It is an open figure. So, it is not a polygon.

Two pairs

The right angles in the given figure are  \(\angle POQ, \angle QOR, \angle ROS, \angle SOP.\)

Given, AB = 8.7 cm, AC = 7 cm and BC = 6 cm
here, AB ≠ BC ≠ CA
i.e. all the sides are of different length.
So, it is a scalene triangle.

(i) Here, AC = AB + BC = 1 + 1 = 2 units [AB = BC = 1unit]
and FG = 1 unit
\(\therefore\) AC > FG is true.
(ii) Here, CD = 1 unit and GH = 1 unit
\(\therefore\) CD = GH is true.
(iii) Here, BC = 1 unit
and EH = EF + FG + GH = 1 + 1 + 1 = 3 units
\(\therefore\) BC < EH is true.

Observing the given angles, we find that \(\angle B > \angle A\) .
On measuring, we find that
Measure of \(\angle A = 40°\)
Measure of \(\angle B =65°\)
∴ \(\angle B > \angle A.\)

\(\angle C\) = Obtuse angle, \(\angle A\) and \(\angle B\) = Acute angles.

On checking the angle moved by an hour hand, while going from 4 to 8 by RA (right angle) tester, it is clear that it is more than 1 right angle.

Yes,it is clear from the figure that, hour hand makes one right angle from 12 to 3 and since 5 lies between 3 and 6. So, revolution of the hour hand is more than 1 right angle.

3 right angle

It is clear from the figure, that the hour hand starts from 10 and turns through 3 right angles. Hence, it will be reach at 7.

Here, the hour hand moves from 5 to 11. It starts from 5 and covers digits 6, 7, 8, 9, 10 and 11. So, hour hand turned by two right angles.

If the hand of a clock starts at 2 and makes \(\frac { 1 }{ 2 } \) of.a revolution, clockwise, i.e. two right angles, it stops at 8.

The situation from 4 to 7 on clock is given in the figure.
Hence, from 4 to 7, hour hand turns through \(\frac { 1 }{ 4 } \) of a revolution, clockwise.

Yes, sum of any two sides in ΔABC is greater than the third side.

\(\overline { AC } \)=2 cm

Line segments \(\overline { AB } ,\overline{ CD } ,\overline{ MN } \)and \(\overline { AB } =\overline { CD } ,\overline{ AB } >\overline { MN } \).

\(\overline{ XY } >\overline{ AB } \).

Given, B is the mid-point of \(\overline{ AC } \)
∴ AB = BC         .....(i)
and C is the mid-point of \(\overline{ BD } \).
∴ BC = CD        .......(ii)
i.e.,
Now, on adding Eqs. (i) and (ii), we get
AB + BC = BC + CD
∴ AB = CD

Let AB be the line segment of length 6 ern and point C lying between A and B.

On measuring the lengths ofline segments AB, AC and CB using ruler, we find that AB = 6 cm, AC = 4 cm and CB= 2 cm
Now, AC + CB= 4+ 2=6 cm = AB
i.e. AC + CB = AB
So, we can say that C lies between A and B.

(a) Since, AC 丄 BD, it means \(\angle E = 90°\)
∴ \(\angle AEB, \angle BEC, \angle CED\) and \(\angle AED\) are right angles.
(b) Since, AE 丄 CE, it means \(\angle E = 90°\)
∴ \(\angle AEC\) is a right angle.
(c) Since, AC 丄 CD, it means \(\angle C= 90°\)
∴ \(\angle ACD\) is a right angle.
(d) Since, OP 丄 AB, it means \(\angle K = 90°\)
\(\therefore\ \ \ \ \ \angle AKO, \angle OKB, \angle BKP\) and \(\angle AKP\) are right angles.

Given, four broom sticks of length 8 cm, 8 cm, 5 cm and 5 cm.
(a) Three types of quadrilaterals can be formed.
(b) Name of the quadrilaterals are rectangle, parallelogram and kite.
(c) The value is scientific, temper and curiosity.

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.

Now, first of all, we draw AB = 2.5 cm and CD = 3.2 cm.

\(\overline{ AB } +\overline { CD } =\overline { EF } \)
Now, we have to draw a line segment

∴ \(\overline { EF } =\overline{ AB } +\overline{ CD } \) = 2.5 +3.2 = 5.7 cm
Hence, new length is 5.7 cm.

(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.

\(\frac { 3 }{ 4 } \) revolution

PQ is not the perpendicular bisector of line segment AB because AO \(\neq\) BO.

\(\angle ACD = \angle ACB = 90°\)

In the figure, D is the mid-point of \(\overline{ BC } \) and \(\overline{ BD } \) = \(\overline{ CD } \).

Here, \(\overline { AB } +\overline { BC } =\overline { AC } \)
Hence, the length of line segment \(\overline{ AB } \) and \(\overline{ BC } \) make the length of line segment \(\overline{ AC } \).