Students can Download Maths Chapter 15 Quadrilaterals Additional Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka State Syllabus Class 8 Maths Chapter 15 Quadrilaterals Additional Questions

Question 1.

Complete the following

a. A quadrilateral has four sides.

b. A quadrilateral has two diagonals

c. A quadrilateral in which one pair of sides are parallel to each other is trapezium.

d. In an isosceles trapezium the base angles are equal.

e. In a rhombus, the diagonals bisect each other in right angles.

f. In a square, all the sides are equal.

Question 2.

Let ABCD be a parallelogram, what special name will you give if.

a. AB = BC

Solution:

Rhombus

b. ∠BAD – 90°

Solution:

Rectangle

c. AB = AD and ∠BAD = 90°

Solution:

Square

Question 3.

A quadrilateral has three acute angles each measuring 70°. What is the measure of the fourth angle?

Solution:

Measure of the fourth angle

= 360 – (70 + 70 + 70) = 360 – 210

= 140°

Question 4.

The difference between two adjacent angles of a parallelogram is 20° . Find the measures of all the angles ‘ of the parallelogram.

Solution:

Let the adjacent angles be ‘x’ and ‘y’

x + y = 180° [Adjacent angles are supplementary]

x = \(\frac{200}{2}\) ; x = 100°

x + y = 180

100 + y = 180°

y = 180 – 100

y = 180°

∴The angles are 100°, 80°, 100° and 80°

Question 5.

The angles of a quadrilateral are in the ratio 1 : 2 : 3 : 4 find all the angles.

Solution:

Ratio of the angles 1 : 2 : 3 : 4

Let the angles be x, 2x, 3x and 4x

x + 2x + 3x + 4x = 360°

[Angle sum property of a quadrilateral]

10x = 360°

x = \(\frac{360}{10}\) ; x = 36°

2x = 2 x 36 = 72°, 3x = 3 x 36 = 108°

4x = 4 x 36 = 144°

∴ The angles are 36°, 72°, 108° and 144°

Question 6.

Let PQRS be a parallelogram with PQ = 10 cm and QR = 6 cm. Calculate the measures of the other two sides and measures of the other two sides and perimter of PQRS.

Solution:

PQ = RS = 10 cm

[opposite sides of a parallelogram]

QR = SP = 6 cm

Perimeter = PQ + QR + RS + SP

= 10 + 6 + 10 + 6

= 32 cm

Question 7.

The perimeter of a square is 60 cm. Find the length of its side.

Solution:

Perimeter of a square = 4a

60 = 4a

∴ a = \(\frac{60}{4}\) a = 15 cm

The length of the side is 15 cm

Question 8.

Let ABCD be a square and let AC = BD = 10 cm let AC and BD intersect at O. Find OC and OD.

Solution:

The diagonals of a square bisect each other

Question 9.

Let PQRS be a rhombus with PR = 15 cm and QS = 8 cm. Find the area of the rhombus.

Solution:

Area of rhombus PQRS = Area of ∆ PQS + Area of ∆ RQS

Area of rhombus = 60 cm^{2}

Question 10.

Let ABCD be a parallelogram and suppose the bisectors of ∠A and ∠B meet at P prove that ∠APB = 90°

Solution:

∠A + ∠B = 180°

[Adjacent angles of a prallelogram]

\(\frac{1}{2}\)∠A + \(\frac{1}{2}\) ∠B = ∠B . 180° [Multiplying by \(\frac{1}{2}\) ]

∠PAB + ∠PBA = 90° [AP and BP are angular bisectors]

∠PAB + ∠PBA +∠APB = 180° [Angle sum property of a triangle]

90 +∠APB = 180°

∠APB = 180 – 90

∠APB = 90°

Question 11.

Let ABCD be a sqaure. Locate points P, Q, R, S on the sides AB, BC, CD, DA respectively such that AP = BQ = CR = DS. Prove that PQRS is a square.

Solution:

Let AP = BQ = CR = DS = x

AP + PB = AB

x + PB = AB

PB = AB – x

AB = BC

∴ QC = AB – x

In ∆ PBQ and ∆ QCR

BQ = CR[= x]

BP = QC [AB – x]

∠B = ∠C [= 90°]

∴ ∆ PBQ = A QCR [SAS postulate]

∴ PQ = QR

[corresponding sides] similarly we can show that QR = SR and SR = SP

i.e., PQ = QR = RS = SP …… (i)

∠BQP = ∠CRQ [corresponding angles]

∠BQP + ∠PQR + ∠RQC = 180° [Angle sum property of a triangle]

∠BQP + ∠PQR + 90 – ∠CRQ = 180°

[∴ ∠CQR = 90 – ∠CRQ]

∠BQP + ∠PQR – ∠BQP = 180 – 90

[∵ ∠DQP = ∠CRQ]

∠PQR = 90°

∴ PQRS is a square.

Question 12.

Let ABCD be a rectangle and let P, Q, R, S be the midpoints of AB, BC, CD, DA respectively prove that PQRS is a Rhombus.

Solution:

In ∆ APS and ∆ CQR

AP = CR [P and R are the midpoints of AB and CD)

AS = CQ [S and Q are the midpoints of AD and BC]

∠SAP = ∠RCQ [=90° angles of a rectangle]

∴ ∆ APS ≅ ∆ CQR [SAS postulate]

∴ SP = QR [corresponding sides] similarly we can show PQ = QR = RS

i.e., PQ = QR = RS = SP

∴ PQRS is a rhombus

Question 13.

Let ABCD be a quadrilateral in which diagonals intersect at ‘O’ perpendicularly. Prove that AB + BC + CD + DA > (AC + BD)

Solution:

In A AOD, AD > AO + OD … (i)

In A AOB, AB > AO + OB … (ii)

In A BOC, BC > BO + OC … (iii)

In A COD, CD > DO + OC … (iv)

By adding (i), (ii), (iii) and (iv)

AD + AB + BC + CD > AO + OD + AO + OB + BO + OC + DO + CO > 2 AO + 2 OC + 2 OB + 2 OD > 2 (AO + OC + OB + OD)

AD + AB + BC + CD > 2 (AC + BD) [∵ AO + OC = AC OB + OD = BD]

∴ AD + AB + BC + CD > AC + BD

Question 14.

Let ABCD be a quadrilateral with diagonals AC and BD. Prove the follwoing statements (compare these with the previous problem)

a. AB + BC + CD > AD

b. AB + BC + CD + DA > 2AC

c. AB + BC + CD + DA > 2BD

d. AB + BC + CD + DA > AC + BD

Solution:

a. AB + BC + CD + DA > AC + BD [From previous problem]

AB + BC + CD > AC + BD – AD

∴ AB + BC + CD > – AD

b. AB + BC + CD + DA > 2 (AC + BD) [From the previous problem]

AB + BC + CD + DA > 2AC + 2BD

∴ AB + BC + CD + DA > 2AC

c. AB + BC + CD + DA > 2AC + 2BD [From the previous problem]

∴ AB + BC + CD + DA > 2BD

d. AB + BC + CD + DA > AC + BD [From the previous problem]

Question 15.

Let PQRS be a kite such that PQ > PS. Prove that ∠PSR > ∠PQR (Hint : Join QS)

Solution:

Join QS

In ∆ PQS, PQ > PS [data]

∴ ∠PSQ > ∠PQS …(i)

In ∆ SRQ, RQ > SR

[∵ PQ = QR and SP = SR]

∴ ∠QSR. > ∠SQR … (ii)

By adding (i) and (ii)

∠PSQ + ∠QSR > ∠PQS + ∠SQR

∠PSR >∠PQR

Question 16.

Let ABCD be a quadrilateral in which AB is the smallest side and CD is a

largest side ∠A > ∠C and ∠B > ∠D ( Hint: join AC and BD)

Solution:

Join AC and BD In ∆ ABD, AB is the smallest side

AD > AB

∠ABD > ∠APB

[Angle opposite to larger side] …(i)

In ∆ CBD, CD is the largest side

∴ CD > CB

∴ [CBD > ∠CDB [Angle opposite to larger side] …. (ii)

By adding (i) and (ii) .

∠ABD + ∠CBD >∠ADB + ∠CDB

∠ABC > ∠ADC

∴ ∠B >∠D

In ∆ ABC, AB is the smallest side] ……. (iv)

∴ BC > AB ∠BAC > ∠ACB [Angle opposite to larger side] … (iii)

In ∆ ADC, CD is the largest side

∴ CD > AD

∠CAD > ∠ACD [Angle opposite to larger side] …(iv)

By adding (iii) and (iv)

∠BAC + ∠CAD > ∠ACB + ∠ACD

∠BAD > ∠ACD

∠A > ∠C

Question 17.

In a triangle ABC, let D be the midpoint of BC. Prove that AB + AC > 2 AD (what property of quadrilateral is needed here)

Solution:

In ∆ ABD, AB + BD > AD … (i)

In ∆ ACD, AC + DC > AD …(ii)

By adding (i) and (ii)

AB + BD + AC + DC > AD + AD

AB + AC + BD + DC > 2AD

AB + AC + BC > 2AD

or

AB + AC > 2AD

Question 18.

Let ABCD be a quadrilateral and let P, Q, R, S be the midpoints of AB, BC, CD, DA respectively prove that PQRS is a parallelogram [what extra result you need to prove this] It can be proved if ABCD is a rectangle

Solution:

In ∆ APS and ∆ CQR.

AP = CR [P and R are the midpoints of AB and CD]

AS = CQ [∵ S and Q are the midpoints of AD and BC]

∠SAP = ∠RCQ [= 90° angles of a rectangle]

∴ ∆ APS ≅ ∆ CQR [SAS posltulate]

∴ SP = QR [corresponding sides]

similarly we can show PQ = QR = RS

i.e., PQ = QR = RS = SP

∴ PQRS is a rhombus

∴ PQRS is a parallelogram.

Question 19.

Prove that the base angles of an isosceles trapezium are equal.

Solution:

In trap ABCD, Draw ⊥^{rs} to base

\(\overline{\mathrm{BE}}\) and \(\overline{\mathrm{CF}}\)

\(\overline{\mathrm{BE}}\) ≅ \(\overline{\mathrm{CF}}\) (∵ Altitudes of trapezium)

\(\overline{\mathrm{AB}}\) = \(\overline{\mathrm{CD}}\) (Given)

\(\overline{\mathrm{AE}}\) = \(\overline{\mathrm{FD}}\) (Sides of Right angle)

From SSS congruence

∠A ≅ ∠D [CPCT]

Question 20.

Suppose in a quadrilateral ABCD, AC = BD and AD = BC. Prove that ABCD is a trapezium.

Solution:

Data: ABCD is a quadrilateral in which

AD = BC and AC = BD

To Prove: ABCD is a trapezium

Proof: In ∆ ADB and ∆ ACB AB = AB (Common side)

AD = BC (data)

BD = AC (data)

∴ ∆ ADB ≅ ∆ ACB (SSS postulate)

∴ ∠DAB = ∠ABC (CPCT) …(i)

In ∆ ADC and ∆ BDC

DC = DC (common side)

AD = BC (data)

AC = BD (data)

∴ ∆ ADC ≅ ∆ BDC (SSS postulate)

∴ ∠ADC = ∠BCD (IPCT) …(ii)

In Quadrilateral ABCD

∠BAD + ∠ABC + ∠DCD + ∠ADC = 360°

∠ABC + ∠ABC + ∠BCD + ∠BCD = 360°

[∠ABC = ∠DAB, ∠ADC = ∠BDC] 2∠ABC + 2∠DCD = 360°

2 (∠ABC + ∠BCD = 360°)

∠ABC +∠BCD = \(\frac{360^{\circ}}{2}\) = 180°

But theSe are co-interior angles AB || DC

∴ ABCD is a trapezium.

Question 21.

Prove logically the diagonals of a parallelogram bisect each other. Show conversely that a quadrilateral in which diagonals bisect each other is a parallelogram.

Solution:

Given: A parallelogram ABCD with diagonal AC and BD intersecting at O.

To prove: AO = OC and BO = OD

Proof: Consider ∆ AOD and ∆ COB.

We observe

∠ADO = ∠CBO (alternate angles)

∠DAO = ∠BCO (alternate angles)

AD = BC (opposite sides of a parallelogram)

By ASA, ∆ AOD ≅ ∆ COB.

Hence AO = OC and OD = OB

Given: a quadrilateral ABCd such that AC and BD bisect each other.

To prove: ABCD is a parallelogram.

Proof: Let AC and BD intersect at K.

In triangles AKD and CKB, we observe

AK = KC (given)

BK = KD (given)

∠AKD = CKB (vertically opposite angles)

Hence by SAS congruency,

∆ AKD ≅ ∆ CKB. lt follows that

∠DAK = ∠BCK and AD = CB.

Thus AD || BC. Similarly, we can prove AB || CD and AB = CD.

Therefore ABCD is a parallelogram.

Question 22.

Prove logically that the diagonal of a rectangle are equal.

Solution:

In rectangle ABCD, AC and BD are diagonal

In ∆ AOD and ∆ BOC

∠AOD = ∠BOC (V.O.A)

∠ADO = ∠OBC (Alternate angle]

∠DAO = ∠BCQ (Alternate angle]

∴ ∆ AOD ≅ ∆ BOC OA = OC, OB = OD

In ∆ AOB and ∆ COD

∠AOB = ∠COD (V.O.A)

∠OAB = ∠PCD (Alternate Angle)

∠QBA = ∠ODC (Alternate Angle)

∆ AOB ≅ ∆ COD

∴ OA = OC, OB = OD

∴ Diagonal AC = Diagonal BD.

Question 23.

In a rhombus PQRS, ZSQR = 40° and PQ = 3 cm. Find ZSPQ, ZQSR and the perimeter of the rhombus.

Solution:

Given PQRS is a rhombus

∠SQK = 40°, PQ = 3cm

Perimeter of rhombus PQRS

= PQ + QR + SR + SP 3 + 3 + 3 + 3 = 12 cm

Question 24.

Let ABCD be a rhombus and ∠ABC = 124°. Calculate. ∠A, ∠D and ∠C.

Solution:

In a rhombus ABCD,

If ∠ABC = 124° then

∠ADC = 124° (Opposite angles are equal)

∠DAD = 180° – 124°

∠DAD = 56°

∠DCD = ∠DAD = 56°

∴ ∠A = 56°, ∠D = 124°, ∠C = 56°

Question 25.

Four congruent rectangles are placed as shown in the figure. Area of the outer square is 4 times that of the inner square is 4 times that of the inner square. What is the ratio of length to breadth of the congruent rectangles?

Solution:

Outer square

ABCD = 4 × Inner square PQRS

Question 26.

Let ABCD be a quadrilateral in which ∠A = ∠C, and ∠B = ∠D. Prove that ABCD is a parallelogram.

Solution:

ABCD is a Quadrilateral in which AC and BD are the diagonals intersecting Each other at O.

OA = OC, and OB = OD.

∵ O is the midpoint of diagonals

∴ Each pair of opposite angles are equal. It is a parallelogram.

Question 27.

In a quadrilateral ABCD, suppose AB = CD and AD = BC. Prove that ABCD is a parallelogram.

Solution:

Given: Quadrilateral ABCD in which AB = DC and AD = BC

To Prove: ABCD is a parallelogram.

Construction: Draw diagonal AC.

Proof:

∴ ABCD is a parallelogram.