Class 9 Maths Ganita Manjari Chapter 12 Solutions — Quadrilaterals

(i) The sides adjacent to AB are BC and DA; its opposite side is CD. (ii) The angles adjacent to ∠A are ∠B and ∠D; its opposite angle is ∠C. (iii) Opposite sides have no common endpoint. Opposite angles occur at vertices not joined by a side; adjacent angles occur at the two endpoints of one side.

At the chosen vertex, draw a sufficiently small circle that meets only the two sides incident at that vertex. Those sides divide the small neighbourhood into two sectors. The internal angle is the angle of the sector lying inside the quadrilateral. At a concave vertex it is the reflex sector, greater than 180°. “Always take the smaller angle” is incorrect. The opposite vertex helps determine the polygon’s interior, but simply choosing the sector that contains that vertex is not a valid rule in every concave quadrilateral.

AB ∥ DC: no. The two parallel sides lie on distinct parallel lines. In a simple boundary ABCD, the two joining sides cannot cross, so the endpoints occur in the order giving a convex trapezium. A reflex corner would force a crossing or a degeneracy.

$\mathrm{AB}=\mathrm{CD}$: yes. Take A(0,0), B(5,0), C(5,6), D(2,2). D lies inside triangle ABC, so ABCD is concave at D; $\mathrm{AB}=5$ and $\mathrm{CD}=\sqrt{3^{2}+4^{2}}=5$.

$\angle A=\angle C$: yes. Take A(−2,0), B(0,1), C(2,0), D(0,3). Reflection in the vertical axis exchanges A and C and fixes B and D. Thus their internal angles are equal, although B is a reflex vertex.

An affine change of coordinates preserves crossings and convexity, so take $A=\left(0,0\right)$, $B=\left(1,0\right)$, $C=\left(0,1\right)$. The lines are $x=0$, $y=0$ and $x+y=1$. For $D=\left(x,y\right)$, excluding these lines, the complete classification is:

Thus one region gives convex figures, four give concave figures, and two give crossed figures. In each concave case the named vertex lies inside the triangle formed by the other three.

No, under the assumption that no three vertices are collinear. If opposite sides intersect, their two intersecting lines determine a plane containing all four endpoints. The entire four-sided boundary therefore lies in that plane.

(i) True. One right angle in a parallelogram makes both adjacent angles 90° and its opposite angle 90°, so it is a rectangle. (ii) False. Every rhombus has perpendicular diagonals, including a rhombus with angles 60°, 120°, 60°, 120°. It need not be a square. (iii) True. In parallelogram ABCD with $\mathrm{AC}=\mathrm{BD}$, triangles DAB and CBA have $\mathrm{DA}=\mathrm{CB}$, AB common and $\mathrm{DB}=\mathrm{CA}$. SSS gives $\angle \mathrm{DAB}=\angle \mathrm{CBA}$. Adjacent angles are supplementary, so both are 90°; ABCD is a rectangle.

Given $\angle \mathrm{DAC}=\angle \mathrm{CAB}$, use AD ∥ BC and AB ∥ DC to get $\angle \mathrm{BCA}=\angle \mathrm{ACD}$. Thus AC also bisects ∠C. Moreover $\angle \mathrm{CAB}=\angle \mathrm{BCA}$, so $\mathrm{AB}=\mathrm{BC}$ in triangle ABC. Opposite sides of a parallelogram are equal; hence all four sides are equal and ABCD is a rhombus.

(i) Yes. Triangles ABC and ADC are congruent by ASA (halves of A and C, AC common), so $\mathrm{AB}=\mathrm{AD}$ and $\mathrm{BC}=\mathrm{CD}$. Similarly triangles BAD and BCD give $\mathrm{AB}=\mathrm{BC}$ and $\mathrm{AD}=\mathrm{CD}$. All four sides are equal.

(ii) Yes. Bisecting diagonals first establish a parallelogram. If they are also perpendicular, right triangles AOB and COB have $\mathrm{AO}=\mathrm{CO}$, OB common and the included angles equal. Thus $\mathrm{AB}=\mathrm{BC}$, making the parallelogram a rhombus.

(iii) No. An isosceles trapezium can have equal diagonals without being a rectangle. For example A(−2,0), B(2,0), C(1,2), D(−1,2) has both diagonals $\sqrt{13}$ and unequal parallel sides. Adding the condition that the diagonals bisect each other makes it a rectangle.

Let E be the intersection of the bisectors of A and B, F of B and C, G of C and D, and H of D and A. Adjacent angles of the original parallelogram sum to 180°, so their halves sum to 90°. In triangle AEB, $\angle \mathrm{AEB}=90^\circ$, and the same reasoning applies cyclically. Therefore EFGH is a rectangle. The condition $\mathrm{AB}\ne \mathrm{BC}$ excludes a rhombus, in which the bisectors coincide with the diagonals and all four named intersections collapse to the same point.

(i) Each of the four small triangles APQ, PBR, QRC and PQR has side lengths $\frac{\mathrm{AB}}{2}$, $\frac{\mathrm{BC}}{2}$ and $\frac{\mathrm{AC}}{2}$, by the midpoint theorem. They are congruent by SSS; in particular △PQR ≅ △QPA with the stated correspondence.

(ii) Given PQR, draw through P a line parallel to QR, through Q a line parallel to PR, and through R a line parallel to PQ. Their pairwise intersections form A, B and C: A is the intersection of the first two, B of the first and third, and C of the second and third. The resulting parallelogram relations show $\mathrm{AP}=\mathrm{PB}$, $\mathrm{AQ}=\mathrm{QC}$ and $\mathrm{BR}=\mathrm{RC}$. Thus PQR is exactly the midpoint triangle of ABC.

Let $X=\mathrm{MN}$ ∩ AD. Since M and N are midpoints, MN ∥ BC, and therefore MX ∥ BD. Apply the converse midpoint theorem in triangle ABD: M bisects AB, so X bisects AD. The endpoint cases $D=B$ or $D=C$ reduce to the given midpoints M or N.

Let E be the midpoint of AD. In triangles ADC and ADB, EG ∥ DC with $\mathrm{EG}=\frac{\mathrm{DC}}{2}$, and EH ∥ AB with $\mathrm{EH}=\frac{\mathrm{AB}}{2}$. Because AB ∥ DC, E,G,H are collinear. Hence GH ∥ AB and:

If $\mathrm{AB}=\mathrm{CD}$, the original quadrilateral is a parallelogram and $G=H$. A single point does not determine the line GH; this is why the unequal-length assumption is needed.

(i) PQRS is a parallelogram, so its diagonals PR and QS bisect each other. (ii) If $\mathrm{AC}=\mathrm{BD}$, adjacent sides $\mathrm{PQ}=\frac{\mathrm{AC}}{2}$ and $\mathrm{QR}=\frac{\mathrm{BD}}{2}$ are equal, so PQRS is a rhombus and PR ⟂ QS. Conversely, if the diagonals PR and QS of this parallelogram are perpendicular, it is a rhombus; hence $\mathrm{PQ}=\mathrm{QR}$ and $\mathrm{AC}=\mathrm{BD}$. The converse is true.

(i)–(ii) Use the original quadrilateral as reference. Copy the position of A relative to the copied, labelled parallelogram by transferring triangle PSA with its lengths and its correct side of PS. Call the corresponding point A′. Reflect A′ through P to obtain B′; reflect B′ through Q to obtain C′; reflect C′ through R to obtain D′. Equivalently, transfer the entire reference construction by the rigid motion used to copy PQRS. This guarantees congruence, rather than merely matching side midpoints.

To check the final side, write point positions as vectors. Since $P+R=Q+S$:

Thus S is the midpoint of A′D′, as required; the other three midpoint conditions follow directly from the construction. Each corresponding triangle is carried by the same rigid motion, so A′B′C′D′ is congruent to ABCD. Different placements, such as a half-turn about the centre with the corresponding cyclic relabelling, give other congruent copies.

Important: the four midpoint positions alone do not uniquely determine the original shape. Choosing A′ arbitrarily and making the three reflections still gives those midpoints, but generally produces a non-congruent quadrilateral. The instruction to recreate a congruent copy requires the original reference measurements and orientation.

(iii) PQRS is a square exactly when its adjacent sides are equal and perpendicular. Since they are respectively half of, and parallel to, AC and BD, this is equivalent to $\mathrm{AC}=\mathrm{BD}$ and AC ⟂ BD. Both directions follow.

At an edge-to-edge vertex we would need an integer k with $108k=360$, which is impossible. Even allowing a tile corner to meet the interior of another tile’s side gives $108k+180l=360$ with positive k and non-negative integer l; neither $l=0$ nor $l=1$ permits an integer k. Thus congruent regular pentagons alone cannot tile the plane without gaps or overlaps. Irregular pentagons are a different question.

Sample: take D(0,0), A(3,0), R(1,1), T(0,3), in that order. R is the concave vertex. Make copies using 180° rotations about side midpoints. The same method works, including around the reflex corner. I prefer the half-turn method because a side midpoint gives an exact placement without estimating an angle. This preference is a sample response.

Rotate a copy of triangle ABC through 180° about the midpoint of one side. The original and the copy make a parallelogram. Translate this parallelogram in the directions of its two sides to fill the plane; each parallelogram remains divided into two copies of the original triangle.

In Fig. 12.33 the two endpoints lie on ruled lines four equal intervals apart. Mark where the drawn segment crosses the ruled line halfway between those endpoint lines—two intervals from either end. Parallel equally spaced lines cut equal successive lengths on any fixed transversal, so the two parts of the segment have equal length. Use the endpoint levels, not the number of all visible rules on the paper.

Let $\theta =\angle \mathrm{AED}=\angle \mathrm{BEC}$ at the crossing E. Triangle AED gives $\angle A+\angle D=180^\circ -\theta$, and triangle BEC gives $\angle B+\angle C=180^\circ -\theta$. Therefore:

To obtain 2°, choose the crossing rays so $\theta =179^\circ$. Place A and D at equal distances from E on those rays, and B and C at equal distances on their opposite rays. Each small triangle then has two base angles of 0.5°, so the four vertex angles total 2°. The angles here are the smaller angles between the incident sides, not the interior angles of a simple region.

Let O be the common midpoint of AC and BD. On segment BD, $\mathrm{DP}=\mathrm{BQ}$ implies that P and Q lie symmetrically about O, so $\mathrm{OP}=\mathrm{OQ}$. Hence the diagonals AC and PQ of APCQ bisect each other and APCQ is a parallelogram. As intended in the figure, P and Q are distinct; if both equal O the named figure degenerates and is not a quadrilateral.

The crease that takes A to B is the perpendicular bisector of AB. Let it meet AB at M and AC at N. Since $\angle \mathrm{ABC}=90^\circ$, both the crease and BC are perpendicular to AB, so MN ∥ BC. M is the midpoint of AB; the converse midpoint theorem then makes N the midpoint of AC.

(i) Let M be the centroid and P,Q,R the side midpoints. Pair the two small triangles adjacent to each side. For the pair MPB and MPC, rotate MPC by 180° about P, taking C to B and M to M′. Now M,P,M′ are collinear, P lies between M and M′, and the union is triangle BMM′. Its side lengths are BM, CM and $2\mathrm{PM}=\mathrm{AM}$. The other two pairs similarly produce triangles with sides AM, BM and CM. All three assembled triangles are congruent by SSS.

(ii) Put the origin at M and write the position vectors of A,B,C as a,b,c, so $a+b+c=0$. One assembled triangle has vertices 0,b,−a. Its median lengths are half the lengths |b−a|, |b−c| and |a−c|, namely $\frac{\mathrm{AB}}{2}$, $\frac{\mathrm{BC}}{2}$ and $\frac{\mathrm{AC}}{2}$. Applying part (i) again makes each new side two-thirds of one of those medians. Consequently all nine final triangles have side lengths:

(iii) One valid alternative construction: choose a longest side as horizontal base and let the altitude be h. Draw the parallel at height $\frac{h}{2}$. Cut off the top triangle and split it along its altitude into two right triangles. Rotate each small right triangle through 180° about the midpoint of its original sloping side. These fill the two missing corners of the lower trapezium, forming a rectangle of base length BC and height $\frac{h}{2}$. Cut this rectangle along a diagonal, subdividing any pieces it crosses. Its two halves are congruent right triangles. This is a finite straight-cut dissection; it need not use the exact hint’s cuts.

(i) Let $M=\mathrm{EF}$ ∩ BD. In triangle DAB, E is the midpoint of DA and EM ∥ AB; hence M bisects DB and $\mathrm{EM}=\frac{\mathrm{AB}}{2}$. In triangle BDC, M bisects BD and MF ∥ DC; hence F bisects BC and $\mathrm{MF}=\frac{\mathrm{CD}}{2}$. E,M,F occur in that order, so:

(ii) If E and F are midpoints, take M as the midpoint of BD. The midpoint theorems in DAB and BDC give EM ∥ AB and FM ∥ DC. Since the bases are parallel, these are the same line through M, so EF ∥ AB. This direct proof also gives the length formula.

In triangles AOP and COQ, $\mathrm{AO}=\mathrm{CO}$, $\angle \mathrm{AOP}=\angle \mathrm{COQ}$ (vertically opposite), and $\angle \mathrm{OAP}=\angle \mathrm{OCQ}$ (alternate angles, AB ∥ CD). ASA gives $\mathrm{OP}=\mathrm{OQ}$. Thus O bisects PQ. The half-turn symmetry of a parallelogram gives the same result immediately.

If the distance between the parallel sides is h, each smaller trapezium has height $\frac{h}{2}$. Therefore:

(i) Ignoring a change of starting vertex or reversal of direction, there are (4−1)!/$2=3$ cyclic boundaries. (ii) If all four points lie on their convex hull, exactly one boundary is convex and the other two are self-intersecting. If one point lies inside the triangle formed by the others, all three boundaries are simple and concave. There is then no convex or crossed quadrilateral. The no-three-collinear assumption excludes boundary degeneracies.

Here $\mathrm{AP}=1$ and $\mathrm{AQ}=5$. Equal divisions on AB cut by parallels to BC give equal corresponding divisions on AC. To justify this without assuming general similarity, draw from a division point on AB a line parallel to AC up to the next parallel; the resulting parallelogram transfers the AC increment, and ASA between successive small triangles makes those increments equal.

(iv) If $\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{m}{n}$ for positive integers m,n, divide AP into m equal steps and PB into n steps of the same length. The parallels through the division points divide AC into m+n equal corresponding steps, of which m form AQ and n form QC. Thus:

This establishes precisely the rational-ratio case requested.

(i) Put $A=0$, $B=u$, $D=v$ and $C=u+v$. Then $M=\frac{u}{2}$ and $N=v+\frac{u}{2}$. At the intersection of DM with AC write $t\left(u+v\right)=v+s\left(\frac{u}{2}-v\right)$. Comparing the two independent directions gives $t=\frac{s}{2}$ and $t=1-s$, hence $t=\frac{1}{3}$. At BN ∩ AC the same calculation gives $t=\frac{2}{3}$. Thus DM and BN trisect AC.

(ii) Method 1: make PQ a diagonal of any parallelogram by choosing B off PQ and taking D so the diagonal midpoints coincide. Use the midpoint construction in (i). Method 2: draw any auxiliary ray from P, mark three equal steps, join its third mark to Q, and draw parallels through the first two marks; Question 11 proves trisection. Method 3: choose B and C with Q as midpoint of BC. In triangle PBC the centroid M lies two-thirds of the way from P to Q. Bisect PM to obtain the first trisection point; M is the second.

(i) The midpoint theorem applies to triangles ABC, ADC, ABD and CBD regardless of whether the four-sided boundary is concave or crossed. It still gives the two opposite pairs of midpoint sides parallel and equal. (ii) The midpoint parallelogram collapses to a segment exactly when the diagonal directions AC and BD are parallel (allowing a further coincident-point collapse only in more degenerate inputs). Under the chapter’s distinct-vertex/no-three-collinear conditions, the segment case can occur for a crossed figure.

(iii) Yes, it also applies to a skew quadrilateral in space. In vector form, $P=\frac{A+B}{2}$, $Q=\frac{B+C}{2}$, $R=\frac{C+D}{2}$ and $S=\frac{D+A}{2}$, so:

The midpoint diagonals have the same midpoint, and PQRS lies in the plane generated by the two diagonal directions. A genuinely non-planar ABCD cannot have parallel AC and BD, since two parallel lines lie in one plane.

No. In square A(0,0), B(1,0), C(1,1), D(0,1), take $P\left(\frac{1}{4},0\right)$, $Q\left(1,\frac{3}{4}\right)$, $R\left(\frac{3}{4},1\right)$, $S\left(0,\frac{1}{4}\right)$. Opposite side vectors of PQRS are equal, so it is a parallelogram, but none of the four points is a midpoint.

(i) Extend PQ to S with $\mathrm{QS}=\mathrm{PQ}$. Since $\mathrm{AQ}=\mathrm{QC}$, $\mathrm{PQ}=\mathrm{QS}$ and the included angles at Q are vertically opposite, △APQ ≅ △CSQ by SAS. Thus $\mathrm{AP}=\mathrm{CS}$ and AP ∥ CS. Since $\mathrm{AP}=\mathrm{PB}$ and A,P,B are collinear, $\mathrm{PB}=\mathrm{CS}$ and PB ∥ CS, making PBCS a parallelogram. Therefore PS ∥ BC and $\mathrm{PS}=\mathrm{BC}$. As $\mathrm{PS}=2\mathrm{PQ}$, the midpoint theorem follows.

(ii) Given P midpoint of AB and PQ ∥ BC, choose S on the ray PQ so $\mathrm{PS}=\mathrm{BC}$. Then PBCS is a parallelogram, so $\mathrm{CS}=\mathrm{PB}=\mathrm{AP}$ and CS ∥ AP. Triangles APQ and CSQ have equal angles at A and C, equal angles at P and S, and $\mathrm{AP}=\mathrm{CS}$. ASA yields $\mathrm{AQ}=\mathrm{QC}$. Thus Q is the midpoint of AC; the first part then also gives $\mathrm{PQ}=\frac{\mathrm{BC}}{2}$.

This is open-ended. Each row states all needed assumptions, avoiding a false converse made by dropping “parallelogram”.

Use AC as a hinge. At A, the smaller spatial angle between AB and AD is strictly less than ∠BAC + ∠CAD; at C it is strictly less than ∠BCA + ∠ACD, because the two triangle planes are distinct. Angles B and D remain the triangle angles. Adding and using the two triangle angle sums gives a total strictly less than 360°. The angle inequality follows by comparing the direct arc between two rays on a small sphere with the broken arc through the diagonal ray.

An exact 2° example (extension): take a skew quadrilateral with four equal sides L and equal opposite diagonals d, chosen so an isosceles triangle with sides L,L,d has vertex angle 0.5°. Such a spatial configuration exists: use A(−a,−t,0), C(−a,t,0), B(a,0,−t), D(a,0,t), with $a>0$ and a suitably small positive t. Here all four side lengths are $\sqrt{4a^2+2t^2}$ and both diagonals are 2t. Choosing $t^2=2a^2(\sec0.5^\circ-1)$ makes every vertex angle 0.5°, hence their sum 2°. This last coordinate verification uses trigonometry beyond the main chapter; the equal-side construction explains the geometry.

(i) The shift vector is $E'-E=O-E$. The four corresponding side quadrilaterals SS′O′O, OO′M′M, MM′E′E and EE′S′S are parallelograms: each has opposite sides equal and parallel under that same shift. Hence $S'-S=O'-O=M'-M=E'-E$, proving that one translation carries every vertex and the whole boundary onto its counterpart. When two corresponding sides are collinear, use the equal directed-segment statement instead of calling a degenerate figure a parallelogram.

(ii) Translate the original shape by integer multiples of its two diagonal vectors. The nine choices with each multiplier −1, 0 or 1 give the requested patch. Fill the intervening regions with half-turned copies. The proof that these are exactly the gaps, rather than an assumption based on a sketch, is given in Question 23.

(i) With c the longest side, circles centred at the endpoints of a length-c segment with radii a and b meet off that segment exactly when $a+b>c$ and |a−b| < c. For positive a ≤ b ≤ c, the second inequality is automatic, so the essential test is a+b > c.

(ii) If the fourth length is x, it must be shorter than $2+5+11=18$, and 11 must be shorter than 2+5+x, giving $x>4$. Thus:

So 100 is impossible, 10 is possible, and 1 is impossible. Endpoints 4 and 18 give a degenerate straight figure, not a quadrilateral.

(iii) For four positive lengths, the longest must be strictly less than the sum of the other three. Necessity follows by joining its endpoints along the other three sides. For sufficiency, choose a diagonal length x common to the two open intervals (|a−b|,a+b) and (|c−d|,c+d); the longest-side condition ensures they overlap. Construct the two triangles on opposite sides of the diagonal. Choose x away from the finitely many collinear cases. Their union has the required simple four-sided boundary.

(i) A diagonal joins two non-adjacent vertices. Its segment need not lie wholly inside a concave polygon.

(ii) From each vertex there are n−3 diagonals; counting over all vertices counts each diagonal twice.

Inserting one new vertex into a side adds n−2 diagonals from that vertex to old vertices, and makes the old side into one further diagonal. Thus the increase is n−1, consistent with:

Any simple planar n-gon can be divided by non-crossing internal diagonals into n−2 triangles. For a convex polygon, draw a fan from one vertex. For a concave polygon, use an internal diagonal and continue recursively; a single fan from an arbitrary vertex may leave the polygon. Adding the triangle angles gives:

For $n=3$,4,5,6,7,8 the sums are 180°, 360°, 540°, 720°, 900°, 1080°. Reflex internal angles are included for concave polygons.

(i) Suppose P lies on AB and Q on AC, PQ ∥ BC and $\mathrm{PQ}=\frac{\mathrm{BC}}{2}$. Extend PQ to S with $\mathrm{PS}=\mathrm{BC}$. PBCS is a parallelogram. Since Q bisects PS, triangles APQ and CSQ have equal angles at P and S, equal vertically opposite angles at Q, and $\mathrm{PQ}=\mathrm{QS}$; ASA gives $\mathrm{AP}=\mathrm{CS}$ and $\mathrm{AQ}=\mathrm{QC}$. But $\mathrm{CS}=\mathrm{PB}$, so P and Q are both midpoints.

(ii) Keep P as the midpoint of AB. The converse reads: if PQ ∥ BC, then Q is the midpoint of AC. In the named conditions, P MID + PRLL imply Q MID + HALF. Exchanging the names of the two sides gives the symmetric version.

(iii) The remaining essentially different choice is P MID + HALF. It does not imply Q MID or PRLL. Take A(0,0), B(4,0), C(4,2), P(2,0), $Q\left(\frac{6}{5},\frac{3}{5}\right)$. Then Q lies three-tenths of the way from A to C, while:

Q is not the midpoint (2,1), and PQ is not parallel to the vertical side BC. A circle centred at P can meet AC in two different points, which is the source of this counterexample.

Challenge proof using translations. Let K be the region of the simple quadrilateral ABCD, let $u=C-A$ and $v=D-B$ be its diagonal vectors, and let H(K) be its half-turn about the midpoint of AB. Consider all regions K + iu + jv and H(K) + iu + jv for integers i,j. The two diagonal vectors are independent for a non-degenerate simple quadrilateral, and their lattice cell has area twice the area of K.

Check the boundary matches. The AB side of K is matched by the reversed AB side of H(K). The BC side is matched by H(K)+u; the CD side by H(K)+u+v; the DA side by H(K)+v. These statements follow by writing the half-turn as X ↦ A+B−X. Thus every edge is paired with an equal, oppositely directed edge. All copies retain the same orientation, so their interiors lie on opposite sides of a shared edge.

For a point away from all edges, count how many copies contain it. The family is locally finite. Crossing a generic edge removes one copy and adds one copy, so the count cannot change; a path can avoid the isolated vertices. The count is therefore constant throughout the plane away from boundaries. Its average over one lattice cell equals the total area of the two representative copies divided by the cell area:

The constant count is therefore exactly one. This proves that there are no uncovered open regions and no overlapping interiors; shared boundaries complete the covering. The argument also handles a concave K, using its ordinary positive area. Method 1 generates this family through its edge half-turns. Method 3 lists the translations of K and then the H(K) gaps. Method 2 uses the equivalent midpoint-parallelogram grid, with the reference shape transferred consistently as in Exercise 12.3 Question 5. All three therefore describe the same tiling. This global argument is stronger than checking only the 360° angle sum at one vertex.

Use a unit square with bottom-left (0,0), bottom-right (1,0), top-right (1,1) and top-left (0,1). The marked points are side midpoints. The four oblique boundary lines are:

Their consecutive intersections are $P\left(\frac{1}{5},\frac{3}{5}\right)$, $Q\left(\frac{2}{5},\frac{1}{5}\right)$, $R\left(\frac{4}{5},\frac{2}{5}\right)$, $S\left(\frac{3}{5},\frac{4}{5}\right)$. Adjacent sides are perpendicular, and all have the same length. For example:

The shaded figure is a square, so its area is $\mathrm{PQ}^{2}=\frac{1}{5}$ of the area of the unit square. The required fraction is one-fifth.