We know that, in a triangle the sum of two sides are always greater than the third side. Consider
\(\triangle POS,OP+OS>PS\)..................1
\(\triangle POQ,OP+OQ>PQ\)..................2
\(\triangle QOR,OQ+OR>QR\).............3
\(\triangle ROS,OR+OS>SR\)................4
On adding Eqs. (i), (ii), (iii) and (iv), we get
2 (OP + OS + OQ + OR) > PS + PQ+ QR + SR
\(\Rightarrow\) 2(PR+ QS) > PQ+ QR+RS+SP

As per the given information in the question, height of a pole is 8 m. Distance between the bottom of the pole and a point on the ground is 6 m. On the basis of given information, we can draw the following figure:

Let the length of the rope be x,
AB = Height of the pole
BC = Distance between the bottom of the pole and a point on the ground where rope was tied.
To find the length of the rope,we will use Pythagoras theorem, such that,
(AC)² =(AB)² + (BC)²
\(\Rightarrow\) (AC)² =(8)² + (6)² \(\Rightarrow\) X² = 64 + 36
\(\Rightarrow\) X² =100
\(\Rightarrow x=\sqrt 100=10m\)

Let one of the angle of a triangle be x. If angles are arranged in descending order. So, angles will be x, (x - 20°) and (x - 40°).
We know that, the sum of all angles in a triangle is equal to 180°.
So, x+(x-20°)+(x-40°)=180°
\(\Rightarrow\) x + x + x-20° - 40° = 180°
\(\Rightarrow\) 3x =180° + 60°
\(\Rightarrow\) 3x = 240°
\(\Rightarrow x=\frac{240°}{3}=80°\)
\(\Rightarrow\) x=80°
So, angles will be 80°, (80° - 20° ) and (80° - 40°)
i.e. 80°, 60°, 40.

As per the given information in the question, one angle of a triangle is 60°. Let the other two angles be x and 3x.
We know that, the sum of all angles in a triangle is equal to 180°.
So, x+3x+60⁰=180°
\(\Rightarrow\) 4x + 60° =180°
\(\Rightarrow\) 4x =180° - 60°
\(\Rightarrow\) 4x =120°
\(\Rightarrow x=\frac{120^0}{4}=30^0\)
x=30⁰
So, angles will be x = 30° and 3x = 3 x 30° = 90°
Hence, the two angles are 90° and 30°.

Let ABCD be the rhombus and AC and BD are its diagonal.
Then, given AC = 30 cm and BD = 16 cm.
We know that, the diagonals of a rhombus bisect each other at right angle.

\(\therefore OA=\frac{1}{2}AC=\frac{1}{2}\times30cm=15cm\)
\(OB=\frac{1}{2}BD=\frac{1}{2}\times16cm=8cm\)
Also, \(\angle AOB\) is right angle.
In \(\triangle AOB\) by using Pythagoras property,
AB² = OA² + OB²
\(\Rightarrow\) AB² = (15)² + (8)²
\(\Rightarrow\) AB² = 225 + 64
\(\Rightarrow\) AB² = 289
\(\Rightarrow AB=\sqrt289=17\)
Perimeter of a rhombus = 4 X Side = 4 X AB
=4 X 17=68 cm
Hence, the perimeter of a rhombus is 68 cm.

\(\angle X=140^0\)

Each interior angles will be equal to 60⁰.

\(\angle A=65^0,yes \angle ACD>\angle A\)

The two interior opposite angles are 48⁰ and 64⁰.

 We know that, On the basis of sides, a triangle is called
(i) scalene triangle, if all three sides of triangle are unequal.
(ii) isosceles triangle, if any rwo sides are equal.
(iii) equilateral triangle, if all three sides are equal.
On the basis of angles, a triangle is called
(i) acute angled triangle, if each angle is less than 90°.
(ii) right angled angled triangle, if one angle is a right angle.
(iii) obtuse angled triangle, if one angle is greater than 90°. Now,
Now,
(i) (a) In \(\triangle ABC\),AC=BC=8 cm
i.e. rwo sides are equal
Therefore,\(\triangle ABC\) is an isosceles triangle.
(b) Also, all the angles of \(\triangle ABC\) are less than 90°.
Therefore,\(\triangle ABC\) is an acute angled triangle.
(ii) (a) In \(\triangle PQR\),\(PQ\neq QR\neq RP \) [given]
i. e. all three sides are unequal.
Therefore,\(\triangle PQR\) is a scalene triangle.
(b) Also,\(\angle R=90^0\)   [given]
Therefore, \(\triangle PQR\) is a right angled triangle.
(iii) (a) In \(\triangle LMN\),LN=MN=7cm
i.e. rwo sides are equal.
Therefore,\(\triangle LMN\) is an isosceles triangle.
(b) Also,\(\angle N>90^0\)     [given]
Therefore,\(\triangle LMN\) is an obtuse angled triangle.
(iv) (a) In \(\triangle RST\),RS = ST = TR = 5.2 cm [given]
i.e. all three sides are equal.
Therefore,\(\triangle RST\) is an equilateral triangle.
(b) Also, all the angles of \(\triangle RST\) are acute.
Therefore,\(\triangle RST\) is an acute angled triangle.
(v) (a) In \(\triangle ABC\),AB = BC = 3 cm
i.e. rwo sides are equal.
Therefore,\(\triangle ABC\) is an isosceles angled triangle.
(b) Also,\(\angle B>90^0\)
Therefore, \(\triangle ABC\) is an obtuse angled triangle.
(vi) (a) In \(\triangle PQR\),PQ = QR = 6 cm    [given]
i.e. rwo sides are equal.
Therefore, \(\triangle PQR\) is an isosceles triangle.
(b) Also,\(\angle Q=90^0\) [given]
Therefore \(\triangle PQR\) is a right angled angled triangle.