecting lines lie in one plane. Since A and B lie on the first line and C and D on the second, all four vertices lie in that plane. The same reasoning applies if AD and BC meet.
Therefore, an actually self-intersecting quadrilateral is planar. A drawing or projection of a non-planar quadrilateral can look crossed even when the spatial sides do not meet.

Class 9 Maths Ganita Manjari Chapter 12 Exercise Set 12.2 Solutions

(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.
Answer:
(i) True.
Adjacent angles of a parallelogram sum to 180° and opposite angles are equal. If one angle is 90°, all four are 90°, so it is a rectangle.

(ii) False.
Every rhombus has perpendicular diagonals, including rhombuses whose angles are 60° and 120°. Such a rhombus is not a square. An additional right angle or equality of its diagonals, would make it a square.

(iii) True.
In parallelogram ABCD, compare △ABC and △BAD. AB is common, BC = AD and AC = BD by the given condition. The triangles are congruent by SSS, so ∠ABC = ∠BAD. These adjacent angles also sum to 180°, hence both are 90°. Therefore ABCD is a rectangle.

Answer:
Given ∠BAC = ∠CAD.
Since AB ∥ CD, ∠BAC = ∠ACD.
Since AD ∥ BC, ∠CAD = ∠BCA.
Therefore ∠ACD = ∠BCA, so AC bisects ∠C.
Also, ∠BAC = ∠BCA in △ABC.

Hence AB = BC, because sides opposite equal angles are equal.
Opposite sides of a parallelogram are equal: AB = CD and BC = AD.
Combining these equalities gives AB = BC = CD = DA. Thus ABCD is a rhombus.

(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.)
Answer:
(i) Yes.
Here the condition means that AC bisects the angles at A and C and BD bisects the angles at B and D.
In △ABC and △ADC, the angles at A and C are equal in pairs and AC is common.
By ASA, the triangles are congruent. Thus AB = AD and BC = CD.
Similarly, comparing △ABD and △CBD using the angle bisectors at B and D gives AB = BC.
Therefore all four sides are equal, so ABCD is a rhombus.

(ii) Yes.
Diagonals that bisect each other make ABCD a parallelogram.
Let their intersection be O. Right triangles AOB and COB have AO = CO and common side OB.
They are congruent by SAS, so AB = BC. A parallelogram with equal adjacent sides is a rhombus.

(iii) No.
A non-rectangular isosceles trapezium has equal diagonals.
For example, ABCD gives equal diagonals AC = BD, but the angles are not all right angles.
Add the condition that ABCD is a parallelogram or equivalently add that its diagonals bisect each other. A parallelogram with equal diagonals is a rectangle, as proved in Question 1(iii).

Answer:
Using the intersections of the bisectors of adjacent angles, as in the figure.
In a parallelogram, ∠A + ∠B = 180°.
In the triangle formed by AB and the bisectors of ∠A and ∠B, the angle at their meeting point is
180° − (∠A/2 + ∠B/2) = 180° − 90° = 90°.

The same argument works for each pair of adjacent vertex-angle bisectors.
Opposite angles are equal and opposite sides are parallel, so the bisector lines of opposite angles are parallel.

Thus the four intersection points form a quadrilateral with four right angles: a rectangle.
If AB = BC, the original parallelogram is a rhombus. Its angle bisectors lie along its two diagonals and all four relevant intersections coincide at the diagonal intersection. There is then no rectangle with four distinct vertices. The condition AB ≠ BC excludes this collapsed case.

Class 9 Maths Ganita Manjari Chapter 12 Exercise Set 12.3 Solutions

Answer:
(i) The Midpoint Theorem gives
PQ = BC/2 = BR = RC,
QR = AB/2 = AP = PB,
PR = AC/2 = AQ = QC.
Therefore, by SSS
△PQR ≅ △QPA ≅ △RBP ≅ △CRQ.

(ii) Yes.
Through P draw a line parallel to QR, through Q a line parallel to PR and through R a line parallel to PQ.
Call the intersections of the first and second lines A, the first and third B, and the second and third C.
For example, APQR is a parallelogram, so AP = RQ. Also PBRQ is a parallelogram, so PB = RQ. Thus AP = PB. Similarly AQ = QC and BR = RC.
Hence, the reconstructed triangle has P, Q, R as its side midpoints and is the required △ABC.

Answer:
Let MN meet AD at X. Since M and N are midpoints, MN ∥ BC by the Midpoint Theorem.

As D is on BC, MX ∥ BD. In △ABD, the line through midpoint M of AB parallel to BD bisects AD, by the converse of the Midpoint Theorem.
Thus AX = XD.
If D coincides with B or C, the conclusion follows directly from M or N being a midpoint.

Answer:
Let E be the midpoint of AD.
In △ADC, EG joins two midpoints, so EG ∥ DC and EG = DC/2.
In △ADB, EH joins two midpoints, so EH ∥ AB and EH = AB/2.

Since AB ∥ DC, EG and EH lie on the same line through E. Consequently E, G, H are collinear and GH ∥ AB.
In this trapezium, G and H lie on the same ray from E, giving GH = |CD − AB|/2.
If AB = CD, the equal-and-parallel-sides test makes ABCD a parallelogram. Its diagonals have the same midpoint, so G = H and GH is not a non-zero segment.
The assumption excludes that case.

(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?
Answer:
(i) By Varignon’s theorem, PQRS is a parallelogram. Its diagonals PR and QS therefore bisect each other.
(ii) The Midpoint Theorem gives PQ = AC/2 and QR = BD/2.
If AC = BD, then PQ = QR.
Thus PQRS is a rhombus, whose diagonals are perpendicular: PR ⟂ QS.
The converse is also true.
If PR ⟂ QS, they are perpendicular diagonals of the parallelogram PQRS.
That makes PQRS a rhombus, so PQ = QR.
Therefore AC/2 = BD/2 or AC = BD.

(i) Copy only PQRS on another 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.
Answer:
(i) Keep the original drawing available as a reference. On the copied PQRS, reproduce triangle SPA: use the lengths SP, PA and SA and place A′ on the corresponding side of SP, preserving the original orientation relative to Q and R.
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′A′. This recreates the required quadrilateral.

(ii) The copied, oriented triangle SPA′ is congruent to the original triangle SPA. The same rigid placement that matches the copied parallelogram to the original therefore matches A′ to A.
Because P is a midpoint, B is uniquely the point obtained by a half-turn of A about P. Our construction makes B′ in exactly that way. Likewise, C′ and D′ are obtained by the corresponding half-turns about Q and R, so they match C and D.
Consequently A′B′C′D′ is congruent to ABCD. Since S was the midpoint of AD, it is also the midpoint of A′D′. Thus A′, S, D′ are collinear and SA′ = SD′. The same midpoint statements hold at P, Q and R by construction.

(iii) PQ ∥ AC, QR ∥ BD, PQ = AC/2 and QR = BD/2.
If PQRS is a square, PQ ⟂ QR and PQ = QR. Therefore AC ⟂ BD and AC = BD.
Conversely, if AC ⟂ BD and AC = BD, adjacent sides PQ and QR of the Varignon parallelogram are perpendicular and equal. Hence PQRS is a square.

Class 9 Maths Ganita Manjari Chapter 12 Exercise Set 12.4 Solutions

(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.)

Answer:
Each internal angle of a regular pentagon is (5 − 2) × 180°/5 = 108°.
At a point where tile corners meet, their angles must add to 360°.
But 360°/108° = 10/3 is not an integer. Three corners give 324°, leaving a 36° gap; four give 432°, causing an overlap.

Even allowing a corner to meet the interior of another tile’s edge does not help. That edge contributes 180°, leaving 180°, which is also not an integer multiple of 108°.

More generally, at a pentagon vertex we would need 108k + 180l = 360 with k a positive integer and l a non-negative integer. Dividing by 36 gives 3k + 5l = 10, which has no such solution.
Therefore congruent regular pentagons cannot tile the plane.

Answer:
One possible DART has D(1, 1), A(0, 3), R(0, 0) and T(3, 0), joined in the order D-A-R-T-D. D lies inside triangle ART, so it is the reflex vertex.
Make congruent copies. Place a half-turned copy across a side by rotating through 180° about that side’s midpoint. Repeat across exposed sides, keeping already placed copies in their existing positions. The copies fit into a repeating tiling; a finite portion is shown in following Figure.

Alternatively, use the Varignon parallelogram grid, place the equally oriented copies at the selected grid positions and fill the intervening gaps with half-turned copies.

Preference: I prefer the half-turn method because each new tile is located by a single rotation about a known midpoint.
The general justification, valid for this non-convex tile too.

Class 9 Maths Ganita Manjari Chapter 12 End-of-Chapter Exercises Solutions

(Hint: Can you use the parallelogram tiling in the introduction?)
Answer:
Take any triangle ABC. Rotate a congruent copy through 180° about the midpoint of BC. The two triangles together form a parallelogram, since the half-turn makes the appropriate opposite sides equal and parallel.
Tile the plane with translated copies of this parallelogram. Keep the diagonal separating its two constituent triangles in every copy. This gives a tiling entirely by copies of the original triangle.

Answer:
The two endpoints in Fig. 12.33 lie on ruled lines four equal spaces apart. Mark the point where the segment crosses the ruled line two spaces from either endpoint. This point is the midpoint.

Justification:
Draw a vertical through one endpoint and a horizontal through the other, forming a right triangle with the given segment as its sloping side. 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.