Question

Two triangles are congruent to each other as given in the figure by RHS criteria, then find the values of BC and DE. [3 Marks]

Answer 1

In the figure,
$△$ABC $\cong$ $△$FED by RHS criteria
AC is hypotenuse of $△$ABC
AB , BC are sides of $△$ABC
$\angle$B = ${90}^{\circ }$
and
DF is hypotenuse of $△$DEF
DE , EF are sides of $△$DEF
$\angle$E = ${90}^{\circ }$

AB = 3 cm, EF = 3 cm
AC = 5 cm, DF = 5 cm
[1 Mark]
Applying pythagoras theorem in $\Delta$DEF,
$D{F}^2=D{E}^2+E{F}^2$
$5^2=D{E}^2+3^2$
$25-9=D{E}^2$
$16=D{E}^2$
$DE=\sqrt{16}=4cm$ [1 Mark]

$\therefore$ AC = DF [Hypotenuse]
$\angle$B = $\angle$E = ${90}^{\circ }$
Now, BC = ED = 4 cm [C.P.C.T] [1 Mark]