Triangles Test

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Triangles Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

It is given that $$AB=BC$$ and $$AD=EC.$$ The $$\triangle ABE\cong \triangle CBD$$ by __________ congruency.

A
$$SSS$$

B
$$ASA$$

C
$$SAS$$

D
$$AAS$$

Solution

Given, $$AD=EC$$
$$\Rightarrow $$ $$AD+DE=DE+EC$$
$$\Rightarrow $$ $$AE=DC$$
Also, $$AB=BC$$
$$\Rightarrow $$ $$\angle BCA=\angle BAC$$          (isos. $$\triangle $$ property)
$$\Rightarrow $$ $$\angle BCD=\angle BAE$$
$$\therefore $$ In $$\triangle $$s $$ABE$$ and $$CBD,$$
$$AB=CB$$          (Given)
$$AE=DC$$          (proved above)
$$\angle BAE=\angle BCD$$ (Proved above)
$$\therefore $$ $$\triangle ABE\cong \triangle CBD$$          (SAS)
Question 2
1 / -0

In the figure, $$\displaystyle \Delta ABC\cong \Delta ADC$$ by the congruence postulate

A
SSS

B
RHS

C
SAS

D
ASA

Solution

According to Right angle hypotenuse side rule, two right-angled triangles are congruent if the hypotenuses and one pair of corresponding sides are equal.

Hence, the given two triangles are congruent by the RHS rule.
Question 3
1 / -0

In the figure, $$\displaystyle AD\parallel BC$$ and $$\displaystyle AD=BC$$. Then $$\displaystyle \Delta ABD\cong \Delta CDB$$ by congruence postulate.

A
SSS

B
ASA

C
AAS

D
SAS

Solution

In $$\displaystyle \Delta ABD$$ and $$\displaystyle \Delta CDB$$
$$\displaystyle AD=BC$$
$$\displaystyle \angle BDA=\angle DBC$$ (Alternate angles)
$$\displaystyle BD=BD$$
$$\displaystyle \therefore \quad \Delta ABD\cong \Delta CDB$$ (By SAS postulate)
Question 4
1 / -0

In the above figure, $$\angle{B}=\angle{C}=65^\circ$$ and $$\angle{D}=30^\circ$$. Then

A
$$BC < CA < CD$$

B
$$BC>CA>CD$$

C
$$BC < CA$$, $$CA>CD$$

D
$$BC>CA$$, $$CA < CD$$

Solution

Here $$\angle{B}=\angle{C}=65^\circ$$ and $$\angle{D}=30^\circ$$
$$\therefore$$ From the triangle ABD $$\angle{A}=85^\circ$$
In triangle ABC $$\angle{A}=50^\circ$$
$$\because BC<CA$$
In triangle ACD $$\angle{A}=85^\circ-50^\circ=35^\circ$$
$$\therefore CD>CA$$
Question 5
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If $$\overline{AC}$$ and $$\overline{BD}$$ intersect at $$O$$ such that $$AO = CO$$ and $$BO = DO$$, then:

A
$$AB = CD$$

B
$$AB \parallel CD$$ and $$AB = CD$$

C
$$AB \parallel CD$$

D
None of these

Solution

In $$\triangle AOB$$ and $$\triangle DOC$$,
side $$OA =$$ side $$OC$$ ... given
side $$OB =$$ side $$OD$$ ... given
$$\angle AOB \cong \angle COD$$ ...Vertically opposite angles
$$\therefore \triangle AOB \cong \triangle COD$$ ...SAS test of congruence.
$$\therefore$$ side $$AB$$ $$\cong$$ side $$CD$$ ...c.s.c.t
$$\therefore \angle BAO \cong \angle DCO$$ ....c.a.c.t
i.e. $$\angle BAC \cong \angle DCA$$ ....$$A-O-C$$ and $$B-O-D$$
$$\therefore AB \parallel CD$$ ....by converse of alternate angles test
So, option B is correct.
Question 6
1 / -0

In the figure, $$\displaystyle PS=QR$$ and $$PQ=SR$$, then choose the correct option.

A
$$\displaystyle \Delta QRP\cong \Delta PSR$$

B
$$\displaystyle \Delta PQR\cong \Delta PRS$$

C
$$\displaystyle \Delta PQR\cong \Delta RSP$$

D
$$\displaystyle \Delta RQP\cong \Delta RSP$$

Solution

In $$\displaystyle \Delta PQR$$ and $$\displaystyle \Delta PSR$$
$$\displaystyle PS=QR,PQ=SR,PR=PR$$
$$\displaystyle \therefore \quad \Delta RQP\cong \Delta RSP$$ (By SSS postulate)
Question 7
1 / -0

In $$\displaystyle \Delta ABC$$ and $$\displaystyle \Delta DBC,AC=BD$$ and $$\displaystyle \angle ABC=\angle BCD={ 90 }^{ o }$$. Then $$\displaystyle \Delta ABC\cong \Delta DCB$$ by the postulate

A
SSS

B
SAS

C
RHS

D
ASA

Solution

In $$\displaystyle \Delta ABC$$ and $$\displaystyle \Delta DBC$$,
$$\displaystyle \angle ABC=\angle BCD$$
$$\displaystyle AC=BD$$ (Givenn)
$$\displaystyle BC=BC$$ (Common Side)
$$\displaystyle \Delta ABC\cong \Delta DCB$$ WCB (By RHS postulate)
Question 8
1 / -0

In $$ \displaystyle \bigtriangleup ABC,AD\perp BC,\angle B=\angle C $$ and AB = AC State by which property $$ \displaystyle \Delta ADB\cong \Delta ADC$$ ?

A
RHS property

B
SSS property

C
SAS property

D
ASA property

Solution
Question 9
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The two quadrilaterals are congruent.
Which angle in quadrilateral WXYZ corresponds to $$\displaystyle \angle BCD$$ in quadrilateral ABCD?

A
$$\displaystyle \angle XYZ$$

B
$$\displaystyle \angle XWZ$$

C
$$\displaystyle \angle WZY$$

D
$$\displaystyle \angle WXY$$

Solution

Clearly, $$\angle WXY$$ corresponds to $$\angle BCD$$ because $$BC$$ faces the angles marked with one arc and three arc and $$XY$$ also faces the angles marked with one arc and three arcs.
So, $$\text{D}$$ is the correct option.
Question 10
1 / -0

In $$\displaystyle \Delta ABC,\angle A={ 12 }^{ o }$$ and $$\angle C={ 97 }^{ o }$$. Arranging the sides (according to lengths) in ascending order, we get:

A
$$BC<AB< AC$$

B
$$AB< AC< BC$$

C
$$BC< AC< AB$$

D
$$AB< BC < AC$$

Solution

In $$\displaystyle \Delta ABC,\angle A={ 12 }^{ o }$$ and $$\angle C={ 97 }^{ o }$$
$$\displaystyle \therefore \quad \angle B={ 71 }^{ o }\quad \left( \because \angle A+\angle B+\angle C={ 180 }^{ o } \right) $$
Now arranging the angles in ascending order $$\displaystyle \angle A<\angle B <\angle C$$.
Sides opposite to greater angle are greater in length and the angles and the opposite sides of a triangle have same relation.
Thus, arranging the sides in ascending order, we get $$BC< AC <AB.$$