NCERT Solutions · Class 9 Mathematics I’m Up and Down, and Round and Round

Construct the triangle, then read off the third angle.Step $\displaystyle 1$ — draw the triangle. Draw \(\displaystyle \mathrm{AB}=5\ \mathrm{cm}\). At A draw a ray making \(\displaystyle 70^\circ\) with AB, and at B a ray making \(\displaystyle 60^\circ\) with AB, both rays on the same side of AB. The two rays meet at C. (They must meet: \(\displaystyle 70^\circ+60^\circ=130^\circ<180^\circ\).)Step $\displaystyle 2$ — draw the circumcircle. The centre has to be equidistant from A, B and C. A point equidistant from A and B lies on the perpendicular bisector of AB; a point equidistant from B and C lies on the perpendicular bisector of BC. So construct those two perpendicular bisectors with compass and straightedge — they cross at a single point O, the circumcentre. With O as centre and OA as radius, draw the circle. It passes through B and C too, because \(\displaystyle \mathrm{OA}=\mathrm{OB}=\mathrm{OC}\).Step $\displaystyle 3$ — inside or outside? Find the third angle from the angle sum of a triangle: \[\angle \mathrm{C}=180^\circ-70^\circ-60^\circ=50^\circ. \] The three angles are \(\displaystyle 70^\circ,\ 60^\circ,\ 50^\circ\). Every one of them is less than \(\displaystyle 90^\circ\), so \(\displaystyle \triangle \mathrm{ABC}\) is an acute-angled triangle. As the chapter says just after Theorem $\displaystyle 1$ (Fig. $\displaystyle 5.5$), the circumcentre of an acute-angled triangle lies inside the triangle.Check your drawing. Measuring should give a radius of about \(\displaystyle 3.3\ \mathrm{cm}\), with \(\displaystyle \mathrm{BC}\approx 6.1\ \mathrm{cm}\) and \(\displaystyle \mathrm{CA}\approx 5.7\ \mathrm{cm}\). If your \(\displaystyle \mathrm{OA},\ \mathrm{OB},\ \mathrm{OC}\) do not all come out the same, the perpendicular bisectors were drawn inaccurately.The centre lies inside the triangle.

Construct the triangle, then look at the largest angle.Step $\displaystyle 1$ — draw the triangle. Draw \(\displaystyle \mathrm{AB}=5\ \mathrm{cm}\). At A draw a ray making \(\displaystyle 100^\circ\) with AB, and mark C on that ray with \(\displaystyle \mathrm{AC}=4\ \mathrm{cm}\). Join BC. Two sides and the angle between them are given, so this fixes the triangle completely (SAS) — everybody's drawing will be the same shape.Step $\displaystyle 2$ — draw the circumcircle. Construct the perpendicular bisector of AB and the perpendicular bisector of AC. Their point of intersection O is equidistant from A, B and C, so it is the circumcentre. Draw the circle with centre O and radius OA.Step $\displaystyle 3$ — inside or outside? Here \(\displaystyle \angle \mathrm{A}=100^\circ\), which is more than \(\displaystyle 90^\circ\), so \(\displaystyle \triangle \mathrm{ABC}\) is an obtuse-angled triangle. As the chapter states just after Theorem $\displaystyle 1$ (Fig. $\displaystyle 5.6$), the circumcentre of an obtuse-angled triangle lies outside the triangle. In your drawing you will find O on the far side of BC from A — that is, across the side opposite the obtuse angle.Check your drawing. \(\displaystyle \mathrm{BC}\) should measure about \(\displaystyle 6.9\ \mathrm{cm}\) and the radius \(\displaystyle \mathrm{OA}=\mathrm{OB}=\mathrm{OC}\) about \(\displaystyle 3.5\ \mathrm{cm}\), so the diameter is about \(\displaystyle 7.0\ \mathrm{cm}\). Notice how close BC comes to being a diameter: a chord that long has to pass very near the centre, which is what squeezes O out to the other side of it.The centre lies outside the triangle.

Construct with compass arcs; all three distances must come out equal.Step $\displaystyle 1$ — draw the triangle. Draw \(\displaystyle \mathrm{AB}=6\ \mathrm{cm}\). With A as centre draw an arc of radius \(\displaystyle 7\ \mathrm{cm}\), and with B as centre another arc of radius \(\displaystyle 7\ \mathrm{cm}\). They cross at C. (They do cross, since \(\displaystyle 7+7=14>6\).) Join AC and BC. The triangle is isosceles: \(\displaystyle \mathrm{CA}=\mathrm{CB}=7\ \mathrm{cm}\), base \(\displaystyle \mathrm{AB}=6\ \mathrm{cm}\).Step $\displaystyle 2$ — draw the circumcircle. Construct the perpendicular bisectors of AB and of BC. They meet at O, the circumcentre. Draw the circle with centre O passing through A.Step $\displaystyle 3$ — measure OA, OB, OC. All three come out equal, about \(\displaystyle 3.9\ \mathrm{cm}\). They have to be equal — that is what "circumcentre" means: O is the one point equidistant from all three vertices, and that common distance is the radius of the circumcircle. Measuring is a check on how accurately you drew, not a discovery.Where the \(\displaystyle 3.9\) comes from. Let M be the midpoint of AB, so \(\displaystyle \mathrm{AM}=3\ \mathrm{cm}\). Since \(\displaystyle \mathrm{CA}=\mathrm{CB}\), the point C is equidistant from A and B, so C lies on the perpendicular bisector of AB; hence CM is perpendicular to AB, and by the Baudhāyana–Pythagoras theorem in \(\displaystyle \triangle \mathrm{CMA}\), \[\mathrm{CM}=\sqrt{7^{2}-3^{2}}=\sqrt{40}\approx 6.32\ \mathrm{cm}. \] O lies on that same line CM. Write \(\displaystyle R=\mathrm{OA}=\mathrm{OB}=\mathrm{OC}\); then \(\displaystyle \mathrm{OM}=\mathrm{CM}-R\), and \(\displaystyle \triangle \mathrm{OMA}\) is right-angled at M, so \[R^{2}=\mathrm{OM}^{2}+\mathrm{AM}^{2}=\left(\sqrt{40}-R\right)^{2}+9 .\] Expanding, \(\displaystyle R^{2}=40-2\sqrt{40}\,R+R^{2}+9\), so \(\displaystyle 2\sqrt{40}\,R=49\) and \[R=\frac{49}{2\sqrt{40}}\approx 3.87\ \mathrm{cm}. \]Since \(\displaystyle \mathrm{OM}=6.32-3.87\approx 2.45\ \mathrm{cm}\) is positive, O sits between M and C — inside the triangle, as it should for this triangle, whose angles are about \(\displaystyle 64.6^\circ,\ 64.6^\circ,\ 50.8^\circ\), all acute.\(\displaystyle \mathrm{OA}=\mathrm{OB}=\mathrm{OC}\approx 3.9\ \mathrm{cm}\).

The centre can be no closer to A and B than the midpoint of AB.Let a circle with centre O and radius \(\displaystyle r\) pass through A and B, and let M be the midpoint of AB.Since \(\displaystyle \mathrm{OA}=\mathrm{OB}=r\), the centre O is equidistant from A and B, so O lies on the perpendicular bisector of AB — the line through M perpendicular to AB (Section $\displaystyle 5.3$).Case $\displaystyle 1$: \(\displaystyle \mathrm{O}=\mathrm{M}\). Then \(\displaystyle r=\mathrm{OA}=\mathrm{MA}=\tfrac{1}{2}\mathrm{AB}\).Case $\displaystyle 2$: O is some other point of the perpendicular bisector. Then \(\displaystyle \triangle \mathrm{OMA}\) is a genuine right-angled triangle, right-angled at M, so by the Baudhāyana–Pythagoras theorem \[r^{2}=\mathrm{OA}^{2}=\mathrm{OM}^{2}+\mathrm{AM}^{2}=\mathrm{OM}^{2}+\left(\tfrac{1}{2}\mathrm{AB}\right)^{2}>\left(\tfrac{1}{2}\mathrm{AB}\right)^{2}, \] so \(\displaystyle r>\tfrac{1}{2}\mathrm{AB}\).Putting the two cases together, every circle through A and B has \(\displaystyle r\ge \tfrac{1}{2}\mathrm{AB}\), and the value \(\displaystyle \tfrac{1}{2}\mathrm{AB}\) is actually reached — take M as centre and \(\displaystyle \tfrac{1}{2}\mathrm{AB}\) as radius; that circle really does pass through A and B, and AB is a diameter of it.This also matches the picture in Fig. $\displaystyle 5.4$: as the centre slides away from M along the perpendicular bisector, the circles through A and B get bigger and flatter, and the smallest one is the one you get by stopping at M.The least possible radius is \(\displaystyle \tfrac{1}{2}\mathrm{AB}\), half the distance between A and B, and that smallest circle has AB as a diameter.