Steps of Construction:

• Draw OR = 4.4 cm.
• Draw ∠ ROX = 105°.
• With O as the center and a radius of 5.8 cm, draw an arc-cutting ray OX at M.
• Draw ∠ ORE = 90°.
• Draw ∠ OMZ = 58°. Angle and extend the ray MZ. Let the rays MZ and RE intersect at point E.

Steps of Construction:

• Draw EF = 6.5 cm.
• With F as centre and radius 7.2 cm, draw an arc.
• With E as centre and radius 4.5 cm, draw an arc cutting the previous arc at D.
• Join ED.
• With D as centre and radius 5.5 cm, draw an arc.
• With E as centre and radius 7.8 cm, draw an arc cutting the previous arc at G.
• Join DG and GF.

Suppose ABCD is a rectangle.

Here, segment AC is a diagonal and segment AD is one side of the rectangle ABCD.

l(AC) = 26 cm and l(AD) = 24 cm.

In right Δ ACD,

l(AC)² = l(AD)² + l(CD)²                ... (Pythagoras theorem)

= l(CD)² = l(AC)² - l(AD)²

= l(CD)² = (26)² - (24)²

l(CD)² = 676 – 576 = 100

l(CD) = \(sqrt{100}\) = 10 cm

Thus, the other side of the rectangle is 10 cm.

Let ABCD be the required rhombus.

l(AC) = 16 cm and l(BD) = 12 cm

Diagonals of rhombus bisect each other

∴ l(AM) = \(\frac{1}{2}\) x l(AC) and l(BM) = \(\frac{1}{2}\) x l(BD)

∴ l(AM) = \(\frac{1}{2}\) x 16 and l(BM) = \(\frac{1}{2}\) x 12

∴ l(AM) = 8 cm and l(BM) = 6 cm

Diagonals of rhombus are perpendicular to each other

∴ m∠ AMB = 90°

In right angled Δ AMB,

by Pythagoras theorem,

l(AB)² = l(AM)² + l(MB)²

∴ l(AB)² = 8² + 6²

∴ l(AB)² = 64 + 36

∴ /(AB)² = 100

∴ l(AB) = \(\sqrt{100}\)

∴ l(AB) = 10 cm

Perimeter of rhombus ABCD = 4 x l(AB) = 4 × 10 = 40 cm

Answer is : Side of rhombus is 10 cm and perimeter of rhombus is 40 cm.

Let ABCD be the required square.

Here, segment AC is a diagonal of square ABCD.

In Δ ABC,

by Pythagoras theorem,

l(AC)² = l(AB)² + l(BC)²

∴ l(AC)² = (8)² + (8)²

∴ l(AC)² = 64 + 64 = 128

l(AC) = \(\sqrt{128}\) cm          ….[Taking square root of both sides]

∴ l(AC) = \(\sqrt{64×2}\) cm

∴ l(AC) = 8\(\sqrt{2}\) cm

Thus, the length of diagonal of square is 8\(\sqrt{2}\) cm.

Let ABCD be the required rhombus.

m∠ ABC = 50°

Opposite angles of rhombus are congruent.

∴ m∠ ADC = m∠ ABC = 50°

Side AD || side BC and line AB is the transversal,

m∠ BAD + m∠ ABC = 180°  ... (Interior angles)

∴ m∠ BAD + 50° = 180°

∴ m∠ BAD = 180 - 50°

∴ m∠ BAD = 130°

Opposite angles of rhombus are equal

∴ m∠ BCD = m∠ BAD = 130°

Answer is : The measures of remaining three angles of a rhombus are 50°, 130° and 130°.

Let □ ABCD be given parallelogram and m∠ A = (3x - 2)° and m∠ C = (50 - x)°.

Opposite angles of parallelogram are of equal measures.

∴ m∠ A = m∠ C

∴ (3x - 2)° = (50 - x)°

∴ 3x – 2 = 50 - x

∴ 3x + x = 50 + 2

∴ 4x = 52

∴ x = 52/4

∴ x = 13

∴ m∠ A = m∠ C = (3x - 2)° = (3 × 13 - 2)° = 37°

side AB || side DC and line AD is the transversal,

m∠ A + m∠ D = 180°                    ….. [Interior angles]

∴ 37° + m∠ D = 180°

∴ m∠ D = 180° - 37°

∴ m∠ D = 143°

m∠ B = m∠ D = 143° ... [Opposite angles of parallelogram are of equal measures]

Answer is : The measure of each angle of a parallelogram are 37°, 143°, 37° and 143°.

WXYZ is a parallelogram.

(1) l(XY) = l(WZ) = 4.5 cm ... (Opposite sides of a parallelogram are congruent)

(2) l(XW) = l(YZ) = 8.2 cm ... (Opposite sides of a parallelogram are congruent)

(3) l(OZ) = l(OX) = 2.5 cm ... (Diagonals of parallelogram bisect each other)

(4) l(WY) = 2 x l(WO) = 2 × 3.3 = 6.6 cm ... (Diagonals of parallelogram bisect each other)

(5) m∠ WXY = m∠ WZY = 120° ... (Opposite angles of a parallelogram are congruent)

Now,

m∠ WZY + m∠ XWZ = 180° ... (Adjacent angles of a parallelogram are  supplementary)

∴ 120° + m∠ XWZ = 180°

m∠ XWZ = 180° - 120° = 60°

l(BC) of 7 cm can be drawn.

On drawing, ∠ ABC = 40°, we can locate point A on it such that l(AB) = 3 cm

Opposite sides of parallelogram are congruent.

∴ l(CD) = l (AB) = 3 cm

l(AD) = l(BC) = 7 cm.

Thus D can be obtained at a distance of 7 cm from A and 3 cm from C.  □ ABCD can thus be constructed.

Let □ ABCD be the given quadrilateral.

The ratio of consecutive angles is 1 : 2 : 3 : 4

Let the common multiple be x

∴ m∠ A =x°, m∠ B = 2x°, m∠ C = 3x° and m∠ D = 4x°

Sum of the measures of all angles of a quadrilateral is 360°

∴ m∠ A + m∠ B + m∠ C + m∠ D = 360°

∴ x° + 2x° + 3x° + 4x° = 360°

∴ 10x = 360

∴ x = 360/10 = 36

m∠ A = x = 36°

m∠ B = 2x = 2 x 36° = 72°

m∠ C = 3x = 3 x 36° = 108°

m∠ D = 4x = 4 x 36° = 144°

m∠ B + m∠ C = 72° + 108° = 180°

∴ side CD || side AB     ... [. Interior angles are equal.]

m∠ A + m∠ B = 36 + 72 = 108

∴ m∠A + m∠B ≠ 180°

∴ Side BC is not parallel to side AD

∴ □ ABCD is a trapezium.

If □ PQRS is a parallelogram, then the information ∠ SRQ = 70° and SR = 3.5 cm or ∠ PQR = 110° and PQ =3.5 cm is unnecessary. …. [ ∵ It is possible to construct a parallelogram if its adjacent sides and the included angle are known.]