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Q1(v):
State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form : (v)
State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form : (v)
Solution :
Given: Two triangles, $\triangle ABC$ and $\triangle DEF$.
In $\triangle ABC$:
- $AB = 2.5$
- $BC = 3$
- $\angle A = 80^\circ$
In $\triangle DEF$:
- $EF = 6$
- $DF = 5$
- $\angle F = 80^\circ$
To Find: Determine if the triangles are similar, state the criterion used, and write the symbolic representation if they are similar.
Step 1: Analyze the sides and angles of the triangles.
We examine the ratios of the sides that include the given angle. In $\triangle ABC$, the angle $80^\circ$ is at vertex $A$. The sides forming this angle are $AB$ and $AC$. However, the length of $AC$ is not provided. In $\triangle DEF$, the angle $80^\circ$ is at vertex $F$. The sides forming this angle are $EF$ and $DF$.
Step 2: Check for similarity criteria.
For two triangles to be similar by the SAS (Side-Angle-Side) criterion, the ratio of the two sides must be equal, and the included angle must be equal.
Let us check the ratios of the given sides:
Ratio 1: $\frac{AB}{DF} = \frac{2.5}{5} = 0.5$
Ratio 2: $\frac{BC}{EF} = \frac{3}{6} = 0.5$
Here, the ratios of the sides are equal ($\frac{AB}{DF} = \frac{BC}{EF} = \frac{1}{2}$).
Step 3: Evaluate the included angle condition.
For SAS similarity, the equal angle must be the included angle between the proportional sides.
In $\triangle ABC$, the sides are $AB$ and $BC$. The included angle is $\angle B$. We are given $\angle A = 80^\circ$.
In $\triangle DEF$, the sides are $DF$ and $EF$. The included angle is $\angle F$. We are given $\angle F = 80^\circ$.
Since the given angle $\angle A$ in $\triangle ABC$ is not the included angle between sides $AB$ and $BC$, and we do not have information about $\angle B$ or $\angle F$ being the included angle for the proportional sides in both triangles, the SAS criterion cannot be applied.
Step 4: Conclusion.
Because the equal angles ($80^\circ$) are not the included angles between the sides that are in the same ratio, the triangles do not satisfy the SAS similarity criterion. There is no other information provided to satisfy AA or SSS criteria.
Final Answer: The triangles are not similar.
More Questions from Class 10 Mathematics Triangles EXERCISE 6.3
- Q1(i): State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form : (i)
- Q1(ii): State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form : (ii)
- Q1(iii): State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form : (iii)
- Q1(iv): State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form : (iv)
- Q1(vi): State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form : (vi)
- Q10(i): $CD$ and $GH$ are respectively the bisectors of $\angle ACB$ and $\angle EGF$ such that $D$ and $H$ lie on sides $AB$ and $FE$ of $\triangle ABC$ and $\triangle EFG$ respectively. If $\triangle ABC \sim \triangle FEG$, show that: (i) $\frac{CD}{GH} = \frac{AC}{FG}$
- Q10(ii): $CD$ and $GH$ are respectively the bisectors of $\angle ACB$ and $\angle EGF$ such that $D$ and $H$ lie on sides $AB$ and $FE$ of $\triangle ABC$ and $\triangle EFG$ respectively. If $\triangle ABC \sim \triangle FEG$, show that: (ii) $\triangle DCB \sim \triangle HGE$
- Q10(iii): $CD$ and $GH$ are respectively the bisectors of $\angle ACB$ and $\angle EGF$ such that $D$ and $H$ lie on sides $AB$ and $FE$ of $\triangle ABC$ and $\triangle EFG$ respectively. If $\triangle ABC \sim \triangle FEG$, show that: (iii) $\triangle DCA \sim \triangle HGF$
- Q11: In Fig. 6.40, $E$ 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 $\triangle ABD \sim \triangle ECF$.
- Q12: Sides $AB$ and $BC$ and median $AD$ of a triangle $ABC$ are respectively proportional to sides $PQ$ and $QR$ and median $PM$ of $\triangle PQR$ (see Fig. 6.41). Show that $\triangle ABC \sim \triangle PQR$.
- Q13: $D$ is a point on the side $BC$ of a triangle $ABC$ such that $\angle ADC = \angle BAC$. Show that $CA^2 = CB \cdot CD$.
- Q14: Sides $AB$ and $AC$ and median $AD$ of a triangle $ABC$ are respectively proportional to sides $PQ$ and $PR$ and median $PM$ of another triangle $PQR$. Show that $\triangle ABC \sim \triangle PQR$.
- Q15: A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
- Q16: If $AD$ and $PM$ are medians of triangles $ABC$ and $PQR$, respectively where $\triangle ABC \sim \triangle PQR$, prove that $\frac{AB}{PQ} = \frac{AD}{PM}$.
- Q2: In Fig. 6.35, $\triangle ODC \sim \triangle OBA$, $\angle BOC = 125^{\circ}$ and $\angle CDO = 70^{\circ}$. Find $\angle DOC$, $\angle DCO$ and $\angle OAB$.
- Q3: Diagonals $AC$ and $BD$ of a trapezium $ABCD$ with $AB \parallel DC$ intersect each other at the point $O$. Using a similarity criterion for two triangles, show that $\frac{OA}{OC} = \frac{OB}{OD}$.
- Q4: In Fig. 6.36, $\frac{QR}{QS} = \frac{QT}{PR}$ and $\angle 1 = \angle 2$. Show that $\triangle PQS \sim \triangle TQR$.
- Q5: $S$ and $T$ are points on sides $PR$ and $QR$ of $\triangle PQR$ such that $\angle P = \angle RTS$. Show that $\triangle RPQ \sim \triangle RTS$.
- Q6: In Fig. 6.37, if $\triangle ABE \cong \triangle ACD$, show that $\triangle ADE \sim \triangle ABC$.
- Q7(i): In Fig. 6.38, altitudes $AD$ and $CE$ of $\triangle ABC$ intersect each other at the point $P$. Show that: (i) $\triangle AEP \sim \triangle CDP$
- Q7(ii): In Fig. 6.38, altitudes $AD$ and $CE$ of $\triangle ABC$ intersect each other at the point $P$. Show that: (ii) $\triangle ABD \sim \triangle CBE$
- Q7(iii): In Fig. 6.38, altitudes $AD$ and $CE$ of $\triangle ABC$ intersect each other at the point $P$. Show that: (iii) $\triangle AEP \sim \triangle ADB$
- Q7(iv): In Fig. 6.38, altitudes $AD$ and $CE$ of $\triangle ABC$ intersect each other at the point $P$. Show that: (iv) $\triangle PDC \sim \triangle BEC$
- Q8: $E$ is a point on the side $AD$ produced of a parallelogram $ABCD$ and $BE$ intersects $CD$ at $F$. Show that $\triangle ABE \sim \triangle CFB$.
- Q9(i): In Fig. 6.39, $ABC$ and $AMP$ are two right triangles, right angled at $B$ and $M$ respectively. Prove that: (i) $\triangle ABC \sim \triangle AMP$
- Q9(ii): In Fig. 6.39, $ABC$ and $AMP$ are two right triangles, right angled at $B$ and $M$ respectively. Prove that: (ii) $\frac{CA}{PA} = \frac{BC}{MP}$
CBSE Solutions for Class 10 Mathematics Triangles
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