Quadrilaterals — Exercise 11(A)

In the given figure, ABCD is a parallelogram in which ∠A = 70°. Calculate ∠B, ∠C, ∠D.

Answer

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°.

In the given figure, ABCD is a parallelogram. Side DC is produced to E and ∠BCE = 105°. Calculate ∠A, ∠B, ∠C and ∠D.

Answer

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°.

If an angle of a parallelogram is two-third of its adjacent angle, find the angles of the parallelogram.

Answer

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

We know that,

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

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

∴ x = 108°.

⇒ $\dfrac{2x}{3} = \dfrac{2}{3} \times 108°$ = 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°.

In the adjoining figure, ABCD is a parallelogram in which ∠BAD = 70° and ∠CBD = 50°. Calculate :

(i) ∠ADB

(ii) ∠CDB.

Answer

(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°.

In the given figure, ABCD is a rhombus in which ∠A = 72°. If ∠CBD = x°, find the value of x.

Answer

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 = $\dfrac{1}{2}$ ∠B [∵ BD bisects ∠B]

⇒ ∠CBD = $\dfrac{108°}{2}$

⇒ ∠CBD = x° = 54°

⇒ x = 54.

Hence, x = 54.

In the adjoining figure, equilateral △ EDC surmounts square ABCD. If ∠DEB = x°, find value of x.

Answer

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.

In the adjoining figure, ABCD is a rhombus whose diagonals intersect at O. If ∠OAB : ∠OBA = 2 : 3, find the angles of △ OAB.

Answer

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 = $\dfrac{90°}{5}$

⇒ x = 18°.

∠OAB = 2x = 2(18°) = 36°.

∠OBA = 3x = 3(18°) = 54°.

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

In the given figure, ABCD is a rectangle whose diagonals intersect at O. Diagonal AC is produced to E and ∠ECD = 140°. Find the angles of △ OAB.

Answer

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°.

In the given figure, ABCD is a kite whose diagonals intersect at O. If ∠DAB = 54° and ∠BCD = 76°, calculate :

(i) ∠ODA

(ii) ∠OBC.

Answer

(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°.

In the given figure, ABCD is an isosceles trapezium in which ∠CDA = 2x° and ∠BAD = 3x°. Find all the angles of the trapezium.

Answer

We know that,

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

∴ ∠A + ∠D = 180°

⇒ 3x + 2x = 180°

⇒ 5x = 180°

⇒ x = $\dfrac{180°}{5}$

⇒ 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°.

In the given figure, ABCD is a trapezium in which ∠A = (x + 25)°, ∠B = y°, ∠C = 95° and ∠D = (2x + 5)°. Find the values of x and y.

Answer

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.

In the given figure, ABCD is rhombus and △EDC is a equilateral. If ∠BAD = 78°, calculate :

(i) ∠CBE

(ii) ∠DBE

Answer

(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 = $\dfrac{102°}{2}$

⇒ ∠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 = $\dfrac{42°}{2}$

⇒ ∠CBE = 21°.

Hence, ∠CBE = 21°.

(ii) From figure,

⇒ ∠DBE = ∠CBD - ∠CBE

⇒ ∠DBE = 51° - 21° = 30°.

Hence, ∠DBE = 30°.

DEC is an equilateral triangle in a square ABCD. If BD and CE intersect at O and ∠COD = x°, find the value of x.

Answer

Given,

ABCD is a square,

∠ADC = 90°

The diagonal BD bisects the ∠ADC at the vertex.

∠BDC = $\dfrac{∠ADC}{2} = \dfrac{90°}{2}$ = 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°.

If one angle of a parallelogram is 90°, show that each of its angles measures 90°.

Answer

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°.

In the adjoining figure, ABCD and PQBA are two parallelograms. Prove that :

(i) DPQC is a parallelogram.

(ii) DP = CQ.

(iii) ΔDAP ≅ ΔCBQ.

Answer

(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.

In the adjoining figure, ABCD is a parallelogram. BM ⟂ AC and DN ⟂ AC. Prove that :

(i) ΔBMC ≅ ΔDNA.

(ii) BM = DN.

Answer

(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.

In the adjoining figure, ABCD is a parallelogram and X is the mid-point of BC. The line AX produced meets DC produced at Q. The parallelogram AQPB is completed. Prove that :

(i) ΔABX ≅ ΔQCX.

(ii) DC = CQ = QP.

Answer

(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.

In the adjoining figure, ABCD is a parallelogram. Line segments AX and CY bisect ∠A and ∠C respectively. Prove that :

(i) ΔADX ≅ ΔCBY

(ii) AX = CY

(iii) AX ∥ CY

(iv) AYCX is a parallelogram

Answer

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 $\Big(\dfrac{1}{2}∠A = \dfrac{1}{2}∠C \Big)$

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 ($\dfrac{1}{2}∠A = \dfrac{1}{2}∠C$)

∴ ∠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.

In the given figure, ABCD is a parallelogram and X, Y are points on diagonal BD such that DX = BY. Prove that CXAY is a parallelogram.

Answer

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.

Show that the bisectors of the angles of a parallelogram enclose a rectangle.

Answer

ABCD is a //gm.

From figure,

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

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

In triangle APD,

∠DAP + ∠PDA + ∠APD = 180°

$\dfrac{1}{2}∠A + \dfrac{1}{2}∠B$ + ∠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°]

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

In triangle BRC,

∠CBR + ∠BCR + ∠CRB = 180°

$\dfrac{1}{2}∠B + \dfrac{1}{2}∠C$ + ∠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°]

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

In triangle ASB,

∠SAB + ∠SBA + ∠ASB = 180°

$\dfrac{1}{2}∠B + \dfrac{1}{2}∠B$ + ∠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.

If a diagonal of a parallelogram bisects one of the angles of the parallelogram, prove that it also bisects the second angle and then the two diagonals are perpendicular to each other.

Answer

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.

In the given figure, ABCD is a parallelogram and E is the mid-point of BC. If DE and AB produced meet at F, prove that AF = 2AB.

Answer

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.