In the given figure- triangles ABC and DCB are right angled A and D respectively and AC - DB- then Delta ABC congDelta DCB of from-AAASASRHSNone of these

Question

Toppr Admin · Accepted Answer

Considering the triangles-xA0-x394-ABC-xA0- and-xA0-x394-DCBAC-DB-xA0- -Given-xA0-x2220-BAC-x2220-CDB-90-x2218-BC-BC-xA0- -Common Hypotenuse-xA0-So- by RHS-xA0- congruency-xA0-x394-ABC-xA0-x2245-xA0- -x394-DCB-xA0-xA0- Corresponding-xA0- sides and angles-xA0- of congruent triangles are also-equal-xA0-

In the given figure, triangles $A B C$ and $D C B$ are right angled at $A$ and $D$ respectively and $A C$ $=$ $D B$ , then $Δ$ $A B C$ $≅ Δ$ $D C B$ of from.

$A A A$
$S A S$
$R H S$
None of these

Considering the triangles
$Δ A B C$ and $Δ D C B$
$A C = D B$ ....Given
$∠ B A C = ∠ C D B = 90^{\circ}$
$B C = B C$ .....Common Hypotenuse
So, by $R H S$ congruency
$Δ A B C$ $≅$ $Δ D C B$
Corresponding sides and angles of congruent triangles are also equal.

In the given figure, triangles ABC and DCB are right angled at A and D respectively and AC = DB, then Δ ABC ≅Δ DCB of from.AAASASRHSNone of these

Considering the triangles ΔABC and ΔDCBAC=DB ....Given ∠BAC=∠CDB=90∘BC=BC .....Common Hypotenuse So, by RHS congruency ΔABC ≅ ΔDCB Corresponding sides and angles of congruent triangles are also equal.

In the given figure, triangles $A B C$ and $D C B$ are right angled at $A$ and $D$ respectively and $A C$ $=$ $D B$ , then $Δ$ $A B C$ $≅ Δ$ $D C B$ of from.

$A A A$
$S A S$
$R H S$
None of these

Considering the triangles
$Δ A B C$ and $Δ D C B$
$A C = D B$ ....Given
$∠ B A C = ∠ C D B = 90^{\circ}$
$B C = B C$ .....Common Hypotenuse
So, by $R H S$ congruency
$Δ A B C$ $≅$ $Δ D C B$
Corresponding sides and angles of congruent triangles are also equal.