Class 10 Mathematics Chapter 11 Question Answer | Constructions | English Medium

Class 10 Mathematics Chapter 11 Question Answer | Constructions | English Medium | ASSEB

Welcome to HSLC Guru

Hello dear student! Welcome to HSLC Guru. In this lesson you will find the complete English-medium solutions for ASSEB (Assam State School Education Board) Class 10 Mathematics, Chapter 11 — Constructions. In Class 9 you learned how to construct certain basic figures using a ruler and a pair of compasses — bisecting line segments and angles, drawing perpendiculars, copying triangles by the SSS, SAS, ASA and RHS criteria. In this chapter you will go a step further and learn three classical constructions that combine those basic skills with the theory of similar triangles and the tangent properties of a circle: (i) dividing a line segment in any given ratio $m:n$ using the parallel-rays method; (ii) constructing a triangle similar to a given triangle with a scale factor that may be a proper fraction (less than 1) or an improper fraction (greater than 1); and (iii) drawing the pair of tangents to a given circle from a point lying outside it. Each construction is presented with numbered steps, a step-by-step SVG diagram showing the arcs and final figure, and a complete geometric justification. Exercises 11.1 and 11.2 from the ASSEB Class 10 prescribed textbook are solved in full, followed by additional questions and a glossary that often appear in HSLC examinations.

Summary

This chapter studies three ruler-and-compass constructions that go beyond the standard SSS, SAS, ASA and RHS triangle constructions of Class 9. The first is the division of a line segment in a given ratio $m:n$ — done by drawing an auxiliary ray from one end, marking $m+n$ equal arcs along it, joining the last mark to the other end of the segment, and drawing a parallel through the $m^{\text{th}}$ mark; the parallel cuts the segment at the required division point, by the Basic Proportionality (Thales) Theorem. The second construction draws a triangle similar to a given triangle with scale factor $\frac{m}{n}$, where $mn$ enlarges it; the same parallel-rays trick on one side of the original triangle produces the new vertex, and a parallel to the third side yields the second new vertex. The third construction draws the pair of tangents to a circle from an external point $P$: the midpoint $M$ of $OP$ is found, and a circle on $OP$ as diameter is drawn; this auxiliary circle meets the original circle at the two tangent points, because the angle in a semicircle is a right angle, which forces the radius–tangent perpendicularity. Mastery of these three constructions, together with their justifications, is sufficient to answer every question in Exercises 11.1 and 11.2.

What You Already Know (Class 9 Recap)

Before starting this chapter, recall the following classical ruler-and-compass constructions from Class 9, because every Class 10 construction is built out of them:

To bisect a given line segment.
To bisect a given angle.
To draw the perpendicular bisector of a given line segment.
To drop a perpendicular from a point to a given line, or erect a perpendicular at a point on a given line.
To copy a given angle at a given vertex on a given ray.
To construct angles of $60°,\ 90°,\ 30°,\ 45°,\ 120°$ etc. directly from $60°$ arcs and bisections.
To construct a triangle from given data using one of the SSS, SAS, ASA or RHS criteria.

The two new tools added in Class 10 are (a) the parallel-rays method for distributing a length into a given number of equal parts, and (b) the use of an auxiliary intermediate circle to find tangent points.

Construction 11.1 — Division of a Line Segment in a Given Ratio

Problem. Given a line segment $AB$ and two positive integers $m$ and $n$, construct the point $P$ on $AB$ such that $AP:PB = m:n$.

Steps of Construction

Draw the line segment $AB$ of the given length.
Through $A$, draw any ray $AX$ making an acute angle with $AB$ (an angle of about $30°$ to $60°$ is convenient).
Locate $m+n$ points $A_1, A_2, A_3, \ldots, A_{m+n}$ on the ray $AX$ such that $AA_1 = A_1A_2 = A_2A_3 = \cdots = A_{m+n-1}A_{m+n}$. (Use the compass set to a fixed small radius and step it along the ray.)
Join $A_{m+n}$ to $B$.
Through the point $A_m$ (the $m^{\text{th}}$ mark from $A$), draw a line parallel to $A_{m+n}B$, meeting $AB$ at the point $P$.
Then $P$ divides $AB$ in the ratio $m:n$, i.e. $AP:PB = m:n$.

Justification

In $\triangle AB A_{m+n}$, the line $PA_m$ is parallel to $BA_{m+n}$ (by construction). By the Basic Proportionality Theorem (Thales’ Theorem) applied to $\triangle AB A_{m+n}$,

$$\frac{AP}{PB} = \frac{AA_m}{A_m A_{m+n}}.$$

But $AA_m$ contains $m$ equal sub-segments and $A_m A_{m+n}$ contains the remaining $n$ equal sub-segments, all of the same length. Therefore

$$\frac{AA_m}{A_m A_{m+n}} = \frac{m}{n} \quad\Longrightarrow\quad \frac{AP}{PB} = \frac{m}{n},$$

which is exactly the required division.

Alternative Method (parallel rays from both ends)

You can also draw two oppositely directed rays — $AX$ from $A$ above $AB$ and $BY$ from $B$ below $AB$. Mark $m$ equal segments $A_1, \ldots, A_m$ along $AX$ and $n$ equal segments $B_1, \ldots, B_n$ along $BY$ (both with the same compass radius). Join $A_m$ to $B_n$; this line meets $AB$ at the required point $P$. The justification uses the AA-similarity of $\triangle PAA_m$ and $\triangle PBB_n$.

Construction 11.2 — Triangle Similar to a Given Triangle

Case I: Scale Factor $\dfrac{m}{n}$ with $m
Problem. Given $\triangle ABC$, construct a triangle similar to it whose sides are $\frac{m}{n}$ of the corresponding sides of $\triangle ABC$.

Steps of Construction

Construct $\triangle ABC$ with the given measurements.

Through $B$, draw a ray $BX$ making an acute angle with $BC$ on the side opposite to vertex $A$.

Mark $n$ points $B_1, B_2, \ldots, B_n$ on $BX$ such that $BB_1 = B_1B_2 = \cdots = B_{n-1}B_n$.

Join $B_n$ to $C$.

Through $B_m$, draw a line parallel to $B_nC$, meeting $BC$ at $C’$.

Through $C’$, draw a line parallel to $CA$, meeting $BA$ at $A’$.

$\triangle A’BC’$ is the required triangle similar to $\triangle ABC$ with scale factor $\frac{m}{n}$.

Justification

By construction, $B_m C’ \parallel B_n C$, so by the BPT in $\triangle BB_nC$,

$$\frac{BC’}{C’C} = \frac{BB_m}{B_m B_n} = \frac{m}{n-m} \quad\Longrightarrow\quad \frac{BC’}{BC} = \frac{m}{n}.$$

Also $C’A’ \parallel CA$, so $\triangle BC’A’ \sim \triangle BCA$ by AA similarity. Therefore the ratios of all corresponding sides equal $\dfrac{BC’}{BC} = \dfrac{m}{n}$, i.e. each side of $\triangle A’BC’$ is $\dfrac{m}{n}$ of the corresponding side of $\triangle ABC$.

Case II: Scale Factor $\dfrac{m}{n}$ with $m>n$ (enlarged triangle)

Problem. Given $\triangle ABC$, construct a similar triangle whose sides are $\frac{m}{n}$ of the corresponding sides of $\triangle ABC$, where $m>n$.

Steps of Construction

Construct $\triangle ABC$ as given.

Draw a ray $BX$ making an acute angle with $BC$ on the side opposite to $A$.

This time mark $m$ points $B_1, B_2, \ldots, B_m$ along $BX$ at equal distances ($m$, not $n$, because we need to extend beyond $C$).

Join $B_n$ to $C$.

Through $B_m$, draw a line parallel to $B_n C$, meeting $BC$ produced at $C’$ (beyond $C$).

Through $C’$, draw a line parallel to $CA$, meeting $BA$ produced at $A’$.

$\triangle A’BC’$ is the required enlarged triangle.

Justification

By construction, $B_m C’ \parallel B_n C$, so $\triangle BB_n C \sim \triangle B B_m C’$, giving $\dfrac{BC’}{BC} = \dfrac{BB_m}{BB_n} = \dfrac{m}{n}$. Also $C’A’ \parallel CA$, so $\triangle BC’A’ \sim \triangle BCA$. Hence each side of $\triangle A’BC’$ is $\dfrac{m}{n}$ times the corresponding side of $\triangle ABC$, as required.

Construction 11.3 — Tangents to a Circle from an External Point

Problem. Given a circle with centre $O$ and a point $P$ outside it, construct the two tangents from $P$ to the circle.

Steps of Construction

Draw the given circle with centre $O$ and the given radius $r$.

Mark the external point $P$ and join $OP$.

Construct the perpendicular bisector of $OP$ to locate its midpoint $M$.

With $M$ as centre and $MP$ (= $MO$) as radius, draw an auxiliary circle. This circle passes through $O$ and through $P$, and intersects the original circle at two points — call them $T_1$ and $T_2$.

Join $PT_1$ and $PT_2$. These two line segments are the required tangents from $P$ to the circle.

Justification

Join $OT_1$. Since $OP$ is a diameter of the auxiliary circle (centre $M$, radius $MP=MO$), the angle in the semicircle at $T_1$ is a right angle:

$$\angle OT_1 P = 90°.$$

Therefore $OT_1 \perp PT_1$. But $OT_1$ is a radius of the original circle, and a line through a point on the circle that is perpendicular to the radius at that point is precisely the tangent at that point. Hence $PT_1$ is a tangent to the given circle at $T_1$. By the same reasoning, $PT_2$ is a tangent at $T_2$. These are the two tangents from the external point $P$.

Length of Tangent Formula

If the radius of the circle is $r$ and $OP = d$, then in the right triangle $OT_1P$,

$$PT_1 = \sqrt{OP^2 – OT_1^2} = \sqrt{d^2 – r^2}.$$

Also $PT_1 = PT_2$, i.e. the two tangents from an external point are equal in length.

Special Case: Tangent at a Point on the Circle

If the point lies on the circle (not outside), the tangent there is unique. To draw the tangent at a point $A$ on a circle with centre $O$:

Join $OA$.

At $A$, construct a line perpendicular to $OA$.

This perpendicular is the required tangent at $A$.

This works because the radius drawn to the point of contact is perpendicular to the tangent at that point.

Exercise 11.1

Q1. Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts.

Answer:

Steps of Construction:

Draw $AB = 7.6$ cm.

Through $A$, draw any ray $AX$ making an acute angle with $AB$.

Mark $5+8 = 13$ equal arcs $A_1, A_2, \ldots, A_{13}$ on $AX$ such that $AA_1 = A_1A_2 = \cdots = A_{12}A_{13}$.

Join $A_{13}B$.

Through $A_5$ (the $5^{\text{th}}$ point), draw a line parallel to $A_{13}B$, meeting $AB$ at $P$.

Then $AP:PB = 5:8$.

Measurement: $AP = \dfrac{5}{13}\times 7.6 \approx 2.92$ cm, $PB = \dfrac{8}{13}\times 7.6 \approx 4.68$ cm.

Justification: By BPT, $\dfrac{AP}{PB} = \dfrac{AA_5}{A_5 A_{13}} = \dfrac{5}{8}$.

Q2. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are $\frac{2}{3}$ of the corresponding sides of the first triangle.

Answer:

Construct $\triangle ABC$ with $BC = 6$ cm, $AB = 5$ cm, $CA = 4$ cm (use SSS).

Through $B$, draw ray $BX$ making an acute angle with $BC$ on the side opposite to $A$.

Mark $3$ points $B_1, B_2, B_3$ on $BX$ with equal spacing.

Join $B_3 C$.

Through $B_2$, draw a line parallel to $B_3 C$, meeting $BC$ at $C’$.

Through $C’$, draw a line parallel to $CA$, meeting $BA$ at $A’$.

$\triangle A’BC’$ is the required triangle with $\dfrac{BC’}{BC} = \dfrac{BA’}{BA} = \dfrac{2}{3}$.

Justification: $B_2 C’ \parallel B_3 C$ gives $\dfrac{BC’}{BC} = \dfrac{BB_2}{BB_3} = \dfrac{2}{3}$. Also $C’A’ \parallel CA$ gives $\triangle BC’A’ \sim \triangle BCA$ by AA, so each side scales by $\dfrac{2}{3}$.

Q3. Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are $\frac{7}{5}$ of the corresponding sides of the first triangle.

Answer:

Construct $\triangle ABC$ with $AB = 5$ cm, $BC = 6$ cm, $CA = 7$ cm.

Draw ray $BX$ making an acute angle with $BC$ below $BC$.

Mark $7$ equal arcs $B_1, B_2, \ldots, B_7$ on $BX$ (because $7>5$, take the larger of $m,n$).

Join $B_5$ to $C$.

Through $B_7$, draw a line parallel to $B_5 C$, meeting $BC$ produced at $C’$.

Through $C’$, draw a line parallel to $CA$, meeting $BA$ produced at $A’$.

$\triangle A’BC’$ has each side $\dfrac{7}{5}$ of the corresponding side of $\triangle ABC$.

Justification: $B_7 C’ \parallel B_5 C$ gives $\dfrac{BC’}{BC} = \dfrac{7}{5}$, and $C’A’\parallel CA$ then gives the same ratio for the other two sides by AA-similarity.

Q4. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are $1\tfrac{1}{2}$ times the corresponding sides of the isosceles triangle.

Answer: Scale factor $1\tfrac{1}{2} = \dfrac{3}{2}$.

Draw $BC = 8$ cm and find its midpoint $D$.

At $D$, erect $DA \perp BC$ with $DA = 4$ cm. Join $AB$ and $AC$. $\triangle ABC$ is the isosceles triangle.

Through $B$, draw ray $BX$ making an acute angle with $BC$, on the side opposite to $A$.

Mark $3$ equal arcs $B_1, B_2, B_3$ on $BX$.

Join $B_2$ to $C$.

Through $B_3$, draw a line parallel to $B_2 C$, meeting $BC$ produced at $C’$.

Through $C’$, draw a line parallel to $CA$, meeting $BA$ produced at $A’$.

$\triangle A’BC’$ is the required enlarged triangle.

Justification: $\dfrac{BC’}{BC} = \dfrac{BB_3}{BB_2} = \dfrac{3}{2}$, and parallels give similar triangles, so each side scales by $\dfrac{3}{2}$.

Q5. Draw a triangle $ABC$ with side $BC = 6$ cm, $AB = 5$ cm and $\angle ABC = 60°$. Then construct a triangle whose sides are $\dfrac{3}{4}$ of the corresponding sides of the triangle $ABC$.

Answer: Scale factor $\dfrac{3}{4}$ ($m

Draw $BC = 6$ cm. At $B$, construct $\angle ABC = 60°$ and along it mark $BA = 5$ cm. Join $AC$ to obtain $\triangle ABC$.

Through $B$, draw ray $BX$ making an acute angle with $BC$ on the side opposite to $A$.

Mark $4$ equal arcs $B_1, B_2, B_3, B_4$ on $BX$.

Join $B_4$ to $C$.

Through $B_3$, draw a line parallel to $B_4 C$, meeting $BC$ at $C’$.

Through $C’$, draw a line parallel to $CA$, meeting $BA$ at $A’$.

$\triangle A’BC’$ is the required triangle.

Justification: $B_3 C’ \parallel B_4 C \Rightarrow \dfrac{BC’}{BC} = \dfrac{3}{4}$; $C’A’ \parallel CA \Rightarrow \triangle BC’A’ \sim \triangle BCA$ with ratio $\dfrac{3}{4}$.

Q6. Draw a triangle $ABC$ with side $BC = 7$ cm, $\angle B = 45°$, $\angle A = 105°$. Then construct a triangle whose sides are $\dfrac{4}{3}$ times the corresponding sides of $\triangle ABC$.

Answer: Since $\angle A + \angle B + \angle C = 180°$, $\angle C = 180° – 105° – 45° = 30°$.

Draw $BC = 7$ cm. At $B$, draw $\angle XBC = 45°$, and at $C$, draw $\angle YCB = 30°$. Let $BX$ and $CY$ meet at $A$. Then $\triangle ABC$ has $\angle A = 105°$.

Through $B$, draw ray $BZ$ making an acute angle with $BC$ on the side opposite to $A$.

Mark $4$ equal arcs $B_1, B_2, B_3, B_4$ on $BZ$.

Join $B_3 C$.

Through $B_4$, draw a line parallel to $B_3 C$, meeting $BC$ produced at $C’$.

Through $C’$, draw a line parallel to $CA$, meeting $BA$ produced at $A’$.

$\triangle A’BC’$ is the required enlarged triangle with each side $\dfrac{4}{3}$ of the corresponding side of $\triangle ABC$.

Justification: $\dfrac{BC’}{BC} = \dfrac{BB_4}{BB_3} = \dfrac{4}{3}$, and the AA-similarity from $C’A’\parallel CA$ yields the same ratio for $BA’$ and $C’A’$.

Q7. Draw a right triangle in which the sides (other than the hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are $\dfrac{5}{3}$ times the corresponding sides of the given triangle.

Answer:

Draw $BC = 4$ cm. At $B$, erect $BA \perp BC$ with $BA = 3$ cm. Join $AC$. Then $\triangle ABC$ is right-angled at $B$ with $AC = \sqrt{3^2 + 4^2} = 5$ cm.

Through $B$, draw ray $BX$ making an acute angle with $BC$, on the side opposite to $A$.

Mark $5$ equal arcs $B_1, B_2, B_3, B_4, B_5$ on $BX$.

Join $B_3 C$.

Through $B_5$, draw a line parallel to $B_3 C$, meeting $BC$ produced at $C’$.

Through $C’$, draw a line parallel to $CA$, meeting $BA$ produced at $A’$.

$\triangle A’BC’$ is the required enlarged right triangle with sides $5$ cm, $\dfrac{20}{3}$ cm, $\dfrac{25}{3}$ cm (i.e. $\dfrac{5}{3}$ of $3,\ 4,\ 5$).

Justification: $B_5 C’ \parallel B_3 C$ gives $\dfrac{BC’}{BC} = \dfrac{5}{3}$, and $C’A’ \parallel CA$ gives $\triangle BC’A’ \sim \triangle BCA$ with the same ratio. Thus $BA’ = \dfrac{5}{3}\cdot 3 = 5$ cm, $BC’ = \dfrac{5}{3}\cdot 4 = \dfrac{20}{3}$ cm, $C’A’ = \dfrac{5}{3}\cdot 5 = \dfrac{25}{3}$ cm.

Exercise 11.2

In each of the following, give the justification of the construction also.

Q1. 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.

Answer:

Draw a circle with centre $O$ and radius $6$ cm.

Mark a point $P$ such that $OP = 10$ cm. Join $OP$.

Bisect $OP$ at $M$.

With $M$ as centre and radius $MP$, draw a circle. It cuts the original circle at $T_1$ and $T_2$.

Join $PT_1$ and $PT_2$. These are the required tangents.

Length of tangent: $PT_1 = PT_2 = \sqrt{OP^2 – r^2} = \sqrt{100 – 36} = \sqrt{64} = 8$ cm.

Justification: $\angle OT_1 P = 90°$ (angle in a semicircle), so $OT_1 \perp PT_1$. Since $OT_1$ is a radius, $PT_1$ is the tangent at $T_1$. Same reasoning for $T_2$.

Q2. 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.

Answer:

Draw two concentric circles with centre $O$ and radii $4$ cm and $6$ cm.

Mark any point $P$ on the larger circle. Then $OP = 6$ cm.

Bisect $OP$ at $M$.

With $M$ as centre and radius $MP$, draw a circle that meets the inner circle at $T$.

Join $PT$. This is the required tangent from $P$ to the smaller circle.

Calculation: $PT = \sqrt{OP^2 – OT^2} = \sqrt{6^2 – 4^2} = \sqrt{36-16} = \sqrt{20} = 2\sqrt{5} \approx 4.47$ cm.

Justification: $\angle OTP = 90°$ (angle in semicircle), so $OT\perp PT$, making $PT$ a tangent to the smaller circle at $T$. The measured length agrees with $\sqrt{20}\approx 4.47$ cm.

Q3. Draw a circle of radius 3 cm. Take two points $P$ and $Q$ on one of its extended diameters each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points $P$ and $Q$.

Answer:

Draw a circle with centre $O$ and radius $3$ cm.

Produce the diameter to both sides and mark $P$ and $Q$ on opposite sides such that $OP = OQ = 7$ cm.

Bisect $OP$ at $M_1$. With $M_1$ as centre and radius $M_1P$, draw a circle meeting the original circle at $T_1, T_2$. Join $PT_1, PT_2$ — these are the tangents from $P$.

Similarly, bisect $OQ$ at $M_2$. With $M_2$ as centre and radius $M_2 Q$, draw a circle meeting the original circle at $T_3, T_4$. Join $QT_3, QT_4$ — these are the tangents from $Q$.

Length: $PT_1 = QT_3 = \sqrt{49-9} = \sqrt{40} = 2\sqrt{10} \approx 6.32$ cm.

Justification: Each angle $\angle OT_i P$ (or $\angle OT_j Q$) is $90°$ because it stands on a diameter of the auxiliary circle. Hence each $PT_i$ (or $QT_j$) is perpendicular to the radius at the point of contact, so it is a tangent.

Q4. Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°.

Answer: If the two tangents from an external point $P$ are inclined at $60°$, the angle $\angle T_1 O T_2$ between the two radii to the points of contact equals $180° – 60° = 120°$ (because $OT_1 \perp PT_1$, $OT_2\perp PT_2$ and the four angles of quadrilateral $OT_1 PT_2$ sum to $360°$).

Draw a circle with centre $O$ and radius $5$ cm.

Draw two radii $OT_1$ and $OT_2$ with $\angle T_1 OT_2 = 120°$.

At $T_1$, draw a perpendicular to $OT_1$. At $T_2$, draw a perpendicular to $OT_2$. The two perpendiculars meet at $P$.

$PT_1$ and $PT_2$ are the required pair of tangents inclined at $60°$.

Verification of $OP$: In the right triangle $OT_1 P$, $\angle OPT_1 = 30°$ (half of $60°$), so $\cos 30° = \dfrac{OT_1}{OP}$ gives $OP = \dfrac{5}{\cos 30°} = \dfrac{10}{\sqrt 3}\approx 5.77$ cm. $\;PT_1 = OP\sin 30° = \dfrac{10}{\sqrt 3}\cdot\dfrac{1}{2}\cdot\sqrt 3 = 5\sqrt 3/\ldots$ — verify: $PT_1 = \sqrt{OP^2 – r^2} = \sqrt{100/3 – 25} = \sqrt{25/3} = 5/\sqrt 3\approx 2.89$ cm.

Justification: Since $\angle OT_1 P = \angle OT_2 P = 90°$ and $\angle T_1 OT_2 = 120°$, the fourth angle $\angle T_1 P T_2 = 360° – 90° – 90° – 120° = 60°$. So the tangents at $T_1$ and $T_2$ are inclined at $60°$ as required.

Q5. Draw a line segment $AB$ of length 8 cm. Taking $A$ as centre, draw a circle of radius 4 cm and taking $B$ as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.

Answer:

Draw $AB = 8$ cm.

With $A$ as centre and radius $4$ cm, draw a circle $C_1$.

With $B$ as centre and radius $3$ cm, draw a circle $C_2$.

Bisect $AB$ at $M$.

With $M$ as centre and radius $MA$ ($=MB= 4$ cm), draw a circle $C_3$. It meets $C_1$ at $T_1, T_2$ and meets $C_2$ at $T_3, T_4$.

Join $BT_1, BT_2$ — tangents from $B$ to circle $C_1$.

Join $AT_3, AT_4$ — tangents from $A$ to circle $C_2$.

Lengths: $BT_1 = \sqrt{AB^2 – AT_1^2} = \sqrt{64 – 16} = \sqrt{48} = 4\sqrt 3$ cm. $AT_3 = \sqrt{AB^2 – BT_3^2} = \sqrt{64-9} = \sqrt{55}$ cm.

Justification: $\angle AT_1 B = 90°$ (angle in a semicircle on diameter $AB$), so $AT_1 \perp BT_1$, and since $AT_1$ is a radius of $C_1$, $BT_1$ is a tangent to $C_1$. The same argument gives $BT_2$, $AT_3$, $AT_4$.

Q6. Let $ABC$ be a right triangle in which $AB = 6$ cm, $BC = 8$ cm and $\angle B = 90°$. $BD$ is the perpendicular from $B$ on $AC$. The circle through $B, C, D$ is drawn. Construct the tangents from $A$ to this circle.

Answer:

Draw $BC = 8$ cm. At $B$, erect $BA \perp BC$ with $BA = 6$ cm. Join $AC$ ($= 10$ cm).

From $B$, drop the perpendicular $BD$ to $AC$. (Step: with $B$ as centre, mark equal arcs on $AC$, etc.)

Find the circumcircle of $\triangle BDC$. Since $\angle BDC = 90°$ (perpendicular foot), $BC$ is the diameter, so the circumcircle has centre at the midpoint $E$ of $BC$ and radius $\dfrac{BC}{2} = 4$ cm. Draw this circle.

Note that $AB$ is already tangent to this circle at $B$, because $AB \perp BC$ and $BC$ is a diameter (so the tangent at $B$ is perpendicular to $BC$, i.e. it is $AB$).

To find the second tangent from $A$: bisect $AE$ at $F$. With $F$ as centre and radius $FA$, draw a circle meeting the small circle at $B$ and at another point $T$. Join $AT$. Then $AT$ is the second tangent.

Justification: $\angle ATE = 90°$ (angle in a semicircle on diameter $AE$), so $ET\perp AT$; since $ET$ is a radius of the small circle, $AT$ is tangent at $T$. Also $AB\perp EB$ at $B$, so $AB$ is tangent at $B$. Hence $AB$ and $AT$ are the two tangents from $A$.

Q7. Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.

Answer: Since the centre is unknown, we must first locate it.

Trace the circle by drawing along the inner edge of a bangle.

Take any two non-parallel chords $PQ$ and $RS$ on the circle.

Draw the perpendicular bisectors of $PQ$ and $RS$. They intersect at $O$, the centre of the circle.

Mark a point $A$ outside the circle. Join $OA$.

Bisect $OA$ at $M$. With $M$ as centre and $MA$ as radius, draw a circle meeting the original at $T_1$ and $T_2$.

Join $AT_1, AT_2$. These are the required tangents.

Justification: The perpendicular bisector of any chord passes through the centre, so two such bisectors meet at $O$. Then the standard construction (Construction 11.3) gives the pair of tangents from $A$, and the angle in the semicircle property at $T_1, T_2$ provides the perpendicularity needed to call them tangents.

Additional Questions and Answers

Q1. To divide a line segment $AB$ in the ratio $4:7$, what is the minimum number of points to be located at equal distances on the ray $AX$?

Answer: $4 + 7 = 11$ points.

Q2. To divide a line segment $AB$ in the ratio $5:6$, points $A_1, A_2, A_3, \ldots$ are marked at equal distances on a ray $AX$. The point $B$ is then joined to which point?

Answer: $B$ is joined to $A_{11}$ (since $5+6=11$). Then a line through $A_5$ parallel to $A_{11}B$ meets $AB$ at the required point.

Q3. To construct a triangle similar to a given $\triangle ABC$ with its sides $\dfrac{3}{7}$ of the corresponding sides of $\triangle ABC$, first $7$ points $B_1, B_2, \ldots, B_7$ are located at equal distances on $BX$. The next step is to join which point to $C$?

Answer: $B_7$ is joined to $C$. Then a line through $B_3$ parallel to $B_7 C$ gives $C’$ on $BC$.

Q4. To draw a pair of tangents to a circle which are inclined to each other at an angle of $60°$, what should be the angle between the two radii drawn to the points of contact?

Answer: $180° – 60° = 120°$. (The four angles of quadrilateral $OT_1 PT_2$ sum to $360°$; two are $90°$ each at the contact points, so the remaining two are supplementary.)

Q5. To draw a pair of tangents to a circle which are inclined at an angle of $35°$, the angle between the two radii at the contact points should be:

Answer: $180° – 35° = 145°$.

Q6. Which of the following is not used while constructing the bisector of an angle?

Answer: A protractor. (Bisection is done with a compass and ruler only.)

Q7. The lengths of the two tangents drawn from an external point to a circle are:

Answer: Equal in length. If the radius is $r$ and the distance of the external point from the centre is $d$, the common length is $\sqrt{d^2 – r^2}$.

Q8. By geometrical construction, can you divide a line segment in the ratio $\sqrt{3} : \dfrac{1}{\sqrt 3}$?

Answer: Yes. The ratio simplifies: $\sqrt 3 : \dfrac{1}{\sqrt 3} = \sqrt 3 \times \sqrt 3 : 1 = 3:1$. So we divide the segment in the ratio $3:1$ using $3+1=4$ equal arcs on the auxiliary ray.

Q9. To construct a triangle similar to a given triangle with scale factor $\dfrac{8}{5}$, the larger of the two integers is:

Answer: $8$. We mark $8$ equal arcs on the ray $BX$ (we always mark the larger of $m,n$). Join $B_5$ to $C$, then through $B_8$ draw a line parallel to $B_5 C$.

Q10. Two tangents drawn from an external point $P$ to a circle of radius $r$ touch the circle at $A$ and $B$. If $\angle APB = 90°$, find the length $PA$.

Answer: $\angle APB = 90°$ implies $\angle AOB = 90°$ (since $\angle APB + \angle AOB = 180°$). $OAPB$ is then a square with side $r$, so $PA = r$.

Q11. To construct an angle of $22.5°$ using ruler and compass, the construction sequence is:

Answer: First construct $90°$, then bisect to get $45°$, then bisect $45°$ to get $22.5°$.

Q12. The number of tangents that can be drawn from a point inside a circle to the circle is:

Answer: Zero. A point inside the circle cannot be the external point of any tangent.

Q13. If a chord and a tangent are drawn from the same external point to a circle, what relation holds between the tangent length and the chord segments?

Answer: By the tangent-secant relation, $PT^2 = PA\cdot PB$, where $PT$ is the tangent length and $PA, PB$ are the distances from $P$ to the two intersection points of the secant with the circle.

Q14. Draw a line segment of length 9 cm and divide it in the ratio $2:5$. Measure the longer part.

Answer: Use $2+5=7$ arcs. Then $AP = \dfrac{2}{7}\times 9 = \dfrac{18}{7}\approx 2.57$ cm and $PB = \dfrac{5}{7}\times 9 = \dfrac{45}{7}\approx 6.43$ cm. Longer part $\approx 6.43$ cm.

Q15. Construct an equilateral triangle of side 5 cm and then a triangle similar to it with sides $\dfrac{3}{5}$ of the corresponding sides.

Answer: Draw $\triangle ABC$ equilateral with $AB = BC = CA = 5$ cm. Mark $5$ equal arcs $B_1\ldots B_5$ on ray $BX$. Join $B_5C$. Through $B_3$, draw parallel to $B_5C$ meeting $BC$ at $C’$. Through $C’$, draw parallel to $CA$ meeting $BA$ at $A’$. Then $\triangle A’BC’$ is equilateral with side $3$ cm.

HSLC-Pattern Multiple Choice Questions (MCQ)

MCQ 1. To divide a line segment $AB$ in the ratio $3:5$, points $A_1, A_2, A_3, \ldots$ are marked at equal distances on a ray $AX$ such that $\angle BAX$ is acute. The minimum number of points to be located on $AX$ is

(a) 3    (b) 5    (c) 8    (d) 15

Answer: (c) 8. Reason: $m+n = 3+5 = 8$ equal segments are required.

MCQ 2. To divide a line segment $AB$ in the ratio $4:9$, the point $B$ should be joined to

(a) $A_4$    (b) $A_9$    (c) $A_{13}$    (d) $A_5$

Answer: (c) $A_{13}$. Reason: We mark $4+9=13$ points and join $B$ to the last one, $A_{13}$.

MCQ 3. A pair of tangents can be constructed from a point $P$ to a circle of radius $4.5$ cm so that the angle between them is $45°$. In this case the distance of $P$ from the centre of the circle has to be

(a) $5$ cm    (b) more than $4.5$ cm    (c) $4.5$ cm    (d) less than $4.5$ cm

Answer: (b) more than $4.5$ cm. The point must be external; for any external point, two tangents can be drawn.

MCQ 4. To draw a pair of tangents to a circle which are inclined to each other at an angle of $35°$, it is required to draw tangents at the end-points of those two radii of the circle, the angle between which is

(a) $105°$    (b) $70°$    (c) $140°$    (d) $145°$

Answer: (d) $145°$. Reason: $180° – 35° = 145°$.

MCQ 5. To construct a triangle similar to $\triangle ABC$ with its sides $\dfrac{8}{5}$ of the corresponding sides of $\triangle ABC$, draw a ray $BX$ such that $\angle CBX$ is acute and $X$ is on the opposite side of $A$ with respect to $BC$. The minimum number of points to be located at equal distances on ray $BX$ is

(a) 5    (b) 8    (c) 13    (d) 3

Answer: (b) 8. Reason: For similar triangles we use the larger of $m,n$.

MCQ 6. To divide a line segment $AB$ in the ratio $5:7$, first a ray $AX$ is drawn so that $\angle BAX$ is acute and then at equal distances points are marked on the ray $AX$. The minimum number of these points is

(a) 12    (b) 7    (c) 5    (d) 35

Answer: (a) 12.

MCQ 7. The locus of all points equidistant from two given points $A$ and $B$ is

(a) a circle through $A$ and $B$    (b) a line parallel to $AB$    (c) the perpendicular bisector of $AB$    (d) the segment $AB$

Answer: (c) the perpendicular bisector of $AB$. This is why we use the perpendicular bisector to find the midpoint $M$ of $OP$ in the tangent construction.

MCQ 8. If a tangent of length $24$ cm is drawn from a point at a distance $25$ cm from the centre of a circle, the radius of the circle is

(a) $7$ cm    (b) $24$ cm    (c) $25$ cm    (d) $1$ cm

Answer: (a) $7$ cm. Reason: $r = \sqrt{d^2 – t^2} = \sqrt{625-576} = \sqrt{49} = 7$ cm.

MCQ 9. The number of tangents that can be drawn to a circle from a point lying on the circle is

(a) zero    (b) one    (c) two    (d) infinitely many

Answer: (b) one — the unique tangent at that point, perpendicular to the radius there.

MCQ 10. To construct a triangle similar to $\triangle ABC$ with sides $\dfrac{3}{7}$ times the corresponding sides, after marking $7$ equal arcs $B_1,\ldots,B_7$ on ray $BX$, we join

(a) $B_3$ to $C$    (b) $B_7$ to $C$    (c) $B_4$ to $C$    (d) $B_{10}$ to $C$

Answer: (b) $B_7$ to $C$. Then a parallel through $B_3$ to $B_7C$ meets $BC$ at $C’$.

MCQ 11. A pair of tangents to a circle of radius $6$ cm is drawn from a point distant $10$ cm from its centre. The length of each tangent is

(a) $4$ cm    (b) $6$ cm    (c) $8$ cm    (d) $10$ cm

Answer: (c) $8$ cm. Reason: $\sqrt{10^2 – 6^2} = \sqrt{64} = 8$ cm.

MCQ 12. To divide a line segment in the ratio $2:5$ by the parallel-rays method using two oppositely directed rays from $A$ and $B$, we mark

(a) $2$ arcs from $A$, $5$ arcs from $B$    (b) $5$ arcs from $A$, $2$ arcs from $B$    (c) $7$ arcs from each    (d) $2$ arcs from each

Answer: (a) $2$ arcs from $A$ and $5$ arcs from $B$. Joining the last marks gives a line that meets $AB$ at the required division point.

Fill in the Blanks

To divide a line segment in the ratio $m:n$, we mark $\underline{\;m+n\;}$ equal arcs on the auxiliary ray.

To construct a similar triangle with scale factor $\dfrac{m}{n}$, we mark $\underline{\;\max(m,n)\;}$ equal arcs on the auxiliary ray.

The angle in a semicircle is $\underline{\;90°\;}$.

The two tangents drawn from an external point to a circle are $\underline{\;\text{equal in length}\;}$.

The radius drawn to the point of contact is $\underline{\;\text{perpendicular}\;}$ to the tangent at that point.

If the angle between two tangents from an external point is $\theta$, then the angle subtended by the two contact points at the centre is $\underline{\;180°-\theta\;}$.

The length of a tangent from a point $P$ at distance $d$ from the centre of a circle of radius $r$ is $\underline{\;\sqrt{d^2-r^2}\;}$.

The construction technique for dividing a line segment uses the $\underline{\;\text{Basic Proportionality}\;}$ theorem as its justification.

To construct an angle of $45°$, we first construct a $\underline{\;90°\;}$ angle and then bisect it.

The number of tangents that can be drawn to a circle from an external point is $\underline{\;\text{two}\;}$.

True or False

By geometrical construction, a line segment can be divided in the ratio $\sqrt{2}:\sqrt{3}$. (False — irrational ratios with no rational simplification cannot be done by ruler-and-compass division-of-segment.) However, $\sqrt{2}:\sqrt{8} = 1:2$ is rational and can be done.

The pair of tangents drawn from an external point to a circle are perpendicular to the radii at the contact points. (True.)

A tangent can be constructed from a point lying inside a circle. (False — no tangent exists from an interior point.)

To construct a similar triangle with scale factor $\dfrac{3}{2}$, we mark $5$ arcs on the auxiliary ray. (False — we mark $\max(3,2)=3$ arcs.)

The two tangents from an external point are always equal in length. (True.)

By geometrical construction, a line segment can be divided in the ratio $5:\sqrt{2}$. (False — one of the parts is irrational and cannot be measured by integer arcs.)

If the angle between two tangents drawn from an external point is $60°$, the angle between the two radii to the contact points is $120°$. (True.)

The midpoint of $OP$ in the tangent construction is found by drawing the perpendicular bisector of $OP$. (True.)

HSLC Long-Answer Practice Questions

LAQ 1. Draw a line segment $PQ$ of length 12 cm. Divide it into three equal parts using ruler and compass only. Justify your construction.

Answer: Dividing into three equal parts means each part is in the ratio $1:1:1$.

Draw $PQ = 12$ cm.

Through $P$, draw any ray $PX$ at an acute angle with $PQ$.

Mark three equal arcs $P_1, P_2, P_3$ on $PX$ with the same compass radius.

Join $P_3 Q$.

Through $P_1$, draw a line parallel to $P_3 Q$, meeting $PQ$ at $A$.

Through $P_2$, draw a line parallel to $P_3 Q$, meeting $PQ$ at $B$.

Then $PA = AB = BQ = 4$ cm.

Justification: Triangles $P A P_1$, $PBP_2$, $PQP_3$ are similar (each has a parallel cut). By BPT, $\dfrac{PA}{PP_1} = \dfrac{PB}{PP_2} = \dfrac{PQ}{PP_3}$, and since $PP_1:PP_2:PP_3 = 1:2:3$, we get $PA:PB:PQ = 1:2:3$, hence $PA = AB = BQ$.

LAQ 2. Construct a $\triangle PQR$ with $PQ = 7$ cm, $QR = 5$ cm, $\angle PQR = 75°$. Then construct another triangle similar to $\triangle PQR$ whose sides are $\dfrac{6}{5}$ of the corresponding sides of $\triangle PQR$. Give justification.

Answer:

Draw $QR = 5$ cm. At $Q$, construct $\angle PQR = 75°$ (since $75° = 60° + 15°$, or $75° = 90° – 15°$, use bisection of $30°$ to get $15°$ then add to $60°$).

Along the $75°$ ray from $Q$, mark $QP = 7$ cm. Join $PR$ to obtain $\triangle PQR$.

Through $Q$, draw ray $QX$ making an acute angle with $QR$ on the side opposite to $P$.

Mark $6$ equal arcs $Q_1, Q_2, \ldots, Q_6$ on $QX$ (since $6 > 5$).

Join $Q_5$ to $R$.

Through $Q_6$, draw a line parallel to $Q_5 R$, meeting $QR$ produced at $R’$.

Through $R’$, draw a line parallel to $RP$, meeting $QP$ produced at $P’$.

$\triangle P’QR’$ is the required enlarged triangle with each side $\dfrac{6}{5}$ of the corresponding side of $\triangle PQR$.

Justification: $Q_6 R’ \parallel Q_5 R$ implies $\triangle QQ_5R \sim \triangle Q Q_6 R’$, so $\dfrac{QR’}{QR} = \dfrac{6}{5}$. Also $R’P’ \parallel RP$ implies $\triangle QR’P’ \sim \triangle QRP$, so all corresponding sides are in the same ratio $\dfrac{6}{5}$.

LAQ 3. Draw a circle of radius 4 cm. Take a point $P$ outside the circle at a distance of 9 cm from the centre. Construct the pair of tangents from $P$. Calculate the length of each tangent.

Answer:

Draw a circle with centre $O$, radius $4$ cm.

Mark $P$ such that $OP = 9$ cm.

Find the midpoint $M$ of $OP$ by drawing its perpendicular bisector.

With $M$ as centre and radius $MP$ ($= 4.5$ cm), draw an auxiliary circle, which meets the original circle at $T_1$ and $T_2$.

Join $PT_1$ and $PT_2$ — these are the required tangents.

Length: $PT_1 = PT_2 = \sqrt{9^2 – 4^2} = \sqrt{81 – 16} = \sqrt{65} \approx 8.06$ cm.

Justification: $\angle OT_1 P = 90°$ (angle in semicircle), so $OT_1 \perp PT_1$. Since $OT_1$ is a radius, $PT_1$ is the tangent at $T_1$. Similarly for $T_2$.

LAQ 4. Draw a line segment $AB$ of length $7$ cm. Using ruler and compass, find a point $P$ on $AB$ such that $\dfrac{AP}{AB} = \dfrac{3}{5}$.

Answer: $\dfrac{AP}{AB} = \dfrac{3}{5}$ implies $\dfrac{AP}{PB} = \dfrac{3}{2}$ (since $AB = AP + PB$, so $PB = AB – AP = \dfrac{2}{5}AB$).

Draw $AB = 7$ cm.

Draw ray $AX$ at acute angle to $AB$.

Mark $5$ equal arcs $A_1, \ldots, A_5$ on $AX$.

Join $A_5$ to $B$.

Through $A_3$, draw a line parallel to $A_5 B$, meeting $AB$ at $P$.

Then $\dfrac{AP}{AB} = \dfrac{AA_3}{AA_5} = \dfrac{3}{5}$.

LAQ 5. Construct a $\triangle ABC$ with $AB = AC = 5$ cm and $BC = 6$ cm (an isosceles triangle). Then construct a triangle similar to $\triangle ABC$ with scale factor $\dfrac{4}{3}$.

Answer:

Draw $BC = 6$ cm. With $B$ and $C$ as centres and radius $5$ cm each, draw arcs intersecting at $A$. Join $AB$ and $AC$.

Through $B$, draw ray $BX$ at acute angle to $BC$ below.

Mark $4$ equal arcs $B_1, B_2, B_3, B_4$ on $BX$.

Join $B_3$ to $C$.

Through $B_4$, draw a line parallel to $B_3 C$, meeting $BC$ produced at $C’$.

Through $C’$, draw a line parallel to $CA$, meeting $BA$ produced at $A’$.

$\triangle A’BC’$ is the required isosceles triangle with $A’B = A’C’ = \dfrac{20}{3}$ cm and $BC’ = 8$ cm.

LAQ 6. Two circles of radii $5$ cm and $3$ cm intersect at two points and the distance between their centres is $6$ cm. Find the length of the common chord. (Although this is not a strict construction problem, it illustrates the geometry that underlies tangent constructions.)

Answer: Let $O_1$ and $O_2$ be the centres with $O_1 O_2 = 6$ cm, radii $5$ cm and $3$ cm. The common chord $AB$ is perpendicular to $O_1 O_2$ at some point $M$. Let $O_1 M = x$, then $MO_2 = 6 – x$. From the right triangles, $AM^2 = 25 – x^2 = 9 – (6-x)^2$. Solving: $25 – x^2 = 9 – 36 + 12x – x^2$, so $25 = -27 + 12x$, $12x = 52$, $x = \dfrac{13}{3}$. Then $AM^2 = 25 – \dfrac{169}{9} = \dfrac{225-169}{9} = \dfrac{56}{9}$, $AM = \dfrac{\sqrt{56}}{3} = \dfrac{2\sqrt{14}}{3}$. So common chord $AB = 2AM = \dfrac{4\sqrt{14}}{3}$ cm $\approx 4.99$ cm.

Worked Example: Tangent Lengths and Angles

Example A. A circle has radius $r=5$ cm. From an external point $P$ at distance $OP = 13$ cm, two tangents $PT_1$ and $PT_2$ are drawn. Find (i) the length of each tangent, (ii) the angle $\angle T_1 P T_2$, and (iii) the area of the quadrilateral $OT_1 PT_2$.

Solution.

(i) $PT_1 = PT_2 = \sqrt{OP^2 – r^2} = \sqrt{169 – 25} = \sqrt{144} = 12$ cm.

(ii) In the right triangle $OT_1 P$, $\sin(\angle OPT_1) = \dfrac{OT_1}{OP} = \dfrac{5}{13}$, so $\angle OPT_1 = \sin^{-1}\!\left(\dfrac{5}{13}\right) \approx 22.62°$. The angle between the two tangents is twice this: $\angle T_1 P T_2 \approx 45.24°$.

(iii) The quadrilateral $OT_1 PT_2$ consists of two congruent right triangles each with legs $5$ cm and $12$ cm. Total area = $2 \times \dfrac{1}{2} \times 5 \times 12 = 60$ cm².

Example B. Construct two tangents to a circle of radius $4$ cm that are inclined to each other at $90°$. Find the distance of the external point from the centre.

Solution. If the angle between tangents is $90°$, the angle between the two radii is $180° – 90° = 90°$. So $OT_1 PT_2$ is a quadrilateral with all four angles equal to $90°$, i.e. a rectangle. Since $OT_1 = OT_2 = r = 4$ cm and $PT_1 = PT_2$ (tangents from external point), it is actually a square. Hence $PT_1 = 4$ cm and

$$OP = \sqrt{OT_1^2 + PT_1^2} = \sqrt{16 + 16} = 4\sqrt{2} \approx 5.66\text{ cm}.$$

Example C. A line segment $AB = 10$ cm is divided into a ratio $3:2$ at $P$. Find $AP$ and $PB$ and verify by construction.

Solution. $AP = \dfrac{3}{5}\times 10 = 6$ cm; $PB = \dfrac{2}{5}\times 10 = 4$ cm. Construction: mark $5$ equal arcs on ray $AX$, join $A_5 B$, draw line through $A_3$ parallel to $A_5 B$; it meets $AB$ at the required $P$ where $AP=6$ cm.

Worked Example: Tangents to Circles with Different Configurations

Example D. A circle has centre $O$ and radius $3$ cm. Find the length of the chord that is the locus of all points $T$ on the circle such that the tangent at $T$ passes through a fixed external point $P$ at distance $OP = 5$ cm.

Solution. The two tangent points $T_1$ and $T_2$ define a chord — the chord of contact. Its length can be computed using the right triangles. $PT_1 = \sqrt{25-9} = 4$ cm. The chord $T_1 T_2$ is perpendicular to $OP$ at some point $N$. By similar-triangle / area arguments, $T_1 N = \dfrac{r\cdot PT_1}{OP} = \dfrac{3\times 4}{5} = \dfrac{12}{5}$ cm, so the chord $T_1 T_2 = \dfrac{24}{5} = 4.8$ cm.

Example E. Construct a circle of radius $3.5$ cm and from a point on the circle, draw a tangent to the circle. Justify.

Solution. Since the point lies on the circle, only one tangent can be drawn at that point.

Draw the circle with centre $O$ and radius $3.5$ cm.

Take a point $A$ on the circle. Join $OA$.

At $A$, construct a perpendicular $\ell$ to $OA$ (using the standard right-angle construction at a point).

$\ell$ is the required tangent at $A$.

Justification: The radius $OA$ is perpendicular to $\ell$ at $A$, and $A$ lies on the circle. By the converse of the radius–tangent theorem, $\ell$ is the unique tangent at $A$.

Example F. A circle has centre $O$ and radius $6$ cm. A line $AB$ is drawn such that $OA = 10$ cm and $OB = 8$ cm. Construct the tangents from $A$ and from $B$ to the circle. Compare their lengths.

Solution. From $A$: $\sqrt{100-36} = \sqrt{64} = 8$ cm. From $B$: $\sqrt{64-36} = \sqrt{28} = 2\sqrt{7} \approx 5.29$ cm. The tangent from the more distant point is longer.

Some Useful Theorems Used in Constructions

Theorem 1 (Basic Proportionality Theorem / Thales’ Theorem). If a line is drawn parallel to one side of a triangle and intersects the other two sides at distinct points, then it divides those two sides in the same ratio.

Symbolically: in $\triangle ABC$, if $DE\parallel BC$ with $D$ on $AB$ and $E$ on $AC$, then $\dfrac{AD}{DB} = \dfrac{AE}{EC}$.

This theorem is the heart of Constructions 11.1 and 11.2 — it is what allows the parallel-rays method to produce the desired ratio.

Theorem 2 (Converse of BPT). If a line divides any two sides of a triangle in the same ratio, then it is parallel to the third side.

Theorem 3 (AA-similarity). If two angles of one triangle are respectively equal to two angles of another, the two triangles are similar.

Theorem 4 (Tangent–Radius Theorem). The tangent at any point of a circle is perpendicular to the radius drawn through the point of contact.

Theorem 5 (Length of two tangents from an external point). The lengths of tangents drawn from an external point to a circle are equal.

Theorem 6 (Angle in a Semicircle). The angle subtended at any point on the circle by a diameter is a right angle.

This is the theorem that makes Construction 11.3 work: the point of contact lies on the auxiliary circle (where the diameter is $OP$), so the angle there is automatically $90°$.

Common Mistakes Students Make

Confusing the number of arcs in division vs similar-triangle problems. For dividing a segment in ratio $m:n$, mark $m+n$ arcs. For a similar triangle with scale factor $\dfrac{m}{n}$, mark only $\max(m,n)$ arcs.

Drawing the auxiliary ray on the wrong side. For the similar-triangle construction, the ray $BX$ must be drawn on the side of $BC$ opposite to vertex $A$, so that the parallel constructions work cleanly.

Forgetting the “produced” extension when the scale factor exceeds $1$. The new vertex $C’$ lies beyond $C$ on ray $BC$ extended, and similarly $A’$ lies beyond $A$ on ray $BA$ extended.

Incorrect angle calculation in the inclined-tangents problem. If the angle between the tangents is $\theta$, the angle between the radii is $180° – \theta$, NOT $\theta$.

Skipping the justification. Most boards mark the justification separately; even a perfect figure without justification loses marks.

Erasing arcs. Construction marks must remain visible. Examiners verify the construction by inspecting the arcs.

Using a protractor. Constructions are ruler-and-compass-only. Use a protractor only to check the answer, never to build it.

Wrong placement of tangent points. The tangent points $T_1, T_2$ are the intersections of the original circle with the auxiliary circle (whose diameter is $OP$), not with any other circle.

Summary Table of the Three Constructions

Construction Auxiliary tool Number of arcs on auxiliary ray Justification theorem

Divide segment in ratio $m:n$ Acute-angle ray from one endpoint $m+n$ Basic Proportionality Theorem (BPT / Thales)

Similar triangle with scale factor $\dfrac{m}{n}$, $m Acute-angle ray from $B$ on opposite side of $A$ $n$ (the larger of $m,n$) BPT and AA-similarity

Similar triangle with scale factor $\dfrac{m}{n}$, $m>n$ Same as above $m$ (the larger of $m,n$) BPT and AA-similarity, with $BC$ produced

Tangents from external point $P$ Auxiliary circle on $OP$ as diameter — Angle in semicircle = $90°$; tangent $\perp$ radius

Tips for HSLC Examination Questions

In the steps of construction, always state explicitly: “Draw a ray $BX$ making an acute angle with $BC$” — examiners look for this exact phrase.

Always mark the larger of $m, n$ on the auxiliary ray for similar-triangle problems. The total number of arcs is the larger integer (not $m+n$, that rule is only for the line-segment-division problem).

Do not erase the construction arcs in your answer sheet — examiners award marks for visible arc traces.

Provide the justification step. A construction without justification typically loses 1–2 marks.

For tangent problems, remember the four key facts: (i) the radius is perpendicular to the tangent at the point of contact; (ii) the two tangents from an external point are equal in length; (iii) the angle in a semicircle is $90°$; (iv) if the angle between two tangents is $\theta$, the angle between the two radii at the contact points is $180°-\theta$.

For the “bangle” problem, the trick is to first locate the centre using perpendicular bisectors of two chords.

Glossary of Key Terms

Term Meaning

Construction A geometric figure drawn using only an unmarked ruler and a pair of compasses, no protractor.

Ratio $m:n$ The proportion in which a quantity is split into two parts whose sizes are in the ratio of the integers $m$ and $n$.

Scale factor The fraction $\dfrac{m}{n}$ that compares each side of a similar triangle to the corresponding side of the original. If $mn$ it is larger.

Basic Proportionality Theorem (BPT) If a line is drawn parallel to one side of a triangle to intersect the other two sides, it divides those two sides in the same ratio. Used to justify the parallel-rays construction.

Similar triangles Triangles with equal corresponding angles and corresponding sides in the same ratio. The new triangles produced in this chapter are always similar to the original by AA-similarity.

Tangent to a circle A straight line that touches the circle at exactly one point. The radius drawn to that point is perpendicular to the tangent.

External point A point lying outside a circle. From every external point, exactly two tangents can be drawn to the circle, and the two tangent lengths are equal.

Length of tangent If $OP=d$ and the radius is $r$, then the tangent length from $P$ is $\sqrt{d^2-r^2}$.

Auxiliary circle (intermediate circle) The circle drawn on $OP$ as diameter, used to locate the points of contact $T_1, T_2$ of the two tangents from $P$. It exploits the fact that the angle in a semicircle is $90°$.

Angle in a semicircle The angle subtended at any point on a circle by a diameter is $90°$. This is the key theorem behind the tangent-from-external-point construction.

Perpendicular bisector The line that cuts a given segment into two equal parts at $90°$. Used to find the midpoint $M$ of $OP$, and to locate the centre of an unknown circle from two of its chords.

Concentric circles Two or more circles sharing the same centre but having different radii. A tangent from any point on the outer circle to the inner one has length $\sqrt{R^2-r^2}$, where $R$ and $r$ are the two radii.

Ruler-and-compass The classical Greek constraint that all constructions must be performed using only an unmarked straight edge (ruler) and a compass — no measurement of angles is allowed except by construction.

This completes the ASSEB Class 10 Mathematics Chapter 11 — Constructions. Practise drawing each of the three constructions at least three times until you can produce the steps fluently from memory. Pay particular attention to the wording of the justifications; they are frequently the difference between full marks and partial credit in HSLC examinations.