M.C.Q - Page 6 - MATHS STD 9 Questions - Vidyadip

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M.C.Q

MCQ 2511 Mark

Two sides of a triangle are oflengths $5\ cm$ and $1.5\ cm$. The length of the third side of the triangle cannot be:

A
$3.6\ cm$
B
$3.8\ cm$
C
$4\ cm$
✓
$3.4\ cm$

Answer

Correct option: D.

$3.4\ cm$

Given that: Two sides of triangle are $5\ cm$ and $1.5\ cm$. We know that the sum of two sides of the triangle is always greater than the third side. Hence, $3.4\ cm$ cannot be the third side. If it is the third side the sum of $3.4\ cm$ and $1.5\ cm$ will be smaller than $5\ cm,$ so, the triangle will not be possible.

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MCQ 2521 Mark

Line sgements $AB$ and $CD$ intersect at $O$ such that $\text{AC}||\text{DB}.$ If $\angle\text{CAB} = 45^\circ$ and $\angle\text{CDB} = 55^\circ,$ then $\angle\text{BOD} =$

✓
$80^\circ $
B
$90^\circ$
C
$100^\circ$
D
$135^\circ$

Answer

Correct option: A.

$80^\circ $

$\text{AC}||\text{DB}$
And, AB is transverse to these parallel lines
So, $\angle\text{CAB} = \angle\text{ABD}$ (Alternate angles)
$\Rightarrow \angle\text{ABD} = 45^\circ$
Now In $\triangle\text{BOD}$
$\angle\text{BOD} + \angle\text{ODB} + \angle\text{DBA} = 180^\circ$
$\angle\text{DBA} = \angle\text{ABD} = 45^\circ, \ \angle\text{ODB} = 55^\circ$
So, $\angle\text{BOD} = 180^\circ - 45^\circ - 55^\circ$
$= 80^\circ$

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MCQ 2531 Mark

In fig, in $\triangle\text{ABC}, AB = AC$, then the value of $x$ is:

A
$100^{\circ}$
B
$80^{\circ}$
C
$120^{\circ}$
✓
$130^{\circ}$

Answer

Correct option: D.

$130^{\circ}$

Triangle $ABC$ is an iscosceles triangle and hence in the triangle other two angles are $50$ and $50.$
Therefore,$X = 180 - 50 = 130$

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MCQ 2541 Mark

In $\triangle\text{ABC},$ if $\angle\text{B} = 30^\circ$ and $\angle\text{C} = 70^\circ,$ then which of the following is the longest side?

A
$AB$ or $AC$
✓
$BC$
C
$AB$
D
$AC$

Answer

Correct option: B.

$BC$

Since the sum of all sides of a triangle is $180^\circ .$ So, $\angle\text{C} = 70^\circ,\ \angle\text{B} = 30^\circ,\ \angle\text{A} = 80^\circ.$
We have a theorem which states that the side opposite to the greatest angle is the longest.
So, the side opposite to angle $A$ is the longest.

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MCQ 2551 Mark

If two acute angles of a right triangle are equal, then each acute is equal to:

A
$30^\circ$
✓
$45^\circ$
C
$60^\circ$
D
$90^\circ$

Answer

Correct option: B.

$45^\circ$

Let the measure of each acute angle of a triangle be $x^\circ $.Then, we have
$x^\circ + x^\circ + 90^\circ = 180^\circ$
$i.e. 2x^\circ = 90^\circ$
$i.e. x^\circ = 45^\circ$

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MCQ 2561 Mark

If $\triangle\text{ABC}\cong\triangle\text{PQR}$ then which of the following is not true?

✓
$BC = PQ$
B
$AC = PR$
C
$BC = QR$
D
$AB = PQ$

Answer

Correct option: A.

$BC = PQ$

Since $ABC$ is not congruent to $RPQ, BC = PQ$ is not true.

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MCQ 2571 Mark

In the adjoining figure, $BC = AD$, $\text{CA}\bot\text{AB}$ and $\text{BD}\bot\text{AB}.$ The rule by which $\triangle\text{ABC}\cong\triangle\text{BAD}$ is:

A
$ASA$
✓
$RHS$
C
$SSS$
D
$SAS$

Answer

Correct option: B.

$RHS$

In $\triangle\text{ABC}$ and $\angle\text{BAG} = \angle\text{ABD}$
$BAD$, we have (Right angles)
$BC = AD$ (Hypotentuses and Given)
$AB = AB$ (conunon in both)
Hence, $\triangle\text{ABC}\cong\triangle\text{BAD}$ by $RHS$ criterion.

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MCQ 2581 Mark

If $\angle\text{OCA}=80^\circ, \ \angle\text{COA}=40^\circ,$ and $\angle\text{BDO}=70^\circ$ then $x^\circ + y^\circ = ?$

A
$270^\circ$
B
$210^\circ$
✓
$230^\circ$
D
$190^\circ$

Answer

Correct option: C.

$230^\circ$

In the given figure, $\angle\text{BOD} = \angle\text{COA}$ (Vertically opposite angles)$\therefore\ \angle\text{BOD} = 40^\circ ... (\text{i})$
In $\triangle\text{ACO}$
$\angle\text{OAE}=\angle\text{OCA}+\angle\text{COA}$ (Exterior angle of a triangle is equal to the sum of two opposite interior angles)
$⇒ \text{x}^\circ = 80^\circ + 40^\circ = 120^\circ ... \ (\text{ii})$
In $\angle\text{BDO},$
$\angle\text{DBF}=\angle\text{BDO}+\angle\text{BOD}$ (Exterior angle of a triangle is equal to the sum of two opposite interior angles)
$⇒\text{y}^\circ=70^\circ+40^\circ=110^\circ\ ...\ (\text{iii})$
Adding $(2)$ and $(3)$ we get
$\text{x}^\circ+\text{y}^\circ=120^\circ+110^\circ=230^\circ$

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MCQ 2591 Mark

In the adjoining figure, $\triangle\text{ABC}\cong\triangle\text{ADC}.$ If $\angle\text{BAC} = 30^\circ$ and $\angle\text{ABC} = 100^\circ$ then $\angle\text{ACD}$ is equal to:

A
$80^{\circ}$
B
$60^{\circ}$
C
$30^{\circ}$
✓
$50^{\circ}$

Answer

Correct option: D.

$50^{\circ}$

In triangle ABC, $\angle\text{BAC} = 30^\circ$and $\angle\text{ABC} = 100^\circ$ (Given)$\angle\text{BAC} + \angle\text{ABC} + \angle\text{BCA} = 180^\circ$
$\angle\text{BAC} = 50^\circ$
Also $\angle\text{ACD} = 50^\circ$(Since, $\triangle\text{ABC}\cong\triangle\text{ADC}$)

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MCQ 2601 Mark

In $\triangle\text{ABC},$ if $\angle\text{B} = 30^\circ$ and $\angle\text{C} = 70^\circ,$ then which of the following is the longest side?

A
$AC$
✓
$BC$
C
$AB$
D
$AB$ or $AC$

Answer

Correct option: B.

$BC$

Since the sum of all sides of a triangle is $180^\circ .$
So, $\angle\text{C}=70^\circ,\ \angle\text{B}=70^\circ,\ \angle\text{A}=80^\circ,$
We have a theorem which states that the side opposite to the greatest angle is the longest.
So, the side opposite to angle A is the longest.

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MCQ 2611 Mark

In fig. which of the following statement is true?

A
$\angle\text{B} = \angle\text{C}$
✓
$\angle\text{B}$ is the smallest angle in the triangle.
C
$\angle\text{B}$ is the greatest angle in the triangle.
D
$\angle\text{A}$ is the smallest angle in the triangle.

Answer

Correct option: B.

$\angle\text{B}$ is the smallest angle in the triangle.

In a triangle angle opposite to smallest side is least $AC$ is least side and hence $B$ is smaller.

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MCQ 2621 Mark

$ABC$ is an isosceles triangle such that $AB = AC$ and $AD$ is the median to base BC. Then, $\angle\text{BAD}=$

✓
$55^\circ$
B
$70^\circ$
C
$35^\circ$
D
$110^\circ$

Answer

Correct option: A.

$55^\circ$

If $AD$ is the median, then $D$ is the mid-point of $BC.$
$\text{BD}=\text{DC}$
So consider $\triangle\text{ADB}$ and $\triangle\text{ADC}$
$\text{AD}=\text{AD}$ (common)
$\text{DB}=\text{DC}$
$\text{BA}=\text{CA}$
So by SSS, $\triangle\text{ADB}\cong\triangle\text{ADC}$
Now $\angle\text{B}=\angle\text{C}=35^\circ$
$\Rightarrow\angle\text{BAD}=\angle\text{DAC}$
So in $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow2\angle\text{BAD}+35^\circ+35^\circ=180^\circ$
$\Rightarrow2\angle\text{BAD}=110^\circ$
$\Rightarrow\angle\text{BAD}=55^\circ$
Hence, correct option is $(a).$

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MCQ 2631 Mark

In $\triangle\text{ABC},$ $\angle\text{A} = 35^\circ$ and $\angle\text{B} = 65^\circ$, then the longest side of the triangle is:

✓
$AB$
B
$BC$
C
$AC$
D
None of these

Answer

Correct option: A.

$AB$

As per angle sum property, $\angle\text{A} +\angle\text{B}+\angle\text{C}= 180^\circ$
Hence, $35^\circ + 65^\circ + \angle\text{C} = 180^\circ$
$\Rightarrow\ \angle\text{C} = 80^\circ$ which is the greatest angle.
We know that the side opposite to the greatest angle i.e $AB$ would be the greatest.
Hence, $AB$ is the longest side.

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MCQ 2641 Mark

$D, E$ and $F$ are the mid points of sides $AB, BC$ and $CA$ of $\triangle\text{ABC}.$ If perimetre of $\triangle\text{ABC}.$ is $16\ cm,$ then perimetre of $\triangle\text{DEF}.$

A
$4\ cm$
✓
$8\ cm$
C
None of these
D
$32\ cm$

Answer

Correct option: B.

$8\ cm$

Using relation,$\text{Perimeter.}\ \triangle\text{DEF}=\frac{1}{2}\text{Perimeter.}\ \triangle\text{ABC}$
$=\frac{1}{2}\times16=18\text{cm}$

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MCQ 2651 Mark

If $a, b, c$ are the lengths of the sides of a triangle, then

A
$A – B > C$
B
$C > A + B$
✓
$C < A + B$
D
$C = A + B$

Answer

Correct option: C.

$C < A + B$

Put the sidesof triangle $a, b, c$
For a possible triangle the following are possible.
$a + b > = c$
$b + c > = a$
$a + c > = b$

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MCQ 2661 Mark

In triangles $ABC$ and $PQR, AB = AC$, $\angle\text{C} = \angle\text{P}$ and $\angle\text{B} = \angle\text{Q}.$ The two triangles are:

A
Congruent but not isosceles.
B
Isosceles and congruent.
✓
Isosceles but not congruent.
D
Neither congruent nor isosceles.

Answer

Correct option: C.

Isosceles but not congruent.

Given: $\triangle\text{ABC}$ and $\triangle\text{PQR}, \ \text{AB = AC}, \ \angle\text{C} = \angle\text{P}$ and $\angle\text{B} = \angle\text{Q}.$

$\text{AB = AC}$
$\Rightarrow\ \angle\text{B} = \angle\text{C}$ (opposite angles to equal sides are equal)
Hence, $\triangle\text{ABC}$ is an isosceles triangle.
$\angle\text{C} = \angle\text{P}$ and $\angle\text{B} = \angle\text{Q}$ (given)
$\Rightarrow \angle\text{P} = \angle\text{Q}\ \ (\therefore \angle\text{B} = \angle\text{C})$
$\Rightarrow \text{QR} = \text{PR}$ (opposite sides to equal angles are equal)
Hence, $\triangle\text{PQR}$ is an isosceles triangle.
So, the two triangles are isosceles but not congruent.
As $AAA$ is not the criterion for a triangle to be congruent.

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MCQ 2671 Mark

Which is true?

✓
A triangle can have two right angles.
B
A triangle can have two obtuse angles.
C
A triangle can have two acute angles.
D
An exterior angle of a triangle is less than either of the interior opposite angles.

Answer

Correct option: A.

A triangle can have two right angles.

We know that, by the angle sum property,
the sum of all the angles of a triangle is $180^\circ ...(i)$
So, if two angles are right angles, then their sum is $180^\circ .$
So, the measure of the third angle is zero, which is not possible.
Thus, $(a)$ is not true.

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MCQ 2681 Mark

In Fig. if $\text{AB}\perp\text{BC},$ then $x =$

A
$18^\circ$
✓
$22^\circ$
C
$25^\circ$
D
$32^\circ$

Answer

Correct option: B.

$22^\circ$

$\text{AB}\perp\text{BC}$$\Rightarrow\angle\text{ABC}=90^\circ$
$\angle\text{CAB}=32^\circ$ (Opposite angles)
Now, in $\triangle\text{ABD}$
$\angle\text{DAB}=\text{x}^\circ+32^\circ$
$\angle\text{ABD}=90^\circ$
$\angle\text{BDA}=\text{x}^\circ+14^\circ$
In a $\triangle,$ sum of all angles $= 180^\circ$
$\Rightarrow\angle\text{DAB}+\angle\text{ABD}+\angle\text{BDA}=180^\circ$
$\Rightarrow\text{x}^\circ+32^\circ+90^\circ+\text{x}^\circ+14^\circ=180^\circ$
$\Rightarrow2\text{x}^\circ=180^\circ-136^\circ$
$\Rightarrow2\text{x}^\circ=44^\circ$
$\Rightarrow\text{x}^\circ=22^\circ$

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MCQ 2691 Mark

If $\triangle\text{ABC}\cong\triangle\text{ACB},$ then $\triangle\text{ABC}$ is isosceles with.

✓
$AB = AC$
B
$AB = BC$
C
$AC = BC$
D
None of these

Answer

Correct option: A.

$AB = AC$

$\triangle\text{ABC}\cong\triangle\text{ACB}$
$\Rightarrow\text{AB}=\text{AC}$
or
$\text{AC}=\text{AB}$
So, in $\triangle\text{ABC}$ is isosceles with $\text{AB}=\text{AC}.$
Hence, correct option $(a).$

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MCQ 2701 Mark

It is not possible to construct a triangle when the lengths of its sides are:

✓
$5.3\ cm, 2.2\ cm, 3.1\ cm$
B
$6\ cm, 7\ cm, 8\ cm$
C
$4\ cm, 6\ cm, 6\ cm$
D
$9.3\ cm, 5.2\ cm, 7.4\ cm$

Answer

Correct option: A.

$5.3\ cm, 2.2\ cm, 3.1\ cm$

Put the sidesof triangle $a, b, c$
For a possible triangle the following are should possible.
$a + b > = c$
$b + c > = a$
$a + c > = b$
Here, $2.2 + 3.1 = 5.3$
So, $a + b = c$
So the triangle becomes a streight line.
So we cannot draw a triangle with sides $5.3\ cm, 2.2\ cm, 3.1\ cm.$

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MCQ 2711 Mark

In $\triangle\text{ABC, }\angle\text{A}=40^{\circ}$ and $\angle\text{B}=60^{\circ}.$ Then the longest side of $\triangle\text{ABC}$ is:

A
$BC$
B
$AC$
✓
$AB$
D
cannot be determined

Answer

Correct option: C.

$AB$

In $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^{\circ}$ ...(Using Angle Sum Property)
$\Rightarrow40^{\circ}+60^{\circ}+\angle\text{C}=180^{\circ}$
$\Rightarrow\angle\text{C}=80^{\circ}$
We know that, the greater angle has the longest side opposite to it.
Since $\angle\text{A}<\angle\text{B}<\angle\text{C, }\text{BC < AC < AB}.$
So, the longest side is $AB.$

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MCQ 2721 Mark

In the above quadrilateral $ACBD$, we have $AC= AD$ and $AB$ bisect the $LA$ .Which of the following is true?

A
$\triangle\text{ABC}\cong\triangle\text{ABD}$
B
$\angle\text{C}=\angle\text{D}$
✓
$\text{All are true}$
D
$\text{BC = BD}$

Answer

Correct option: C.

$\text{All are true}$

$\text{AC = AD}$$\angle\text{AB}= \angle\text{BAD}$
$\text{AB = AB}$
By SAS, we have
$\triangle\text{ABC}\cong\triangle\text{ABD}$
Hence, we have $\text{BC = BD}$ and $\angle\text{C}= \angle\text{D}.$
So, all the given options are true.

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MCQ 2731 Mark

In the given figure, $AB > AC$. Then, which of the following is true?

A
$AB < AD$
B
Cannot be determined
✓
$AB > AD$
D
$AB = AD$

Answer

Correct option: C.

$AB > AD$

$AB > AC$ [given.]
$∴ \ \angle\text{ACB} > \angle\text{ABC}$
Now, $\angle\text{ADB} > \angle\text{ACD}$ (exterior angle is always greater than each interior angle)
$⇒ \angle\text{ADB} > \angle\text{ACB} > \angle\text{ABC}$
$\Rightarrow\ \angle\text{ADB} > \angle\text{ADB}$
$⇒\text{AB > AD}$

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