Solution:

 In right \(\triangle\)ABC,
 AC² = AB² + BC² ... (Pythagoras’ Theorem)
∴ AC² = 24² + 7²
∴ AC² = 576 + 49
∴ AC² = 625
∴ AC = \(\sqrt{625}\)
∴ AC = 25 cm
∴ The length of the diagonal is 25 cm.

Solution:

In right \(\triangle\)ABC,

 AC² = AB² + BC² ... (Pythgoras’ Theorem)

∴ 13² = AB² + BC²

∴ 169 = AB² + AB²

∴ 169 = 2AB²

∴ \(\displaystyle \frac{169}{2}\) = AB²

i.e. AB² = \(\displaystyle \frac{169}{2}\)

∴ AB = \(\displaystyle \sqrt{\frac{169}{2}}\)

∴ AB = \(\displaystyle \frac{\sqrt{169}}{\sqrt{2}}\)

∴ AB = \(\displaystyle \frac{13}{\sqrt{2}}\)

Rationalizing the denominator:

∴ AB = \(\displaystyle \frac{13}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}\)

∴ AB = \(\displaystyle \frac{13\sqrt{2}}{2}\)

∴ AB = \(\displaystyle 6.5\sqrt{2}\) cm ... (i)

∴ The side of the square is \(\displaystyle 6.5\sqrt{2}\) cm

Solution:

Let the two adjacent sides of the parallelogram be 3x cm and 4x cm respectively.

From the given information,

∴ 3x + 4x + 3x + 4x = 112

∴ 14x = 112

∴ \(\displaystyle x = \frac{112}{14}\)

∴ \(\displaystyle x = 8\) cm ... (i)

∴ 3x = 3 × 8 = 24 cm

and 4x = 4 × 8 = 32 cm

∴ The lengths of the sides of that parallelogram are 24 cm, 32 cm, 24 cm and 32 cm respectively.

Solution:

Let \(\Box\)PQRS be the given rhombus.
Let PR = 20 cm and QS = 48 cm.

The diagonals of a rhombus are perpendicular bisectors of each other. ... (Theorem)
∴ \(\angle\) POQ = 90\(^\circ\)

Also, PO = \(\displaystyle \frac{1}{2}\) PR

∴ PO = \(\displaystyle \frac{1}{2} \times 20\)

∴ PO = 10 cm ... (i)

And, QO = \(\displaystyle \frac{1}{2}\) QS

∴ QO = \(\displaystyle \frac{1}{2} \times 48\)

∴ QO = 24 cm ... (ii)

In right \(\triangle\)POQ,

PQ² = PO² + QO² ... (Pythagoras’ Theorem)

∴ PQ² = 10² + 24²

∴ PQ² = 100 + 576

∴ PQ² = 676

∴ PQ = \(\sqrt{676}\)

∴ PQ = 26 cm

∴ The length of side PQ is 26 cm.

Solution:

Let \(\Box\)PQRS be the given rectangle.
Let the diagonals PR and QS intersect at M.

Now, \(\angle\)SMP \(\cong\) \(\angle\)QMR ... (Vertically opposite angles)
But, \(\angle\)QMR = 50\(^{\circ}\) ... (Given)
∴ \(\angle\)SMP = 50\(^{\circ}\) ... (i)

Also, the diagonals of a rectangle are congruent and bisectors of each other. ... (Theorem)
∴ SQ = RP
∴ \(\frac {1}{2}\)SQ = \(\frac {1}{2}\)RP
∴ MS = MP
∴ In \(\triangle\)SMP,
\(\angle\)MSP = \(\angle\)MPS ... (Theorem of isosceles triangle) ... (ii)

And, in \(\triangle\)SMP,
\(\angle\)MPS + \(\angle\)MSP + \(\angle\)SMP = 180\(^{\circ}\) ... (Theorem)
∴ \(\angle\)MPS + \(\angle\)MPS + 50\(^{\circ}\) = 180\(^{\circ}\) ... [From (ii)]
∴ 2\(\angle\)MPS + 50\(^{\circ}\) = 180\(^{\circ}\) − 50\(^{\circ}\)
∴ 2\(\angle\)MPS = 130\(^{\circ}\)

∴ \(\angle\)MPS = \(\displaystyle \frac{130^{\circ}}{2}\)

∴ \(\angle\)MPS = 65\(^{\circ}\)

Proof:

In \(\Box\)ABQP,
 seg AB | | seg PQ ... (Given) and
 seg AB \(\cong\) seg PQ ... (Given)
∴, \(\Box\)ABQP is a parallelogram. ... (Theorem)
∴ seg AP | | seg BQ ... (i)
  ... (Opposite sides of a | |^m).
and seg AP \(\cong\) seg BQ ... (ii)
  ... (Opposite sides of a | |^m)

Also, in \(\Box\)ACRP,
 seg AC | | seg PR ... (Given) and
 seg AC \(\cong\) seg PR ... (Given)
∴, \(\Box\)ACRP is a parallelogram. ... (Theorem)
∴ seg AP | | seg CR ... (iii)
  ... (Opposite sides of a | |^m).
and seg AP \(\cong\) seg CR ... (iv)
  ... (Opposite sides of a | |^m)

And, in \(\Box\)BQRC,
 seg BQ | | seg CR ... [From (i) and (iii)] and
 seg BQ \(\cong\) seg CR ... [From (ii) and (iv)]
∴, \(\Box\)BQRC is a parallelogram. ... (Theorem)
∴ seg BC | | seg QR ... (v)
  ... (Opposite sides of a | |^m).
and seg BC \(\cong\) seg QR ... (vi)
  ... (Opposite sides of a | |^m)

Construction:

Draw seg AQ. Extend segments DC and AQ such that they meet in point E.

Proof:

In \(\triangle\)ABQ and \(\triangle\)ECQ,
 \(\angle\)AQB \(\cong\) \(\angle\)EQC ... (Vertically opposite angles)
 seg BQ \(\cong\) seg CQ ... (Given)
 \(\angle\)ABQ \(\cong\) \(\angle\)ECQ ... (Alternate angles)
∴ \(\triangle\)ABQ \(\cong\) \(\triangle\)ECQ ... (A S A test)
∴ seg AB \(\cong\) seg EC ... (c s c t) ... (i)
and seg AQ \(\cong\) seg EQ ... (c s c t)
∴ Q is the midpoint of seg AE.
But, P is the midpoint of seg AD. ... (Given)
∴ In \(\triangle\)ADE,
PQ | | DE ... (Midpoint theorem)
i.e. PQ | | DC ... (∵ D – C – E)
But, DC | | AB ... (Given)
∴ PQ | | AB ... (ii)

Also, in \(\triangle\)ADE, P is the midpoint of seg AD and Q is the midpoint of seg AE.

∴ \(\displaystyle \text{PQ} = \frac{1}{2} \text{DE}\) ... (Midpoint theorem)

∴ \(\displaystyle \text{PQ} = \frac{1}{2} \text{(DC + CE)}\) ... (∵ D – C – E)

But, CE = AB ... [From (i)]

∴ \(\displaystyle \text{PQ} = \frac{1}{2} \text{(DC + AB)}\)

i.e. \(\displaystyle \text{PQ} = \frac{1}{2} \text{(AB + DC)}\)

Construction:

Draw seg MN. Draw seg DM and extend it to meet side AB in point E.

Proof:

In \(\triangle\)DMC and \(\triangle\)EMA,
 seg CM \(\cong\) seg AM ... (Given)
 \(\angle\)CDM \(\cong\) \(\angle\)AEM ... (Alternate angles)
 \(\angle\)CMD \(\cong\) \(\angle\)AME ... (Vertically opposite angles)
∴ \(\triangle\)DMC \(\cong\) \(\triangle\)EMA ... (S A A test)
∴ seg DM \(\cong\) seg EM ... (c s c t)
∴ M is the mid-point of seg DE. ... (i)

Now, in \(\triangle\)DEB,
M and N are the midpoints of seg DE and seg DB respectively.
 ... [From (i) and Given]
∴ MN | | EB ... (Midpoint theorem)
i.e. MN | | AB ... (∵ A – E – B)