State which pairs of triangles in the figure 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.
(N/A) In $\Delta DEF$,the sum of angles is $180^{\circ}$.
Given $\angle D = 70^{\circ}$ and $\angle E = 80^{\circ}$.
Therefore,$\angle F = 180^{\circ} - (70^{\circ} + 80^{\circ}) = 180^{\circ} - 150^{\circ} = 30^{\circ}$.
In $\Delta PQR$,the sum of angles is $180^{\circ}$.
Given $\angle Q = 80^{\circ}$ and $\angle R = 30^{\circ}$.
Therefore,$\angle P = 180^{\circ} - (80^{\circ} + 30^{\circ}) = 180^{\circ} - 110^{\circ} = 70^{\circ}$.
Comparing the two triangles:
$\angle D = \angle P = 70^{\circ}$
$\angle E = \angle Q = 80^{\circ}$
$\angle F = \angle R = 30^{\circ}$
Since all corresponding angles are equal,by the $AAA$ (Angle-Angle-Angle) similarity criterion,the triangles are similar.
Symbolic form: $\Delta DEF \sim \Delta PQR$.
Given $\angle D = 70^{\circ}$ and $\angle E = 80^{\circ}$.
Therefore,$\angle F = 180^{\circ} - (70^{\circ} + 80^{\circ}) = 180^{\circ} - 150^{\circ} = 30^{\circ}$.
In $\Delta PQR$,the sum of angles is $180^{\circ}$.
Given $\angle Q = 80^{\circ}$ and $\angle R = 30^{\circ}$.
Therefore,$\angle P = 180^{\circ} - (80^{\circ} + 30^{\circ}) = 180^{\circ} - 110^{\circ} = 70^{\circ}$.
Comparing the two triangles:
$\angle D = \angle P = 70^{\circ}$
$\angle E = \angle Q = 80^{\circ}$
$\angle F = \angle R = 30^{\circ}$
Since all corresponding angles are equal,by the $AAA$ (Angle-Angle-Angle) similarity criterion,the triangles are similar.
Symbolic form: $\Delta DEF \sim \Delta PQR$.
Explore More
Similar Questions
In the figure,$D$ is a point on the hypotenuse $AC$ of $\Delta ABC,$ such that $BD \perp AC,$ $DM \perp BC,$ and $DN \perp AB.$ Prove that $DM^{2} = DN \cdot MC.$
Difficult
View SolutionUsing Theorem $6.1,$ prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.
Medium
View Solution$ABC$ is an isosceles triangle with $AC = BC$. If $AB^2 = 2 AC^2$,prove that $ABC$ is a right triangle.
Medium
View Solution$CD$ 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 $\Delta EFG$ respectively. If $\Delta ABC \sim \Delta FEG,$ show that:
$(i) \frac{CD}{GH} = \frac{AC}{FG}$
$(ii) \Delta DCB \sim \Delta HGE$
$(iii) \Delta DCA \sim \Delta HGF$
$(i) \frac{CD}{GH} = \frac{AC}{FG}$
$(ii) \Delta DCB \sim \Delta HGE$
$(iii) \Delta DCA \sim \Delta HGF$
Difficult
View Solution$ABC$ is an isosceles triangle right-angled at $C$. Prove that $AB^{2} = 2AC^{2}$.
Difficult
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo