Triangles Test

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Triangles Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

In the given figure, triangles $$ABC$$ and $$DCB$$ are right angled at $$A$$ and $$D$$ respectively and $$AC$$ $$=$$ $$DB$$, then $$\Delta$$ $$ABC$$ $$\cong\Delta$$ $$DCB$$ of from.

A
$$AAA$$

B
$$SAS$$

C
$$RHS$$

D
None of these

Solution

In $$\Delta ABC$$ and $$\Delta DCB,$$
$$AC = DB$$ [Given]
$$ \angle BAC = \angle CDB = 90^{\circ}$$
$$BC=BC$$ [Common Hypotenuse ]

So, by $$RHS$$ rule of congruence,
$$\Delta ABC $$ $$\cong $$ $$\Delta DCB$$

Hence, option $$C$$ is correct.
Question 2
1 / -0

The ________ criterion is used to construct a triangle congruent to another triangle whose length of three sides are given.

A
$$SAS$$

B
$$SSS$$

C
$$RHS$$

D
$$ASA$$

Solution

When the length of all three sides of a triangle are given, then by Side-Side-Side i.e. $$SSS$$ criterion we can say that the sides of the other triangle will be equal to that by $$CPCT$$.

Hence, the answer is $$SSS$$.
Question 3
1 / -0

In $$\triangle ABC$$, if $$AB = 7$$ cm, $$\angle A= 40^o$$ and $$\angle B = 70^o$$, which criterion can be used to construct this triangle?

A
ASA

B
SSS

C
SAS

D
RHS

Solution

Given : In a triangle $$ABC, AB = 7 cm, \angle A=40^{o}$$ and $$\angle B=70^{o}$$
So, here we know two of the angles and a side including these angles.
Hence, Angle Side Angle i.e ASA criterion can be used to construct this triangle.
Question 4
1 / -0

If $$ \triangle ABC$$ and $$\triangle DBC$$ are on the same base BC,AB=DC and AC=DB,then which of the following gives a CORRECT congruence relationship?

A
$$\triangle ABC=\triangle DBC$$

B
$$\triangle ABC=\triangle CBD$$

C
$$\triangle ABC=\triangle DCB$$

D
$$\triangle ABC=\triangle BCD$$

Solution

In triangle $$\Delta ABC$$ and $$\Delta DCB$$ we have
$$AB=DC$$ , $$AC=DB$$ and $$BC=BC$$
then from $$SSS$$ congruence ,$$\Delta ABC=\Delta DCB$$
Question 5
1 / -0

Which of the following triangles is congruent to the given triangle?

A

B

C

D

Solution

Triangle in option (C) is congruent to given triangle by SAS congruene criteria.
Question 6
1 / -0

Two triangles are congruent, if two angles and the side included between them in one triangle is equal to the two angles and the side included between them of the other triangle.This is known as

A
RHS congruence criterion

B
ASA congruence criterion

C
SAS congruence criterion

D
SSS congruence criterion

Solution

When two angles and side included in between these angles of one triangle are same as another one, both these triangle are Congruent.
This criteria of congruency is knows as ASA criterion as a side included between two angles are all same.
Question 7
1 / -0

In a triangle ABC , if AB , BC and AC are the three sides of the triangle , then which of the following statements is necessarily true ?

A
$$\displaystyle AB + BC < AC$$

B
$$\displaystyle AB + BC > AC$$

C
$$\displaystyle AB + BC = AC$$

D
$$\displaystyle AB^2 + BC^2 = AC^2$$

Solution

The sum of any two sides of a triangle is greater than the third side .

In $$\triangle ABC, AB, BC$$ and $$AC$$ are the three sides ,

Now ,

$$AB + BC > AC$$
Question 8
1 / -0

In the given figure, $$B < A$$ and $$D > C$$, then:

A
$$AD > BC$$

B
$$AD = BC$$

C
$$AD < BC$$

D
$$AD = 2BC$$

Solution

In $$\triangle AOB$$, $$\angle {A}>\angle {B}$$
Hence, $$OB>OA$$..... (i)
In $$\triangle ODC$$, $$\angle {D}>\angle {C}$$
Hence, $$OC>OD$$.... (ii)
Adding inequalities (i) and (ii), we get
$$OB+OC>OA+OD$$
$$BC>AD$$
Question 9
1 / -0

If $$D$$ and $$E$$ are the mid-points of the sides $$AB$$ and $$AC$$ respectively of $$\triangle ABC$$. $$DE$$ is produced to $$F$$. To prove that $$CF$$ is equal and parallel to $$DA$$, we need an additional information which is:

A
$$\triangle DAE$$ = $$\triangle EFC$$

B
$$AE = EF$$

C
$$DE = EF$$

D
$$\triangle ADE = \triangle ECF$$

Solution

Given $$CF$$ is equal and parallel to $$DA$$
$$D$$ is the midpoint of $$ AB$$
So, $$ AD $$ =$$ DB$$= $$ CF$$= $$\cfrac { c }{ 2 } $$
Similarly $$ AE$$ =$$ EC $$=$$\cfrac { b }{ 2 } $$
Let $$\angle AED=\alpha $$ $$ADE=\beta \quad $$and$$\quad DAE=\gamma$$
Since $$AD $$ is parallel to $$ FC$$
$$\angle FEC = \alpha \angle EFC=\beta $$and$$ \angle ECF=\gamma$$
Applying $$S.A.S$$ Congruency for triangle$$ ADE$$ and $$ CFE$$, $$AD= CF$$, $$AE= AC$$
$$\angle DAE=\angle ECF$$
So $${ \Delta }^{ k } DAE$$ and$${ \Delta }^{ k } FCE$$ are congruent
So $$ DE = EF$$
It is given that $$ DE= EF$$,then by $$S.S.S. $$ congruency all angles will be equal and $$ CF$$ and parallel to $$DA$$
$$\therefore $$ Additionsl information needed is $$ DE= EF$$.
Question 10
1 / -0

$$AB$$ and $$CD$$ are the smallest and largest sides of a quadrilateral $$ABCD$$.
Out of $$\angle B$$ and $$\angle D$$, decide which is greater.

A
$$\angle B$$ will be greater.

B
$$\angle D$$ will be greater.

C
Both the angles are equal.

D
Cannot be determined.

Solution

In the fig. $${ABCD}$$ is a quadrilateral with $${AB}$$ as the smallest side, and $${CD}$$ as the largest side.
We join the diagonal $${BD}$$.
We label the angles as shown in the figure as $$\theta 1$$ , $$\theta 2$$, $$\theta 3$$, $$\theta 4$$ as shown in figure.
Now in $$\Delta {ABD}$$,
$${AB}<{AD}$$
$$\implies \theta 2 < \theta 1 \rightarrow {(1)}$$
Similarly in $$\Delta {BCD}$$,
$${BC}<{CD}$$
$$\implies \theta 4 < \theta 3 \rightarrow {(2)}$$
Adding $${(1)} and {(2)}$$,
$$ \theta 2 + \theta 4 < \theta 1 + \theta 3$$
which is nothing but,
$$\angle{D} < \angle{B}$$
Thus $$\angle(B)$$ will be greater.