Class 9 Maths Ganita Manjari Chapter 5 Solutions — I’m Up and Down, and Round and Round

Detailed construction: draw $\mathrm{AB}=5\,\mathrm{cm}$. At A draw a ray making 70° with AB; at B draw a ray making 60° with BA on the same side of AB. Their intersection is C. For each of two sides, draw equal-radius compass arcs from its endpoints (radius greater than half the side), and join the arc intersections to make its perpendicular bisector. Let the two bisectors meet at O. Set the compass to OA and draw the circle.

Draw $\mathrm{AB}=5\,\mathrm{cm}$. At A construct a 100° ray and mark C on it with $\mathrm{AC}=4\,\mathrm{cm}$. Join BC. Construct perpendicular bisectors of AB and AC; their intersection O is outside the triangle. Draw the circle with centre O and radius OA.

Draw $\mathrm{AB}=6\,\mathrm{cm}$; use arcs of radius 7 cm from A and B to locate C. Join AC,BC and construct two perpendicular bisectors. For an elementary exact check, let M be the midpoint of AB. Then $\mathrm{CM}=\sqrt{7^{2}-3^{2}}=2\sqrt{10}$. Write $\mathrm{OM}=t$ toward C. $\mathrm{OA}^{2}=t^{2}+9$ and $\mathrm{OC}^{2}=\left(2\sqrt{10}-t\right)^{2}$. Equating gives $4\sqrt{10}\cdot t=31$, so $R=\mathrm{OC}=\frac{49}{4\sqrt{10}}\approx 3.874\,\mathrm{cm}$. Measurements on paper should be close to this, rather than treated as exact.

This assumes AB is not a diameter; a diameter and the centre are collinear and do not form a non-degenerate triangle.

The triangles must belong to the same circle, or to circles with equal radii; equal bases alone do not suffice across unequal circles.

(i) No. With X,Y on the circle and on the same side of AB, the same-segment theorem gives $\angle \mathrm{AXB}=\angle \mathrm{AYB}$. The vertices must be distinct from A,B.

(ii) Interpreting the printed “same side of the circle” as “same side of AB”: not always. If AB is a diameter, points on opposite semicircles both subtend 90°. For a non-diameter chord, equal angles at points of the same circle do require the same side: opposite-side angles are supplementary and can be equal only when both are 90°.

(iii) Not in general. If X,Y lie on the same side of AB and the angles are non-degenerate, the converse theorem guarantees concyclicity. Without that condition, consider A(−1,0), B(1,0), X(0,2), Y(0,−2). Reflection gives equal angles at X,Y. The circle through A,B,X has centre $\left(0,\frac{3}{4}\right)$, radius $\frac{5}{4}$, but Y is $\frac{11}{4}$ from that centre, so Y is not on it. X and Y are also both outside the original unit circle through A,B.

The vertex must lie on the remaining part of the circle, outside the 70° arc and away from its endpoints. This condition is needed for the half-angle rule.

For the kite reasoning, label consecutive sides $\mathrm{AB}=\mathrm{BC}=5$ and $\mathrm{CD}=\mathrm{DA}=12$. Triangles ABD and CBD are congruent by SSS, so $\angle A=\angle C$; cyclicity makes each 90°. Each area is $\left(\frac{1}{2}\right)\cdot 5\cdot 12=30$. If the side order is instead 5,12,5,12, the cyclic parallelogram is a $5\times 12$ rectangle, with the same area 60.

Construct the perpendicular bisectors of two non-parallel sides. Their intersection is the circumcentre O; you do not need to draw the circle. Check whether O is inside the quadrilateral, outside it, or on a side. For a convex quadrilateral, O is inside exactly when it lies strictly on the inward side of every edge.

An angle-only claim that every interior angle must be less than 90° is false: their sum is 360°, and many cyclic quadrilaterals with obtuse angles contain their centre. If a side is a diameter, O lies on the boundary; this is a third case, not “inside” or “outside”.

Let equal chords AB and CD meet at P inside the circle. Let E,F be their midpoints. Since the chords are equal, $\mathrm{OE}=\mathrm{OF}$; also OE⊥AB and OF⊥CD. Right triangles OEP and OFP have common hypotenuse OP and equal legs OE,OF, so $\mathrm{PE}=\mathrm{PF}$ by RHS.

Put $\mathrm{AE}=\mathrm{BE}=\mathrm{CF}=\mathrm{DF}=h$ and $\mathrm{PE}=\mathrm{PF}=t$. Each chord is split by P into lengths h+t and h−t. Thus the two segment lengths agree as an unordered pair, with the correspondence determined by the endpoint labels. If both chords are diameters, $P=O$ and all four segments equal the radius.

For an exact compass construction, first draw $\mathrm{AB}=6\,\mathrm{cm}$ and bisect it at M. At M erect a perpendicular and mark O with $\mathrm{OM}=3\,\mathrm{cm}$. Set the compass directly to OA (rather than the rounded decimal 4.24), then draw the circle centred at O. $\mathrm{OA}=\mathrm{OB}=3\sqrt{2}\,\mathrm{cm}$.

For a fixed chord length $0<L<2r$, every midpoint M satisfies $\mathrm{OM}=\sqrt{r^{2}-\frac{L^{2}}{4}}=d$. Conversely, through any point at distance d from O, the line perpendicular to OM cuts a chord of length L. Thus the full locus is the concentric circle of radius d.

If $L=2r$, the chords are diameters and all midpoints coincide at O, so the locus is a single point. If $L>2r$, no such chords exist.

In triangles OAB and OAC, OA is common, $\mathrm{OB}=\mathrm{OC}$ (radii) and $\mathrm{AB}=\mathrm{AC}$ (given). By SSS, ΔOAB≅ΔOAC. Hence $\angle \mathrm{BAO}=\angle \mathrm{OAC}$, so AO bisects ∠BAC. This directly identifies the internal angle bisector.

The requested angles ∠MOP and ∠MNP both subtend chord MP. Since N,O lie on the same side of MP in the stated cyclic order, $\angle \mathrm{MOP}=\angle \mathrm{MNP}$. Their common value is not determined by the information given.

The diameter condition instead gives $\angle \mathrm{MON}=90^\circ$ and $\angle \mathrm{MPN}=90^\circ$. In triangle MOP, $\angle \mathrm{MOP}=90^\circ -\angle \mathrm{NOP}$; in triangle MNP, $\angle \mathrm{MNP}=90^\circ -\angle \mathrm{NMP}$, and $\angle \mathrm{NOP}=\angle \mathrm{NMP}$ because they subtend NP. These facts confirm the equality but do not make the two requested angles right angles.

Wording correction: for ∠CDE to be the exterior angle at D, E must lie on AD extended beyond D. If CD is extended beyond D as the textbook says, ∠CDE is a straight angle and the stated equality does not generally hold.

With A,D,E collinear in that order, $\angle \mathrm{CDE}+\angle \mathrm{CDA}=180^\circ$. Also $\angle \mathrm{ABC}+\angle \mathrm{CDA}=180^\circ$ because ABCD is cyclic. Subtracting gives $\angle \mathrm{CDE}=\angle \mathrm{ABC}$.

To justify the maximum distance, let M be the foot from O to any chord line through A. Then $\mathrm{OA}^{2}=\mathrm{OM}^{2}+\mathrm{AM}^{2}$, so $\mathrm{OM}\le \mathrm{OA}$, with equality exactly when $M=A$. If $A=O$, every chord through A is a diameter of the same length; there is no unique shortest chord and OA has no direction.

Name the unlabelled diameter endpoints in Fig. 5.30 L and R, keeping the top vertex A as in the textbook. Since $\mathrm{OL}=\mathrm{OA}=\mathrm{OR}$, triangles OLA and ORA are isosceles. The angles at L and R are a and b, so the corresponding parts at A are also a and b. In triangle LAR, $a+b+\left(a+b\right)=180^\circ$; hence $\angle \mathrm{LAR}=a+b=90^\circ$.

Take diameter AB as the x-axis. The perpendicular chords have endpoints C(u,c), C′(u,−c), D(v,d), D′(v,−d). Their midpoints are $M\left(\frac{u+v}{2},\frac{c+d}{2}\right)$ and $M'\left(\frac{u+v}{2},-\frac{c+d}{2}\right)$. Their x-coordinates are equal, so MM′ is vertical and hence perpendicular to AB, provided $M\ne M'$.

The usual diagram labels C,D on the same side of AB, giving c,$d>0$ and distinct midpoints. If an alternative endpoint labelling makes $c+d=0$, $M=M'$ and no joining line is defined; that degenerate case must be excluded.

Use the labels in Fig. 5.31. Since $\mathrm{OA}=\mathrm{OB}=\mathrm{OC}=\mathrm{OD}$, the base-angle pairs are p in triangle AOB, q in BOC, u in COD and v in DOA. The four quadrilateral angles are p+v, p+q, q+u, u+v. Their sum is $2\left(p+q+u+v\right)=360^\circ$, so $p+q+u+v=180^\circ$. Each opposite pair sums to exactly this value.