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

As we know, the measure of exterior angle is equal to the sum of opposite two interior angles.

In $\triangle\text{PQR},$ $\angle\text{x}$ is the exterior angle.
So, $\angle\text{x}=\angle\text{P}+\angle\text{Q}$
$=60^{\circ}+40^{\circ}=100^{\circ}$

In any Triangle, the sum of any two sides is always greater than the
other side.in $\triangle\text{ABC},\text{AB}+\text{BC}>\text{CA}....\text{eqn(1)}$
$\text{BC + CA > AB}.....\text{eqn}(2)$
$\text{CA + AB > BC}...\text{eqn}(3)$ In equation $(2),$ adding $AB$ both side we
get, $\text{AB + BC + CA > 2AB}$

As we know, sum of the lengths of any two sides of a triangle is always greater than the length of the third side.

$\text{In} \ \triangle\text{PQR,}$ $PR + QR > PQ$
$\Rightarrow PR > PQ - QR$
$\Rightarrow PQ - QR < PR$

$x + y = 180^\circ - 40^\circ = 140^\circ x - y = 30^\circ .$
$x = 85^\circ .$

Theorem: In any triangle, sum of any two sides is greater than the third side. Also, the difference of any two sides must be smaller than the third side.
Therefore, in an $\triangle\text{ABC} AB - BC < CA$ and $AB + BC > CA.$

As we know, sum of any two sides of a triangle is always greater than the third side.
Hence, option (c) satisfies the given condition.
Verification$18 + 14 > 5$
$18 + 5 > 14$
$5 + 14 > 18.$

In right traingle $BOC,$
$BC^2 = OC^2 + OB^2$
$\Rightarrow (15)^2 = (9)^2 +OB^2$
$\Rightarrow 225 = 81 + OB^2$
$\Rightarrow OB^2 = 144$
$\Rightarrow OB^2 =(12)^2$
$\Rightarrow OB = 12m$
Hence, the correct answer is option $(d).$

In $\triangle \text{ABD}$
$\angle \text{ADB}+\angle \text{BAD}+\angle \text{ABD}=180^\circ$ [Angle sum property of triangle]
$\Rightarrow 61^\circ+59^\circ+\angle \text{ABD}=180^\circ$
$\Rightarrow \angle \text{ABD}=60^\circ$
$\angle \text{ABD}+\angle \text{DBC}=180^\circ$ [Linear pair angles]
$\Rightarrow 60^\circ+\text{y}^\circ=180^\circ$
$\Rightarrow \text{y}=120$
Now, $\angle \text{ADB}=\angle \text{GDE}=61^\circ$ [Vertically opposite angles]
Now, In $\triangle \text{GDE},$
$\angle \text{GDE}+\angle \text{DGE}+\angle \text{GED}=180^\circ$ [Angle sum property of triangle]
$\Rightarrow 61^\circ+69^\circ+\angle \text{GED}=180^\circ$
$\Rightarrow \angle \text{GED}=50^\circ$
Now, $\angle \text{GED}+\angle \text{GEF}=180^\circ$ [Linear pair angles]
$\Rightarrow 50^\circ+\text{x}^\circ=180^\circ$
$\Rightarrow \text{x}=130$
Hence, the correct answer is option $(a).$

$\angle\text{x}+\angle\text{y}=120^\circ,\angle\text{x}-\angle\text{y}=10^\circ$
$\angle\text{x}=65^\circ,\angle=55^\circ$

Since, $AB \| CD$
$\angle \text{ABD}+\angle \text{CDB}=180^\circ$ [Angles on the same side of a transversal line are supplementary]
$\Rightarrow \text{x}^\circ+27^\circ+112^\circ=180^\circ$
$\Rightarrow \text{x}^\circ=41^\circ$
$\Rightarrow \text{x}=41$
Now, In $\triangle \text{ABC},$
$\angle \text{A}+\angle \text{B}+\angle \text{C}=180^\circ$ [Angle sum property of triangle]
$\Rightarrow 49^\circ+41^\circ+\text{y}^\circ=180^\circ$
$\Rightarrow \text{y}^\circ=90^\circ$
$\Rightarrow \text{y}=90$
Hence, the correct answer is option $(a).$

Since, $∠\text{D}$ and $∠\text{F}$ are complementary angles.
In $\triangle\text{DEF},$
$∠\text{D}+\angle\text{E}+\angle\text{F}=180^{\circ}$ [angle sum property of a triangle]
$\Rightarrow \ ∠\text{D}+90^{\circ}+\angle\text{F}=180^{\circ}$$[\because\angle\text{E}=90^{\circ},\text{given}]$
$\Rightarrow \ ∠\text{D}+\angle\text{F}=180^{\circ}-90^{\circ}$
$\Rightarrow \ ∠\text{D}+\angle\text{F}=90^{\circ}$
Note: Two angles whose measures add to $180^\circ $ are known as supplementry angles and two angles whose measures add to $90^\circ $ are known as complementary angles.

$AAA$ is not a congruency criterion, because if all the three angles of two triangles are equal; this does not imply that both the triangles fit exactly on each other.

Since, $AB = DC$ [given]
and $AC = DB [common base]$
$BC = BC$
By $SSS$ congruence criterion, $\triangle\text{ABC}\cong\triangle\text{DCB}$

$\angle \text{TRS}+\angle \text{TRQ}=180^\circ$ [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 $(b).$

Value of exterior angle is equal to the sum of two opposite angles.
Thus, here, $110 = 50 + x$
so, $x = 60$ degree

Suppose $BC$ is the ladder which is placed againts the wall $OA.$ The foot of the ladder $C$ is $15m$ away from the foot $O$ of the wall and its top reaches the window which is $20m$ above the ground.

In right traingle $BOC,$
$B C^2=O C^2+O B^2 $
$\Rightarrow B C^2=(15)^2+(20)^2 $
$\Rightarrow B C^2=225+400 $
$\Rightarrow B C^2=625 $
$\Rightarrow B C^2=(25)^2 $
$\Rightarrow B C=25 m$
Hence, the correct answer is option $(b).$

In $(c)$
$BC^2 = AC^2 + AB^2$
$\Rightarrow (17)^2 = (15)^2 + (8)^2$
$\Rightarrow 289 =225 + 64$
$\Rightarrow 289 = 289$
Since, the sum of the square of two smallest side is equal to the square of largest side.
Hence, $ABC$ is a right angle triangle at $A.$
Hence, the correct answer is option $(c).$

As we know, the measure of any exterior angle is equal to the sum of two opposite interior angles.
Let the interior angle be $x.$
Given that, interior opposite angles are equal.
$\therefore \ 130^{\circ}=\text{x}+\text{x}$
$\Rightarrow \ 130^{\circ}=2\text{x}$
$\Rightarrow \ \text{x}=\frac{130^{\circ}}{2}$
$\Rightarrow \ \text{x}=65^{\circ}$
Hence, the interior angle is $= 65^\circ .$

$\angle \text{A}+\angle \text{B}+\angle \text{C}=180^\circ$ [Angle sum property of triangle]
$\Rightarrow 150^\circ+\angle \text{C}=180^\circ$
$\Rightarrow \angle \text{C}=30^\circ$
Now, $\angle \text{B}+\angle \text{C}=75^\circ$
$\Rightarrow \angle \text{B}+30^\circ=75^\circ$
$\Rightarrow \angle \text{B}=45^\circ$
Hence, the correct answer is option $(b).$

Consider $\triangle ABC$ in which $AD$ divides $BC$ in the ratio $1:1.$

Now, $BD : DC = 1 : 1$
$\Rightarrow \ \frac{\text{BD}}{\text{DC}}=\frac{1}{1}$
$\therefore \ \text{BD}=\text{DC}$
Since, Ad divides BC into two equal parts. Hence, $AD$ is the median.

Let the smallest angle be x, then the other angle be $4x.$
Now,
$x + 4x + 90^\circ = 180^\circ $
$\Rightarrow 5x = 90^\circ $
$\Rightarrow x = 18^\circ $
Thus, the measure of the angles are $18^\circ $, and $4(18)^\circ = 72^\circ $
Hence, the correct answer is option $(d).$

Let the third angle be $x$ Sum of angles $= 180^\circ 40^\circ + 60^\circ + x = 180^\circ 100^\circ + x = 180^\circ $
$\Rightarrow x = 80^\circ $

$\angle \text{ACD}+\angle \text{ACB}=180^\circ$ [Linear angles]
$\Rightarrow 91^\circ+\angle \text{ACB}=180^\circ$
$\Rightarrow \angle \text{ACB}=89^\circ$
Since, $AB \| DE$
$\angle \text{DEC}=\angle \text{CAB}=46^\circ$ [Alternate angles]
Now,
$\angle \text{ACB}+\angle \text{CAB}+\angle \text{ABC}=180^\circ$ [Angle sum property of triangle]
$\Rightarrow 89^\circ + 46^\circ + \text{x}^\circ = 180^\circ$
$⇒ \text{x}^\circ = 45^\circ$
$\Rightarrow \text{x} = 45$
Hence, the correct answer is option $(d).$

Suppose $BC$ is the ladder which is placed againts the wall $OA$. The foot of the ladder $C$ is $15m$ away from the foot $O$ of the wall and its top reaches the window which is $20m$ above the ground.
In right traingle $ABC$
$AB^2 = BC^2 + AC^2$
$\Rightarrow AB^2 = (5)^2 + (5)^2$
$\Rightarrow AB^2 =25 + 25$
$\Rightarrow AB^2 = 50$
$\Rightarrow \text{AB}^2=(5\sqrt{2})^2$
$\Rightarrow \text{AB}=5\sqrt{2}\text{cm}$
Hence, the correct answer is option (b).

Given that, $\triangle\text{PQR}\cong\triangle\text{STU}$
$\Rightarrow \ \text{PQ} =\text{ST} $
$\Rightarrow \ \text{QR} =\text{TU} $
$\Rightarrow \ \text{PR} =\text{SU} $
Hence, $\text{TU}=\text{QR}=6\text{cm}.$

As we know, sum of any two sides in a triangle is always greater than the third side.
So, only $4$ is the minimum value that satisfies as a side in triangle.
$\begin{cases}10<6.5+4\\6.5<10+4\\4<10+6.5\end{cases}\Bigg\}$