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

Cleary sum of any two given sides of is greater than the third side,
so a triangle can be formed. Now all the sides of triangle are unequal, so the triangle formed is a scalene $\triangle$

Since, $AB \| DE$
$\angle \text{DCB}=\angle \text{CBA}=3\text{x}^\circ$ [Alternate angles]
Now,
$\angle \text{ACB}+\angle \text{CAB}+\angle \text{CBA}=180^\circ$ [Angle sum property of triangle]
$\Rightarrow 55^\circ + 2\text{x}^\circ + 3\text{x}^\circ= 180$
$\Rightarrow 5\text{x}^\circ= 125^\circ$
$\Rightarrow \text{x} = 25$
Hence, the correct answer is option $(a).$

In the given triangle $\triangle\text{ABC}$

$AB = AC$
$AD$ is perpendicular
So in $\triangle\text{ADC},\triangle\text{ADB}$
$AD = AD$ Common
$\triangle\text{ADC}=\triangle\text{ADB}=90^\circ$
$\triangle\text{ADC}\cong\triangle\text{ADB}$
$\Rightarrow\angle\text{B}=\angle\text{C}$

As we know, sum of all the interior angles of a triangle is $180^\circ .$
$\text{In} \ \triangle\text{ABC}, \ \angle\text{A}+\angle\text{B}+\angle\text{C}=180^{\circ}$ $[\text{interior angles of }\triangle\text{ABC}].....(\text{i})$
$\text{In} \ \triangle\text{DEF}, \ \angle\text{D}+\angle\text{E}+\angle\text{F}=180^{\circ}$ $[\text{interior angles of }\triangle\text{DEF}].....(\text{ii})$
On adding Equation. (i) and (ii), we get.
$\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}+\angle\text{E}+\angle\text{F}\\ =180^{\circ}+180^{\circ}$
$=360^{\circ}$

The exterior angle is equal to the sum of two opposite interior angle.
$80 = x + 30,$
$x = 50 degree$

In $\triangle\text{PQR}$ and $\triangle\text{PQS},$
$PR = PS = a cm$
$RQ = SQ = b cm$
$PQ = PQ =$ Common line segment
By $SSS$ congruence criterion,
$\triangle\text{PQR}\cong\triangle\text{PQS}$

$\text{In} \ \angle\text{ABC},$ $\angle\text{A}+\angle\text{B}+ \angle\text{C}=180^{\circ}$ [angle sum property of a triangle]
$\Rightarrow \ \angle\text{A}+60^{\circ}+60^{\circ}=180^{\circ}$ ${[\because\angle\text{B}=\angle{\text{C}}=60^{\circ},\text{given}]}$
$\Rightarrow \ \angle\text{A}=120^{\circ}-80^{\circ}$
$\Rightarrow \ \angle\text{A}=60^{\circ}$

Since, all the angles are of $60^\circ $.
so, it is an equilateral triangle.

Given, $\angle\text{BAC}=90^{\circ},$ $\text{AD}\bot\text{BC}$ and $\angle\text{ BAD}=50^{\circ}$
In $\triangle\text{ABD},$
$\angle\text{ABD}+\angle\text{DAB}+\angle\text{ADB}=180^{\circ}$ [angle sum property of a triangle]
$\Rightarrow \ \angle\text{ABD}+50^{\circ}+90^{\circ}=180^{\circ}$
$\Rightarrow \ \angle\text{ABD}+140^{\circ}=180^{\circ}$
$\Rightarrow \ \angle\text{ABD}=180^{\circ}-140^{\circ}$
$\Rightarrow \ \angle\text{ABD}=40^{\circ}$
Now In$\triangle\text{ABD},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^{\circ}$ [angle sum property of a triangle]
$\Rightarrow \ 90^{\circ}+40^{\circ}+\angle\text{C}=180^{\circ}$
$\Rightarrow \ \angle\text{C}=180^{\circ}-130^{\circ}$
$\Rightarrow \ \angle\text{C}=50^{\circ}$
$\therefore\ \angle\text{ACD}=50^{\circ}$

$x + 45 + 45^\circ = 180^\circ$
$\Rightarrow x = 90^\circ .$

$\angle\text{ABC}=60^\circ$
$\therefore\angle\text{ABD}=180^\circ-60^\circ=120^\circ$
$\therefore\text{x}=180^\circ-(120^\circ+30^\circ)$
$=30^\circ$

$\angle\text{PQR}=50^\circ$
$\therefore\angle\text{QPR}-180^\circ-(50^\circ+45^\circ)$
$=85^\circ$