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  • IIt is given that triangleABC cong triangleFDE and AB = 5 cm, angleB = 40° and angleA = 80°. Then which of the following is true?(A) DF = 5 cm,angleF = 60°(B) DF = 5 cm, angleE = 60°(C) DE = 5 cm, angleE = 60°(D) DE = 5 cm, \angleD = 40°

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?
(A) $D F=5 \mathrm{~cm}, \angle F=60^{\circ}$
(B) $D F=5 \mathrm{~cm}, \angle E=60^{\circ}$
(C) $\mathrm{DE}=5 \mathrm{~cm}, \angle \mathrm{E}=60^{\circ}$
(D) $D E=5 \mathrm{~cm}, \angle \mathrm{D}=40^{\circ}$

Answers (1)

Solution

In $\triangle \mathrm{ABC}, \angle \mathrm{A}=80^{\circ}, \angle \mathrm{B}=40^{\circ}$ and $\angle C=x$ (assume)

$\because \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{x}=180^{\circ}$ (angle sum property of a triangle)

$
\begin{aligned}
& \Rightarrow 80^{\circ}+40^{\circ}+x=180^{\circ} \\
& \Rightarrow x=180^{\circ}-120^{\circ} \\
& \Rightarrow x=60^{\circ} \\
& \Rightarrow D C=60^{\circ} \\
& \because \triangle A B C \cong \triangle F D E
\end{aligned}
$

By ASA congruence relation

$
\mathrm{AB}=\mathrm{FD}, \angle \mathrm{~A}=\angle \mathrm{F}, \angle \mathrm{~B}=\angle \mathrm{D}
$

$
\begin{aligned}
& \Rightarrow D F=5 \mathrm{~cm}, \angle F=80^{\circ}, \angle B=40^{\circ}=\angle D \\
& \because \text { Also, } \angle C=\angle E \\
& \Rightarrow \angle E=60^{\circ}
\end{aligned}
$

Therefore, $\mathrm{DF}=5 \mathrm{~cm}, \angle \mathrm{E}=60^{\circ}$

Hence option (B) is correct.

 

Posted by

infoexpert24

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