10th Maths Chapter 4 Geometry

10th Maths Chapter 4 Geometry solutions, with the chapter's formulas and every exercise worked step by step, as per the 2026-27 Samacheer Kalvi textbook.

This chapter looks at similar triangles and circles. You learn when two triangles are similar, using the AA, SAS and SSS tests, and how similarity links their sides, perimeters and areas. It also covers tangents to a circle and their properties. In the exercises you prove results, find unknown lengths, and solve height, shadow and distance problems from real situations.

Exercises in this chapter

Key formulas

Similar triangles — equal corresponding angles
$\triangle ABC \sim \triangle PQR \Rightarrow \angle A = \angle P,\ \angle B = \angle Q,\ \angle C = \angle R$
Similar triangles — proportional corresponding sides
$\triangle ABC \sim \triangle PQR \Rightarrow \dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{CA}{RP}$
AA (same as AAA) similarity criterion
$\angle A = \angle P,\ \angle B = \angle Q \Rightarrow \triangle ABC \sim \triangle PQR$
SAS similarity criterion
$\angle A = \angle P \text{ and } \dfrac{AB}{PQ} = \dfrac{AC}{PR} \Rightarrow \triangle ABC \sim \triangle PQR$
SSS similarity criterion
$\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{CA}{RP} \Rightarrow \triangle ABC \sim \triangle PQR$
Ratio of perimeters of similar triangles
$\dfrac{\text{Perimeter of } \triangle ABC}{\text{Perimeter of } \triangle PQR} = \dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{CA}{RP}$
Ratio of areas of similar triangles
$\dfrac{\text{Area of } \triangle ABC}{\text{Area of } \triangle PQR} = \dfrac{AB^2}{PQ^2} = \dfrac{BC^2}{QR^2} = \dfrac{CA^2}{RP^2}$
Tangent perpendicular to radius at the point of contact
$OP \perp PT,\ \text{i.e. } \angle OPT = 90^\circ$
Number of tangents from an exterior point
$\text{From an exterior point } P,\ \text{exactly } 2 \text{ tangents can be drawn to a circle}$
Lengths of tangents from an exterior point are equal
$PA = PB$
Direct common tangents to two circles
$AB = CD \quad (\text{the two direct common tangents are equal in length})$
Median relation (Apollonius type) — part (i)
$b^{2} = p^{2} + ax + \dfrac{a^{2}}{4}$
Median relation (Apollonius type) — part (ii)
$c^{2} = p^{2} - ax + \dfrac{a^{2}}{4}$
Median relation (Apollonius type) — part (iii)
$b^{2} + c^{2} = 2p^{2} + \dfrac{a^{2}}{2}$

Chapter 4 one-mark practice

If in triangles $ABC$ and $EDF$ , $\dfrac{AB}{DE} = \dfrac{BC}{FD}$ then they will be similar, when
1. (1) $\angle B = \angle E$
2. (2) $\angle A = \angle D$
3. (3) $\angle B = \angle D$
4. (4) $\angle A = \angle F$
Show answer
Answer: (3) $\angle B = \angle D$
1. $\dfrac{AB}{DE}=\dfrac{BC}{FD}$
2. In $\Delta ABC$ the sides $AB$ , $BC$ include $\angle B$ .
3. In $\Delta EDF$ the sides $DE$ , $FD$ include $\angle D$ .
4. By SAS similarity, the included angles must be equal.
5. So $\angle B=\angle D$ .
Question 1 of 10
In $\Delta LMN$ , $\angle L = 60^\circ$ , $\angle M = 50^\circ$ . If $\Delta LMN \sim \Delta PQR$ then the value of $\angle R$ is
1. (1) $40^\circ$
2. (2) $70^\circ$
3. (3) $30^\circ$
4. (4) $110^\circ$
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Answer: (2) $70^\circ$
1. In $\Delta LMN$ , $\angle N = 180^\circ-(60^\circ+50^\circ)=70^\circ$ .
2. $\Delta LMN \sim \Delta PQR$ gives $\angle R = \angle N$ .
3. $\angle R = 70^\circ$ .
Question 2 of 10
If $\Delta ABC$ is an isosceles triangle with $\angle C = 90^\circ$ and $AC = 5$ cm, then $AB$ is
1. (1) $2.5$ cm
2. (2) $5$ cm
3. (3) $10$ cm
4. (4) $5\sqrt{2}$ cm
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Answer: (4) $5\sqrt{2}$ cm
1. $\angle C=90^\circ$ , so $AB$ is the hypotenuse and the equal sides are $AC=BC=5$ cm.
2. $AB^2=AC^2+BC^2=5^2+5^2=50$
3. $AB=\sqrt{50}=5\sqrt{2}$ cm.
Question 3 of 10
In a given figure $ST \parallel QR$ , $PS = 2$ cm and $SQ = 3$ cm. Then the ratio of the area of $\Delta PQR$ to the area of $\Delta PST$ is
1. (1) $25 : 4$
2. (2) $25 : 7$
3. (3) $25 : 11$
4. (4) $25 : 13$
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Answer: (1) $25 : 4$
1. $PQ=PS+SQ=2+3=5$ cm.
2. $ST\parallel QR \Rightarrow \Delta PST \sim \Delta PQR$ .
3. $\dfrac{[\Delta PQR]}{[\Delta PST]}=\dfrac{PQ^2}{PS^2}=\dfrac{5^2}{2^2}=\dfrac{25}{4}$ .
Question 4 of 10
The perimeters of two similar triangles $\Delta ABC$ and $\Delta PQR$ are $36$ cm and $24$ cm respectively. If $PQ = 10$ cm, then the length of $AB$ is
1. (1) $6\dfrac{2}{3}$ cm
2. (2) $\dfrac{10\sqrt{6}}{3}$ cm
3. (3) $66\dfrac{2}{3}$ cm
4. (4) $15$ cm
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Answer: (4) $15$ cm
1. In similar triangles, ratio of perimeters = ratio of corresponding sides.
2. $\dfrac{AB}{PQ}=\dfrac{36}{24}$
3. $AB=10\times\dfrac{36}{24}=15$ cm.
Question 5 of 10
If in $\Delta ABC$ , $DE \parallel BC$ . $AB = 3.6$ cm, $AC = 2.4$ cm and $AD = 2.1$ cm then the length of $AE$ is
1. (1) $1.4$ cm
2. (2) $1.8$ cm
3. (3) $1.2$ cm
4. (4) $1.05$ cm
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Answer: (1) $1.4$ cm
1. $DE\parallel BC$ , so by Thales theorem $\dfrac{AD}{AB}=\dfrac{AE}{AC}$ .
2. $\dfrac{2.1}{3.6}=\dfrac{AE}{2.4}$
3. $AE=\dfrac{2.1\times 2.4}{3.6}=1.4$ cm.
Question 6 of 10
In a $\Delta ABC$ , $AD$ is the bisector of $\angle BAC$ . If $AB = 8$ cm, $BD = 6$ cm and $DC = 3$ cm. The length of the side $AC$ is
1. (1) $6$ cm
2. (2) $4$ cm
3. (3) $3$ cm
4. (4) $8$ cm
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Answer: (2) $4$ cm
1. $AD$ bisects $\angle BAC$ , so $\dfrac{AB}{AC}=\dfrac{BD}{DC}$ .
2. $\dfrac{8}{AC}=\dfrac{6}{3}=2$
3. $AC=\dfrac{8}{2}=4$ cm.
Question 7 of 10
In the adjacent figure $\angle BAC = 90^\circ$ and $AD \perp BC$ then
1. (1) $BD \cdot CD = BC^2$
2. (2) $AB.AC = BC^2$
3. (3) $BD \cdot CD = AD^2$
4. (4) $AB \cdot AC = AD^2$
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Answer: (3) $BD \cdot CD = AD^2$
1. $\angle BAC=90^\circ$ and $AD\perp BC$ , so $AD$ is the altitude to the hypotenuse.
2. Then $\Delta ABD \sim \Delta CAD$ .
3. $\dfrac{BD}{AD}=\dfrac{AD}{CD}$
4. $BD\cdot CD = AD^2$ .
Question 8 of 10
Two poles of heights 6 m and 11 m stand vertically on a plane ground. If the distance between their feet is 12 m, what is the distance between their tops?
1. (1) $13$ m
2. (2) $14$ m
3. (3) $15$ m
4. (4) $12.8$ m
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Answer: (1) $13$ m
1. Difference of heights $=11-6=5$ m.
2. Distance between feet $=12$ m.
3. Distance between tops $=\sqrt{12^2+5^2}=\sqrt{144+25}=\sqrt{169}$
4. $=13$ m.
Question 9 of 10
In the given figure, $PR = 26$ cm, $QR = 24$ cm, $\angle PAQ = 90^\circ$ , $PA = 6$ cm and $QA = 8$ cm. Find $\angle PQR$
1. (1) $80^\circ$
2. (2) $85^\circ$
3. (3) $75^\circ$
4. (4) $90^\circ$
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Answer: (4) $90^\circ$
1. In right $\Delta PAQ$ , $PQ^2=PA^2+QA^2=6^2+8^2=100$ , so $PQ=10$ cm.
2. Now $PQ^2+QR^2=10^2+24^2=100+576=676$ .
3. $PR^2=26^2=676$ .
4. Since $PQ^2+QR^2=PR^2$ , by converse of Pythagoras theorem $\angle PQR=90^\circ$ .
Question 10 of 10

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