SolveItClass 9 · NCERT

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

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Exercise Set 5.2 1–2 (part 2 of 9)

  1. Exercise 1

    Show that the triangle formed by a chord and the centre of the circle is isosceles.

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    This solution has not been cross-checked against the answer printed in NCERT.

    Two of its sides are radii of the same circle.Let the circle have centre C and radius \(\displaystyle r\), and let AB be a chord of it. Joining the two ends of the chord to the centre gives \(\displaystyle \triangle \mathrm{CAB}\).A and B are points on the circle. By the very definition of a circle — the set of points at a fixed distance from the centre — every point of the circle is at distance \(\displaystyle r\) from C. So \[\mathrm{CA}=r \quad\text{and}\quad \mathrm{CB}=r, \qquad\text{hence}\qquad \mathrm{CA}=\mathrm{CB}. \]A triangle with two equal sides is isosceles.Because the two equal sides are CA and CB, the base is the chord AB, and the base angles are equal: \[\angle \mathrm{CAB}=\angle \mathrm{CBA}. \]One thing to be careful about: if the chord AB happens to be a diameter, then C lies on AB, the three points A, C, B are collinear and there is no triangle at all. So the statement is about a chord that is not a diameter.Every chord of a circle, together with the centre, forms an isosceles triangle — the two radii are its equal sides and the chord is its base.

  2. Exercise 2

    Show that if two such isosceles triangles (occurring in the previous question) have equal base length, they are congruent to each other.

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    This solution has not been cross-checked against the answer printed in NCERT.

    SSS congruence — the four radii give two pairs of equal sides for free.Let AB and DE be two chords of the same circle, with centre C and radius \(\displaystyle r\). By the previous question, \(\displaystyle \triangle \mathrm{CAB}\) and \(\displaystyle \triangle \mathrm{CDE}\) are both isosceles, with bases AB and DE. "Equal base length" means \[\mathrm{AB}=\mathrm{DE}. \]Now compare \(\displaystyle \triangle \mathrm{CAB}\) and \(\displaystyle \triangle \mathrm{CDE}\) side by side:
    In \(\displaystyle \triangle \mathrm{CAB}\) and \(\displaystyle \triangle \mathrm{CDE}\)Reason
    \(\displaystyle \mathrm{CA}=\mathrm{CD}\)both are radii, each equal to \(\displaystyle r\)
    \(\displaystyle \mathrm{CB}=\mathrm{CE}\)both are radii, each equal to \(\displaystyle r\)
    \(\displaystyle \mathrm{AB}=\mathrm{DE}\)given (equal bases)
    All three pairs of corresponding sides are equal, so by the SSS congruence criterion, \[\triangle \mathrm{CAB}\cong\triangle \mathrm{CDE}. \]Since corresponding parts of congruent triangles are equal, this also gives \(\displaystyle \angle \mathrm{ACB}=\angle \mathrm{DCE}\) — which is exactly Theorem $\displaystyle 2$, that equal chords subtend equal angles at the centre.A word of caution: the argument uses \(\displaystyle \mathrm{CA}=\mathrm{CD}\) and \(\displaystyle \mathrm{CB}=\mathrm{CE}\), and that needs the two chords to belong to the same circle (or to two circles of the same radius). Equal chords in circles of different sizes give triangles that are certainly not congruent.The two isosceles triangles are congruent, by SSS.