Chapter 8: Quadrilaterals

(D) 120º

Explanation:

According to the question,

Three angles of quadrilateral are 75°, 90° and 75°

Consider the fourth angle to be x.

We know that,

Sum of all angles of a quadrilateral = 360°

⇒ 75° + 90° + 75° + x = 360°

⇒ 240° + x = 360°

⇒ x = 360° – 240°

⇒ x = 120°

Hence, the fourth angle is 120°.

Therefore, option (D) is the correct answer.

(B) 50º

Explanation:

According to the question,

A diagonal of a rectangle is inclined to one side of the rectangle at 25º

i.e., Angle between a side of rectangle and its diagonal = 25°

Consider the acute angle between diagonals to be = x

We know that diagonals of a rectangle are equal in length i.e.,

AC = BD

Dividing RHS and LHS by 2,

⇒ ½ AC = ½ BD

Since, O is mid-point of AC and BD

⇒ OD = OC

Since, angles opposite to equal sides are equal

⇒ ∠y = 25°

We also know that,

Exterior angle is equal to the sum of two opposite interior angles.

So, ∠BOC = ∠ODC + ∠OCD

⇒ ∠x = ∠y + 25°

⇒ ∠x = 25° + 25°

⇒ ∠x = 50°

Hence, the acute angle between diagonals is 50°.

Therefore, option (B) is the correct answer.

(C) 50º

Explanation:

According to the question,

ABCD is a rhombus

∠ACB = 40°

∵ ∠ACB = 40°

⇒ ∠OCB = 40°

∵ AD ∥ BC

⇒ ∠DAC = ∠BCA = 40° [Alternate interior angles]

⇒ ∠DAO = 40°

Since, diagonals of a rhombus are perpendicular to each other

We have,

∠AOD = 90°

We know that,

Sum of all angles of a triangle = 180°

⇒ ∠AOD + ∠ADO + ∠DAO = 180°

⇒ 90° + ∠ADO + 40° = 180°

⇒ 130° + ∠ADO = 180°

⇒ ∠ADO = 180° – 130°

⇒ ∠ADO = 50°

⇒ ∠ADB = 50°

Hence, ∠ADB = 50°

Therefore, option (C) is the correct answer.

(C) diagonals of PQRS are perpendicular

Explanation:

Let the rectangle be ABCD,

We know that,

Diagonals of rectangle are equal

∴ AC = BD

⇒ PQ = QR

∴ PQRS is a rhombus

Diagonals of a rhombus are perpendicular.

Hence, diagonals of PQRS are perpendicular

Therefore, option (C) is the correct answer.

(D) diagonals of PQRS are equal.

Explanation:

Since, ABCD is a rhombus

We have,

AB = BC = CD = DA

Now,

Since, D and C are midpoints of PQ and PS

By midpoint theorem,

We have,

DC = ½ QS

Also,

Since, B and C are midpoints of SR and PS

By midpoint theorem

We have,

BC = ½ PR

Now, again, ABCD is a rhombus

∴ BC = CD

⇒ ½ QS = ½ PR

⇒ QS = PR

Hence, diagonals of PQRS are equal

Therefore, option (D) is the correct answer.

(C) trapezium

Explanation:

As angles A, B, C and D of the quadrilateral ABCD, taken in order, are in the ratio 3: 7: 6: 4,

We have the angles A, B, C and D = 3x, 7x, 6x and 4x.

Now, sum of the angle of a quadrilateral = 360^o.

3x + 7x + 6x + 4x = 360^o

⇒20x = 360^o

⇒ x = 360 ÷ 20 =18^o

So, the angles A, B, C and D of quadrilateral ABCD are,

∠A = 3×18^o = 54^o,

∠B = 7×18^o = 126^o

∠C = 6×18^o = 108^o

∠D = 4×18^o = 72^o

AD and BC are two lines cut by a transversal CD

Now, sum of angles ∠C and ∠D on the same side of transversal,

∠C +∠D =108^o + 72^o =180

Hence, AD|| BC

So, ABCD is a quadrilateral in which one pair of opposite sides are parallel.

Hence, ABCD is a trapezium.

Therefore, option (C) is the correct answer.

(D) quadrilateral whose opposite angles are supplementary

Explanation:

We know that,

Sum of all angles of a quadrilateral = 360°

⇒ ∠A + ∠B + ∠C + ∠D = 360°

Dividing LHS and RHS by 2,

⇒ ½ (∠A + ∠B + ∠C + ∠D) = ½ × 360° = 180°

Since, AP, PB, RC and RD are bisectors of ∠A, ∠B, ∠C and ∠D

⇒ ∠PAB + ∠ABP + ∠RCD + ∠RDC = 180° … (1)

We also know that,

Sum of all angles of a triangle = 180°

∠PAB + ∠APB + ∠ABP = 180°

⇒ ∠PAB + ∠ABP = 180° – ∠APB …(2)

Similarly,

∴ ∠RDC + ∠RCD + ∠CRD = 180°

⇒ ∠RDC + ∠RCD = 180° – ∠CRD …(3)

Substituting the value of equations (2) and (3) in equation (1),

180° – ∠APB + 180° – ∠CRD = 180°

⇒ 360° – ∠APB – ∠CRD = 180°

⇒ ∠APB + ∠CRD = 360° – 180°

⇒ ∠APB + ∠CRD = 180° …(4)

Now,

∠SPQ = ∠APB [vertically opposite angles]

∠SRQ = ∠DRC [vertically opposite angles]

Substituting in equation (4),

⇒ ∠SPQ + ∠SRQ = 180°

Hence, PQRS is a quadrilateral whose opposite angles are supplementary.

Therefore, option (D) is the correct answer.