MATHS M.C.Q. [1 Marks Each]

Here,

$\ \angle1+\angle\text{P}=180^{\circ}$ $[\text{linear pair}]$
$​​​​\Rightarrow\ \angle1+\angle\text{140}^{\circ}=180^{\circ}$
$​​​​\Rightarrow\ \angle1=180^{\circ}-\angle\text140^{\circ}$
$​​​​\Rightarrow\ \angle1=40^{\circ}$
Since, $PQ = PR$
$\therefore \ \angle\text{Q}=\angle\text{R}=\text{x} \ $ ​$[\text{say}]$​​​​​​
In$ \ \triangle\text{PQR},$ $\ \angle\text{P}+\angle\text{Q}+\angle\text{R}=180^{\circ}$ [angle sum property of a triangle]
$\Rightarrow \ 40^{\circ}+\text{x}+\text{x}=180^{\circ}$
$\Rightarrow \ 2\text{x}=180^{\circ}-40^{\circ}$
$\Rightarrow \text{2x}=140^{\circ}$
$\Rightarrow\text{x}=70^{\circ}$
So, $\angle\text{Q}=\angle\text{R}=70^{\circ}$
Given that, $RS = RQ$
$\therefore \ \angle2=\angle3=70^{\circ}$
$\text{In} \ \triangle\text{SQR},$ $\angle2+\angle3+\angle4=180^{\circ}$ [angle sum property of a triangle]
$\Rightarrow \ 70^{\circ}+70^{\circ}+\angle4=180^{\circ}$
$\Rightarrow \ \angle4=180^{\circ}-140^{\circ}$
$\Rightarrow \ \angle4 = 40^{\circ}$
Also, $ST || QR$ [given]
Now, $\angle4=\angle6=40^{\circ}$ [alternate interior angles]
$\therefore \ \angle\text{TSR}=40^{\circ}$

Let one angle be $\angle\text{x}$
$\therefore$ other angle is $\angle\text{x}+25^\circ$
Now
Sum of all the angles of triangle $= 180^\circ $
$⟹ x + x + 25^\circ + 65^\circ = 180^\circ $
$⟹ 2x = 180^\circ - 90^\circ $
$⟹ x = 45^\circ $

Given, length of rectangle $= 60cm$ and its diagonal $= 61cm.$

Let the breadth of a rectangle be $xcm.$
In right angled $\triangle\text{ABC},$
$⇒ (AC)^2 = (AB)^2 + (BC)^2$
$⇒ (BC)^2 = (AC)^2 + (AB)^2$ [by pythagoras theoram]
$⇒ x^2 = (61)^2 - (60)^2 = 3721 - 3600 = 121$
$\Rightarrow \ \text{x}=\sqrt{121}=11\text{cm}$
$\therefore$ Breadth of rectangle $= 11cm$ and length of rectangle $= 60cm.$
Now, perimeter of rectangle $= 2(l + b)$
$= 2(60 + 11) = 2 × 71$
$= 142cm.$

$\text{Given}, \ \text{BC}=\text{CA},$
$\therefore\angle\text{B}=\angle\text{A}=40^{\circ}$ $[\because$ opposite angles of two equal sides are equal$]$
As we know, the measure of any exterior angles of a triangle is equal to the sum of the measure of its two interior opposite angles.
So, $\angle\text{ACD}=\angle\text{A}+\angle\text{B}=40^{\circ}+40^{\circ}$
$\angle\text{ACD}=80^{\circ}$.

As we know, the sum of the interior angles of a triangle is $180^\circ .$​​​​​​​

$\text{In} \ \angle\text{ABC},$ $\angle\text{A}+\angle\text{B}+ \angle\text{C}=180^{\circ}$ [angle sum property of a triangle]
$\Rightarrow \ \angle\text{A}+40^{\circ}+40^{\circ}=180^{\circ}$
$\Rightarrow \ \angle\text{A}=180^{\circ}-80^{\circ}$
$\Rightarrow \ \angle\text{A}=100^{\circ}$ $[\text{obtuse angle }]$
Therefore, it is an obtuse angled triangle. Since, it has one angle which is greater than $90^\circ .$

$\angle \text{TRS}+\angle \text{TRQ}=180^\circ$ [Linear angles]
$\Rightarrow 5\text{x}^\circ+\angle \text{TRQ}=180^\circ$
$\Rightarrow \angle \text{TEQ}=180^\circ-5\text{x}^\circ$
Now, $\angle \text{QTR}+\angle \text{TRQ}=\angle \text{PQT}$ [Exterior angle property of triangle]
$\Rightarrow 3\text{x}^\circ+180^\circ-5\text{x}^\circ=120^\circ$
$\Rightarrow 2\text{x}^\circ=60^\circ$
$\Rightarrow \text{x}=30$
Hence, the correct answer is option $(b).$

$3 + 4 > 5; 4 + 5 > 3; 5 + 3 > 4.$

In $\triangle\text{ADC},$
$\angle\text{ADC}+\angle\text{DAC}+\angle\text{ACD}=180^{\circ}$ [angle sum property of a triangle]
$\Rightarrow \ \angle\text{ADC}+50^{\circ}+\angle\text{ACD}=180^{\circ}$ $[\because\angle\text{DAC}=50^{0}]$
$\Rightarrow \ \angle\text{ACD}=130^{\circ}-\angle\text{ACD}......(\text{i})$
In $\triangle\text{ DBC},$ $\angle\text{ADC}=\angle\text{DBC}+\angle\text{BCD}$
$[\because$ exterior angle is equal to sum of opposite interior angles$]$
$\Rightarrow \ \angle\text{ADC}=70^{\circ}+\angle\text{ACD}$ $[\because\angle\text{ACD}=\angle\text{BCD}]$
$\Rightarrow \ \angle\text{ADC}=70^{\circ}+130^{\circ}-\angle\text{ADC}$ [from equation (i)]
$\Rightarrow \ \angle\text{ADC}=200^{\circ}-\angle\text{ADC}$
$\Rightarrow 2\ \angle\text{ADC}=200^{\circ}$
$\Rightarrow \ \angle\text{ADC}=\frac{200^{\circ}}{2}$
$\Rightarrow \ \angle\text{ADC}=100^{\circ}$

In $\triangle \text{DEF}$
$\angle \text{DEF}+\angle \text{DFE}+\angle \text{EDF}=180^\circ$ [Angle sum property of triangle]
$\Rightarrow 110^\circ+40^\circ+\angle \text{EDF}=180^\circ$
$\Rightarrow \angle \text{EDF}=30^\circ$
Now, $\angle \text{EDF}+\angle \text{FDA}=180^\circ$ [Linear pair angles]
$\Rightarrow 30^\circ+\text{x}^\circ=180^\circ$
$\Rightarrow \text{x}=150$
Now, $\angle \text{EDF}=\angle \text{ADB}=30^\circ$ [Vertically opposite angles]
Now, In $\triangle \text{ABD},$
$\angle \text{ADB}+\angle \text{DAB}+\angle \text{ABD}=180^\circ$ [Angle sum property of triangle]
$\Rightarrow 30^\circ+90^\circ+\angle \text{ABD}=180^\circ$
$\Rightarrow \angle \text{ABD}=60^\circ$
Now, $\angle \text{ABD}+\angle \text{DBC}=180^\circ$ [Linear pair angles]
$\Rightarrow 60^\circ+\text{y}^\circ=180^\circ$
$\Rightarrow \text{y}=120$
Hence, the correct answer is option $(c).$

In $\triangle\text{ADB} \ \text{and} \ \triangle\text{ADC}, BD = DC [D$ is the mid-point]
$AB = AC$ [given]
$AD = AD$ [common side]

By $SSS$ congruence criterion, $\triangle\text{ABD}\cong\triangle\text{ACD}$
$\therefore \ \triangle\text{ADB}\cong\triangle\text{ADC}$ [by $CPCT]$
We know that, $\angle\text{ADB}+\angle\text{ADC}=180^{\circ}$ [linear pair]
$\Rightarrow2\angle\text{ADC}=180^{\circ}$ $[\because\angle\text{ADB}=\angle\text{ADC}]$
$\Rightarrow \ \angle\text{ADC}=90^{\circ}$

Two figures are said to be congruent, if the trace copy of figure $1$ fits exactly on that of:

Now, if $\triangle\text{ABC}$ and $\triangle\text{DEF}$ are congruent, then
$AB = DF, BC = EF$
$AC = DE$, $\angle\text{A}=\angle\text{D}$
$\angle\text{B}=\angle\text{F},$ $\angle\text{ C}=\angle\text{E}$
Hence, option (b) is not true.

If all the angles of a triangle measure less than $90^\circ $, then such a triangle is called acute angled triangle.
For eg: An equilateral triangle with all the angles equal to $60^\circ $ is acute angled triangle.

Largest angle $=\frac{4}{2+3+4}\times180^\circ=80^\circ$

$\angle \text{AFE}+\angle \text{EFG}=180^\circ$ [Linear pair angles]
$\Rightarrow 124^\circ+\angle\text{EFG}=180^\circ$
$\Rightarrow \angle\text{EFG}=56^\circ$
Since, $AB || CD$
$\therefore \angle \text{EFG}=\angle \text{FHK}=56^\circ$ [Corresponding angles]
$\Rightarrow \text{x}=56$
Now, $\angle \text{QKH}+\angle \text{GKH}=180^\circ$ [Linear pair angles]
$\Rightarrow 103^\circ+\angle \text{GKH}=180^\circ$
$\Rightarrow \angle \text{GKH}=77^\circ$
Since, $AB || CD$
$\therefore \angle \text{EGF}=\angle \text{GKH}=77^\circ$ [Corresponding angles]
$\Rightarrow \text{y}=77$
In $\triangle \text{EHK},$
$\angle \text{EHK}+\angle \text{EKH}+\angle \text{HEK}=180^\circ$ [Angle sum property of triangle]
$\Rightarrow 56^\circ+77^\circ+\text{z}^\circ=180^\circ$
$\Rightarrow \text{z}=47$
Hence, the correct answer is option $(d).$

In $\triangle\text{PQS},$
$110^{\circ}+\angle\text{1}=180^{\circ}$ [linear pair of angles]
$\Rightarrow \ ∠\text{1}=180^{\circ}-110^{\circ}$
$\Rightarrow\angle\text{1}=70^{\circ}$

Also, $ \ ∠\text{1}=\angle\text{2}=70^{\circ}$ $[\because\text{PQ}=\text{PS}]$
As we know, the measures of any eterior angle of a triangle is equal to the sum of the measures of its two interior opposite angles.
$\therefore \ \angle2=\text{x}+25^{\circ}$
$\Rightarrow 70^{\circ}=\text{x}+25^{\circ}$ $\because\angle2=70^{0}$
$\Rightarrow\text{x}=70^{\circ}-25^{\circ}$
$\Rightarrow \ \text{x}=45^{\circ}$

Let $BE$ be the smaller tree and $AD$ be the bigger tree.
Now, we have to find $AB$ (i.e. the distance between their tops).

By observing,
$ED = BC = 4m$ and $BE = CD = 4m$
$\text{In} \ \triangle\text{ABC},$
$BC = 4m$
and $AC = AD - CD = (7 - 4)m = 3m$
In right angled $ \ \triangle\text{ABC},$
$AB = AC^2 + BC^2 = 4^2 + 3^2$ [by pythagoras theoram]
$= 16 + 9$
$\Rightarrow AB^2 = 25$
$\Rightarrow \text{AB} = \sqrt{25}$
$\Rightarrow \ \text{AB}=5\text{m}$
Therefore, distance between thier tops is $5m.$

As we know, sum of any two sides of a triangle is always greater than the third side.
So, option (d) satisfy this rule.
Verification$6 + 6 > 10$
$6 + 10 > 6$
$10 + 6 > 6.$

A triangle has $3$ altitudes.

$(2x)^\circ + (3x - 5)^\circ + (4x - 13)^\circ = 180^\circ $ [Angle sum property of triangle]
$\Rightarrow 2x + 3x - 5 + 4x - 13 = 180^\circ $
$\Rightarrow 9x - 18 = 180$
$\Rightarrow 9x = 198$
$\Rightarrow x = 22$
Hence, the correct answer is option $(a).$