Quadrilaterals

Given,

∠A = 70°

ABCD is a parallelogram.

⇒ ∠C = ∠A = 70° [∵ opposite angles of a parallelogram are equal]

⇒ ∠A + ∠B = 180° [∵ AD ∥ BC and sum of Co-interior angles is 180°]

⇒ ∠B = 180° - ∠A = 180° - 70°

⇒ ∠B = 110°

⇒ ∠D = ∠B = 110° [∵ opposite angles of a parallelogram are equal]

Hence, ∠C = 70°, ∠B = 110° and ∠D = 110°.

Given,

∠BCE = 105°

ABCD is a parallelogram.

⇒ ∠BCE + ∠BCD = 180° [Linear pairs]

⇒ ∠BCD = 180° - ∠BCE

⇒ ∠BCD = 180° - 105°

⇒ ∠BCD = 75°.

⇒ ∠A = ∠BCD = 75° [∵ opposite angles of a parallelogram are equal]

⇒ ∠A + ∠B = 180° [∵ AD ∥ BC and sum of Co-Interior angles is 180°]

⇒ ∠B = 180° - ∠A

⇒ ∠B = 180° - 75°

⇒ ∠B = 105°

⇒ ∠D = ∠B = 105° [∵ opposite angles of a parallelogram are equal]

Hence, ∠A = 75°, ∠C = 75°, ∠B = 105° and ∠D = 105°.

Let the measure of the adjacent angle be x. Then, the other angle will be $\frac{2x}{3}$.

We know that,

Sum of its adjacent angles of a //gm = 180°.

$$\begin{aligned}\Rightarrow x + \frac{2x}{3} = 180^{\circ} \\[1em] \Rightarrow \frac{3x + 2x}{3} = 180^{\circ} \\[1em] \Rightarrow \frac{5x}{3} = 180^{\circ} \\[1em] \Rightarrow 5x = 540^{\circ} \\[1em] \Rightarrow x = \frac{540^{\circ}}{5} \\[1em] \Rightarrow x = 108^{\circ}\end{aligned}$$

∴ x = 108°.

⇒ $\frac{2x}{3} = \frac{2}{3} \times 108°$ = 72°

Since opposite angles in a parallelogram are equal, the four angles are : 72°, 108°, 72°, and 108°.

Hence, angles of the parallelogram are 72°, 108°, 72°, and 108°.

(i) Given,

∠BAD = 70°

∠CBD = 50°

ABCD is a parallelogram.

⇒ ∠ADB = ∠DBC = 50° [Alternate angles are equal,as AD ∥ BC and DB is transversal]

Hence, ∠ADB = 50°.

(ii) ∠BAD + ∠ABC = 180° [∵ AD ∥ BC and sum of Co-interior angles is 180°]

⇒ ∠ABC = 180° - ∠BAD

⇒ ∠ABC = 180° - 70°

⇒ ∠ABC = 110°.

From figure,

⇒ ∠ABC = ∠DBA + ∠CBD

⇒ 110° = ∠DBA + 50°

⇒ ∠DBA = 110° - 50°

⇒ ∠DBA = 60°

⇒ ∠CDB = ∠DBA = 60°.

⇒ ∠CDB = ∠DBA = 60° [Alternate angles are equal, as DC ∥ AB and DB is transversal]

Hence, ∠CDB = 60°.

Given,

∠A = 72°

ABCD is a rhombus. Thus, opposite sides are parallel.

⇒ ∠A + ∠B = 180° [∵ AD ∥ BC and sum of Co-interior angles is 180°]

⇒ 72° + ∠B = 180°

⇒ ∠B = 180° - 72°

⇒ ∠B = 108°

We know that,

In a rhombus diagonals bisects the vertex angle.

⇒ ∠CBD = $\frac{1}{2}$ ∠B [∵ BD bisects ∠B]

⇒ ∠CBD = $\frac{108°}{2}$

⇒ ∠CBD = x° = 54°

⇒ x = 54.

Hence, x = 54.

Given,

△EDC is an equilateral triangle, so all sides are equal.

ED = DC = EC ......(1)

In square all sides are equal.

AB = CB = DC = AD .....(2)

From (1) and (2) we get,

DC = EC = CB

⇒ EC = CB.

In △ECB,

EC = CB

⇒ ∠BEC = ∠CBE = a (let) (Angles opposite to equal sides are equal in isosceles triangle)

From figure,

⇒ ∠C = ∠ECD + ∠DCB

⇒ ∠ECD = 60° (As each angle of a equilateral triangle = 60°)

⇒ ∠DCB = 90° (As each angle of a square = 90°)

⇒ ∠ECB = ∠ECD + ∠DCB = 60° + 90° = 150°.

In triangle BEC,

⇒ ∠BEC + ∠CBE + ∠ECB = 180°

⇒ a + a + 150° = 180°

⇒ 2a = 180° - 150°

⇒ 2a = 30°

⇒ a = 15°.

From figure,

x° = ∠DEC - ∠BEC = 60° - 15° = 45°.

Hence, x = 45.

Let ∠OAB = 2x and ∠OBA = 3x.

The diagonals of rhombus are perpendicular to each other.

∴ ∠AOB = 90°

In △AOB,

⇒ ∠AOB + ∠OAB + ∠OBA = 180°

⇒ 90° + 2x + 3x = 180°

⇒ 2x + 3x = 180° - 90°

⇒ 5x = 90°

⇒ x = $\frac{90°}{5}$

⇒ x = 18°.

∠OAB = 2x = 2(18°) = 36°.

∠OBA = 3x = 3(18°) = 54°.

Hence, ∠OAB = 36°, ∠OBA = 54°, ∠AOB = 90°.

Given,

∠ECD = 140°.

ABCD is a rectangle.

⇒ ∠DCO + ∠DCE = 180°

⇒ ∠DCO = 180° - 140°

⇒ ∠DCO = 40°.

∠CAB = ∠DCA = 40° [Alternate angles are equal, as CD ∥ AB and AC is transversal]

From figure,

∠OAB = ∠CAB = 40°

OB = OA [∵ diagonals of a rectangle are equal and bisect each other]

∠OAB = ∠OBA = 40° [Angles opposite to equal sides in a triangle are equal.]

In △AOB,

⇒ ∠AOB + ∠OAB + ∠OBA = 180°

⇒ ∠AOB + 40° + 40° = 180°

⇒ ∠AOB = 180° - 80°

⇒ ∠AOB = 100°.

Hence, ∠OAB = 40°, ∠ABO = 40°, ∠AOB = 100°.

(i) In kite ABCD,

AB = AD [Adjacent sides of kite are equal]

In triangle ABD,

⇒ ∠BDA = ∠ABD [Angles opposite to equal sides in a triangle]

In △ADB,

⇒ ∠BDA + ∠ABD + ∠DAB = 180° [∵ Angle sum property]

⇒ 2∠BDA + 54° = 180° [∵ ∠ODA = ∠OBA]

⇒ 2∠ODA = 180° - 54°

⇒ 2∠ODA = 126°

⇒ ∠ODA = 63°.

Hence, ∠ODA = 63°.

(ii) DC = CB [Adjacent sides of kite are equal]

∠BDC = ∠CBD [Angles opposite to equal sides in a triangle are equal]

In △CDB,

⇒ ∠BDC + ∠DCB + ∠CBD = 180°

⇒ 2∠CBD + 76° = 180°

⇒ 2∠OBC = 180° - 76°

⇒ 2∠OBC = 104°

⇒ ∠OBC = 52°.

Hence, ∠OBC = 52°.

We know that,

Sum of adjacent co-interior angles of a trapezium = 180°

∴ ∠A + ∠D = 180°

⇒ 3x + 2x = 180°

⇒ 5x = 180°

⇒ x = $\frac{180°}{5}$

⇒ x = 36°.

∠A = 3x = 3(36°) = 108°

∠D = 2x = 2(36°) = 72°

∠B = ∠A = 108° [∵ base angles of an isosceles trapezium are equal]

∠C = ∠D = 72° [∵ base angles of an isosceles trapezium are equal]

Hence, ∠A = 108°, ∠C = 72°, ∠B = 108° and ∠D = 72°.

We know that,

Sum of adjacent co-interior angles of a trapezium = 180° (As AB || DC)

∴ ∠A + ∠D = 180°

⇒ (x + 25)° + (2x + 5)° = 180°

⇒ 3x° + 30° = 180°

⇒ 3x° = 180° - 30°

⇒ 3x° = 150°

⇒ x° = 50°

⇒ x = 50.

Sum of adjacent co-interior angles of a trapezium = 180° (As AB || DC)

∴ ∠C + ∠B = 180°

⇒ 95° + y° = 180°

⇒ y° = 180° - 95°

⇒ y° = 85°

⇒ y = 85.

Hence, x = 50, y = 85.

(i) In rhombus ABCD,

⇒ ∠C = ∠A = 78° (Opposite angles of rhombus are equal)

The sum of adjacent angles of a rhombus is always 180°.

⇒ ∠A + ∠B = 180°

⇒ 78° + ∠B = 180°

⇒ ∠B = 180° - 78° = 102°.

In △ DBC,

⇒ DC = CB (Sides of rhombus are equal in length) .....(1)

⇒ ∠CDB = ∠CBD = x (let) (In a triangle angles opposite to equal sides are equal.)

By angle sum property of triangle,

⇒ ∠CDB + ∠CBD + ∠BCD = 180°

⇒ x + x + ∠BCD = 180°

⇒ 2x + 78° = 180°

⇒ 2x = 180° - 78°

⇒ 2x = 102°

⇒ x = $\frac{102°}{2}$

⇒ ∠CDB = ∠CBD = 51°.

Given,

DEC is an equilateral triangle, so all the sides of triangle are equal.

∴ DC = EC .......(2)

From equations (1) and (2), we get :

⇒ CB = EC

In triangle CEB,

⇒ ∠CEB = ∠CBE = y (let) [Angles opposite to equal sides in a triangle are equal]

By angle sum property of triangle,

⇒ ∠CEB + ∠CBE + ∠ECB = 180°

⇒ y + y + (∠ECD + ∠BCD) = 180°

⇒ 2y + (60° + 78°) = 180°

⇒ 2y + 138° = 180°

⇒ 2y = 180° - 138°

⇒ 2y = 42°

⇒ y = $\frac{42°}{2}$

⇒ ∠CBE = 21°.

Hence, ∠CBE = 21°.

(ii) From figure,

⇒ ∠DBE = ∠CBD - ∠CBE

⇒ ∠DBE = 51° - 21° = 30°.

Hence, ∠DBE = 30°.

Given,

ABCD is a square,

∠ADC = 90°

The diagonal BD bisects the ∠ADC at the vertex.

∠BDC = $\frac{∠ADC}{2} = \frac{90°}{2}$ = 45°

DEC is an equilateral triangle.

∠DCE = ∠CDE = 60°

From figure,

∠OCD = ∠DCE = 60°

∠ODC = ∠BDC = 45°

In △COD,

⇒ ∠COD + ∠ODC + ∠OCD = 180°

⇒ x° + 45° + 60° = 180°

⇒ x° + 105° = 180°

⇒ x° = 180° - 105°

⇒ x° = 75°

Hence, x = 75°.

Let ABCD be a parallelogram where ∠A = 90°.

Sum of adjacent angles of a parallelogram = 180°.

∠A + ∠B = 180°

90° + ∠B = 180°

∠B = 180° - 90°

∠B = 90°.

In a parallelogram, opposite angles are equal.

∠C = ∠A = 90°

∠D = ∠B = 90°.

∴ ∠A = ∠B = ∠C = ∠D = 90°

Hence, proved that all angles of parallelogram measures 90°.

(i) Given,

ABCD and PQBA are two parallelograms.

DC ∥ AB ....(1)

AB ∥ PQ .....(2)

From (1) and (2) we have,

∴ DC ∥ PQ

Opposite sides of a parallelogram are equal.

Thus, in //gm ABCD

DC = AB .....(3)

Thus, in //gm PQBA

AB = PQ ......(4)

From (3) and (4) we have,

∴ DC = PQ

Since, the pair of opposite sides DC and PQ are equal and parallel. Therefore, DPQC is a parallelogram.

Hence, proved that DPQC is a parallelogram.

(ii) We know that,

DPCQ is a parallelogram.

Opposite sides of a parallelogram are equal.

∴ DP = CQ

Hence, proved that DP = CQ.

(iii) In triangle DAP and CBQ,

DA = CB [opposite sides of parallelogram ABCD]

AP = BQ [opposite sides of parallelogram PQBA]

DP = CQ [opposite sides of parallelogram DPQC]

∴ ΔDAP ≅ ΔCBQ [By SSS rule]

Hence, proved that DP = CQ.

(i) In triangle DNA and BMC,

∠DNA = ∠BMC = 90° (Given)

∠DAN = ∠MCB [Alternate interior angles BC and AD are parallel, AC acts as a transversal]

BC = AD [opposite sides of parallelogram]

∴ ΔBMC ≅ ΔDNA.[By A.A.S. rule]

Hence, proved that ΔBMC ≅ ΔDNA.

(ii) We know that,

ΔBMC ≅ ΔDNA

∴ BM = DN [Corresponding sides of Congruent Triangles]

Hence, proved that BM = DN.

(i) Considering △ABX and △QCX we have,

⇒ ∠XBA = ∠XCQ (Alternate angles are equal)

⇒ XB = XC (As X is mid-point of BC)

⇒ ∠AXB = ∠CXQ (Vertically opposite angles are equal)

Hence, △ABX ≅ △QCX by ASA axiom.

(ii) Since, △ABX ≅ △QCX

∴ AB = CQ (By C.P.C.T.C.) .......(1)

AB = CD and AB = QP (Opposite sides of parallelogram are equal) ........(2)

From (i) and (ii) we get,

⇒ AB = DC = CQ = QP

⇒ DC = CQ = QP

Hence, proved that DC = CQ = QP.

Given,

ABCD is a parallelogram.

AX bisects ∠A

CY bisects ∠C

(i) In a parallelogram opposite angles are equal, thus ∠A = ∠C.

In ΔADX and ΔCBY,

∠D = ∠B [Opposite angles of a parallelogram are equal]

∠DAX = ∠BCY $\Big(\frac{1}{2}∠A = \frac{1}{2}∠C \Big)$

AD = BC [Opposite sides of a parallelogram are equal]

∴ ΔADX ≅ ΔCBY [By A.S.A. rule]

Hence, proved that ΔADX ≅ ΔCBY.

(ii) We know that,

ΔADX ≅ ΔCBY

AX = CY [By C.P.C.T.C.]

Hence, proved that AX = CY.

(iii) ∠DCY = ∠CYB [Alternate interior angles, AB ∥ DC, CY is transversal]

∠BAX = ∠DCY ($\frac{1}{2}∠A = \frac{1}{2}∠C$)

∴ ∠BAX = ∠CYB

These are corresponding angles formed by lines AX and CY with the transversal AB.

Since corresponding angles are equal, the lines AX and CY are parallel.

Hence, proved that AX ∥ CY.

(iv) We know that,

AX ∥ CY and AX = CY.

If one pair of opposite sides of a quadrilateral are equal and parallel, then the quadrilateral is a parallelogram.

Thus, AYCX is a parallelogram.

Hence, proved that AYCX is a parallelogram.

Given,

ABCD is a parallelogram

DX = BY

Diagonals of ∥ gm bisect each other.So,

OA = OC and OD = OB

Also,

OD - DX = OB - BY [DX = BY]

OX = OY

∴ Diagonals of quadrilateral CXAY bisect each other, CXAY is parallelogram

Hence, proved that CXAY is a parallelogram.

ABCD is a //gm.

From figure,

∠A + ∠D = 180° [Sum of Co-Interior angles in a // gm is 180°]

⇒ $\frac{1}{2}∠A + \frac{1}{2}∠D$ = 90°

In triangle APD,

∠DAP + ∠PDA + ∠APD = 180°

$\frac{1}{2}∠A + \frac{1}{2}∠B$ + ∠APD = 180°

90° + ∠APD = 180°

∠APD = 90°

∠SPQ = ∠APD = 90° [Vertically opposite angles are equal]

∠P = 90°

From figure,

∠B + ∠C = 180° [Sum of Co-Interior angles in a // gm is 180°]

⇒ $\frac{1}{2}∠B + \frac{1}{2}∠C$ = 90°

In triangle BRC,

∠CBR + ∠BCR + ∠CRB = 180°

$\frac{1}{2}∠B + \frac{1}{2}∠C$ + ∠CRB = 180°

90° + ∠CRB = 180°

∠CRB = 90°

∠SRQ = ∠CRB = 90° [Vertically opposite angles are equal]

∠R = 90°

From figure,

∠A + ∠B = 180° [Sum of Co-Interior angles in a // gm is 180°]

⇒ $\frac{1}{2}∠A + \frac{1}{2}∠B$ = 90°

In triangle ASB,

∠SAB + ∠SBA + ∠ASB = 180°

$\frac{1}{2}∠B + \frac{1}{2}∠B$ + ∠ASB = 180°

90° + ∠ASB = 180°

∠ASB = 90°

∠S = 90°

In quadrilateral PQRS,

By angle sum property of quadrilateral,

∠P + ∠Q + ∠R + ∠S = 360°

90° + ∠Q + 90° + 90° = 360°

∠Q + 270° = 360°

∠Q = 360° - 270° = 90°.

Since, all the angles of PQRS = 90°

∴ PQRS is a rectangle.

Hence, proved that PQRS is a rectangle.

Let ABCD be a parallelogram, and let the diagonal AC bisect ∠BAD.

Since ABCD is a parallelogram:

AB ∥ DC

AD ∥ BC

∠DAC = ∠CAB [AC bisects ∠DAB]

In △ABC and △ADC:

AC = AC (common side)

∠BAC = ∠DCA (Alternate interior angles are equal)

∠DAC = ∠BCA (Alternate interior angles are equal)

△ABC ≅ △ADC (A.S.A. congruence)

So, ∠DCA = ∠BCA (By C.P.C.T.C.)

That means AC also bisects the opposite angle ∠DCB.

Now,

AD = AB (From congruence)

AB = CD and AD = BC (Opposite sides of parallelogram)

∴ AB = BC = CD = DA

Thus, ABCD is a Rhombus.

We know that,

Diagonals of a rhombus are perpendicular to each other.

Thus, AC ⊥ BD.

Hence, proved that the diagonal bisects the second angle and AC ⊥ BD.

Given,

ABCD is a parallelogram.

E is the midpoint of BC.

DE meets AB produced at F.

AB ∥ DC

AB = DC (opposite sides of a parallelogram are equal)

Since E is the midpoint of BC :

BE = EC

In △DEC and △FEB:

EC = EB (E is mid-point of BC)

∠DEC = ∠FEB (Vertically opposite angles are equal)

∠DCE = ∠FBE (Alternate interior angles are equal)

△DEC ≅ △FEB (By A.S.A. axiom)

DC = BF [By C.P.C.T.C.]

DC = AB

∴ BF = AB

From figure,

AF = AB + BF

AF = AB + AB

AF = 2AB.

Hence, proved that AF = 2AB.

Steps of constructions :

1. Draw PQ = 4 cm.

2. Draw ∠QPX = 90°.

3. With P as centre and radius = 3.2 cm, draw an arc, cutting PX at S.

4. With Q as centre and radius = 4.3 cm, draw an arc.

5. With S as centre and radius = 3.6 cm, draw another arc, cutting the previous arc at R.

6. Join QR and RS.

Hence, PQRS is the required quadrilateral.

Steps of constructions:

1. Draw AB = 4.5 cm.

2. Draw ∠ABX = 75°.

3. With B as centre and radius = 5.2 cm, draw an arc, cutting BX at C.

4. With A as centre and radius = 4.7 cm, draw an arc.

5. With C as centre and radius = 5 cm, draw another arc, cutting the previous arc at D.

6. Join DC and DA.

Hence, ABCD is the required quadrilateral.

Steps of constructions :

1. Draw CD = 5.1 cm.

2. Draw ∠DCX = 60°.

3. With C as centre and radius = 4.7 cm, draw an arc, cutting CX at B.

4. With D as centre and radius = 4.2 cm, draw an arc.

5. With B as centre and radius = 5.1 cm, draw another arc, cutting the previous arc at A.

6. Join DA and BA.

Hence, ABCD is the required quadrilateral.

Steps of construction:

1. Draw UV = 6 cm.

2. Construct ∠VUX = 60°.

3. With U as centre and radius = 5 cm, draw an arc, cutting UX at Z.

4. Construct ∠UVY = 75°.

5. With V as centre and radius = 5.3 cm,draw an arc, cutting arc VY at W.

6. Join VW and ZW.

Hence, UVWZ is required quadrilateral.

Steps of construction:

1. Draw QR = 5.6 cm.

2. Construct ∠RQX = 60°.

3. With Q as centre and radius = 4.5 cm, draw an arc, cutting QX at P.

4. Construct ∠QRY = 120°.

5. With R as centre and radius = 4.1 cm, draw an arc, cutting arc RY at S.

6. Join SP.

Hence, PQRS is required quadrilateral.

Steps of construction :

1. Draw AB = 5.3 cm.

2. Taking A as center, draw an arc of radius 5.3 cm (= AD) and taking B as center, draw one more arc of radius 6.4 cm (= diagonal BD). Let the two arcs intersect at point D.

3. Taking D as center, draw an arc of radius 5 cm (= CD) and taking B as center, draw one more arc of radius 5cm (= BC). Let the two arcs intersect at point C.

4. Join DC and BC.

Hence, ABCD is the required quadrilateral.

Steps of construction:

1. Draw YZ = 4.5 cm.

2. With Y as centre and radius 5.7 cm draw an arc.

3. With Z as centre and radius 5.7 cm draw another arc, cutting previous arc at X.

4. With X as centre and radius 5 cm, draw an arc.

5. With Z as centre and radius 4.6 cm draw another arc, cutting previous arc at T.

6. Join XT and ZT.

Hence, XYZT is required quadrilateral.

Steps of construction:

1. Draw BC = 6 cm.

2. Construct ∠BCX = 60°.

3. With C as centre and radius = 4 cm, draw an arc, cutting CX at D.

4. With D as centre and radius = 6 cm, draw an arc.

5. With B as centre and radius = 4 cm, draw another arc,cutting the previous arc at A.

6. Join AB and DA.

Hence, ABCD is the required parallelogram.

Steps of construction :

1. Draw QR = 4.7 cm.

2. Construct ∠RQX = 120°.

3. With Q as centre and radius = 2.8 cm, draw an arc,cutting QX at P.

4. With P as centre and radius = 4.7 cm, draw an arc.

5. With R as centre and radius = 2.8 cm, draw another arc, cutting the previous arc at S.

6. Join RS and PS.

Hence, PQRS is the required parallelogram.

Steps of construction:

1. Draw BC = 5 cm.

2. With B as centre and radius = 6 cm, draw an arc.

3. With C as centre and radius = 3 cm, draw another arc, cutting previous arc at D.

4. Join BD and CD.

5. With B as centre and radius = 3 cm draw an arc.

6. With D as centre and radius = 5 cm, draw another arc,cutting previous arc at A.

7. Join AD and AB.

Hence, ABCD is the required parallelogram.

Steps of construction :

1. Draw AB = 4.8 cm.

2. With A as centre and radius = 5.4 cm, draw an arc.

3. With B as centre and radius = 3.5 cm, draw another arc, cutting previous arc at C.

4. Join BC and AC.

5. With A as centre and radius = 3.5 cm draw an arc.

6. With C as centre and radius = 4.8 cm, draw another arc,cutting previous arc at D.

7. Join AD and CD.

Hence, ABCD is the required parallelogram.

Steps of construction:

1. Draw AC = 5.6 cm.

2. Find mid-point of AC.

3. Construct ∠AOX = 60°, produce XO to Y.

4. With O as centre and radii = $\frac{1}{2}$ (6.2) = 3.1 cm, draw arcs, cutting OX at D and OY at B.

5. Join AB, BC, CD and DA.

Hence, ABCD is required parallelogram.

Steps of construction :

1. Draw AB = 4.7 cm.

2. Construct ∠ABX = 90°.

3. With B as centre and radius equal to 3.2 cm, draw an arc, cutting BX at C.

4. With A as centre and radius = 3.2 cm, draw an arc.

5. With C as centre and radius = 4.7 cm, draw an arc, cutting previous arc at D.

6. Join DC and DA.

Hence, ABCD is required rectangle.

Steps of construction :

1. Draw AB = 4.5 cm.

2. Construct ∠ABX = 90°.

3. With A as centre and radius equal to 6 cm, draw an arc, cutting BX at C. Join BC

4. With C as centre and radius = 4.5 cm, draw an arc.

5. With A as centre and radius = BC, draw an arc, cutting previous arc at D.

6. Join DC and DA.

Hence, ABCD is required rectangle.

Steps of construction :

1. Draw AC = 6.3 cm.

2. Find mid-point O of AC.

3. Construct ∠AOX = 50°.

4. Produce XO to Y.

5. Set off OB = 3.15 cm and OD = 3.15 cm.

6. Join AB, BC, CD and DA.

Hence, ABCD is required rectangle.

Steps of construction:

1. Draw AC = 6 cm.

2. Draw a perpendicular bisector XY of AC, meeting AC at O.

3. On XY, setoff OD = 3 cm and OB = 3 cm.

4. Join AD, CD, AB and BC.

Hence, ABCD is the required square.

Steps of construction :

1. Draw AB = 4.3 cm.

2. Construct ∠BAX = 45°.

3. On AX, cut off AD = 4.3 cm.

4. With B as centre and radius = 4.3 cm, draw an arc.

5. With D as centre and radius = 4.3 cm, draw another arc, cutting the previous arc at C.

6. Join CB and CD.

Hence, ABCD is the required rhombus.

Steps of construction:

1. Draw AB = 5 cm.

2. With A as centre and radius = 6.8 cm, draw an arc.

3. With B as centre and radius = 5 cm, draw another arc, cutting the previous arc at C.

4. Join BC and AC.

5. With A as centre and radius = 5 cm, draw an arc.

6. With C as centre and radius = 5 cm, draw another arc,cutting the previous arc at D.

7. Join CD and DA.

Hence, ABCD is required rhombus.

Steps of construction :

1. Draw AC = 5.8 cm.

2. Draw the perpendicular bisector XOY of AC.

3. Set off OB = 3.2 cm on OY and OD = 3.2 cm on OX.

4. Join AB, BC, CD and DA.

Hence, ABCD is the required rhombus.

Steps of construction :

1. Draw AC = 5.6 cm.

2. Draw a perpendicular bisector XY of AC, meeting AC at O.

3. On XY, setoff OD = 2.8 cm and OB = 2.8 cm.

4. Join AD, CD, AB and BC.

Hence, ABCD is the required square.

We know that each angle in a regular hexagon = 120°.

Steps of construction :

1. Draw a line segment AB = 2 cm.

2. At A and B, draw ∠ABC = 120° each and cut off AF = BC = 2 cm.

3. At F and C, draw rays making angle of 120° each and cut off EF = CD = 2 cm.

4. Join ED. Hence, ABCDEF is the required hexagon.

Hence, ABCDEF is required regular hexagon.

We know that each angle in a regular hexagon = 120°.

Steps of construction :

1. Draw a line segment AB = 3.5 cm.

2. At A and B, draw ∠ABC = 120° each and cut off AF = BC = 3.5 cm.

3. At F and C, draw rays making angle of 120° each and cut off EF = CD = 3.5 cm.

4. Join ED. Hence, ABCDEF is the required hexagon.

Hence, ABCDEF is required regular hexagon.

Let the fourth angle be x.

Given the three angles are 56°, 115° and 84°:

The sum of the angles of a quadrilateral is 360°

56°+ 115° + 84° + x = 360°

255° + x = 360°

x = 360° - 255°

x = 105°

Hence, option 2 is the correct option.

We know that,

The sum of the angles of a quadrilateral is 360°

Given the ratio of the angles is 2 : 4 : 5 : 7.

Let the angles be 2x, 4x, 5x, and 7x.

2x + 4x + 5x + 7x = 360°

18x = 360°

x = 20°

2x = 40°

4x = 80°

5x = 100°

7x = 140°

Hence, option 3 is the correct option.

In a parallelogram, consecutive angles are supplementary and opposite angles are equal.

∠A + ∠B = 180°

72° + ∠B = 180°

∠B = 180° - 72°

∠B = 108°

∠A = ∠C = 72° (Opposite angles of parallelogram are equal.)

∠D = ∠B = 108° (Opposite angles of parallelogram are equal.)

Hence, option 3 is the correct option.

In a parallelogram, consecutive angles are supplementary and opposite angles are equal.

∠A = 2x + 25° and ∠B = 3x − 5°

∠A + ∠B = 180°

2x + 25° + 3x − 5° = 180°

5x + 20° = 180°

5x = 180° - 20°

5x = 160°

x = 32°.

Hence, option 4 is the correct option.

Let the smallest angle of the parallelogram be denoted as x.

Given,

One angle of a parallelogram is 30° less than twice the smallest angle = 2x - 30°.

In a parallelogram, consecutive angles are supplementary and opposite angles are equal.

x + 2x - 30° = 180°

3x - 30° = 180°

3x = 180° + 30°

3x = 210°

x = 70°.

The two smaller angles are both 70°.

The two larger angles are both 2(70°) - 30° = 110°.

Hence, option 2 is the correct option.

In a parallelogram, opposite sides are equal in length.

AB = CD

BC = DA

Perimeter = 30 cm

We know that,

Perimeter of Parallelogram = 2(Length + Breadth)

30 = 2(9.5 + BC)

15 = (9.5 + BC)

BC = 15 - 9.5

BC = 5.5 cm

BC = AD = 5.5 cm.

AB = CD = 9.5 cm

Hence, option 1 is the correct option.

Let ABCD be the rhombus and the diagonals intersect at point O.

Let diagonal AC = 16 cm.

We know that,

Diagonals of rhombus bisect each other at right angles.

∴ AO = OC = $\frac{AC}{2} = \frac{16}{2}$ = 8 cm and BO = OD = x cm (let).

In right angle triangle AOB,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + Base²

⇒ AB² = AO² + OB²

⇒ 10² = 8² + x²

⇒ 100 = 64 + x²

⇒ x² = 100 - 64

⇒ x² = 36

⇒ x = $\sqrt{36}$ = 6 cm.

From figure,

⇒ BD = BO + OD = 6 + 6 = 12 cm.

Hence, option 2 is the correct option.

The sum of consecutive angles in a rhombus is 180°.

∠A + ∠D = 180°

70° + ∠D = 180°

∠D = 180° - 70°

∠D = 110°.

The diagonal BD bisects ∠D.

∠CDB = $\frac{∠D}{2}$

∠CDB = $\frac{110°}{2}$

∠CDB = 55°.

Hence, option 3 is the correct option.

In a parallelogram, adjacent angles are supplementary.

Let one of the angle of parallelogram be x.

The adjacent angle is given as $\frac{4x}{5}$.

$$\begin{aligned}\Rightarrow x + \frac{4x}{5} = 180° \\[1em] \Rightarrow \frac{5x + 4x}{5} = 180° \\[1em] \Rightarrow \frac{9x}{5} = 180° \\[1em] \Rightarrow 9x = 180° \times 5 \\[1em] \Rightarrow 9x = 900° \\[1em] \Rightarrow x = 100°.\end{aligned}$$

The adjacent angle = $\frac{4 \times 100°}{5}$ = 80°.

The angles opposite to 100° is also 100°.

The angle opposite to 80° is also 80°.

Hence, option 1 is the correct option.

The diagonals of a rhombus are 18 cm and 24 cm.

AC = 18 cm

Then, OA = OC = $\frac{18}{2}$ = 9 cm.

And, BD = 24 cm

Then, OB = OD = $\frac{24}{2}$ = 12 cm.

Since the diagonals of a rhombus bisect at 90°.

Applying pythagoras theorem in triangle AOB, we get :

⇒ AB² = OA² + OB²

⇒ AB² = (9)² + (12)²

⇒ AB² = 81 + 144

⇒ AB² = 225

⇒ AB = $\sqrt{225}$

⇒ AB = 15 cm.

Hence, option 2 is the correct option.

Given,

∠BAD = 60° and ∠BAC = 30°.

From figure,

∠CAD = ∠BAD - ∠BAC = 60° - 30° = 30°.

If diagonal of a parallelogram bisects a vertex angle, then it is a rhombus.

Thus, ABCD is a rhombus.

In rhombus adjacent angles are supplementary.

Thus,

∠BAD + ∠CBA = 180°

60° + ∠CBA = 180°

∠CBA = 180° - 60° = 120°.

Since, diagonals of rhombus bisect the vertex angle,

∴ ∠CBD = $\frac{∠CBA}{2} = \frac{120°}{2}$ = 60°.

Hence, option 2 is the correct option.

In a rectangle all angles are equal to 90°.

∠ABC = ∠OBC + ∠ABO

90° = 64° + ∠ABO

∠ABO = 90° - 64°

∠ABO = 26°.

The diagonals of a rectangle are equal in length and bisect each other. This means :

⇒ AC = BD

⇒ AO = BO

In an isosceles triangle AOB, the angles opposite the equal sides are equal.

∠OAB = ∠ABO = 26°.

Hence, option 3 is the correct option.

In rhombus opposite sides are parallel. The diagonal BD acts as a transversal line.

Therefore, the alternate interior angles are equal.

∠DBC = ∠ADB = 30°

From figure,

∠OBC = ∠DBC = 30°

∠BOC = 90° [Diagonals of rhombus cut at right angles]

In triangle BOC,

∠OBC + ∠BOC + ∠OCB = 180°

30° + 90° + ∠OCB = 180°

120° + ∠OCB = 180°

∠OCB = 180° - 120°

∠OCB = 60°.

Hence, option 2 is the correct option.

Given the ratio ∠A : ∠B = 3 : 5.

Let ∠A = 3x and ∠B = 5x.

Sum of adjacent angles in a parallelogram = 180°.

3x + 5x = 180°

8x = 180°

x = 22.5°

In a parallelogram, opposite angles are equal.

∠C = ∠A = 3x

∠C = 3 × (22.5°)

∠C = 67.5°

Hence, option 1 is the correct option.

In a parallelogram, consecutive angles are supplementary.

∠A + ∠B = 180°

In triangle AEB,

∠AEB + ∠BAE + ∠EBA = 180°

∠AEB + $\frac{1}{2}$ ∠A + $\frac{1}{2}$ ∠B = 180°

∠AEB + $\frac{1}{2}$ (∠A + ∠B) = 180°

∠AEB + $\frac{1}{2} \times$ (180°) = 180°

∠AEB + 90° = 180°

∠AEB = 180° - 90°

∠AEB = 90°.

Hence, option 2 is the correct option.

In parallelogram ABCD, we know that AB ∥ DC. The line AP is a transversal.

∠PAB = ∠APD (Alternate interior angles are equal) ....(1)

Since AP is the bisector of ∠A:

∠PAB = ∠PAD .......(2)

From equation (1) and (2), we get :

∴ ∠PAD = ∠APD

In triangle APD, since two angles are equal, the triangle is isosceles.

AD = DP

ABCD is a parallelogram, AD = BC = 3 cm.

Thus, DP = 3 cm.

Similarly, for the bisector BP and transversal BP:

∠PBA = ∠BPC (Alternate interior angles are equal) .......(3)

Since BP bisects ∠B:

∠PBA = ∠PBC ..........(4)

From equation (3) and (4), we get :

∴ ∠BPC = ∠PBC

In triangle BCP, since two angles are equal, the triangle is isosceles.

PC = BC = 3 cm.

Length of CD = DP + PC = 3 + 3 = 6 cm.

AB = CD = 6 cm [Opposite sides of parallelogram are equal]

Hence, option 3 is the correct option.

Let the rhombus be ABCD and its diagonals AC and BD intersect at point O.

We know that,

Diagonals of a rhombus bisect each other.

AC = 24 cm

OA = $\frac{24}{2}$ = 12 cm.

BD = 10 cm

OB = $\frac{10}{2}$ = 5 cm.

In triangle AOB,

OA² + OB² = AB²

12² + 5² = AB²

144 + 25 = AB²

169 = AB²

AB = $\sqrt{169}$

AB = 13 cm.

Hence, option 4 is the correct option.

If the opposite angles of a quadrilateral are equal, then it will definitely be a parallelogram.

Hence, option 4 is the correct option.

A rectangle is defined as a parallelogram with equal diagonals. If the diagonals of a quadrilateral bisect each other and are also equal, it must be a rectangle.

Hence, option 1 is the correct option.

In triangle PQS,

PQ = PS = 6 cm (because all sides of a rhombus are equal).

Therefore, the base angles are equal:

∠PQS = ∠PSQ.

∠PQR = ∠PSR = 120° [Opposite angles of a rhombus are equal]

∠PQR + ∠QPS = 180° [Consecutive angles are supplementary]

∠QPS = 180° - 120°

∠QPS = 60°.

In triangle PQS,

∠QPS + ∠PQS + ∠PSQ = 180°

60° + 2∠PQS = 180°

2∠PQS = 180° - 60°

2∠PQS = 120°

∠PQS = 60°.

Since all three angles (∠QPS, ∠PQS, and ∠PSQ) are 60°, △PQS is an equilateral triangle.

PQ = PS = QS = 6 cm.

Hence, option 1 is the correct option.

The diagonals of a rectangle are equal in length and bisect eah other, but not necessarily at right angles.

∴ Assertion (A) is false.

A square is a special type of rectangle where all sides are equal, but a rectangle is not necessarily a square.

A rectangle only requires four right angles; it does not require all four sides to be equal.

Every square is a rectangle, but not every rectangle is a square.

∴ Reason (R) is false.

Both A and R are false.

Hence, option 4 is the correct option.

For a quadrilateral to be a rectangle, the diagonals must be equal and they must bisect each other.Having equal diagonals is not enough to say a quadrilateral to be a rectangle.

Assertion (A) is false.

This is a fundamental property of rectangles.

If we have a rectangle ABCD, we can prove the diagonals AC and BD are equal using the SAS congruence rule on △ABC and △DCB.

AB = DC (Opposite sides)

∠B = ∠C = 90°

BC = BC (Common side)

Therefore, AC = BD.

Reason (R) is true.

A is false, R is true

Hence, option 2 is the correct option.

By definition, a parallelogram is a quadrilateral with two pairs of parallel opposite sides.

A square has all the properties of a parallelogram:

Opposite sides are parallel.

Opposite sides are equal.

Opposite angles are equal.

Diagonals bisect each other.

Because a square meets all these criteria, it is considered a special type of parallelogram.

Assertion (A) is true.

A square has equal diagonals, it is not true for a general parallelogram.

In a standard parallelogram (like a rhombus or a slanted parallelogram), one diagonal is typically longer than the other.

Reason (R) is false.

A is true, R is false.

Hence, option 1 is the correct option.

∠A + ∠D = 180° [sum of Co-Int. angles in ∥gm is 180°]

⇒ $\frac{1}{2}∠A + \frac{1}{2}∠D$ = 90°

In triangle ASD,

∠DAS + ∠SDA + ∠ASD = 180°

$\frac{1}{2}∠A + \frac{1}{2}∠D$ + ∠ASD = 180°

90° + ∠ASD = 180°

∠ASD = 90°

∠PSR = ∠ASD = 90° [Vertically opposite angles]

∠S = 90°

∠B + ∠C = 180° [sum of Co-Int. angles in ∥gm is 180°]

⇒ $\frac{1}{2}∠B + \frac{1}{2}∠C$ = 90°

In triangle BQC,

∠QCB + ∠QBC + ∠CQB = 180°

$\frac{1}{2}∠C + \frac{1}{2}∠B$ + ∠CQB = 180°

90° + ∠CQB = 180°

∠CQB = 90°

∠PQR = ∠CQB = 90° [Vertically opposite angles]

∠Q = 90°

∠A + ∠B = 180° [sum of Co-Int. angles in ∥gm is 180°]

⇒ $\frac{1}{2}∠A + \frac{1}{2}∠B$= 90°

In triangle APB,

$\frac{1}{2}∠A + \frac{1}{2}∠B$ + ∠APB = 180°

90° + ∠APB = 180°

∠APB = 90°

∠P = 90°

In quadrilateral PQRS,

By angle sum property of quadrilateral:

∠P + ∠Q + ∠R + ∠S = 360°

90° + 90° + ∠R + 90° = 360°

∠R + 270° = 360°

∠R = 360° - 270°

∠R = 90°

Since, all angles of PQRS = 90°

∴ PQRS is a rectangle

Hence, option 1 is the correct option.

A quadrilateral whose diagonals bisect each other is a Parallelogram. ​

Since ∠A and ∠B are adjacent angles they are supplementary,

∠A + ∠B = 180°

35° + ∠B = 180°

∠B = 180° - 35°

∠B = 145°.

Hence, option 1 is the correct option.

The Mid-point Theorem, which states that the line segment joining the mid-points of two sides of a triangle is parallel to the third side and is half of its length.

In triangle ADC,

M is the mid-point of AD (let)

P is the mid-point of AC

MP ∥ CD

MP = $\frac{1}{2}$ CD

In triangle ABD,

M is the mid-point of AD

Q is the mid-point of BD

MQ ∥ AB

MQ = $\frac{1}{2}$ AB

Since: MP ∥ CD, MQ ∥ AB and AB ∥ CD,

∴ MP ∥ MQ

Both pass through point M. So MP and MQ are the same straight line.

Hence, M, P and Q all lie on same line.

MQ = MP + PQ

PQ = MQ - MP

PQ = $\frac{1}{2}$ AB - $\frac{1}{2}$ CD

PQ = $\frac{1}{2}$ [AB - CD]

Hence, option 3 is the correct option.

A kite is defined by two pairs of adjacent sides being equal. Because of this specific symmetry, the longer diagonal acts as a mirror line, while the shorter diagonal does not.

Kite has exactly 1 axis of symmetry.

Hence, option 3 is the correct option.

No, we cannot construct a unique quadrilateral with only the measures of all four angles and the length of just one side.

In any quadrilateral, the sum of interior angles is always 360°. So even if all four angles are given, they only tell you the shape’s turning, not its exact size or proportions.

Knowing the length of only one side fixes the scale along that side, but the other sides can still vary.

Therefore, the given information is insufficient, and a unique quadrilateral cannot be constructed.

Rhombus:

Diagonals of a rhombus are perpendicular bisectors of each other.

Knowing both diagonal lengths completely determines the rhombus.

Square:

A square has equal diagonals that are perpendicular. Knowing the diagonal length fixes the side length.

Hence, square and rhombus can be constructed if we know the length of its diagonals.

In a rectangle, all angles are 90°.

Given that diagonal bisects ∠B

∠ABD = ∠DBC = 45°

This implies that the diagonal makes equal angles with sides BA and BC, which can happen only if AB = BC.

A rectangle with adjacent sides equal is a square.

Yes, ABCD is a square.