Triangles Test

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Triangles Test - 43

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Weekly Quiz Competition

Question 1
1 / -0

Which of Which of the following statements is not correct?

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

B
A triangle can have all its angles acute

C
A right-angled triangle cannot be equilateral

D
Difference of any to sides of a triangle is greater than the third side

Solution

Since the length of the third side of the triangle is always greater than the difference between two sides.
Question 2
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If $$\Delta ABC$$ and $$\Delta DBC$$ are on the same base $$BC, AB = DC$$ and $$AC = DB$$, then which of the following gives a congruence relationship?

A
$$\Delta ABC\cong \Delta DBC$$

B
$$\Delta ABC \cong \Delta CBD$$

C
$$\Delta ABC\cong \Delta DCB$$

D
$$\Delta ABC\cong \Delta BCD$$

Solution

In $$\triangle ABC$$ and $$\triangle DCB,$$
$$AC = DB$$ [Given]
$$AB = DC$$ [Given]
$$BC = CB$$ [Common]

So, by $$SSS$$ rule of congruence,
$$\Delta ABC = \Delta DCB$$

Hence, option $$C$$ is correct.
Question 3
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The length of two sides of a triangle is $$7 \,cm$$ and $$9 \,cm$$. the length of the third side may lie between

A
$$1 \,cm$$ and $$10 \,cm$$

B
$$2 \,cm$$ and $$8 \,cm$$

C
$$3 \,cm$$ and $$16 \,cm$$`

D
$$1 \,cm$$ and $$16 \,cm$$

Solution

Length of two sides is greater than third side in a trinagle.
Hence, $$7 + 9 = 16 \,cm$$ and $$9 - 7 = 2$$
The length of the third side is lie between $$2$$ and $$16$$.
Question 4
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By which congruency criterion, the two triangles in Fig. are congruent?

A
$$RHS$$

B
$$ASA$$

C
$$SSS$$

D
$$SAS$$

Solution

In triangle $$PQR$$ and $$PQS$$,
$$PQ = PQ$$ ...Common
$$PR = PS = a \,cm$$ ...Given
$$QR = QS = b \,cm$$ ...Given
$$\therefore \Delta PSQ = \Delta PRQ$$ ...$$SSS$$ criterion
AS, all three sides are same so, it is congruent by $$SSS$$ test.
So, option $$C$$ is correct.
Question 5
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If length of two sides of a triangle are $$6 \,cm$$ and $$10 \,cm$$, then the length of the third side can be

A
$$3 \,cm$$

B
$$4 \,cm$$

C
$$2 \,cm$$

D
$$6 \,cm$$

Solution

Let $$y \, cm$$ be the third side of triangle.
Sum of two sides of triangle is always greater than $$3^{rd}$$ side
Hence there are three possibilities,
$$\Rightarrow 6+y > 10\cdots (1)$$
$$\Rightarrow 10+y > 6\cdots(2)$$
$$\Rightarrow 10+6 > y\cdots(3)$$
From $$1,2,3$$
$$y>4, y>-4, y<16$$
As, length cannot be negative
So, $$4 < y >16$$
Therefore, from option third side of a triangle can be $$6 \, cm$$.
Hence, option $$D$$ is correct.
Question 6
1 / -0

Which of the following can be the length of the third side of triangle whose two sides measure $$18 cm$$ and $$14 cm $$?

A
$$4 \,cm$$

B
$$3 \,cm$$

C
$$5 \,cm$$

D
$$32 \,cm$$

Solution

Let $$y \,cm$$ be the third side of the triangle.
Hence,
$$\Rightarrow 18+14>y$$
$$\Rightarrow 32>y$$

$$18-14 <y$$
$$y>4 \ and\ 32>y$$
$$4<y<32$$

Therefore, the third side of a triangle can be $$5 cm$$.
Question 7
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$$ \ln \Delta \mathrm{TPQ}, \angle \mathrm{T}=65^{\circ}, \angle \mathrm{P}=95^{\circ} $$ which of the following is a true statement?

A
$$ \mathrm{PQ}<\mathrm{TP} $$

B
$$ P Q< T Q $$

C
$$ \mathrm{TQ}<\mathrm{TP}<\mathrm{PQ} $$

D
$$ \mathrm{PQ}<\mathrm{TP}<\mathrm{TQ} $$

Solution

$$ \ln \Delta T P Q, \angle T=65^{\circ}, \angle P=95^{\circ} $$
A triangle can have a maximum one obtuse angle.
$$ \angle T<\angle P $$
$$ \mathrm{PQ}<\mathrm{TQ} \quad$$(The side opposite to the greater angle is greater than the side opposite to the smaller angle)
Hence, the correct answer is $$ \mathrm{PQ}<\mathrm{TQ} $$.
Question 8
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Choose the correct alternative answer for the following question. $$ \ln \Delta \mathrm{PQR}, $$ If $$ \angle \mathrm{R}>\angle \mathrm{Q} $$ then $$ \ldots \ldots $$

A
$$ \mathrm{QR}>\mathrm{PR} $$

B
$$ P Q>P R $$

C
$$ \mathrm{PQ}<\mathrm{PR} $$

D
$$ \mathrm{QR}<\mathrm{PR} $$

Solution

$$ \ln \Delta \mathrm{PQR}, $$
If $$ \angle \mathrm{R}>\angle \mathrm{Q} $$
$$ \therefore \mathrm{PQ}>\mathrm{PR} \quad $$ (The sides opposite to the greater angle is greater than side opposite to the smaller angle)
Hence, the correct answer is $$ \mathrm{PQ}>\mathrm{PR} $$.
Question 9
1 / -0

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

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

B
$$\triangle CBA\cong \triangle PRQ$$

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

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

Solution

In$$\triangle ABC$$ and $$\triangle QRP$$
AB=QR
BC=PR
and CA=PQ
By SSS congruency,
$$\triangle ABC\cong \triangle QRP$$
or $$\triangle CBA\cong \triangle PRQ$$
Question 10
1 / -0

In triangles $$ABC$$ and $$DEF$$, $$AB=FD$$ and $$\angle A=\angle D$$. Two triangles are congruent by SAS criterion if

A
$$BC=EF$$

B
$$AC=DE$$

C
$$AC=EF$$

D
$$BC=DE$$

Solution

In $$\triangle ABC$$ and $$\triangle DEF$$
$$AB=DF$$
$$\angle A=\angle D$$
for congruence by SAS rule $$AC=DE$$

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Re-Attempt Weekly Quiz Competition

Triangles Test - 43

Result

Triangles Test - 43

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SHARING IS CARING

Question 1

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

A triangle can have all its angles acute

A right-angled triangle cannot be equilateral

Difference of any to sides of a triangle is greater than the third side

Solution

Question 2

$$\Delta ABC\cong \Delta DBC$$

$$\Delta ABC \cong \Delta CBD$$

$$\Delta ABC\cong \Delta DCB$$

$$\Delta ABC\cong \Delta BCD$$

Solution

Question 3

$$1 \,cm$$ and $$10 \,cm$$

$$2 \,cm$$ and $$8 \,cm$$

$$3 \,cm$$ and $$16 \,cm$$`

$$1 \,cm$$ and $$16 \,cm$$

Solution

Question 4

$$RHS$$

$$ASA$$

$$SSS$$

$$SAS$$

Solution

Question 5

$$3 \,cm$$

$$4 \,cm$$

$$2 \,cm$$

$$6 \,cm$$

Solution

Question 6

$$4 \,cm$$

$$3 \,cm$$

$$5 \,cm$$

$$32 \,cm$$

Solution

Question 7

$$ \mathrm{PQ}<\mathrm{TP} $$

$$ P Q< T Q $$

$$ \mathrm{TQ}<\mathrm{TP}<\mathrm{PQ} $$

$$ \mathrm{PQ}<\mathrm{TP}<\mathrm{TQ} $$

Solution

Question 8

$$ \mathrm{QR}>\mathrm{PR} $$

$$ P Q>P R $$

$$ \mathrm{PQ}<\mathrm{PR} $$

$$ \mathrm{QR}<\mathrm{PR} $$

Solution

Question 9

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

$$\triangle CBA\cong \triangle PRQ$$

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

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

Solution

Question 10

$$BC=EF$$

$$AC=DE$$

$$AC=EF$$

$$BC=DE$$

Solution

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