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Quadrilaterals – Maharashtra Board Class 8 Solutions for Mathematics

Quadrilaterals – Maharashtra Board Class 8 Solutions for Mathematics

MathematicsGeneral ScienceMaharashtra Board Solutions

Exercise – 14

Solution 1:

quadrilaterals-maharashtra-board-class-8-solutions-mathematics-1.1
quadrilaterals-maharashtra-board-class-8-solutions-mathematics-1.2
quadrilaterals-maharashtra-board-class-8-solutions-mathematics-1.3
quadrilaterals-maharashtra-board-class-8-solutions-mathematics1.4

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:

quadrilaterals-maharashtra-board-class-8-solutions-mathematics-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:

quadrilaterals-maharashtra-board-class-8-solutions-mathematics-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:

quadrilaterals-maharashtra-board-class-8-solutions-mathematics-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:

quadrilaterals-maharashtra-board-class-8-solutions-mathematics-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:

quadrilaterals-maharashtra-board-class-8-solutions-mathematics-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:

quadrilaterals-maharashtra-board-class-8-solutions-mathematics-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:

quadrilaterals-maharashtra-board-class-8-solutions-mathematics-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:

quadrilaterals-maharashtra-board-class-8-solutions-mathematics-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:

quadrilaterals-maharashtra-board-class-8-solutions-mathematics-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:

quadrilaterals-maharashtra-board-class-8-solutions-mathematics-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:

quadrilaterals-maharashtra-board-class-8-solutions-mathematics-11

Solution 2:

quadrilaterals-maharashtra-board-class-8-solutions-mathematics-12
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°.

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