M.C.Q

Question 511 Mark

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

Answer

AB > AD
Solution:
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)
$⇒ \angle\text{ADB} > \angle\text{ACB} > \angle\text{ABC}$
$\Rightarrow\ \angle\text{ADB} > \angle\text{ADB}$
$⇒\text{AB > AD}$

Question 521 Mark

In Fig. ABC ids an isosceles triangle whose side AC is produced to E. Through C, CD is drawn parallel to BA. The value of x is:

Answer

104º
Solution:

$\triangle\text{ABC}$ is isosceles
$\angle\text{ABC}=\angle\text{ACB}=52^\circ$
then $\angle\text{BAC}=180^\circ-52^\circ-52^\circ=76^\circ$
If $\text{AB}\parallel\text{CD},$ AC is transversal
then $\angle\text{BAC}=\angle\text{ACD}$ (alternate angles)
$\Rightarrow\angle\text{ACD}=76^\circ$
Now from figure,
$\angle\text{ACD}+\text{x}^\circ=180^\circ$
$\Rightarrow\text{x}^\circ=180^\circ-76^\circ$
$\Rightarrow\text{x}^\circ=104^\circ$
Hence, correct option is (d).

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Question 531 Mark

In an isosceles triangle, if the vertex angle is twice the sum of the base angles, then the measure of vertex angle of the triangle is:

Answer

120º
Solution:
Let the base angles be x each,
Vertex angle = 2(x + x) = 4x
Now, since the sum of all the angles of a triangle is 180º
x + x + 4x = 180º
6x = 180º
x = 30º
Therefore, vertex angle = 4x = 120º

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Question 541 Mark

In Figure, AB and CD are parallel lines and transversal EF intersects them at P and Q respectively. If $\angle\text{APR} = 25^\circ,\ \angle\text{RQC} = 30^\circ$ and $\angle\text{CQF} = 65^\circ,$ then

Answer

x = 55º, y = 40º
Solution:
$\angle\text{OQP} = 180^\circ - \angle\text{OQF}$
= 180º - (30º + 65º)
$⇒ \angle\text{OQP} = 85^\circ\ ...\ (\text{i})$
$\angle\text{APQ}=\angle\text{CQF}$ (Corresponding angles)
⇒ 25º + yº = 65º
⇒ yº = 65º - 25º
⇒ yº = 40º
Now in $\triangle\text{OPQ}$
$\angle\text{O}+\angle\text{OPQ}+\angle\text{PQO} = 180^\circ$
⇒ xº + 40º + 85º = 180º
xº = 180º - 85º - 40º = 55º
⇒ x = 55º, y = 40º

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Question 551 Mark

$\angle\text{x}$ and $\angle\text{y}$ are exterior angles of a triangle ABC at the points B and C respectively, Also, $\angle\text{B} >\angle\text{C},$ then the relation between $\angle\text{x}$ and $\angle\text{y}$ is:

Answer

$\angle\text{x} >\angle\text{y},$
Solution:
Since interior $\angle\text{B} >\angle\text{C},$
Hence x < y.

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Question 561 Mark

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

Answer

130º
Solution:
$\triangle\text{ABC}$ is an iscosceles triangle and hence in the triangle other two angles are 50 and 50.
Therefore,
X = 180 - 50= 130º

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Question 571 Mark

Which of the following pairs of angle are supplementary?

Answer

45º, 135º
Solution:
We know that the sum of supplementary angle is 180º
and the sum of 45º, 135º is also 180º.

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Question 581 Mark

D, E, F are the mid-point of the sides BC, CA and AB respectively of $\triangle\text{ABC}.$ Then $\triangle\text{DEF}$ is congruent to triangle.

Answer

AFE, FBD, EDC
Solution:
DEF is the median triangle and it divides the given triangle into 4 identical triangles including median triangle.

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Question 591 Mark

In $\triangle\text{ABC}$ and $\triangle\text{PQR},$ AB = PR and $\angle\text{A} = \angle\text{P}.$ Then, the two triangles will be congruent by SAS axiom if:

Answer

AC = PQ
Solution:
$\angle\text{A}$ is included between AB and AC and $\angle\text{P}$ is included between PQ and PR and corresponding sides must be equal. Since AB = PR, hence AC = PQ for the given triangles to be congruent by SAS axiom.

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Question 601 Mark

If $\triangle\text{ABC} \cong\triangle\text{PQR}$ by SSS congruence rule, then:

Answer

BC = QR
Solution:
If $\triangle\text{ABC} \cong\triangle\text{PQR}$ by SSS congruence rule, then the corresponding sides must be equal i.e AB = PQ, BC = QR and AC = PR.

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Question 611 Mark

If $\triangle\text{PQR}\cong\triangle\text{EFD},$ then ED =

Answer

PR
Solution:
$\triangle\text{PQR}\cong\triangle\text{EFD},$
$\Rightarrow\text{ED}=\text{PR}$ (congruent sides of congruent triangles)
Hence, correct option is (c)

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Question 621 Mark

The perimeter of a triangle is 36cm and its sides are in the ratio a : b : c = 3 : 4 : 5 then a, b, c are respectively:

Answer

9cm, 12cm, 15cm
Solution:
Let the three sides a, b, c be 3x, 4x and 5x respectively.
Then according to the conditions given in the question, we have
3x + 4x + 5x = 36
12x = 36
x = 3cm
Thus, the three sides are:
a = 3 × 3 = 9cm, b = 4 × 3 = 12cm and c = 5 × 3 = 15cm

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Question 631 Mark

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

Answer

5.3cm, 2.2cm, 3.1cm
Solution:
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 5.3cm, 2.2cm, 3.1cm.

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Question 641 Mark

In Fig. The measure of $\angle\text{B}'\text{A}'\text{C}'$ is:

Answer

60º
Solution:

In $\triangle\text{ABC}$ and $\triangle\text{A}'\text{B}'\text{C},$
$\text{AB}=\text{A}'\text{B}'$
$\text{BC}=\text{B}'\text{C}'$
$\angle\text{ABC}=\angle\text{A}'\text{B}'\text{C}'$
So $\triangle\text{ABC}\cong\triangle\text{A}'\text{B}'\text{C}'$ by SAS creterion
$\Rightarrow\angle\text{BAC}=\angle\text{B}'\text{A}'\text{C}'$
$\Rightarrow3\text{x}=2\text{x}+20$
$\text{x}=20^\circ$
$2\text{x}+20=2\times20+60^\circ=\angle\text{B}'\text{A}'\text{C}'$
Hence, correct option is (b).

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Question 651 Mark

An exterior angle of a triangle is equal to 100º and two interior opposite angles are equal. Each of these angles is equal to:

Answer

50º
Solution:
Let the two interior opposite angles be xº each.
Now, the exterior angle is equal to the sum of the two interior opposite angles.
xº + xº = 180º
⇒ 2xº = 100º
⇒ xº = 50º

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Question 661 Mark

If all the altitudes from the vertices to the opposite sides of a triangle are equal, then the triangle is:

Answer

Equilateral
Solution:
In an equilateral triangle all the altitudes, sides, angles, perpendicular bisectors, medians and angular bisectors are equal.

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Question 671 Mark

If all the three angles of a triangle are equal, then each one of them is equal to:

Answer

60º
Solution:
Let the measure of each angle be x°.
Now, the sum of all angles of any triangle is 180°.
Thus, x° + x° + x° = 180°
i.e. 3x° = 180°
i.e. x° = 60°

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Question 681 Mark

In the given figure, lines AB and CD intersect at a point O. The sides CA and OB have been produced to E and repectively such that $\angle\text{OAE}=\text{x}^\circ$ and $\angle\text{DBF}=\text{y}^\circ.$

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

Answer

230º
Solution:
In $\triangle\text{OAC},$ by angle sum property
$\angle\text{OCA}+\angle\text{COA}+\angle\text{CAO}=180^\circ $
$\Rightarrow80^\circ+40^\circ+\angle\text{CAO}=180^\circ$
$\Rightarrow\angle\text{CAO}=60^\circ$
$\angle\text{CAO}+\angle\text{OAE}=180^\circ$ (linear pair)
$\Rightarrow60^\circ+\text{x}=180^\circ$
$\Rightarrow\text{x}=120^\circ$
$\angle\text{COA}=\angle\text{BOD}$ (vertically opposite angles)
$\Rightarrow\angle\text{BOD}=40^\circ$
In $\triangle\text{OBD},$ by angle sum property
$\angle\text{OBD}+\angle\text{BOD}+\angle\text{ODB}=180^\circ$
$\Rightarrow\angle\text{OBD}+40^\circ+70^\circ=180^\circ$
$\Rightarrow\angle\text{OBD}=70^\circ$
$\angle\text{OBD}+\angle\text{DBF}=180^\circ$ (linear pair)
$\Rightarrow70^\circ+\text{y}=180^\circ$
$\Rightarrow\text{y}=110^\circ$
$\therefore\text{x}+\text{y}=120^\circ+110 ^\circ=230^\circ$

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Question 691 Mark

In Fig. for which value of $x$ is $l_1 || l_2$?

Answer

Let if$ l_1 || l_2$ and$ AB$ is tranverse to it.
Then,
$\angle\text{PBA}$ should be equal $\angle\text{BAS} ($Alternate angles$)$
So if $l_1 || l_2,$ then $\angle\text{BAS}=70^\circ$
$\Rightarrow\angle\text{BAC}=78^\circ-35^\circ=43^\circ\dots(1)$
Now, in $\triangle\text{ABC}$
$\text{x}^\circ+\angle\text{C}+\angle\text{BAC}=180^\circ$
$\Rightarrow\text{x}^\circ+90^\circ+43^\circ=180^\circ$
$\Rightarrow\text{x}^\circ=180^\circ-90^\circ-43^\circ=47^\circ$
$\Rightarrow\text{x}^\circ=47^\circ$
So if $x^\circ = 47^\circ$ then $l_1 || l_2$

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Question 701 Mark

In a $\triangle\text{ABC},$ if $\angle\text{A}=60^\circ,\angle\text{B}=80^\circ$ and the bisectors of $\angle\text{B}$ and $\angle\text{C}$ meet at O, then $\angle\text{BOC}=$

Answer

120º
Solution:

O is point where bisectors of $\angle\text{C }\& \angle\text{B}$ meets.
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$60^\circ+80^\circ+\angle\text{C}=108^\circ$
$\angle\text{C}=40^\circ$
$\frac{\angle\text{C}}{2}=20^\circ$
$\frac{\angle\text{C}}{2}=20^\circ=\angle\text{BCO}\dots(1)$
$\frac{\angle\text{B}}{2}=\frac{80^\circ}{2}=40^\circ=\angle\text{OBC}\dots(2)$
In $\triangle\text{BOC}$
$\angle\text{BCO}+\angle\text{OBC}+\angle\text{BOC}=180^\circ$
from (1) and (2)
$20^\circ+40^\circ+\angle\text{BOC}=180^\circ$
$\Rightarrow\angle\text{BOC}=180^\circ-60^\circ=120^\circ$

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Question 711 Mark

In a $\triangle\text{ABC}, \ \angle\text{A} = 50^\circ$ and BC is produced to a point D. If the bisectors $\angle\text{ACD}$ meet at E, then $\angle\text{E} =$

Answer

25º
Solution:

BE and CE are bisectors of $\angle\text{B}$ and $\angle\text{ACD}.$
in $\triangle\text{ABC}$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\ \angle\text{B}+\angle\text{C}=180^\circ-50^\circ=130^\circ\ ...\ \text{(i)}$
Now,in $\triangle\text{BEC}$
$\angle\text{CBE}+\angle\text{BEC}+\angle\text{ECB}=180^\circ$
$\angle\text{CBE}=\frac{\angle\text{B}}{2},\ \angle\text{BEC}=\angle\text{E},\ \angle\text{ECB}=\angle\text{C}+\angle\text{ACE}$
Now, $\angle\text{ACD}=180^\circ-\angle\text{C}$
$\angle\text{ACE}=\frac{\angle\text{ACD}}{2}=\frac{180^\circ-\angle\text{C}}{2}=90^\circ-\frac{ \angle\text{C}}{2}$
So, $\angle\text{ECB}=\angle\text{C}+90^\circ-\frac{\angle\text{C}}{2}$
$\Rightarrow\ \angle\text{ECB}=90^\circ+\frac{\angle\text{C}}{2}$
Now putting all values in eq. (ii)
$\frac{\angle\text{B}}{2}+\angle\text{E}+90^\circ+\frac{\angle\text{C}}{2}=180^\circ$
$\Rightarrow\ \angle\text{E}=180^\circ-90^\circ-\Big(\frac{\angle\text{B}+\angle\text{C}}{2}\Big)$
$=90^\circ-\Big(\frac{\angle\text{B}+\angle\text{C}}{2}\Big)$
$=90^\circ-\frac{130^\circ}{2}$
$\Rightarrow\ \angle\text{E}=25^\circ$

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Question 721 Mark

Which of the following statements is true?

A triangle can have two acute angles.
A triangle can have two right angles.
A triangle can have two obtuse angles.
All are true.

Answer

A
Solution:
A triangle can have two or even all three acute angles (in case of an equilateral triangle) but it cannot have two right angles or two obtuse angles as the sum of the interior angles of a triangle is 180º and two right angles or two obtuse angles would sum up to 180º or more leaving the no space for the third angle.

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Question 731 Mark

In Fig. if$ l_1 || l_2,$ the value of $x$ is:

Answer

From figure,
$\angle\text{ACS}=180^\circ-2\text{b}^\circ$
also $\angle\text{ACS}=\angle\text{PAC}=2\text{a}^\circ ($alternate angles$)$
$\Rightarrow2\text{a}^\circ=180^\circ-2\text{b}^\circ$
$\Rightarrow\text{a}^\circ+\text{b}^\circ=90^\circ$
Now, in $\triangle\text{ABC}$
$\text{a}^\circ+\text{b}^\circ+\angle\text{ABC}=180^\circ$
$\angle\text{ABC}=180^\circ-2\text{x}^\circ$
$\Rightarrow\text{a}^\circ+\text{b}^\circ+180^\circ-2\text{x}^\circ=180^\circ$
$\Rightarrow2\text{x}^\circ=\text{a}^\circ+\text{b}^\circ=90^\circ$
$\Rightarrow\text{x}^\circ=45^\circ$

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Question 741 Mark

In the given fig., if ABCD is a quadrilateral in which AD = BC, AB = CD, and $\angle\text{D} = \angle\text{B},$ then $\angle\text{CAB}$ is equal to:

Answer

$\angle\text{ACD}$
Solution:
$\angle\text{ACD}$ is the right answer;
In $\triangle\text{ACD}$ and $\triangle\text{CAB}$
AD = BC (Given)
$\angle\text{D}=\angle\text{B}$ (Given)
AB = CD (Given)
$\therefore\triangle\text{ACD}≅\triangle\text{CAB}$ (by SAS congruence criteria),
$⇒ \angle\text{CAB} = \angle\text{ACD}$ (cpct).

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Question 751 Mark

In figure, ABC is a triangle in which $\angle\text{B} = 2\angle\text{C}.$ D is a point on side BC such that AD bisects $\angle\text{BAC}$ and AB = CD. BE is the bisector of $\angle\text{B}.$ The measure of $\angle\text{BAC}$ is:

Answer

72º
Solution:
Given that $\triangle\text{ABC}$
BE is bisector of $\angle\text{B}$ and AD is bisector of $\angle\text{BAC}$
$\angle\text{B} = 2\angle\text{C}$
By exterior angle theorem in triangle ADC
$\angle\text{ADB} = \angle\text{DAC} + \angle\text{C} ...\ \text{(i)}$
In $\triangle\text{ADB},$
$\angle\text{ABD} + \angle\text{BAD} + \angle\text{ADB} = 180^\circ$
$2\angle\text{C} + \angle\text{BAD} + \angle\text{DAC} + \angle\text{C} = 180^\circ$ [From (i)]
$3\angle\text{C} + \angle\text{BAC} = 180^\circ$
$\angle\text{BAC} = 180^\circ - 3\angle\text{C} ...\ \text{(ii)}$
Therefore,
$\text{AB = CD}$
$\angle\text{C} = \angle\text{DAC}$
$\angle\text{C} = \frac{1}{2}\angle\text{BAC}\ ...\ \text{(iii)}$
Putting value of Angle C in (ii), we get
$\angle\text{BAC} = 180^\circ - \frac{1}{2} \angle\text{BAC}$
$\angle\text{BAC} +\frac{3}{2} \angle\text{BAC} =180^\circ$
$\frac{5}{2} \angle\text{BAC} =180^\circ$
$\angle\text{BAC} = \frac{180\times2}{2}$
$=72^\circ$
$\angle\text{BAC} = 72^\circ$

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Question 761 Mark

In the given figure AB > AC. If BO and CO are the bisectors of $\angle\text{B}$ and $\angle\text{C}$ respectively then,

Answer

OB > OC
Solution:
AB > AC (given)
$\therefore\ \angle\text{C}>\angle\text{B}$
Or, $\frac{1}{2}\angle\text{C}>\frac{1}{2}\angle\text{B}$
Therefore, OB > OC

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Question 771 Mark

If $\triangle\text{ABC}≅\triangle\text{LKM},$ then side of $\triangle\text{LKM}$ equal to side AC of $\triangle\text{ABC}$ is:

Answer

LM
Solution:
Since, by corresponding part of congruent triangle AC of $\triangle\text{ABC}$ is equal to the LM of $\triangle\text{LKM}.$

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Question 781 Mark

In the adjoining figure, AB = AC and AD is bisector of $\angle\text{A}.$ The rule by which $\triangle\text{ABD}\cong\triangle\text{ACD}.$

Answer

SAS
Solution:
In $\triangle\text{ABD}$ and $\triangle\text{ADC},$ we have
AB = AC (Given)
$\angle\text{BAD} = \angle\text{DAC}$ (Since AD, bisects $\angle\text{A}$)
AD = AD (common in both)
Hence, $\triangle\text{ABD}\cong\triangle\text{ACD}$ by SAS.

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Question 791 Mark

In figure, ABCD is a quadrilateral in which AB = BC and AD = DC. The measure of $\angle\text{BCD}$ is:

Answer

105º
Solution:
Join AC. We get two isosceles triangles, $\triangle\text{ABC}$ and $\triangle\text{ACD}$
In $\triangle\text{ABC},\ \angle\text{ABC}= 108^\circ$
$\therefore\ \angle\text{BAC} = \angle\text{BCA} = \Big(\frac{180^\circ - 108^\circ}{2}\Big) = \frac{72^\circ}{2} = 36^\circ$
In $\triangle\text{ACD},\ \angle\text{ADC}=42^\circ$
$\therefore\ \angle\text{DAC} = \angle\text{DCA} = \Big(\frac{180^\circ - 42^\circ}{2}\Big) = \frac{138^\circ}{2} = 69^\circ$
Now, $\angle\text{BCD} = \angle\text{BCA} + \angle\text{DCA} = 36^\circ + 69^\circ = 105^\circ$

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Question 801 Mark

If $\triangle\text{ABC}\cong\triangle\text{PQR}$ and $\triangle\text{ABC}$ is not congruent to $\triangle\text{RPQ},$ then which of the following is not true:

Answer

$\text{BC}=\text{PQ}$
Solution:
$\triangle\text{ABC}\cong\triangle\text{PQR}$
$\Rightarrow\text{AB}=\text{PR},\text{AC}=\text{PR},\text{BC}=\text{QR}$
$\triangle\text{ABC}\not\cong\triangle\text{RQP}$
$\Rightarrow\text{AB}\not=\text{QR},\text{AC}\not=\text{RP},\text{BC}\not=\text{PQ}$
Hence, correct option (a).

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Question 811 Mark

In the adjoining figure, the rule by which $\triangle\text{ABC}\cong\triangle\text{ADC}\ ?$

Answer

SSS
Solution:
In $\triangle\text{ABC}$ and $\triangle\text{ADC}$ we have,
AB = AD (4cm)
BC= DC (2.7cm)
AC = AC (commom in both)
Hence, $\triangle\text{ABC}\cong\triangle\text{ADC},$ by SSS criterion.

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Question 821 Mark

In Fig. ABC is a triangle in which $\angle\text{B}=2\angle\text{C}.$ D is a point on side BC such that AD bisects $\angle\text{BAC}$ and $\text{AB}=\text{CD}.$ Be is the bisector of $\angle\text{B}.$ The measure of $\angle\text{BAC}$ is:

Answer

72º
Solution:

$\angle\text{ABE}=\angle\text{EBC}$ $($EBC is bisector of $\angle\text{B})$
and $\angle\text{C}=\frac{\angle\text{B}}{2}$
$\Rightarrow\angle\text{EBC}=\angle\text{ECB}$
So $\triangle\text{EBC}$ is isosceles triangle.
$\Rightarrow\text{EB}=\text{EC}\ ....(1)$
Now Consider $\triangle\text{ABE}$ and $\triangle\text{DCE}$
$\text{AB}=\text{DC}$ (Given)
$\text{BE}=\text{CE}$ [From (1)]
$\angle\text{ABE}=\angle\text{DCE}$ (From above data)
So $\triangle\text{ABE}\cong\triangle\text{DCE}$ by SAS property
$\Rightarrow\text{AE}=\text{DE}$
$\angle\text{BAE}=\angle\text{CDE}=\angle\text{A}$
Now consider $\triangle\text{AED},$
$\text{AE}=\text{DE}$ (above proved)
$\Rightarrow\triangle\text{AED}$ is isosceles triangle
$\Rightarrow\angle\text{EAD}=\angle\text{EDA}=\frac{\angle\text{A}}{2}$ $($AD is Bisector of $\angle\text{A})\ ....(2)$
Now, consider $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\text{C}=180^\circ$
$\Rightarrow\angle\text{A}+2\angle\text{C}+\angle\text{C}=180^\circ(\angle\text{B}=2\angle\text{C)}$
$\Rightarrow\angle\text{A}+3\angle\text{C}=180^\circ\ .....(3)$
Consider $\triangle\text{ADE},$
$\Rightarrow\frac{\angle\text{A}}{2}+\angle\text{ADC}+\angle\text{}\text{C}=180^\circ$
$\Rightarrow\frac{\angle\text{A}}{2}+(\angle\text{EDA}+\angle\text{CDE})+\angle\text{C}=180^\circ$
$\Rightarrow\frac{\angle\text{A}}{2}+\frac{\angle\text{A}}{2}+\angle\text{A}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{}A+\angle\text{A}+\angle\text{C}=180^\circ$
$\Rightarrow2\angle\text{A}+\angle\text{C}=180^\circ\ .....(4)$
Right hand side of equations (3) and (4) are equal, hence Left hand side.
$\Rightarrow\angle\text{A}+3\angle\text{C}=2\angle\text{A}+\angle\text{C}$
$\Rightarrow\angle\text{A}=2\angle\text{C}$
Substituting in equation (3),
$2\angle\text{C}+3\angle\text{C}=180^\circ$
$\Rightarrow5\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{C}=36^\circ$
$\Rightarrow\angle\text{A}=2\times36^\circ=72^\circ$
Hence, correct option is (a).

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Question 831 Mark

In $\triangle\text{ABC, BC = AB}$ and $\angle\text{B}=80^{\circ}.$ Then, $\angle\text{A = ?}$

Answer

50°
Solution:
In $\triangle\text{ABC,}$
$\text{BC = AB}$
$\Rightarrow\angle\text{A}=\angle\text{C}$ (angles opposite to equal sides are equal)
Now, $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^{\circ}$
$\Rightarrow\angle\text{A}+80^{\circ}+\angle\text{A}=180^{\circ}$
$\Rightarrow2\angle\text{A}+100^{\circ}$
$\Rightarrow\angle\text{A}=50^{\circ}$

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Question 841 Mark

An exterior angle of a triangle is 108º and its interior opposite angles are in the ratio 4 : 5. The angles of the triangle are:

Answer

48º, 60º, 72º
Solution:

From figure, we have
$\angle\text{A}+ \angle\text{B} = \angle\text{ACD}$
$⇒ 4\text{x}^\circ + 5\text{x}^\circ = 180^\circ$
$⇒ 9\text{x}^\circ = 108^\circ$
$⇒ \text{x} = 12^\circ$
So, $\angle\text{A} = 48^\circ,\ \angle\text{B} = 60^\circ$
$\Rightarrow \angle\text{C} = 180^\circ - 48^\circ - 60^\circ = 72^\circ$

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Question 851 Mark

In $\triangle\text{ABC},\ \angle\text{A} = 50^\circ,\ \angle\text{B} = 60^\circ.$ Find the longest side of the triangle?

Answer

AB
Solution:
By angle sum property, we have,
$\angle\text{A} + \angle\text{B} + \angle\text{C} = 180^\circ$
$\Rightarrow 50^\circ + 60^\circ + \angle\text{C} = 180^\circ$
$\Rightarrow\ \angle\text{C} = 180^\circ - (50^\circ+ 60^\circ) = 70^\circ$
Therefore, $\angle\text{C}$ is the largest angle in the triangle and the side opposite to it i.e. AB is the longest side.

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Question 861 Mark

In $\triangle\text{ABC},$ if AB = AC and $\angle\text{ACD}=1200,$ find $\angle\text{A}.$

Answer

60°
Solution:
In $\triangle\text{ABC}, \ \text{AB} = \text{AC}$
$⇒ \angle\text{ABC} = \angle\text{ACB}$
Also, $\angle\text{ACD}=120^\circ$
$⇒ \angle\text{ACB} = 180^\circ- \angle\text{ACD}$ (Linear pair)
$⇒ \angle\text{ACB} = 180^\circ- 120^\circ = 60^\circ$
$⇒ \angle\text{ABC} = 60^\circ$
By using angle sum property, we have
$\angle\text{ABC} + \angle\text{ACB} + \angle\text{BAC} = 180^\circ$
$60^\circ + 60^\circ+ \angle\text{A} = 180^\circ$
or, $\angle\text{A} = 180^\circ - 120^\circ= 60^\circ$

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Question 871 Mark

In the given figure, two rays BD and CE intersect at a point A. The side BC of $\triangle\text{ABC}$ have been produced on both sides to points F and G respectively. If $\angle\text{ABF}=\text{x}^\circ, \ \angle\text{ACG}=\text{y}^\circ$ and $\angle\text{DAE}=\text{z}^\circ$ then $\text{z} = ?$

Answer

x + y - 180
Solution:
In the given figure, $\angle\text{ABF}+\angle\text{ABC}=180^\circ$ (Linear pair of angles)
$∴\text{x}^\circ+\angle\text{ABC}=180^\circ$
$⇒\angle\text{ABC}=180^\circ−\text{x}°...\ (\text{i})$
Also, $\angle\text{ACG}+\angle\text{ACB}=180^\circ$ (Linear pair of angles)
$∴\text{y}′+\angle\text{ACB}=180^\circ$
$⇒\angle\text{ACB}=180^\circ−\text{y}′\ ...\ \text{(ii)}$
Also, $\angle\text{BAC}=\angle\text{DAE}=\text{z}° \ ....\ \text{(iii)}$ (Vertically opposite angles)
In $\triangle\text{ABC}$
$\angle\text{BAC}+\angle\text{ABC}+\angle\text{ACB}=180^\circ$ (Angle sum property)
$∴\text{z}^\circ+180−\text{x}^\circ+180^\circ−\text{y}^\circ=180^\circ$ [Using (1), (2) and (3)]
$⇒ \text{z} = \text{x} + \text{y} – 180$

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Question 881 Mark

In $\triangle\text{ABC},$ if $\angle\text{C}>\angle\text{B},$ then

Answer

AB > AC
Solution:
Greater angle has greater side opposite to it.

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Question 891 Mark

In fig, if AD = BC and $\angle\text{BAD} = \angle\text{ABC},$ then $\angle\text{ACB}$ is equal to:

Answer

$\angle\text{BDA}$
Solution:
The two triangles are congruent according to (SAS CONGRUENCY) as AD = BC (given), $\angle\text{BAD} = \angle\text{ABC},$ (given) and AB = AB (common) and hence corresponding angles are equal (cpct).

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Question 901 Mark

In Fig. x + y =

Answer

230º
Solution:
$\triangle\text{ACO}$
$\angle\text{ACO}+\angle\text{COA}+\angle\text{COA}=180^\circ$
Now, $\angle\text{OAC}=180^\circ-\text{x}^\circ$
$\Rightarrow80^\circ+40^\circ+180^\circ-\text{x}^\circ=180^\circ$
$\Rightarrow\text{x}^\circ=120^\circ$
$\angle\text{BOD}=\angle\text{COA}=40^\circ$ (Opposite angles)
$\angle\text{BDO}=70^\circ$
In $\triangle\text{OBD},$
$\angle\text{OBD}=180^\circ-40^\circ-70^\circ=70^\circ$
Also, $\text{y}^\circ=180^\circ-\angle\text{OBD}=180^\circ-70^\circ=110^\circ$
$\Rightarrow\text{x}^\circ+\text{y}^\circ=120^\circ+110^\circ=230^\circ$

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Question 911 Mark

If the altitudes from two vertices of a triangle to the opposite sides are equal, then the triangle is:

Answer

Isosceles
Solution:

A $\triangle\text{ABC}$ is given in which $\text{BL}\perp\text{AC}$ and $\text{CM}\perp\text{AB}$ such that $\text{BL = CM.}$
TO prove: $\text{AB = AC}$
In $\triangle\text{ABL}$ and $\triangle\text{AMC,}$
$\text{BL = CM}$ ...(Given)
$\angle\text{ALB}=\angle\text{AMC}$ ...(Each is 90°)
$\angle\text{LAB}=\angle\text{MAC}$ ...(Common angle)
$\therefore\triangle\text{ABL}$ and $\triangle\text{AMC}$ ...(AAS congruence criterion)
$\Rightarrow\text{AB = AC}$ ...(C.P.C.T.)
Hence, the $\triangle\text{ABC}$ is an Isosceles triangle.

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Question 921 Mark

In the given figure, side $BC$ of $\triangle\text{ABC}$ has been produced to a point $D$. If $\angle\text{A}=3\text{y}^\circ \angle\text{B}=\text{x}^\circ, \ \angle\text{C}=5\text{y}^\circ$ and $\angle\text{CBD}=7\text{y}^\circ.$ Then, the value of $x$ is:

Answer

In the given figure $\angle\text{ACB}+\angle\text{ACD}=180^\circ ($Linear pair of angles$)$
$\therefore 5\text{y}′+7\text{y}′=180^\circ$
$\Rightarrow 12y^\circ = 180^\circ$
$\Rightarrow y = 15 ...(i)$
In $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\angle\text{ACB}=180^\circ ($angle sum property$)$
$\therefore 3\text{y}′+\text{x}′+5\text{y}′=180^\circ$
$\Rightarrow x^\circ + 8y^\circ = 180^\circ$
$\Rightarrow x^2+ 8 \times 15^\circ = 180^\circ [$using $(1)]$
$\Rightarrow x^\circ + 120^\circ = 180^\circ$
$\Rightarrow x^\circ = 180^\circ − 120^\circ = 60^\circ$
Thus, the value of $x$ is $60.$

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Question 931 Mark

The bisector of exterior angles at B and C of $\triangle\text{ABC}$ meet at O. If $\angle\text{A} = \text{x}^\circ,$ then $\angle\text{BOC}.$

Answer

$90^\circ-\frac{\text{x}^\circ}{2}$
Solution:
In $\triangle\text{ABC}$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\angle\text{B}+\angle\text{C}=180^\circ-\text{x}^\circ\ ...\ \text{(i)}$
$\angle\text{CBD}=180^\circ-\angle\text{B}\ ...\ \text{(ii)}$
$\angle\text{ECB}=180^\circ-\angle\text{C}\ ...\ \text{(iii)}$
$\Rightarrow\ \frac{\angle\text{CBD}}{2}=\angle\text{OBC}=90^\circ-\frac{\angle\text{B}}{2}\ ...\ \text{(iv)}$
$\Rightarrow\ \frac{\angle\text{ECB}}{2}=\angle\text{OCB}=90^\circ-\frac{\angle\text{C}}{2}\ ...\ \text{(v)}$
Now, in $\triangle\text{BOC}$
$\angle\text{OBC}+\angle\text{OCB}+\angle\text{BOC}=180^\circ$
$\angle\text{BOC}=180^\circ-(\angle\text{OBC}\ +\ \angle\text{OCB})$
From eq. (iv) and (v)
$\angle\text{BOC}=180^\circ-\Big(90^\circ-\frac{\angle\text{B}}{2}+90^\circ-\frac{\angle\text{C}}{2}\Big)$
$=\ 180^\circ-\Big(180^\circ-\frac{\angle\text{B}}{2}-\frac{\angle\text{C}}{2}\Big)$
$=\frac{\angle\text{B}+\angle\text{C}}{2}$
$=\frac{180^\circ+\text{x}^\circ}{2}$
$\angle\text{BOC}=90^\circ-\frac{\text{x}^\circ}{2}$

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Question 941 Mark

D is a point on the side BC of a $\triangle\text{ABC}$ such that AD bisects $\triangle\text{BAC}$ then:

Answer

BA > BD
Solution:
Since, $\triangle\text{BAC}$ is bisected by AD, then $\triangle\text{BAD}$ is less than $\triangle\text{ABC},$ hence the side opposite $\triangle\text{ABC},$ i.e., BA is greater than the side opposite to $\triangle\text{BAD}$ i.e., BD.

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Question 951 Mark

The base BC of triangle ABC is produced both ways and the measure of exterior angles formed are 94º and 126º. Then, $\angle\text{BAC} =$

Answer

40º
Solution:

$\angle\text{ABC} = 180^\circ - 126^\circ = 54^\circ$
$\angle\text{ACB} = 180^\circ - 94^\circ = 86^\circ$
Now, in $\triangle\text{ABC}$
$\angle\text{BAC} + \angle\text{ABC} + \angle\text{ACB} = 180^\circ$
$⇒\angle\text{BAC} = 180° - \angle\text{ABC} - \angle\text{ACB}$
$= 180^\circ - 54^\circ - 86^\circ$
$⇒\angle\text{BAC} = 40^\circ$

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Question 961 Mark

D, E, F are the mid-point of the sides BC, CA and AB respectively of $\triangle\text{ABC}.$ Then $\triangle\text{DEF}$ is congruent to triangle.

Answer

AFE, BFD, CDE
Solution:
In anytriangle, a line joining the mid-points of any two sides is parallel to the third side.
⇒ EF || BC EF || DC and BD
Similiarly DF || EC
⇒DF || AE and EC
Also DE || AB.
⇒ DE || AF and BF
From this information it is clear that EFDC, EFBD, EAFD
are the parallelogram by property.
Now consider one parallelogram EFDC
Consider $\triangle\text{DEF}$ and $\triangle\text{EDC}$
$\text{DE}=\text{ED}$ (common)
$\angle\text{DEF}=\angle\text{EDC}$
$\angle\text{EDF}=\angle\text{DEC}$ (ASA property)
$\Rightarrow\triangle\text{DEF}\cong\triangle\text{EDC}$
Similiarly in parallelogram EAFD,
$\triangle\text{DEF}\cong\triangle\text{AFC}$
And in parallelogram EFBD
$\triangle\text{DEF}\cong\triangle\text{FBD}$
Hence, correct option is (d).
Note: Option (d) modified.

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Question 971 Mark

An exterior angle of a triangle is 108º and its interior opposite angles are in the ratio 4 : 5 The angles of the triangle are:

Answer

48º, 60º, 72º
Solution:

From figure, we have
$\angle\text{A}+\angle\text{B}=\angle\text{ACD}$
$\Rightarrow4\text{x}^\circ+5\text{x}^\circ=108^\circ$
$\Rightarrow9\text{x}^\circ=108^\circ$
$\Rightarrow\text{x}=12^\circ$
So, $\angle\text{A}=48^\circ,\angle\text{B}=60^\circ$
$\Rightarrow\angle\text{C}=180^\circ-48^\circ-60^\circ=72^\circ$

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Question 981 Mark

Find the measure of angles which is equal to its supplement.

Answer

90º
Solution:
We know that the sum of supplementary angle is 180º
Let the angle be x the other angle is 180° - x and the two angles are equal,
⇒ x = 180º - x
⇒ 2x = 180º
$\Rightarrow\ \text{X}=\frac{180^\circ}{2}$
$\Rightarrow\ \text{X}=90^\circ$

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Question 991 Mark

In fig, if AD = BC and $\angle\text{BAD} = \angle\text{ABC},$ then $\angle\text{ACB}$ is equal to:

Answer

$\angle\text{BDA}$
Solution:
The two triangles are congruent according to (SAS CONGRUENCY) as AD = BC (given), $\angle\text{BAD} = \angle\text{ABC},$ (given) and AB = AB (common) and hence corresponding angles are equal (cpct).

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Question 1001 Mark

The sum of the interior angles of a triangle is:

Answer

180º
Solution:
For a triangle,
Number of sides (n) = 3
Sum of interior angles = (n - 2) × 180º
= (3 - 2) × 180º
= 1 × 180º
= 180º

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MATHS M.C.Q