**Quadrilaterals – Maharashtra Board Class 8 Solutions for Mathematics**

MathematicsGeneral ScienceMaharashtra Board Solutions

**Exercise – 14**

**Solution 1:**

**Exercise – 15**

**Solution 1:**

Length of the diagonal, i.e. l(PR) = 8 cm …..(Given)

Since diagonals of a square are congruent, they are of equal length.

∴ l(PR) = l(QS)

∴ l(QS) = 8 cm

Thus, the length of the diagonal QS of square PQRS is 8 cm.

**Solution 2:**

All the sides of a square are congruent.

Thus, in square ABCD, we have

l(AB) = l(BC) = l(CD) = l(DA)

Now, l(AB) = 4.5 cm ….(Given)

∴ l(BC) = l(CD) = l(DA) = 4.5 cm

Thus, the length of each side of a square is 4.5 cm.

**Solution 3:**

The diagonals of a square bisect each other.

∴ l(DM) = l(FM)

Now, l(DM) = 7 cm … (Given)

∴ l(FM) = 7 cm

l(DF) = l(DM) + l(MF) .…. (D – M – F)

= 7 + 7

= 14 cm

∴ l(DF) = 14 cm ….. (3)

The diagonals of a square are congruent.

∴ l(EG) = l(DF) = 14 cm

∴ l(EG) = 14 cm

**Solution 4:**

∠XMY is the angle at the point of intersection of the diagonals seg XZ and seg YW.

Since, the diagonals of a square are perpendicular bisectors of each other, m∠XMY = 90°.

**Solution 5:**

Seg HF and seg CD are the diagonals of square HDFC.

Now, the diagonals of a square are congruent.

∴ l(CD) = l(HF)

l(HF) = 5 cm …(Given)

∴ l(CD) = 5 cm

**Exercise – 16**

**Solution 1:**

The opposite sides of a rectangle are congruent.

∴ seg PS ≅ seg QR

∴ l(PS) = l(QR)

Now, l(PS) = 9 cm ….(Given)

∴ l(QR) = 9 cm

Similarly, seg PQ ≅ seg SR

∴ l(PQ) = l(SR)

l(PQ) = 7 cm … (Given)

∴ l(SR) = 7 cm

Thus, l(QR) = 9 cm and l(SR) = 7 cm.

**Solution 2:**

Each angle of a rectangle is a right angle.

∴ m∠DAB = 90°.

m∠DAB = m∠DAC + m∠CAB

∴ 90° = m∠DAC + 25° … (Given: m∠CAB = 25°)

∴ m∠DAC = 90 – 25 = 65°.

The opposite sides of a rectangle are parallel.

∴ seg AB || seg DC and AC is the transversal.

∴ ∠CAB ≅ ∠ACD (Alternate angles)

m∠CAB = 25° (Given)

∴ m∠ACD = 25°

Thus, m∠DAC = 65° and m∠ACD = 25°.

**Solution 3:**

The diagonals of a rectangle bisect each other.

Since point K is the point of intersection of diagonals AC and BD, we have

l(AK) = l(KC)

l(AK) = 3.5 cm … (Given)

∴ l(KC) = 3.5cm

l(AC) = l(AK) + l(KC) = 3.5 + 3.5 = 7 cm.

Thus, l(KC) = 3.5 cm and l(AC) = 7 cm.

**Solution 4:**

**Solution 5:**

Given: m∠L = m∠M = m∠N = 90°

The sum of the measures of the angles of a quadrilateral

is 360°.

∴ m∠L + m∠M + m∠N + m∠P = 360°

∴ 90 + 90 + 90 + m∠P = 360

∴ 270 + m∠P = 360

∴ m∠P = 360 – 270

∴ m∠P = 90°

Now, each angle of □LMNP is a right angle.

∴ □LMNP is a rectangle.

Thus, m∠P = 90° and □LMNP is a rectangle.

**Exercise – 17**

**Solution 1:**

Length of one side of a rhombus = 7.5 cm …..(Given)

All the sides of a rhombus are congruent.

Thus, the length of each of the remaining sides of the rhombus is 7.5 cm.

**Solution 2:**

Point P is the point of intersection of the diagonals XZ and YW.

∴ P is the midpoint of diagonal XZ.

Now, the diagonals of a rhombus bisect each other.

∴ l(XZ) = 2 × l(XP)

= 2 × 8 ……[l(XP) = 8 cm]

= 16 cm

∴ l(XZ) = 16 cm.

**Solution 3:**

The opposite angles of a rhombus are congruent.

∴ ∠QPS ≅ ∠QRS

∴ m∠QPS = m∠QRS

But, m∠QPS = 65° …(Given)

∴ m∠QRS = 65°

**Solution 4:**

The diagonals of a rhombus are perpendicular bisectors of each other.

Now, point O is the point of intersection of the diagonals AC and BD.

∴ m∠AOD = m∠BOC = 90°

Thus, m∠AOD = 90° and m∠BOC = 90°.

**Solution 5:**

The opposite angles of a rhombus are congruent.

∴∠N ≅ ∠K and ∠G ≅ ∠I

∴ m∠N = m∠K = 70° and m∠G = m∠I = 110°

Thus, m∠N = 70° and m∠G = 110°.

**Exercise – 18**

**Solution 1:**

**Solution 2:**

The opposite angles of a parallelogram are congruent.

∴ ∠ S ≅ ∠ Q

∴ m∠S = m∠Q

But, m∠Q = 130° …(Given)

∴ m∠S = 130°

Also, the opposite sides of a parallelogram are parallel to each other.

∴ Side PS || side QR and side PQ is the transversal.

∴ m∠P + m∠ Q = 180° ….(Pair of interior angles is supplementary)

∴ m∠P + 130° = 180°

∴ m∠P = 180° – 130°

∴ m∠P = 50°

Now, m∠R = m∠P ….(Opposite angles of a parallelogram)

∴ m∠R = 50° …[m∠P = 50°]

Thus, m∠P = 50°, m∠R = 50° and m∠S = 130°.

**Solution 3:**

Since the opposite angles of a parallelogram are congruent, they are of equal measures.

∴ (3x – 2)° = (50 – x)°

∴ 3x – 2 = 50 – x

∴ 3x + x = 50 + 2

∴ 4x = 52

∴ x = 13

Thus, 3x – 2 = 3 × 13 – 2 = 39 – 2 = 37

∴ (3x – 2)° = 37°

∴ (50 – x)° = 37°

Now, the adjacent angles of a parallelogram are supplementary angles.

Let the angle supplementary to the angle measuring 37° be y.

Then, we have

y + 37 = 180

∴ y = 180 – 37

∴ y = 143°

∴ The angle opposite to y will be also 143°.

Thus, the measures of the angles of the parallelogram are 37°, 143°, 37°, 143°.