Question : In $\triangle$ABC and $\triangle$DEF, $\angle$A = $55^{\circ}$, AB = DE, AC = DF, $\angle$E = $85^{\circ}$ and $\angle$F = $40^{\circ}$. By which property are $\triangle$ABC and $\triangle$DEF congruent?
Option 1: SAS property
Option 2: ASA property
Option 3: RHS property
Option 4: SSS property
Latest: SSC CGL Tier 1 Result 2024 Out | SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL Tier 1 Scorecard 2024 Released | SSC CGL complete guide
Suggested: Month-wise Current Affairs | Upcoming Government Exams
Correct Answer: SAS property
Solution : In $\triangle$ DEF, $\angle$ D + $\angle$ E + $\angle$ F = $180^{\circ}$ ⇒ $\angle$ D + $85^{\circ}$ + $40^{\circ}$ = $180^{\circ}$ ⇒ $\angle$ D = $180^{\circ} - 85^{\circ} - 40^{\circ}$ = $55^{\circ}$ In $\triangle$ ABC and $\triangle$ DEF, AB = DE AC = DF $\angle$ A = $\angle$ D = $55^{\circ}$ By the SAS property, $\triangle$ABC and $\triangle$DEF are congruent. Hence, the correct answer is SAS property.
Candidates can download this ebook to know all about SSC CGL.
Result | Eligibility | Application | Selection Process | Preparation Tips | Admit Card | Answer Key
Question : In $\triangle$ABC and $\triangle$PQR, AB = PQ and $\angle$B = $\angle$Q. The two triangles are congruent by SAS criteria if:
Question : ABC is an isosceles triangle having AB = AC and $\angle$A = 40$^\circ$. Bisector PO and OQ of the exterior angle $\angle$ ABD and $\angle$ ACE formed by producing BC on both sides, meet at O, then the value of
Question : If it is given that for two right-angled triangles $\triangle$ABC and $\triangle$DFE, $\angle$A = 25°, $\angle$E = 25°, $\angle$B = $\angle$F = 90°, and AC = ED, then which one of the following is TRUE?
Question : I is the incenter of a triangle ABC. If $\angle$ ABC = 65$^\circ$ and $\angle$ ACB = 55$^\circ$, then the value of $\angle$ BIC is:
Question : In $\triangle \mathrm{ABC}, \angle \mathrm{A}=50^{\circ}$, BE and CF are perpendiculars on AC and AB at E and F, respectively. BE and CF intersect at H. The bisectors of $\angle \mathrm{HBC}$ and $\angle H C B$ intersect at P. $\angle B P C$ is equal to:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile