Answer: Option A. -> True
:
A
In a right angled triangle, square of the hypotenuse (the side opposite to the right angle) is equal to the sum of the squares of the other two sides. This is Pythagoras' theorem.​

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 C. -> 60∘
:
C
The sum of all the angles of a triangle is ${180}^{\circ }$​.
Therefore,
$\text{Third angle}={180}^{\circ }-({50}^{\circ }+{70}^{\circ })={180}^{\circ }-\left({120}^{\circ }\right)={60}^{\circ }$​

:
Answer: 1 Mark
Justification: 1 Mark
Yes.

The median AD will divide the ∆ABC into 2 triangles ∆ADB and ∆ADC. These two triangles, as shown in the figure will have the same altitude AN as the altitude is the perpendicular distance from the vertex to the base of the triangle.The foot of the perpendicular can be inside as well as outside the triangle as shown in the figure. (Note: The median of the triangle divides it into 2 triangles with equal area.)

:
Properties : 1 Mark
Steps : 2 Marks
Result : 1 Mark

In triangle ABC, AC < AB + BC ------------------- 1
In triangle ADE, AD < AE + DE ------------------- 2
Adding 1 and 2,
AC + AD < AB + BC + AE + DE ----------------3
In triangle ADC, CD < AD + AC ------------------- 4
From 3 and 4, we can write
CD < AB + BC + AE + DE
Adding CD to both sides,
2CD < AB + BC + CD + DE + EA
Or, 2 CD < Perimeter of ABCDE

:
(a) Answer: 1 Mark
(b) Answer: 1 Mark
Reason: 1 Mark
(a) In a right-angled isosceles triangle, sides other than the hypotenuse are equal in length.
In ABC, AB = BC $\Rightarrow \angle 1=\angle 3$ -------------------I
By angle sum property of triangle,
$\angle 1+\angle 2+\angle 3={180}^{\circ }$
$\angle 1+90+\angle 1={180}^{\circ }$
$2\angle 1={90}^{\circ }$
$\Rightarrow \angle 1={45}^{\circ }=\angle 2$
So, angles are ${45}^{\circ },{45}^{\circ }and{90}^{\circ }$
(b)A triangle ABC with the given side lengths of AB = 12 cm, BC = 8 cm andAC = 20 cm is not possible becausein a trianglethe sum of any two sides should be greater than the third side but in this triangle, AB + BC = 20 cm which is equal to the third side AC = 20 cm. AB + BC should have been greater than AC.

:
(a) Steps: 1 Mark
Proof: 1 Mark
(b) Steps: 1 Mark
Result: 1 Mark
(a)

$\angle 1$, $\angle 2$ and $\angle 3$ are angles of triangle ABCand 4 is the exterior angle when BC is extended to D.
$\angle 1$ + $\angle 2$ = $\angle 4$ (Exterior Angle Property)
$\angle 1$ + $\angle 2$ + $\angle 3$ = $\angle 4$ + $\angle 3$ (Adding $\angle 3$ to both sides)
Also, $\angle 3$ + $\angle 4$ = 180${}^o$ (Linear Pair of angles)
$\therefore$ $\angle 1$ + $\angle 2$ + $\angle 3$ = 180${}^o$
(b)

$\angle 1$ + $\angle 2$ = $\angle 4$ (Exterior Angle Property)
$\angle 1+{60}^{\circ }={140}^{\circ }$
$\angle 1={140}^{\circ }-{60}^{\circ }={80}^{\circ }$
Since AE is the angular bisector therefore $\angle EAC=\frac{\angle 1}{2}$
$\Rightarrow x=\frac{80}{2}={40}^{\circ }$

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

:
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 = ${360}^{\circ }$​

:
Each angle: 1 Mark
The three angles in the given triangle are x, 2x and 3x.
For any triangle,
Sum of all the angles = 180${}^o$
Or, x + 2x + 3x = 180${}^o$
Or, 6x = 180${}^o$
Or x = $\frac{{180}^o}{6}$ = 30${}^o$
The angles of the triangle will be 30${}^o$(x), 60${}^o$(2x) and 90${}^o$(3x).