This page provides comprehensive Triangles – Exercise 6.3. Free NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6-3. Step-by-step explained answers for CBSE Board exams. Download PDF and practice now.
Master Similarity Criteria (AAA, SSS, SAS) with step-by-step solutions
1. State which pairs of triangles are similar. Write the similarity criterion used and the symbolic representation.
(i) Triangles ABC and PQR
(ii) Triangles DEF and LMN
(i) Analysis:
$\angle A = \angle P = 40°$
$\angle B = \angle Q = 60°$
$\angle C = \angle R = 80°$
Result: $\Delta ABC \sim \Delta PQR$ by AAA Criterion.
(ii) Analysis of Ratios:
$\frac{DE}{LM} = \frac{2}{4} = \frac{1}{2}$
$\frac{EF}{MN} = \frac{2.5}{5} = \frac{1}{2}$
$\frac{DF}{LN} = \frac{3}{6} = \frac{1}{2}$
Result: $\Delta DEF \sim \Delta LMN$ by SSS Criterion.
2. In the figure, $\Delta OD C \sim \Delta OBA$, $\angle BOC = 110°$ and $\angle CDO = 60°$. Find $\angle DOC$, $\angle DCO$ and $\angle OAB$.
1. Find $\angle DOC$:
$\angle DOC + \angle BOC = 180°$ (Linear Pair)
$\angle DOC = 180 - 110 = 70°$.
2. Find $\angle DCO$:
In $\Delta ODC$: $\angle D + \angle DOC + \angle C = 180°$
$60 + 70 + \angle C = 180 \Rightarrow \angle DCO = 50°$.
3. Find $\angle OAB$:
Since $\Delta ODC \sim \Delta OBA$: $\angle OAB = \angle OCD = 50°$.
3. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that $\frac{OA}{OC} = \frac{OB}{OD}$.
In $\Delta OAB$ and $\Delta OCD$:
1. $\angle AOB = \angle COD$ (Vertically Opposite)
2. $\angle OAB = \angle OCD$ (Alt. Interior angles, AB || DC)
By AA Criterion, $\Delta OAB \sim \Delta OCD$.
Therefore, $\frac{OA}{OC} = \frac{OB}{OD}$ (Corresponding sides).
4. In the figure, $\frac{QR}{QS} = \frac{QT}{PR}$ and $\angle 1 = \angle 2$. Show that $\Delta PQS \sim \Delta TQR$.
Step 1: In $\Delta PQR$, $\angle 1 = \angle 2 \Rightarrow PQ = PR$.
Step 2: Given $\frac{QR}{QS} = \frac{QT}{PR}$. Replace $PR$ with $PQ$: $\frac{QR}{QS} = \frac{QT}{PQ}$.
Inverting, $\frac{QS}{QR} = \frac{PQ}{QT}$.
Step 3: In $\Delta PQS$ and $\Delta TQR$:
Ratio of sides is equal and $\angle Q$ is common.
By SAS Criterion: $\Delta PQS \sim \Delta TQR$.
5. S and T are points on sides PR and QR of $\Delta PQR$ such that $\angle P = \angle RTS$. Show that $\Delta RPQ \sim \Delta RTS$.
In $\Delta RPQ$ and $\Delta RTS$:
1. $\angle R = \angle R$ (Common Angle)
2. $\angle P = \angle RTS$ (Given)
$\therefore \Delta RPQ \sim \Delta RTS$ (AA Criterion).
6. In the figure, if $\Delta ABE \cong \Delta ACD$, show that $\Delta ADE \sim \Delta ABC$.
Given $\Delta ABE \cong \Delta ACD \Rightarrow AB = AC$ and $AE = AD$.
So, $\frac{AD}{AB} = \frac{AE}{AC}$.
In $\Delta ADE$ and $\Delta ABC$:
1. $\frac{AD}{AB} = \frac{AE}{AC}$
2. $\angle A$ is common.
Thus, $\Delta ADE \sim \Delta ABC$ (SAS Criterion).
7. In the figure, altitudes AD and CE of $\Delta ABC$ intersect at P. Show that:
(i) $\Delta AEP \sim \Delta CDP$
(ii) $\Delta ABD \sim \Delta CBE$
(iii) $\Delta AEP \sim \Delta ADB$
(iv) $\Delta PDC \sim \Delta BEC$
All parts are proved using AA Similarity (one 90° angle and one common/vertically opposite angle).
(i) $\angle AEP = \angle CDP = 90°$, $\angle APE = \angle CPD$.
(ii) $\angle ADB = \angle CEB = 90°$, $\angle B$ common.
(iii) $\angle AEP = \angle ADB = 90°$, $\angle A$ common.
(iv) $\angle PDC = \angle BEC = 90°$, $\angle C$ common.
8. E is a point on side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that $\Delta ABE \sim \Delta CFB$.
1. $\angle A = \angle C$ (Opposite angles of parallelogram).
2. $AE || BC \Rightarrow \angle AEB = \angle CBF$ (Alt. Interior Angles).
$\therefore \Delta ABE \sim \Delta CFB$ (AA Criterion).
9. In the figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
(i) $\Delta ABC \sim \Delta AMP$
(ii) $\frac{CA}{PA} = \frac{BC}{MP}$
(i) In $\Delta ABC$ and $\Delta AMP$:
$\angle B = \angle M = 90°$ and $\angle A$ is common.
$\therefore \Delta ABC \sim \Delta AMP$ (AA Criterion).
(ii) Since triangles are similar, corresponding sides are proportional: $\frac{CA}{PA} = \frac{BC}{MP}$.
10. CD and GH are respectively the bisectors of $\angle ACB$ and $\angle EGF$ such that D and H lie on sides AB and FE of $\Delta ABC$ and $\Delta EFG$. If $\Delta ABC \sim \Delta FEG$, show that:
(i) $\frac{CD}{GH} = \frac{AC}{FG}$
(ii) $\Delta DCB \sim \Delta HGE$
(iii) $\Delta DCA \sim \Delta HGF$
Given $\Delta ABC \sim \Delta FEG \Rightarrow \angle C = \angle G$.
Halves are equal: $\angle ACD = \angle FGH$ and $\angle DCB = \angle HGE$.
(i) & (iii) In $\Delta DCA$ and $\Delta HGF$:
$\angle A = \angle F$ (Given) and $\angle ACD = \angle FGH$.
$\therefore \Delta DCA \sim \Delta HGF$ (AA). This implies $\frac{CD}{GH} = \frac{AC}{FG}$.
(ii) Similarly for $\Delta DCB$ and $\Delta HGE$ using $\angle B = \angle E$.
11. In the figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD $\perp$ BC and EF $\perp$ AC, prove that $\Delta ABD \sim \Delta ECF$.
1. $AB = AC \Rightarrow \angle B = \angle C$.
2. In $\Delta ABD$ and $\Delta ECF$:
$\angle ABD = \angle ECF$ (from 1)
$\angle ADB = \angle EFC = 90°$ (Given)
$\therefore \Delta ABD \sim \Delta ECF$ (AA Criterion).
12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of $\Delta PQR$. Show that $\Delta ABC \sim \Delta PQR$.
Given $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM}$.
Since medians bisect base, $BC=2BD, QR=2QM$.
$\Rightarrow \frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM}$.
$\Delta ABD \sim \Delta PQM$ (SSS). Thus $\angle B = \angle Q$.
In $\Delta ABC$ and $\Delta PQR$:
$\frac{AB}{PQ} = \frac{BC}{QR}$ and $\angle B = \angle Q$.
$\therefore \Delta ABC \sim \Delta PQR$ (SAS).
13. D is a point on the side BC of a triangle ABC such that $\angle ADC = \angle BAC$. Show that $CA^2 = CB \cdot CD$.
In $\Delta ABC$ and $\Delta DAC$:
1. $\angle BAC = \angle ADC$ (Given)
2. $\angle C = \angle C$ (Common)
$\therefore \Delta ABC \sim \Delta DAC$ (AA).
Ratio: $\frac{CA}{CD} = \frac{CB}{CA} \Rightarrow CA^2 = CB \cdot CD$.
14. Sides AB and AC and median AD of a triangle ABC are proportional to sides PQ and PR and median PM of another triangle PQR. Show $\Delta ABC \sim \Delta PQR$.
Construction: Extend AD to E such that AD=DE. Join BE, CE.
1. Prove $\Delta ABD \cong \Delta ECD \Rightarrow AB=CE$.
2. Use proportionality to show $\Delta ACE \sim \Delta PRN$ (where N is ext point for PQR).
3. This gives $\angle CAD = \angle RPM$. Similarly $\angle BAD = \angle QPM$.
4. Total $\angle A = \angle P$. With proportional sides, $\Delta ABC \sim \Delta PQR$ (SAS).
15. A vertical pole of length 5 m casts a shadow 3 m long. At the same time, a tower casts a shadow 24 m long. Find the height of the tower.
Similar triangles by AA (90° and sun's angle).
$\frac{h}{5} = \frac{24}{3}$
$h = \frac{24}{3} \times 5 = 8 \times 5 = 40$ m.
16. If AD and PM are medians of $\Delta ABC$ and $\Delta PQR$ respectively, where $\Delta ABC \sim \Delta PQR$, prove that $\frac{AB}{PQ} = \frac{AD}{PM}$.
1. $\Delta ABC \sim \Delta PQR \Rightarrow \angle B = \angle Q$ and $\frac{AB}{PQ} = \frac{BC}{QR}$.
2. Medians bisect BC and QR, so $\frac{BC}{QR} = \frac{2BD}{2QM} = \frac{BD}{QM}$.
3. In $\Delta ABD$ and $\Delta PQM$:
$\frac{AB}{PQ} = \frac{BD}{QM}$ and $\angle B = \angle Q$.
$\therefore \Delta ABD \sim \Delta PQM$ (SAS).
Hence $\frac{AB}{PQ} = \frac{AD}{PM}$.