Question : It is given that $\triangle \mathrm{ABC} \cong \triangle \mathrm{FDE}$ and $\mathrm{AB}=5 \mathrm{~cm}, \angle \mathrm{B}=40^{\circ}$ and $\angle \mathrm{A}=80^{\circ}$. Then which of the following is true?
Option 1: $\mathrm{DE}=5 \mathrm{~cm} \text{ and }\angle \mathrm{D}=40^{\circ}$
Option 2: $\mathrm{DE}=5 \mathrm{~cm}\text{ and }\angle \mathrm{E}=60^{\circ}$
Option 3: $\mathrm{DF}=5 \mathrm{~cm}\text{ and } \angle \mathrm{F}=60^{\circ}$
Option 4: $\mathrm{DF}=5 \mathrm{~cm}\text{ and }\angle \mathrm{E}=60^{\circ}$
New: SSC CHSL Tier 2 answer key released | SSC CHSL 2024 Notification PDF
Recommended: How to crack SSC CHSL | SSC CHSL exam guide
Don't Miss: Month-wise Current Affairs | Upcoming government exams
Correct Answer: $\mathrm{DF}=5 \mathrm{~cm}\text{ and }\angle \mathrm{E}=60^{\circ}$
Solution :
Given, $\triangle ABC \cong \triangle FDE$ and $AB=5 \mathrm{~cm}, \angle B=40^{\circ}, \angle A=80^{\circ}$
Since, $\triangle FDE \cong \triangle ABC$
So, $DF=AB$ [by corresponding parts of Congruent Triangle]
$D F=5 \mathrm{~cm}$
and $\angle E=\angle C$ [by corresponding parts of Congruent Triangle]
$\Rightarrow\angle E=\angle C=180^{\circ}-(\angle A+\angle B)$
[by angle sum property of a $\triangle A B C$ ]
$\Rightarrow \angle E=180^{\circ}-\left(80^{\circ}+40^{\circ}\right)$
$\Rightarrow \angle E=60^{\circ}$
Hence, the correct answer is $\mathrm{DF}=5 \mathrm{~cm}\text{ and } \angle \mathrm{E}=60^{\circ}$.
Related Questions
Know More about
Staff Selection Commission Combined High ...
Result | Eligibility | Application | Admit Card | Answer Key | Preparation Tips | Cutoff
Get Updates Brochure
Your Staff Selection Commission Combined Higher Secondary Level Exam brochure has been successfully mailed to your registered email id “”.
