Answer: Option A. ->
:
(a) Solution: 1 Mark
     Type of triangle: 1 Mark
(b) Solution: 1 Mark
     Correct answer: 1 Mark
(a) For a triangle, sum of all angles = 180${}^o$
Or, (2x - 15${}^o$) + (x + 20${}^o$) + (x + 15${}^o$) = 180${}^o$
Or, 2x + x + x - 15${}^o$ + 20${}^o$ + 15${}^o$ = 180${}^o$
Or, 4x + 20${}^o$ = 180${}^o$
Or, 4x = 160${}^o$
Or, x = $\frac{160}{4}$ = 40${}^o$
The angles of the triangle will be, (2x - 15${}^o$ = 65${}^o$), (x + 20${}^o$ = 60${}^o$) and (x + 15${}^o$ = 55${}^o$)
Since all angles are less than 90${}^o$, it's an 'acute angled triangle' and also all the angles are of different values therefore all the sides will be of different length, hence it is a 'scalene triangle'.
(b
             
In triangle ABD
$\Rightarrow 2x+4+90+30=180$
$\Rightarrow 2x=180-124$
$\Rightarrow x=\frac{56}{2}$
$\Rightarrow x=28$
In triangle ADC
$\Rightarrow 4x-k+90+10=180$
$\Rightarrow 4x-k=180-100$
$\Rightarrow 4\left(28\right)-k=80$ (Substituting the value of $x=28$)
$\Rightarrow k=4\left(28\right)-80$
$\Rightarrow k=112-80$
$\Rightarrow k=32$
$\therefore$ The angles are ${40}^{\circ },{60}^{\circ }and{80}^{\circ }$ for the given triangle.

Answer: Option A. ->
:

A median is line joining a vertex of a triangle to the mid-point of the opposite side.

Answer: Option B. -> False
:
B

A median connects a vertex of a triangle to the mid point of the opposite side.
An altitude is a perpendicular on one of the sides from the opposite vertex.
As OB is the altitude, B may or may not be the midpoint of CE.

Answer: Option B. -> False
:

Let the exterior angles of the triangle be x, y and z respectively.
Then by exterior angle property, interior angles of the triangle will be $(180-x{)}^{\circ }$​, $(180-y{)}^{\circ }$​​, $(180-z{)}^{\circ }$​​

Using angle sum property of a triangle,

(180 - x) + (180 - y) + (180 - z) = 180

540 - (x + y + z) = 180

Therefore, x + y + z = ${360}^{\circ }$​

Answer: Option B. -> False
:
B

A triangle with all the three angles less than ${60}^{\circ }$ is not possible, as the sum of all the angles of a triangle is ${180}^{\circ }$​.

Answer: Option B. -> False
:
B

Sum of lengths of any two sides of a triangle should always be greater than the third side.

Here, 2 + 3 = 5
Therefore, this triangle is not possible.

Answer: Option B. -> False
:
B

The square of the hypotenuse is equal to the sum of the squares of the other two sides. This relation holds only in right-angled triangles.

Answer: Option A. -> Fig 1 and Fig 3 only
:
A

An obtuse angled triangle is the triangle in which one of the angles is greater than ${90}^{\circ }.$ 
An isosceles triangle is the triangle in which two sides are equal.
1. Fig 1:
$\Delta PQR$  is isosceles $[\because PQ=PR]$
         $\Rightarrow \angle Q=\angle R$
[$\because$ angles opposite to equal sides of a triangle are equal]
$\angle P+\angle Q+\angle R={180}^{\circ }$ 
[angle sum property of a triangle]
$\angle P+{25}^{\circ }+{25}^{\circ }={180}^{\circ }$
$\Rightarrow \angle P={180}^{\circ }-{50}^{\circ }={130}^{\circ }$ 
$\Delta PQR$ is an obtuse angled triangle as one of the angles measures 130°.
2. Fig 2:
$\Delta ABC$ is isosceles $[\because AB=AC]$
Similarly as above, we can find the angles of this triangle.
$\angle A={35}^{\circ },\angle B=\angle C={72.5}^{\circ }$ 
Since all angles are less than ${90}^{\circ }$, $\Delta ABC$ is an acute angled triangle.
3. Fig 3:
$\Delta XYZ$ is an isosceles as well as an obtuse angled triangle as angle Y measures 110°.
4. Fig 4:
$\Delta MNO$ is an isosceles as well as a right angled triangle.
Hence, only Fig 1 and Fig 3 are isosceles as well as obtuse angled triangles.

Answer: Option A. -> Fig 1 and Fig 3 only
:

By Exterior angle property of triangles,
x + y = $\angle$A + $\angle$B
         = ${50}^{\circ }$​ + ${70}^{\circ }$
         = ${120}^{\circ }$

Answer: Option A. -> True
:
A

The third side of a triangle must be greater than the difference between the other two sides.