Free Congruence of Triangles 03 Practice Test - 7th grade 

Criteria for congruence of triangles are SSS, SAS, ASA, AAS, RHS.

SSS- Two  triangles are congruent if all the 3 corresponding sides of the given triangles are equal. 

SAS- Two triangles are congruent if 2 corresponding sides of the given triangles and the corresponding angle between those sides are equal to each other. 

ASA- Two triangles are congruent if two angles and the included side of one triangle are equal to the corresponding angles and sides of other triangles. 

AAS- Two triangles are congruent if two pairs of corresponding angles and a pair of opposite sides are equal.

RHS- If the hypotenuse and a side of a right angled triangle are congruent with the hypotenuse and the corresponding side
of the other right angled triangle, then the two triangles are congruent with each other.

Consider $\Delta BOC$  and $\Delta AOD$
1) AD = BC  ( Given ) 
2) $\angle CBO=\angle DAO$= 90°
3) $\angle BOC=\angle AOD$ ......(vertically opposite angles)

$\therefore \Delta BOC\cong \Delta AOD$ ....[AAS Criterion]

$\Rightarrow$ OC = OD ....(congruent parts of congruent triangle)

Since, $\Delta BOC\cong \Delta AOD$, then $\angle BOC={30}^o$. From angle sum property of triangle in $\Delta BOC$, the measure of $\angle BCO$ is ${60}^o$.

Since, $\Delta BOC\cong \Delta AOD$,

Corresponding angles of the triangles are equal. It gives,
$\angle BOC={30}^o$ 
$\angle BOC+\angle BCO+\angle OBC={180}^o$ [Angle sum property]  
$\angle BCO$​​​​​​​ = 180 - (90 + 30) = 180 - 120 = ${60}^o$

In two triangles,

If a pair of corresponding angles and the included side are equal, then they are congruent [ASA congruence criterion].

If a pair of corresponding angles and a non-included side are equal, then they are congruent [AAS congruence criterion].

Therefore, given statement is true.

If all the side lengths of one triangle are equal to the side lengths of another triangle, then the triangles are congruent. This is called SSS criterion.

Given, $AB=AC$
$⟹94=6x+4⟹6x=90⟹x=15$
 

In $△ABD$ and $△ACD$,
(i) $\angle ADB=\angle ADC={90}^{\circ }$ ... (given)
(ii) AB = AC ... (given)
(iii) AD = AD ... (common side)
$\Rightarrow$$△ABD\cong △ADC$ ... (RHS congruence rule)
Then, $BD=CD$ ... (CPCT)
$\Rightarrow 2y–7=11\Rightarrow 2y=18\Rightarrow y=9$

In $△ABD$ and $△ADC$

(i) $\angle ADB=\angle ADC={90}^{\circ }$ .......(given)

(ii) AD = AD ....... (common)

(iii) AB = AC ....... (given)

(iv) $△ABD\cong △ADC$....... (RHS Postulate)

(v) BD = DC …… (cpct)

(vi) $\angle ABD$ = $\angle ACD={60}^{\circ }$ .......(cpct)

Hence (A) and (C)

In $△ABO$ and $△CBO$

(i) $\angle BAO=\angle CAO={90}^{\circ }$ ........ (given)

(ii) BO = BO .........(common side)

(iii) AB = BC........ (given)

$\Rightarrow △ABO\cong △CBO$    ........ (RHS Postulate)
$\Rightarrow$ OA = OC.......(cpct)
$\Rightarrow$ $\angle ABO=\angle CBO={60}^{\circ }$ ........(cpct)

Consider $△ABC$ and $△ADC$



(i) AB = CD ...... (given)
(ii) BC = DA ...... (given)
(iii) AC = AC ...... (common)

$\therefore △ABC\cong △CDA$ ... (SSS Postulate)

$\Rightarrow \angle ABC=\angle CDA$ ..... (CPCT)

$\therefore x={48}^{\circ }$

 $\Rightarrow \angle BCA=\angle DAC$ ......(CPCT)

$\therefore y={56}^{\circ }$

In $△ABD$ and $△ACD$

(i) AB = AC .........(given)

(ii) BD = CD .........(given)

(iii) AD = AD ..........(common)

(iv)$△ABD\cong △ACD$ ......(SSS Postulate)

(v) $\angle BAD=\angle CAD$ .....(cpct)

$\therefore$ AD bisects $\angle$ BAC

(vi) $\angle ABD=\angle ACD$ .....(cpct)

If $\angle BAC={80}^{\circ }$, then $\angle ABD=\angle ACD={50}^{\circ }$ ( not ${80}^{\circ }$)