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

Take ∆ABC. Rotate a copy through 180° about the midpoint of BC. Then B goes to C and C goes to B, and A goes to a new point D. The diagonals AD and BC of ABDC bisect each other at the midpoint of BC, so ABDC is a parallelogram (Theorem 4).

So two copies of the triangle make one parallelogram.

From the introduction, any fixed parallelogram tiles the plane (a grid of sticks pushed over). Each parallelogram of that tiling is cut by a diagonal into two copies of ∆ABC.

The orange line X to Y starts on the 6th line and ends on the 2nd line, so it crosses the lines 6, 5, 4, 3, 2. That is 4 equal gaps, and the middle ruled line (the 4th) lies halfway between the ends.

Mark the midpoint where the orange line crosses the 4th ruled line.

1. Let X be on the lower end and Y on the upper end. Let X' be the point on the 2nd ruled line directly above X.
2. XX' crosses the lines 6, 5, 4, 3, 2 and the gaps are equal, so the 4th ruled line meets XX' at its midpoint N.
3. In ∆XX'Y, the 4th ruled line passes through the midpoint N of XX' and is parallel to X'Y (which lies on the 2nd ruled line).
4. By the converse of the Midpoint Theorem (Theorem 7), this line bisects the third side XY.

E lies on AB and on CD, so the angles of ABCD at the four vertices are also angles of two triangles: ∠A = ∠DAE, ∠D = ∠ADE in ∆AED, and ∠B = ∠CBE, ∠C = ∠BCE in ∆BEC.

The angles ∠AED and ∠BEC are vertically opposite, so they are equal; call each x.

Let the diagonals AC and BD of ABCD meet at O. They bisect each other, so OB = OD and OA = OC.

P is between D and O, and Q is between B and O, with DP = BQ. So:

So O is the midpoint of PQ, and also the midpoint of AC. In quadrilateral APCQ the diagonals AC and PQ bisect each other, so by Theorem 4, APCQ is a parallelogram.

1. When A is folded onto B, the crease is the perpendicular bisector of AB. So the crease meets AB at its midpoint M, and the crease is perpendicular to AB.
2. The right angle is at B, so BC ⊥ AB. Two lines perpendicular to AB are parallel, so the crease is parallel to BC.
3. In ∆ABC, M is the midpoint of AB and the crease through M is parallel to BC. By the converse of the Midpoint Theorem (Theorem 7), the crease bisects AC.

Take ∆MPB and ∆MPC. Since PB = PC, we can place them along this side, with P on P and C on B. The angles ∠MPB and ∠MPC add up to 180° (B, P, C are collinear), so after the move M, P and the new position M' of the other M are in a line.

The new figure is ∆BMM' with MM' = 2MP. The centroid divides each median in the ratio 2 : 1, so MP = 13AP and MB = 23BQ, MC = 23CR:

The other four pieces make triangles in the same way, from the pairs at Q and at R. Each has sides 23AP, 23BQ, 23CR, so all three are congruent by SSS.

In ∆BMM' above, P is the midpoint of MM', so BP is a median of this triangle, and BP = BC2. The pair at Q and the pair at R give triangles congruent to it with medians CA2 and AB2. So the assembled triangle has medians AB2, BC2 and CA2.

Now apply Part (i) to this triangle: the new triangles have sides 23 of its medians:

Each of the 9 triangles has sides AB/3, BC/3 and AC/3. I checked this with coordinates.

1. Cut ∆ABC along the median AP, where P is the midpoint of BC. Then cut ∆APB along its median PR, where R is the midpoint of AB.
2. Turn ∆PBR through 180° about R: B goes to A, and P goes to P', with R the midpoint of PP'. Now ∆PAR and ∆P'AR together make ∆APP'.
3. In ∆APP': AP is common, AP' = PB = PC (the half-turn keeps lengths) and PP' = 2PR = AC (PR is half of AC by the Midpoint Theorem).

So ∆APP' has sides AP, PC, AC, the same three lengths as ∆APC. Hence ∆APP' ≅ ∆APC by SSS. The pieces form two congruent triangles: ∆APC (kept as it is) and ∆APP'.

Let EF meet the diagonal AC at G.

• In ∆ADC, E is the midpoint of AD and EG ∥ DC. By Theorem 7, G is the midpoint of AC and EG = DC2.
• In ∆ABC, G is the midpoint of AC and GF ∥ AB. By Theorem 7, F is the midpoint of BC and GF = AB2.

In ∆ABD, E and M are midpoints of AD and BD, so EM ∥ AB. In ∆BCD, F and M are midpoints of BC and BD, so FM ∥ DC, and DC ∥ AB.

EM and FM are two lines through M, both parallel to AB. So they are the same line, which means E, M, F are collinear on a line parallel to AB. Hence EF ∥ AB.

The direct way (with M) is simple because it needs only the Midpoint Theorem twice. Using (i) needs the same idea as the second proof of Theorem 7 (a unique parallel through a point). I find the direct way simpler.

Compare ∆OAP and ∆OCQ:

• OA = OC, because the diagonals of a parallelogram bisect each other.
• ∠AOP = ∠COQ (vertically opposite angles).
• ∠OAP = ∠OCQ (alternate angles, since AB ∥ CD with transversal AC).

So ∆OAP ≅ ∆OCQ by ASA, and OP = OQ.

The half-turn about O maps the parallelogram to itself, A to C and B to D, so AB goes to CD. It maps the line PQ (through O) to itself, so P on AB goes to a point on CD and on PQ, which is Q. So O is the midpoint of PQ.

By Question 7, EF ∥ AD and EF = AD + BC2 = 3 + 52 = 4 cm.

EF ∥ AD ∥ BC, and E is the midpoint of AB, so EF is halfway between AD and BC. If the height of the trapezium is h, both parts have height h2.

A quadrilateral is a way of joining the 4 points in a cycle. Starting at A, there are 3 choices for the next point, 2 for the next and 1 for the last: 3 × 2 × 1 = 6 orders. But going round the other way gives the same quadrilateral, so there are 62 = 3 different quadrilaterals:

In the first case, joining the points round their outside gives the convex quadrilateral. The other two orders cross, because each uses both diagonals of the convex 4-gon as sides.

In the second case, every order has the inner point as a dent, so all three quadrilaterals are non-convex and none crosses itself.

Suppose parallel lines cut equal parts on one line AB. Then they cut equal parts on AC as well. Reason: let X1 be the midpoint of AX2 on AB, with lines through X1 and X2 parallel to BC meeting AC at Y1 and Y2. In ∆AX2Y2, X1 is the midpoint of AX2 and X1Y1 ∥ X2Y2, so by Theorem 7, Y1 is the midpoint of AY2. Repeating this gives equal parts on AC for any number of equal parts on AB.

AP = 1 and PB = 2, so AB is 3 equal parts of length 1 (cut PB in two, as the dotted line shows). Lines parallel to BC cut AC into 3 equal parts. AQ is one part and equals √5, so each part is √5.

Cut AP = 1 into 3 parts of 13. Then AP is 3 parts and PB is 1 part, so AB is 4 parts. AQ = 3 parts = √5, so each part is √53.

Cut both AP and PB into parts of 13: AP is 3 parts and PB is 2 parts, so AB has 5 parts. AQ = 3 parts = √5, so each part is √53.

Let APPB = mn with whole numbers m and n. Cut AP into m equal parts and PB into n equal parts, each of length d, so AP = md and PB = nd. Lines parallel to BC through the cutting points divide AC into m + n equal parts, with AQ being m parts and QC being n parts.

AB ∥ DC and AB = DC, so MB = DN with MB ∥ DN. So MBND is a parallelogram and DM ∥ BN.

Let DM meet AC at X and BN meet AC at Y.

• In ∆ABY, M is the midpoint of AB and MX ∥ BY. By Theorem 7, X is the midpoint of AY, so AX = XY.
• In ∆CDX, N is the midpoint of CD and NY ∥ DX. By Theorem 7, Y is the midpoint of CX, so CY = YX.

• Using (i): draw any parallelogram with PQ as a diagonal (take any point B off PQ and complete the parallelogram PBQD). Join D to the midpoint of PB and B to the midpoint of QD. These cut PQ into three equal parts.
• Using Question 11: draw any ray from P. With a compass mark 3 equal steps along it, ending at Z. Join Z to Q. Through the other two marks draw lines parallel to ZQ. They meet PQ at the trisection points.
• Using the Centroid Theorem: take any point U off PQ and let V be the point such that Q is the midpoint of UV. In ∆PUV, PQ is a median. The medians from U and V meet PQ at the centroid G, which divides PQ in the ratio 2 : 1, so PG = 23PQ. The midpoint of PG is the second trisection point.

The proof used only the Midpoint Theorem in ∆ABC and ∆ADC (giving PQ ∥ AC ∥ SR) and in ∆BCD and ∆BAD (giving QR ∥ BD ∥ PS). It never used that ABCD is convex or that its sides do not cross; it only needs the midpoints of the four sides. So PQRS is a parallelogram for every quadrilateral.

PQ ∥ AC and QR ∥ BD. If the lines AC and BD are parallel (or the same line), then PQ ∥ QR, and the four points P, Q, R, S lie on one line.

Example: A(0, 0), B(0, 1), C(2, 0), D(2, 1). Then AC ∥ BD, the sides BC and DA cross, and the midpoints are P(0, 0.5), Q(1, 0.5), R(2, 0.5), S(1, 0.5), all on one line. For a convex quadrilateral the diagonals meet, so this cannot happen.

Any three points lie in a plane, so the Midpoint Theorem still applies to ∆ABC, ∆ADC, ∆BCD and ∆BAD. We still get PQ ∥ AC and SR ∥ AC with PQ = SR = AC2, so PQ and SR are equal and parallel. Two parallel lines lie in one plane, so P, Q, R, S are in a plane and form a parallelogram. The Varignon parallelogram is planar even when ABCD is not.

Take the square A(0, 0), B(4, 0), C(4, 4), D(0, 4) and choose P(1, 0) on AB, Q(4, 1) on BC, R(3, 4) on CD, S(0, 3) on DA.

PQ and SR are equal and parallel, so PQRS is a parallelogram (Theorem 5). But P is not the midpoint of AB, because AP = 1 and AB = 4.

R is the half-turn image of P about the centre (2, 2), and S is the half-turn image of Q. So the diagonals PR and QS bisect each other at the centre, whatever the positions of P and Q.

Extend PQ beyond Q to S so that QS = PQ. Then Q is the midpoint of PS. Join CS.

1. In ∆APQ and ∆CSQ: AQ = CQ (Q is the midpoint of AC), PQ = SQ and ∠AQP = ∠CQS (vertically opposite). So ∆APQ ≅ ∆CSQ by SAS.
2. Hence CS = AP = BP, and ∠APQ = ∠CSQ, which are alternate angles, so CS ∥ AP, that is CS ∥ BP.
3. So BP and CS are equal and parallel, and BPSC is a parallelogram (Theorem 5).
4. Then PS ∥ BC and PS = BC. Since PQ is half of PS, PQ ∥ BC and PQ = BC2.

The reason we wanted ∆APQ ≅ ∆CSQ: it gives CS equal and parallel to BP.

P is the midpoint of AB and PQ ∥ BC. Extend PQ to S, where S is the point on line PQ with CS ∥ BA.

1. BP ∥ CS (given) and PS ∥ BC, so BPSC is a parallelogram, and CS = BP = AP.
2. In ∆APQ and ∆CSQ: ∠APQ = ∠CSQ (alternate angles, AB ∥ CS), ∠AQP = ∠CQS (vertically opposite) and AP = CS. So ∆APQ ≅ ∆CSQ by AAS.
3. Hence AQ = CQ: Q is the midpoint of AC. Also PQ = QS, so PQ = PS2 = BC2.

Example for the last "false": A(0, 0), C(4, 0), B(1, 2), D(1, −2). The diagonals AC and BD are both 4 long and perpendicular, but they do not bisect each other, so ABCD is not a square.

Use the diagonal AC. ∆ABC and ∆ADC have angle sums 180° each:

At A there is a corner of 3-dimensional space made by the rays AB, AC, AD. In such a corner, one face angle is less than the sum of the other two: ∠BAD < ∠BAC + ∠CAD, as long as the corner is not flat. In the same way ∠BCD < ∠BCA + ∠ACD.

So ∠A + ∠B + ∠C + ∠D < 360°. The sum is exactly 360° only when ABCD is flat.

Slide SOME along EO by the length EO. This is a translation by the vector EO, so E goes to E' = O, and O goes to O', with O' on the line EO and OO' = EO.

Every vertex moves by the same vector EO. So the four segments EE', SS', OO', MM' are equal and parallel. Each side of SOME, together with its translated copy and the two connecting segments, forms a parallelogram:

In a parallelogram, opposite sides are equal and parallel, so the translated vertices S', O', M', E' are joined in the same way as S, O, M, E: SO = S'O', OM = O'M', ME = M'E' and ES = E'S', with the same angles. So the cutout fits S'O'M'E' exactly.

Let u = EO and w = SM be the two diagonal vectors of SOME. Sliding along EO in both directions gives the copies SOME + u and SOME − u, and sliding along SM gives SOME + w and SOME − w. That is 4 new copies.

Sliding these four along the other diagonal gives the copies at ±u ± w, which are 4 more. All together we get copies at the 9 positions iu + jw, with i and j equal to −1, 0 or 1: a 3 by 3 grid. Neighbours meet only at vertices.

The blank spaces: the translations by u and w repeat a pattern of cell area |u × w|, which is twice the area of SOME (the area of a 4-gon is half the product of its diagonals times the sine of the angle between them). So each gap has the same area as SOME. Turning SOME through 180° about the midpoint of a side fills a gap exactly, as in Method 1, so the gaps are 4-gons congruent to SOME.

Draw a segment of length c. Draw a circle of radius a about one end and a circle of radius b about the other end. A triangle exists exactly when these circles meet. They meet when a + b > c (and b − a < c, which is automatic since b ≤ c).

If a + b ≤ c the circles do not meet (or just touch), and we get a flat figure, not a triangle.

In a 4-gon, the longest side must be shorter than the sum of the other three. (The other three sides form a path between the ends of the longest side, so the path is longer than the straight side.)

If x is the longest side we need x < 2 + 5 + 11 = 18. If 11 is the longest we need 11 < 2 + 5 + x, that is x > 4.

A 4-gon with sides a, b, c, d exists exactly when the longest side is less than the sum of the other three.

Why it is enough: say d is the longest, with d < a + b + c. We want a diagonal t that makes two triangles, one with sides a, b, t and one with sides c, d, t. Both exist when t lies strictly between the larger of |a − b|, |c − d| and the smaller of a + b, c + d. That range is not empty: the only comparison that is not obvious is |c − d| < a + b, which is d − c < a + b, which is exactly d < a + b + c. Build the two triangles on opposite sides of the diagonal t.

A diagonal is a segment joining two vertices that are not adjacent.

The increases are 2, 3, 4, 5, 6, so the guesses are 9 + 5 = 14 for n = 7 and 14 + 6 = 20 for n = 8. Systematic counting agrees.

From each vertex we can draw a diagonal to every vertex except itself and its 2 neighbours, which is n − 3 diagonals. There are n vertices, but each diagonal is counted from both of its ends, so:

Check: n = 7 gives 7 × 42 = 14 and n = 8 gives 8 × 52 = 20.

Add a new vertex between two old neighbours. It is joined to the other n − 2 old vertices as diagonals, and the old side between its two neighbours becomes a diagonal as well.

Check with the formula: (n + 1)(n − 2)2 − n(n − 3)2 = 2n − 22 = n − 1.

Each extra side adds 180°. A 5-gon (convex) splits into 3 triangles from one vertex, a 6-gon into 4, and so on.

Proof by induction. For n = 3 it is 180°. For n > 3, a polygon always has a diagonal that lies inside it. This diagonal splits the n-gon into two polygons with n₁ and n₂ vertices, where n₁ + n₂ = n + 2 (the two ends of the diagonal are counted twice).

By the induction assumption the angle sums are (n₁ − 2) × 180° and (n₂ − 2) × 180°. The angles of the two parts together make up the angles of the n-gon, so:

Statement: In ∆ABC, let P be on AB and Q on AC with PQ ∥ BC and PQ = BC2. Then P and Q are the midpoints of AB and AC.

1. Draw the line through C parallel to BA, meeting line PQ at R.
2. BC ∥ PR and CR ∥ BP, so BPRC is a parallelogram. So PR = BC and CR = BP.
3. PQ = BC2 = PR2, so Q is the midpoint of PR.
4. ∆APQ ≅ ∆CRQ by AAS (∠APQ = ∠CRQ alternate angles, ∠AQP = ∠CQR vertically opposite, PQ = QR).
5. So AQ = QC (Q MID) and AP = CR = BP (P MID).

(Compare Chapter 9 Propositions and their Converses Solutions.) Write the Midpoint Theorem as: "Suppose P is the midpoint of AB and Q is a point on AC. If Q is the midpoint of AC then PQ ∥ BC." Keep the first sentence and reverse the second:

"Suppose P is the midpoint of AB and Q is a point on AC. If PQ ∥ BC then Q is the midpoint of AC."

That is Theorem 7, and PQ = BC2 follows. In terms of the named conditions: (P MID) and (PRLL) give (Q MID) and (HALF).

(1) No overlaps: two copies never share an interior point. (2) No gaps: every point of the plane lies in some copy. Both follow if copies fit exactly along each edge and the angles around every vertex add up to exactly 360°.

• A half-turn about the midpoint of a side maps that side onto itself, with its ends swapped, and puts the image on the other side of the side's line. So the new copy shares exactly that side with the old one, with no overlap.
• At any vertex, going around it one meets the copies obtained by repeated half-turns. Each half-turn about the midpoint of an edge at that vertex brings the next angle of the 4-gon. After four of them, all four angles appear once: 360°, so there is no gap and no overlap.
• Doing two half-turns one after the other is a translation, so the pattern repeats in the plane in two directions, and nothing is left uncovered.

The sides of the Varignon parallelogram are AC2 and BD2, that is half of the diagonal vectors u and w. Placing copies of the 4-gon on every second cell of the grid, in both directions, puts them at the positions iu + jw. These are the same positions as in Method 3, so the same reasoning applies: the copies meet only at vertices, and the gaps are filled with half-turn copies, as in Method 1.

Sliding by the diagonal vectors u = EO and w = SM gives copies at all positions iu + jw. These copies meet only at vertices, by the four parallelograms of Question 18. The gaps have the same area as the 4-gon and are half-turn copies of it, so they are filled as in Method 1.

Let the big square have side 2, so its area is 4. Put the corners at (0, 0), (2, 0), (2, 2), (0, 2). The four lines join a corner to a midpoint:

L1 ∥ L2 and L3 ∥ L4. The directions (1, −2) and (2, 1) are perpendicular, so the shaded region is a square.

Its side is the distance between L1 and L2, which is 4 − 2√(2² + 1²) = 2√5. Check with L3 and L4: the distance between x − 2y = −2 and x − 2y = 0 is also 2√5.