NCERT Solutions Class 10 Maths Chapter 11 Constructions Exercise 11.2

The Geometry and Logic Behind Tangent Construction

To successfully construct tangents, we must understand the underlying geometric theorems from Chapter 10 (Circles). A tangent at any point on a circle is perpendicular to the radius through the point of contact. This means if we draw a tangent from an external point $P$ to a circle with center $O$ touching at point $Q$, then $\angle OQP = 90^\circ$.

But how do we locate this exact point $Q$ where the radius and tangent meet at a right angle? We use a classic property of circles: any angle inscribed in a semicircle is a right angle ($90^\circ$). If we construct a second circle with the line segment $OP$ as its diameter, this circle will intersect our original circle at two points, say $Q$ and $R$. Since $OP$ is the diameter of this new circle, the angles $\angle OQP$ and $\angle ORP$ are inscribed in semicircles, making them exactly $90^\circ$. Consequently, $PQ$ and $PR$ are the required tangents to the circle. This elegant logic makes your construction mathematically sound and easy to justify to board examiners.

Step-by-Step Guide: Tangents from an External Point

1. Draw the Given Circle and External Point — Using a sharp pencil and a compass, draw a circle of the given radius with center $O$. Mark the external point $P$ at the given distance from the center $O$. Join $OP$ with a straight line using a ruler.
2. Construct the Perpendicular Bisector of OP — Place the compass needle on point $O$. Set the width to more than half the length of $OP$. Draw arcs above and below the line $OP$. Keeping the same compass width, place the needle on $P$ and draw arcs intersecting the previous ones. Join the points of intersection to find the midpoint $M$ of $OP$.
3. Draw the Intersecting Circle — Place the compass needle at the midpoint $M$. Set the radius equal to $MO$ (or $MP$). Draw a circle (or a semi-circle if preferred) that passes through both $O$ and $P$. Let this circle intersect the original circle at points $Q$ and $R$.
4. Join and Complete the Tangents — Using a ruler, draw a straight line from point $P$ through $Q$, and another from point $P$ through $R$. The lines $PQ$ and $PR$ are the two required tangents. You can measure their lengths; they will always be equal!

CBSE Board Exam Tips & Common Pitfalls

Based on common mistakes highlighted by CBSE board examiners, keep these crucial tips in mind when attempting constructions ex 11 2 class 10 ncert:

• Keep Arcs Thin and Distinct: Use a well-sharpened lead pencil. Double-drawn, thick, or overlapping arcs lead to deduction of marks for neatness and precision.
• Write the Steps of Construction: Even if the question does not explicitly ask for them, always write brief, clear steps of construction (3 to 5 lines). This ensures you get partial marking even if your final drawing has minor measurement deviations.
• Provide Justification When Asked: If the question states "Give the justification of the construction", you must prove why the line is a tangent. Simply join $OQ$ and show that $\angle OQP = 90^\circ$ (angle in a semicircle), hence $PQ \perp OQ$. Since $OQ$ is a radius, $PQ$ must be a tangent.
• Verify Measurement Mathematically: Use Pythagoras theorem to double-check your construction! In the right-angled triangle $OQP$, $OP^2 = OQ^2 + PQ^2$. For example, if radius $OQ = 6\text{ cm}$ and distance $OP = 10\text{ cm}$, then tangent $PQ = \sqrt{10^2 - 6^2} = 8\text{ cm}$. Measure your drawn tangent with a ruler to ensure it matches this calculation exactly.

Practice Questions with Solutions

• Q: Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths. A: Step 1: Draw a circle with center $O$ and radius $6\text{ cm}$ using your compass. Step 2: Locate point $P$ such that $OP = 10\text{ cm}$ from the center $O$. Join $OP$. Step 3: Draw the perpendicular bisector of $OP$. Mark its intersection with $OP$ as point $M$. Step 4: Place the compass on $M$ with radius $MO = 5\text{ cm}$ and draw a circle that intersects the original circle at points $Q$ and $R$. Step 5: Join $PQ$ and $PR$. Measuring them with a ruler gives $PQ = PR = 8\text{ cm}$. Final answer: The pair of tangents $PQ$ and $PR$ are successfully constructed, each of length $8\text{ cm}$.
• Q: Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation. A: Step 1: Draw two concentric circles with a common center $O$, one of radius $4\text{ cm}$ and the other of radius $6\text{ cm}$. Step 2: Mark any point $P$ on the outer circle and join $OP$. Step 3: Bisect the line segment $OP$ to find its midpoint $M$. Step 4: With $M$ as center and radius $MO$, draw an arc/circle cutting the inner circle at points $Q$ and $R$. Step 5: Join $PQ$ (and $PR$). On measuring, $PQ \approx 4.47\text{ cm}$. Step 6: Verification by Calculation: In right-angled triangle $OQP$, $OP = 6\text{ cm}$ (radius of outer circle) and $OQ = 4\text{ cm}$ (radius of inner circle). By Pythagoras theorem, $PQ = \sqrt{OP^2 - OQ^2} = \sqrt{6^2 - 4^2} = \sqrt{36 - 16} = \sqrt{20} \approx 4.47\text{ cm}$. Final answer: The construction is verified mathematically and measured length matches the calculated value of $4.47\text{ cm}$.
• Q: Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60 degrees. A: Step 1: Remember that the angle between the tangents and the angle between the radii at the center are supplementary. Angle between tangents = $60^\circ$, so angle between radii = $180^\circ - 60^\circ = 120^\circ$. Step 2: Draw a circle of radius $5\text{ cm}$ with center $O$. Step 3: Draw any radius $OA$. Using a protractor or compass, construct an angle of $120^\circ$ at center $O$ to get another radius $OB$ such that $\angle AOB = 120^\circ$. Step 4: Draw perpendiculars ($90^\circ$ angles) at point $A$ and point $B$ to the radii $OA$ and $OB$ respectively. Step 5: Let these two perpendiculars intersect at point $P$. $PA$ and $PB$ are the required tangents. Final answer: Tangents $PA$ and $PB$ are constructed meeting at point $P$ with an inclination angle of $60^\circ$.
• Q: Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q. A: Step 1: Draw a circle of radius $3\text{ cm}$ with center $O$. Extend its diameter on both sides. Step 2: Mark points $P$ and $Q$ on this extended line such that $OP = 7\text{ cm}$ and $OQ = 7\text{ cm}$. Step 3: Bisect segment $OP$ to find its midpoint $M_1$. Bisect segment $OQ$ to find its midpoint $M_2$. Step 4: Draw a circle with center $M_1$ and radius $M_1O$ cutting the main circle at $A$ and $B$. Draw another circle with center $M_2$ and radius $M_2O$ cutting the main circle at $C$ and $D$. Step 5: Join $PA$, $PB$ (tangents from $P$) and $QC$, $QD$ (tangents from $Q$). Final answer: Pairs of tangents $(PA, PB)$ and $(QC, QD)$ are successfully drawn from points $P$ and $Q$ respectively.

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