Question

Question 135

In each of the given pairs of triangles in given figures, using only RHS congruence criterion, determine which pairs of triangles are congruent. In case of congruence, write the result in symbolic form.

Answer 1

a) In $\Delta ABD$ and $\Delta ACD$, AB = AC [given]
AD = AD [common side]
$\angle ADB=\angle ADC={90}^{\circ }$
By RHS congruence criterion, $\Delta ABD\cong \Delta ACD$

b) In $\Delta XYZ$ and $\Delta UZY$, $\angle XYZ=\angle UZY={90}^{\circ }$
XZ = YU [given]
ZY = ZY [common side]
By RHS congruence criterion, $\Delta XYZ\cong \Delta UZY$

c) In $\Delta AEC$ and $\Delta BED$, CE = DE [given]
AE = BE [given]
$\angle ACE=\angle BDE={90}^{\circ }$
By RHS congruence criterion, $\Delta AEC\cong \Delta BED$

d) Here, CD = BD - BC = 14 - 8 = 6 cm
In right angled $\Delta ABC$,
$AC=\sqrt{A{B}^2+B{C}^2}=\sqrt{6^2+8^2}=\sqrt{36+64}$ [By Pythagoras theorem]
$=\sqrt{100}=10cm$
In right angled $\Delta CDE$,
$DE=\sqrt{C{E}^2-C{D}^2}=\sqrt{{10}^2-6^2}=\sqrt{100-36}=\sqrt{64}=8cm$
In $\Delta ABC$ and $\Delta CDE,AC=CE=10cm$
BC = DE = 8 cm
$\angle ABC=\angle CDE={90}^{\circ }$
By RHS congruence criterion, $\Delta ABC\cong \Delta CDE$

e) Not possible, because there is not any right angle.

f) In $\Delta LOM$ and $\Delta LON$, LM = LN = 8 cm
LO = LO [common side]
$\angle LOM=\angle LON={90}^{\circ }$
By RHS congruence criterion, $\Delta LOM\cong \Delta LON$