Triangles Test

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Triangles Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

In the adjoining figure, $$AB=AC$$ and $$AP\bot BC$$. Then:

A
$$AB=AP$$

B
$$AB < AP$$

C
$$AB>AP$$

D
$$AB\le AP$$

Solution

Given: $$AB = AC$$, $$AP \perp BC$$
In $$\triangle ABP$$, we have
$$\angle APB = 90^{\circ}$$
$$\angle ABP = 90 - \angle BAP$$ ....(Angle sum property)
Thus, $$\angle APB > \angle ABP$$
$$AB > AP$$ .... (Side opposite to larger angle is large)
Question 2
1 / -0

Which pair of triangles is not congruent by $$SAS$$ congruence criterion

A

B

C

D
None of these

Solution

In option (3), we have

$$\angle BAC = \angle DFE = 120^{\circ}$$

$$BC = DE = 5$$

$$AC = DF = 3$$

For SAS to hold true corresponding angle included by two congruent sides must be congruent

here, angle A and F have not included angles by the congruent sides.
Hence, $$\triangle ABC \not\cong \triangle DEF$$
Question 3
1 / -0

If $$AB=QR$$, $$BC=PQ$$ and $$CA=PR$$, then:

A
$$\triangle ABC\cong \triangle PQR$$

B
$$\triangle CBA\cong \triangle PQR$$

C
$$\triangle BAC\cong \triangle RPQ$$

D
$$\triangle PQR\cong \triangle BCA$$

Solution

In $$\triangle ABC$$ and $$\triangle PQR$$
$$AB=QR$$
$$BC=PQ$$
$$CA=PR$$
Thus, $$\triangle CBA \cong \triangle PQR$$ (SSS rule)
Question 4
1 / -0

State the property by which $$\Delta ADB\, \cong\, \Delta ADC$$ in the following figure.,

A
SAS property

B
SSS property

C
RHS property

D
ASA property

Solution

From figure, we have
$$AB = AC $$ ....Given
$$\angle BAD\, =\, \angle CAD$$
$$AD = AD$$ ....Common sides
$$\therefore$$ By SAS property, $$\Delta ADB \cong \Delta ADC$$.
Question 5
1 / -0

Ankita wants to prove $$\Delta ABC\cong \Delta DEF$$ using $$SAS$$. She knows $$AB=DE$$ and $$AC=DF$$. What additional piece of information does she need?

A
$$\;\angle A=\angle D$$

B
$$\;\angle C=\angle F$$

C
$$\;\angle B=\angle E$$

D
$$\;\angle A=\angle B$$

Solution

By SAS, two corresponding sides and their included angles should be congruent.
Since $$ AB=DE $$ and $$ AC=DF $$,
The included angles of $$ AB$$ and $$AC = \angle A $$
and included angle of $$ DE$$ and $$DF = \angle D.$$
$$ \Rightarrow \angle A = \angle D $$ is needed.
Question 6
1 / -0

By which congruency property, the two triangles connected by the following figure are congruent.

A
SAS property

B
SSS property

C
RHS property

D
ASA property

Solution

Separate the given figure into two triangles.

$$\Delta ABC$$ and $$\Delta ABD$$

We can observe that,

$$AC = AD$$ {Both are $$x \,\,cm$$}

$$BC = BD$$ {Both are $$y\,\,cm$$}

$$AB = AB$$ {Common Side}

So $$\Delta ABC\, \cong\, \Delta ABD$$ by SSS property.
Question 7
1 / -0

If $$4$$ cm and $$3$$ cm are the lengths of two sides of a triangle then the length of the third side may be_____.

A
$$11$$ cm

B
$$3$$ cm

C
$$5$$ cm

D
$$12$$ cm

Solution

In a triangle, sum of two sides $$ > $$ third side.

So, if two sides are $$ 4 cm $$ and $$ 7 cm $$, then considering the given options,

A. $$ 11 cm $$

$$ 4 + 7 \ngtr 11 $$

Hence, this is not the correct option.

B. $$ 6 cm $$
We see that
$$ 4 + 7 > 6 $$
$$ 4 + 6 > 7 $$
$$ 7 + 6 > 4 $$
Hence, $$ 6 cm $$ is the correct option.
Question 8
1 / -0

Two sides of a triangle are $$7$$ and $$10$$ units, which of the following length can be the length of the third side?

A
$$19$$ units

B
$$17$$ units

C
$$13$$ units

D
$$3$$ units

Solution

The sum of any two sides is greater than the third side.
and difference between two sides should be lesser than the third side.
For option $$A,$$ $$7+10\ngtr 19$$
For option $$B,$$ $$7+10\ngtr 17$$
For option $$C,$$ $$7+10 > 13$$
Also, $$ |10-7| < 13$$
For option $$D,$$ $$7+10 > 3$$
but $$|7-10| \nless 3$$
Only, $$C$$ is correct.
Question 9
1 / -0

In the given fig. if AD = BC and AD || BC, then:

A
AB = AD

B
AB = DC

C
BC = CD

D
None

Solution

In the give fig. AD = BC, AC = AC,
$$\angle ADC\, =\, \angle ABC$$ ($$\because $$ AD || BC)
By SAS theorem $$\Delta ABC\, \cong\, \Delta CDA$$
So AB = DC.
Question 10
1 / -0

ABC is an isosceles triangle in which the altitudes $$BE$$ and $$CF$$ are drawn to the equal sides $$AC$$ and $$AB$$ respectively. Then

A
$$BE = CF$$

B
$$BE = AB$$

C
$$AB = BC$$

D
$$AC = BC$$

Solution

In $$\triangle ABE$$ and $$\triangle ACF$$,

we have $$\angle BEA=\angle CFA$$ (Each $${ 90 }^{ 0 }$$)

$$\angle A= \angle A$$ (Common angle)

$$AB=\,AC$$ (Given)

$$\therefore \triangle ABE \cong \triangle ACF$$ (By SAS congruence criteria)

$$\therefore BF=\,CF$$ [C.S.C.T]