NCERT Solutions · Class 9 Mathematics Quadrilaterals

(i) Cut \(\displaystyle PB\) into two parts of length $\displaystyle 1$, so \(\displaystyle AB\) has $\displaystyle 3$ equal parts. Parallels to \(\displaystyle BC\) through the cut points cut \(\displaystyle AC\) into $\displaystyle 3$ equal parts, each equal to \(\displaystyle AQ\).\[\text{equal parts on } AB \;\Rightarrow\; \text{equal parts on } AC \quad \text{(Theorem 7, Exercise 7)} \]\[QC = 2\,AQ = 2\sqrt5 \](ii) Cut \(\displaystyle AP\) into $\displaystyle 3$ parts of length \(\displaystyle \tfrac13\): \(\displaystyle AB = \tfrac43\) has $\displaystyle 4$ equal parts, so \(\displaystyle AC\) has $\displaystyle 4$, and \(\displaystyle AQ\) is $\displaystyle 3$ of them.\[3u = \sqrt5 \;\Rightarrow\; u = \frac{\sqrt5}{3}, \qquad QC = u = \frac{\sqrt5}{3} \](iii) Use parts of length \(\displaystyle \tfrac13\): \(\displaystyle AP\) is $\displaystyle 3$ parts and \(\displaystyle PB\) is $\displaystyle 2$, so \(\displaystyle AC\) has $\displaystyle 5$ equal parts and \(\displaystyle AQ\) is $\displaystyle 3$ of them.\[3u = \sqrt5, \qquad QC = 2u = \frac{2\sqrt5}{3} \](iv) Let \(\displaystyle \frac{AP}{PB} = \frac mn\). Cut \(\displaystyle AP\) into \(\displaystyle m\) and \(\displaystyle PB\) into \(\displaystyle n\) equal parts of one common length. Parallels through the \(\displaystyle m + n\) cut points cut \(\displaystyle AC\) into \(\displaystyle m + n\) equal parts \(\displaystyle v\); \(\displaystyle Q\) is the \(\displaystyle m\)th cut.\[AQ = mv, \quad QC = nv \;\Rightarrow\; \frac{AQ}{QC} = \frac mn = \frac{AP}{PB} \]Answer: (i) \(\displaystyle 2\sqrt5\); (ii) \(\displaystyle \dfrac{\sqrt5}{3}\); (iii) \(\displaystyle \dfrac{2\sqrt5}{3}\); (iv) \(\displaystyle \dfrac{AP}{PB} = \dfrac{AQ}{QC}\).

(i) In the given figure, let \(\displaystyle DM\) and \(\displaystyle BN\) meet \(\displaystyle AC\) at \(\displaystyle X\) and \(\displaystyle Y\).\[MB = \tfrac12 AB = \tfrac12 DC = DN, \qquad MB \parallel DN \]\[MBND \text{ is a parallelogram} \quad \text{(Theorem 5)} \;\Rightarrow\; DM \parallel BN \]\[\triangle ABY:\ AM = MB,\ MX \parallel BY \;\Rightarrow\; AX = XY \quad \text{(Theorem 7)} \]\[\triangle CDX:\ CN = ND,\ NY \parallel DX \;\Rightarrow\; CY = YX \quad \text{(Theorem 7)} \]\[AX = XY = YC \](ii) 1. Complete a parallelogram \(\displaystyle PBQD\), \(\displaystyle B\) off \(\displaystyle PQ\); \(\displaystyle M, N\) midpoints of \(\displaystyle PB, QD\). By (i), \(\displaystyle DM\) and \(\displaystyle BN\) trisect \(\displaystyle PQ\). 2. Ray from \(\displaystyle P\), steps \(\displaystyle PA_1 = A_1A_2 = A_2A_3\); parallels to \(\displaystyle A_3Q\) through \(\displaystyle A_1, A_2\) meet \(\displaystyle PQ\) at \(\displaystyle T_1, T_2\). \[PT_1 : T_1Q = PA_1 : A_1A_3 = 1:2, \qquad PT_2 : T_2Q = PA_2 : A_2A_3 = 2:1 \quad \text{(Exercise 11)} \] 3. Take \(\displaystyle B\) off \(\displaystyle PQ\); extend \(\displaystyle BQ\) to \(\displaystyle C\) with \(\displaystyle QC = BQ\); \(\displaystyle E\) midpoint of \(\displaystyle PC\); \(\displaystyle BE\) meets \(\displaystyle PQ\) at \(\displaystyle G\); \(\displaystyle H\) midpoint of \(\displaystyle PG\). \[PG : GQ = 2 : 1 \quad \text{(Centroid Theorem, medians } PQ, BE \text{ of } \triangle PBC) \] \[PH = HG = \tfrac13 PQ = GQ \]Answer: (i) \(\displaystyle AX = XY = YC\). (ii) Parallelogram \(\displaystyle PBQD\) with (i); equal steps on a ray with parallels; centroid \(\displaystyle G\) and midpoint \(\displaystyle H\) of \(\displaystyle PG\).

Yes, in both cases.(i) The proof uses only the Midpoint Theorem in triangles \(\displaystyle ABC, ADC, BCD, BAD\). Any three of the four points form a triangle, whatever the shape of \(\displaystyle ABCD\), so convexity is never used.\[PQ \parallel AC \parallel SR \quad (\triangle ABC,\ \triangle ADC) \]\[QR \parallel BD \parallel PS \quad (\triangle BCD,\ \triangle BAD) \]Both pairs of opposite sides are parallel, so \(\displaystyle PQRS\) is a parallelogram.(ii) Since \(\displaystyle PQ \parallel AC\) and \(\displaystyle QR \parallel BD\), the parallelogram collapses exactly when \(\displaystyle AC \parallel BD\):\[AC \parallel BD \;\Longleftrightarrow\; PQ,\ QR \text{ on one line} \;\Longleftrightarrow\; P, Q, R, S \text{ collinear} \]With \(\displaystyle A, C\) on one line and \(\displaystyle B, D\) on a parallel line, two opposite sides always cross: \(\displaystyle ABCD\) is self-intersecting.(iii) Yes. Each triangle is planar even if \(\displaystyle ABCD\) is not, so \(\displaystyle PQ \parallel AC \parallel SR\) and \(\displaystyle QR \parallel BD \parallel PS\) still hold. Lines parallel to a third are parallel in space, and parallel lines are coplanar: \(\displaystyle PQRS\) is a planar parallelogram.Answer: (i) Only the Midpoint Theorem in four triangles is used, never convexity. (ii) When \(\displaystyle AC \parallel BD\). (iii) Yes.

Converse of Theorem $\displaystyle 9$: if \(\displaystyle P, Q, R, S\) on \(\displaystyle AB, BC, CD, DA\) form a parallelogram, they are the midpoints.False. In a square \(\displaystyle ABCD\) of side $\displaystyle 4$ take\[AP = BQ = CR = DS = 1, \qquad PB = QC = RD = SA = 3 \]\[\triangle APS \cong \triangle BQP \cong \triangle CRQ \cong \triangle DSR \quad \text{(SAS, right angles)} \]\[PQ = QR = RS = SP \;\Rightarrow\; PQRS \text{ is a rhombus, hence a parallelogram} \]Yet \(\displaystyle AP = 1 \neq PB = 3\), so \(\displaystyle P\) is not a midpoint. The same happens for any \(\displaystyle AP = BQ = CR = DS \neq 2\).Answer: No. In a square, points \(\displaystyle \tfrac14\) of the way along each side from \(\displaystyle A, B, C, D\) form a parallelogram (a square) but are not midpoints.

(i) Extend \(\displaystyle PQ\) by its own length, so \(\displaystyle QS = PQ\).\[AQ = CQ \ \text{(} Q \text{ midpoint of } AC \text{)}, \quad \angle AQP = \angle CQS \ \text{(vertically opposite)}, \quad PQ = SQ \]\[\triangle APQ \cong \triangle CSQ \quad \text{(SAS)} \]\[\angle APQ = \angle CSQ \;\Rightarrow\; AB \parallel CS \quad \text{(alternate angles)} \]\[CS = AP = PB, \quad CS \parallel PB \;\Rightarrow\; PBCS \text{ is a parallelogram} \quad \text{(Theorem 5)} \]\[PS \parallel BC, \quad PS = BC \;\Rightarrow\; PQ \parallel BC, \quad PQ = \tfrac12 PS = \tfrac12 BC \](ii) Given \(\displaystyle AP = PB\), \(\displaystyle PQ \parallel BC\), \(\displaystyle Q\) on \(\displaystyle AC\). Extend \(\displaystyle PQ\) to \(\displaystyle S\) with \(\displaystyle PS = BC\).\[PS \parallel BC, \quad PS = BC \;\Rightarrow\; PBCS \text{ is a parallelogram} \quad \text{(Theorem 5)} \]\[CS \parallel BP, \quad CS = BP = AP \]\[\angle APQ = \angle CSQ \ \text{(alternate, } AB \parallel CS\text{)}, \quad \angle AQP = \angle CQS \ \text{(vertically opposite)}, \quad AP = CS \]\[\triangle APQ \cong \triangle CSQ \quad \text{(AAS)} \;\Rightarrow\; AQ = QC, \quad PQ = QS = \tfrac12 BC \]Answer: (i) Extend \(\displaystyle PQ\) by \(\displaystyle QS = PQ\). (ii) Extend \(\displaystyle PQ\) to \(\displaystyle S\) with \(\displaystyle PS = BC\); then \(\displaystyle AQ = QC\).

Six converses, each with its verdict. This list is not exhaustive; other converses are equally valid.1. Diagonals bisect each other at right angles \(\displaystyle \Rightarrow\) rhombus: True.\[\text{parallelogram (Theorem 4)}, \quad \triangle AOB \cong \triangle BOC \cong \triangle COD \cong \triangle DOA \ \text{(SAS)} \;\Rightarrow\; AB = BC = CD = DA \]2. Diagonals bisect the angles \(\displaystyle \Rightarrow\) rhombus: True.\[\triangle ABC \cong \triangle ADC \ \text{(ASA)} \Rightarrow AB = AD; \qquad \triangle BAD \cong \triangle BCD \ \text{(ASA)} \Rightarrow AB = CB \]3. Diagonals perpendicular \(\displaystyle \Rightarrow\) rhombus: False. Kite:\[A(-1,0),\ B(0,-1),\ C(3,0),\ D(0,1): \quad AC \perp BD, \quad AB = \sqrt2 \neq BC = \sqrt{10} \]4. Diagonals equal \(\displaystyle \Rightarrow\) rectangle: False. Isosceles trapezium:\[(0,0),\ (4,0),\ (3,2),\ (1,2): \quad \text{both diagonals } \sqrt{13}, \quad \text{no right angle} \]5. Parallelogram with equal diagonals \(\displaystyle \Rightarrow\) rectangle: True.\[\triangle ABC \cong \triangle DCB \ \text{(SSS)} \Rightarrow \angle B = \angle C, \quad \angle B + \angle C = 180^\circ \Rightarrow \angle B = 90^\circ \]6. Diagonals equal and bisecting each other at right angles \(\displaystyle \Rightarrow\) square: True.\[\text{bisect} \Rightarrow \text{parallelogram}; \quad \text{equal} \Rightarrow \text{rectangle (5)}; \quad \text{perpendicular} \Rightarrow \text{rhombus (1)} \]Answer: Converses $\displaystyle 1$, $\displaystyle 2$, $\displaystyle 5$, $\displaystyle 6$ are true; $\displaystyle 3$ (kite) and $\displaystyle 4$ (isosceles trapezium) are false.

Turn triangle \(\displaystyle ADC \) about the hinge \(\displaystyle AC \) through the fold angle \(\displaystyle \psi \). \(\displaystyle \angle B,\angle D \) sit in rigid triangles, so they never change.Let \(\displaystyle F \) be the foot of \(\displaystyle D \) on \(\displaystyle AC \), \(\displaystyle \pi \) the plane through \(\displaystyle F \) perpendicular to \(\displaystyle AC \), \(\displaystyle B' \) the foot of \(\displaystyle B \) on \(\displaystyle \pi \), and \(\displaystyle D_0 \) the flat position of \(\displaystyle D \) (in plane \(\displaystyle ABC \), opposite \(\displaystyle B \)).\[BD^2=BB'^2+B'D^2,\qquad B'D\le B'F+FD=B'D_0 \]\[D\ne D_0\ \Rightarrow\ BD<BD_0 \]With two sides fixed, a shorter \(\displaystyle BD \) gives a smaller angle, so \(\displaystyle \angle A,\angle C \) shrink as \(\displaystyle \psi \) falls:\[\angle BAD<\angle BAD_0\le\angle BAC+\angle CAD,\qquad \angle BCD<\angle BCD_0\le\angle BCA+\angle ACD \]\[\angle A+\angle B+\angle C+\angle D<(\angle BAC+\angle ABC+\angle BCA)+(\angle CAD+\angle CDA+\angle ACD)=360^\circ \]For \(\displaystyle 2^\circ \), take congruent isosceles triangles with \(\displaystyle \angle B=\angle D=0.5^\circ \):\[\text{flat }(\psi=180^\circ):\ 360^\circ,\qquad \text{closed }(D\text{ on }B):\ 0.5^\circ+0.5^\circ+0^\circ+0^\circ=1^\circ \]The sum changes continuously and \(\displaystyle 1^\circ<2^\circ<360^\circ \), so some fold gives \(\displaystyle 2^\circ \); it is neither flat nor closed, so non-planar.Answer: the sum is always \(\displaystyle <360^\circ \); yes; \(\displaystyle \angle B,\angle D \) stay, \(\displaystyle \angle A,\angle C \) shrink.

(i) \(\displaystyle S' \) and \(\displaystyle S \) lie on the same side of \(\displaystyle EO' \), and \(\displaystyle S'O'M'E' \) is a copy of \(\displaystyle SOME \) with \(\displaystyle E'=O \). Corresponding angles are equal, so:\[\angle SEO=\angle S'E'O' \Rightarrow ES\parallel E'S',\ \ ES=E'S' \Rightarrow ESS'E' \text{ is a parallelogram} \]\[\angle MEO=\angle M'E'O' \Rightarrow EM\parallel E'M',\ \ EM=E'M' \Rightarrow EMM'E' \text{ is a parallelogram} \]\[\angle SOE=\angle S'O'E' \Rightarrow OS\parallel O'S',\ \ OS=O'S' \Rightarrow SOO'S' \text{ is a parallelogram} \]\[\angle MOE=\angle M'O'E' \Rightarrow OM\parallel O'M',\ \ OM=O'M' \Rightarrow OMM'O' \text{ is a parallelogram} \]\[\overrightarrow{SS'}=\overrightarrow{MM'}=\overrightarrow{EE'}=\overrightarrow{OO'} \]So the slide by \(\displaystyle \overrightarrow{EE'} \) takes \(\displaystyle S,O,M,E \) to \(\displaystyle S',O',M',E' \): the cutout matches \(\displaystyle S'O'M'E' \) exactly.(ii) Write \(\displaystyle \vec u=\overrightarrow{EO} \), \(\displaystyle \vec w=\overrightarrow{SM} \). Slides by \(\displaystyle \pm\vec u,\pm\vec w \) give $\displaystyle 4$ copies; four more slides give \(\displaystyle \pm\vec u\pm\vec w \): the grid \(\displaystyle Q+i\vec u+j\vec w \), \(\displaystyle |i|,|j|\le1 \), with \(\displaystyle Q=SOME \).\[E+\vec u=O,\quad S+\vec w=M,\quad E+\vec u+\vec w=P,\quad S+\vec u+\vec w=R \qquad (P=O+\vec w,\ R=M+\vec u) \]Four copies enclose \(\displaystyle OMPR \):\[\rho(X)=O+M-X:\qquad \rho(S)=P,\quad \rho(E)=R,\quad \rho(O)=M,\quad \rho(M)=O \]So the gap is \(\displaystyle SOME \) turned through \(\displaystyle 180^\circ \) about the midpoint of \(\displaystyle OM \): congruent to \(\displaystyle SOME \).\[\text{around }O:\quad \underbrace{\angle O}_{Q}+\underbrace{\angle E}_{Q+\vec u}+\underbrace{\angle M}_{\text{gap }OMPR}+\underbrace{\angle S}_{\text{other gap}}=360^\circ \]so nothing overlaps.Answer: (i) the slide by \(\displaystyle \overrightarrow{EO} \) carries \(\displaystyle SOME \) onto \(\displaystyle S'O'M'E' \); (ii) the gaps are \(\displaystyle SOME \) turned through \(\displaystyle 180^\circ \).

(i) Draw \(\displaystyle PQ=c \); arcs of radius \(\displaystyle a \) about \(\displaystyle P \) and \(\displaystyle b \) about \(\displaystyle Q \) meet at \(\displaystyle R \) exactly when \(\displaystyle a+b>c \).\[a\le b\le c:\quad a+b>c\ \Rightarrow\ a+c>b,\ \ b+c>a \]\[(4,5,7):\ 4+5=9>7\ \text{(triangle)};\qquad (2,3,5):\ 2+3=5\ \text{(arcs only touch: no triangle)} \](ii) By the rule in (iii), the longest side must be less than the sum of the other three.\[x=100:\ 100<2+5+11=18\ \text{false: no} \]\[x=10:\ 11<2+5+10=17\ \text{true: yes} \]\[x=1:\ 11<2+5+1=8\ \text{false: no} \]\[x\le11:\ 11<7+x\Rightarrow x>4;\qquad x>11:\ x<18\qquad\Rightarrow\qquad 4<x<18 \](iii) Necessary: a straight segment is the shortest path. Sufficient: take a diagonal \(\displaystyle x \) in the range below; triangles \(\displaystyle (a,b,x) \) and \(\displaystyle (c,d,x) \) exist by (i); glue them on opposite sides of \(\displaystyle x \).\[d=AD<AB+BC+CD=a+b+c\qquad(d\text{ longest; strict, as no three vertices are collinear}) \]\[\max(|a-b|,\ d-c)<x<\min(a+b,\ c+d)\ \text{ is non-empty: } d-c<a+b,\ \ |a-b|<d<c+d \]\[\max(a,b,c,d)<\tfrac12(a+b+c+d) \]Answer: (i) a triangle exists iff \(\displaystyle a+b>c \); (ii) $\displaystyle 100$: no, $\displaystyle 10$: yes, $\displaystyle 1$: no, and \(\displaystyle 4<x<18 \); (iii) the longest side must be less than the sum of the other three.

(i) A diagonal joins two non-adjacent vertices; \(\displaystyle D(n) \) counts them. Each vertex gives \(\displaystyle n-3 \) diagonals (not itself, not its two neighbours), and each diagonal is counted from both ends.\[\begin{array}{c|cccccc} n & 3&4&5&6&7&8\\ \hline D(n) & 0&2&5&9&14&20 \end{array}\qquad \text{steps: } +2,+3,+4,+5,+6 \]\[D(7)=\frac{7\cdot4}{2}=14,\qquad D(8)=\frac{8\cdot5}{2}=20 \](ii) Insert a new vertex \(\displaystyle V \) between adjacent vertices \(\displaystyle A,B \): \(\displaystyle AB \) becomes a diagonal and \(\displaystyle V \) joins the other \(\displaystyle n-2 \) vertices, so \(\displaystyle n-1 \) diagonals are added.\[D(n+1)=D(n)+(n-1) \]\[\frac{n(n-3)}{2}+(n-1)=\frac{n^2-n-2}{2}=\frac{(n+1)(n-2)}{2}=\frac{(n+1)\big((n+1)-3\big)}{2} \]With \(\displaystyle D(3)=0 \), the formula holds for every \(\displaystyle n \).Answer: \(\displaystyle D(5)=5,\ D(6)=9,\ D(7)=14,\ D(8)=20 \); \(\displaystyle D(n)=\dfrac{n(n-3)}{2} \), and \(\displaystyle n-1 \) diagonals are added going from \(\displaystyle n \) to \(\displaystyle n+1 \) sides.