Class 9 Mathematics · Ganita Manjari Part 1

Class 9 Maths Chapter 5 – Exercise Set 5.1 Solutions

These solutions cover the four construction and reasoning questions in Exercise Set 5.1. Construction measurements may vary slightly depending on the accuracy of the drawing.

Question 1 — Construct a triangle and its circumcircle

Given: $AB=5$ cm, $\angle A=70^\circ$, $\angle B=60^\circ$.

Construction:

  1. Draw $AB=5$ cm.
  2. At $A$, construct a $70^\circ$ angle.
  3. At $B$, construct a $60^\circ$ angle.
  4. The two rays meet at $C$.
  5. Draw perpendicular bisectors of two sides. Their intersection is the circumcentre $O$.
  6. With centre $O$ and radius $OA$, draw the circumcircle.

Now $\angle C=180^\circ-70^\circ-60^\circ=50^\circ$. All three angles are less than $90^\circ$, so $\triangle ABC$ is acute-angled.

Answer: The circumcentre lies inside the triangle.

Question 2 — Circumcircle of an obtuse triangle

Given: $AB=5$ cm, $AC=4$ cm and $\angle A=100^\circ$.

Draw $AB=5$ cm, construct $100^\circ$ at $A$, mark $AC=4$ cm on the second ray and join $BC$. Construct the perpendicular bisectors of two sides to locate the circumcentre $O$, then draw the circle with centre $O$.

Since $\angle A=100^\circ>90^\circ$, the triangle is obtuse-angled.

Answer: The circumcentre lies outside the triangle.

Question 3 — Circumradius by construction

Given: $AB=6$ cm, $BC=7$ cm and $CA=7$ cm.

  1. Draw $AB=6$ cm.
  2. With centres $A$ and $B$ and radius $7$ cm, draw arcs meeting at $C$.
  3. Join $AC$ and $BC$.
  4. Construct perpendicular bisectors to locate $O$.
  5. Draw the circumcircle.

Because $O$ is the circumcentre, $OA=OB=OC$. For this construction the measured value is approximately $4$ cm.

Answer: $OA=OB=OC\approx4$ cm. The exact measured value may vary slightly with drawing accuracy.

Question 4 — Least possible radius through two fixed points

For every circle through fixed points $A$ and $B$, its centre lies on the perpendicular bisector of $AB$. The radius is smallest when the centre is the midpoint of $AB$.

Then $AB$ is the diameter, so

$$r_{\min}=\frac{AB}{2}.$$

Answer: The least possible radius is half the distance $AB$.