(a)No
(b)Yes
(c)Yes
(d)Yes
(e)Yes 
(f)Yes.

\(\angle ABC = 45° + 70° = 115°\)

AD = 4 cm

BC = 4 cm, CA = 4 cm

Name of the angles are as follows
(a) \(\angle 1=\angle CBD\)
(b) \(\angle 2 =\angle DBE\)
(c) \(\angle 3=\angle EBA\)
(d) \(\angle 1+\angle 2=\angle CBE\)
(e) \(\angle 2+\angle 3=\angle DBA\)
(e) \(\angle 1+\angle 2+\angle 3=\angle CBA\)
(g) Put the value of \(\angle CBA\)
\([\therefore \angle CBA =\angle 1+\angle 2+\angle 3]\)Now, \(\angle CBA-\angle 1=\angle 1+\angle 2+\angle 3-\angle 1\)
\(=\angle 2+\angle 3=\angle DBA\)
\([\because \angle DBA=\angle 2+\angle 3]\)

The 3-D shape formed is a square pyramid

(i) Vertices are A, B, C, D and E.
(ii) Edges are AB, AC, AD, AE, BC, CD, DE and EB.
(iii) Faces are: square BCDE,  ΔABC, ACD, ΔADE and ΔABE.

(a) Given, \(\angle A = \angle B = \angle C = 60°\)

Hence, it is an equilateral triangle.
(b) Given, \(\angle B =\angle C = 50°\)

hence, it is an isosceles triangle.
(c) Given, \(\angle A = 45°, \angle B = 45°, \angle C = 90°\)

Hence, it is an isosceles right angled triangle.
(d) Given, \(\angle A = 50°, \angle B = 60°, \angle C = 70°\)

Hence, it is a scalene triangle.

This polygon has four sides, so it is a quadrilateral. e.g.

This polygon has three sides, so it is a triangle. e.g.

This polygon has five sides, so it is a pentagon. e.g.

This polygon has eight sides, so it is an octagon. e.g.

(a) If we starts facing East and makes \(\frac { 1 }{ 2 } \) of a revolution clockwise i.e. two right angles, we will face West.

(b) If we starts facing East and makes 1\(\frac { 1 }{ 2 } \)of a revolution clockwise i.e. one complete revolution and two right angles, we will face West.

(c) If we starts facing West and makes \(\frac { 3 }{ 4 } \) of a revolution anti-clockwise i.e. three right angles, we will face North.

(d) We have,

If we starts facing South and makes one full revolution either clockwise or anti-clockwise, we will face South again.
There is no need to specify ciockwise or anti-clockwise because one full revolution will bring us back to the original position.

(i) Here, we draw a ΔABC.
In which, AB = 3 cm,
BC = 3.5 cm and AC = 3.9 cm
Now, by adding two sides of Δ ABC
i.e. AB + BC = 3 + 3.5
= 6.5 cm
Clearly, AB + BC > AC

(ii) Here, we draw a ΔABC in which AB =3 cm, BC = 4 cm and AC = 5 cm.
Now, by adding two sides of ΔABC
i.e. AB + BC = 3 + 4 = 7 cm
Clearly, AB + BC > AC.

(iii) Here, we draw a ΔABC in which AB = BC = AC = 2.5 cm
Now, by adding two sides of ΔABC
i.e. AB + BC = 2.5 + 2.5 = 5 cm
Clearly, AB + BC > AC

(iv) Here, we draw a ΔABC in which
AB = 3 cm, BC = 35 cm and AC = 4 cm
Now, by adding two sides of ΔABC.
i.e. AB + BC = 3 + 35 = 65 cm
Clearly, AB + BC > AC

(v) Here, we draw a ΔABC in which AB = 2 cm, BC = 4 cm and AC = 45 cm.
Now, by adding two sides of ΔABC
i.e. AB + BC = 2 + 4 = 6 cm
Clearly, AB + BC > AC.
From all these conclusions, we observe that in each case, the sum of the lengths of any two sides is greater than the third side. So, the sum of the lengths of any two sides can never be less than the third side.