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Q5:
$S$ and $T$ are points on sides $PR$ and $QR$ of $\triangle PQR$ such that $\angle P = \angle RTS$. Show that $\triangle RPQ \sim \triangle RTS$.
Solution :
Given: A triangle $\triangle PQR$ where $S$ is a point on side $PR$ and $T$ is a point on side $QR$. It is given that $\angle P = \angle RTS$.
To Prove: $\triangle RPQ \sim \triangle RTS$.
Step 1: Identifying the triangles to be compared.
We are required to prove the similarity between $\triangle RPQ$ and $\triangle RTS$. Let us examine the vertices of these two triangles:
- Triangle 1: $\triangle RPQ$
- Triangle 2: $\triangle RTS$
Step 2: Analyzing the angles of the triangles.
To prove that two triangles are similar, we can use the Angle-Angle (AA) similarity criterion, which states that if two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.
Consider $\triangle RPQ$ and $\triangle RTS$:
1. In $\triangle RPQ$, we have $\angle R$ (which is $\angle PRQ$).
2. In $\triangle RTS$, we have $\angle R$ (which is $\angle TRS$).
Since both triangles share the same vertex $R$, we can state:
$\angle PRQ = \angle TRS$ [Common angle to both triangles]
Step 3: Utilizing the given information.
It is explicitly given in the problem statement that:
$\angle P = \angle RTS$
In the context of our triangles:
$\angle RPQ = \angle RTS$ [Given]
Step 4: Applying the AA Similarity Criterion.
We have established two correspondences between the angles of $\triangle RPQ$ and $\triangle RTS$:
1. $\angle R = \angle R$ [Common angle]
2. $\angle P = \angle RTS$ [Given]
By the Angle-Angle (AA) similarity criterion, if two angles of one triangle are equal to two angles of another triangle, the triangles are similar.
Therefore, $\triangle RPQ \sim \triangle RTS$.
Final Answer: Since $\angle R = \angle R$ (common) and $\angle P = \angle RTS$ (given), by the AA similarity criterion, $\triangle RPQ \sim \triangle RTS$.
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(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)
- 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 $\triangle PQS \sim \triangle TQR$.
- 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
Chapters in CBSE - Class 10 Mathematics
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