In the kite ABCD BCD 80 BAD 70 the remaining two angles CBA and CDA are equal find each equal angle

A closed figure having four sides formed by joining four points, no three of which are collinear, in an order is called quadrilateral.

For example:

Quadrilateral is the union of four line segments that join four coplanar points; no three of which are collinear and each segment meet exactly two other lines, each at their end point.

A quadrilateral is convex if each of the internal angle is less than 180˚. Otherwise it is concave quadrilateral.

3.5.2 Properties of quadrilateral

Let ABCD be a quadrilateral.

1. Then points A, B, C and D are called the vertices of the quadrilateral.
2. The segments AB, BC, CD and DA are the four sides of the quadrilateral.
3. The angles ∠DAB, ∠ABC, ∠BCD, and ∠CDA are the four angles of the quadrilateral.
4. The segments AC and BD are called as the diagonals of the quadrilateral.

Adjecent sides and opposite sides:

1. The sides of a quadrilateral are said to be adjacent sides or consecutive sides. If they have a common end point. In the adjoining figure, AB and AD are adjacent or consecutive sides. Identify the other pair of adjacent sides.
2. Two sides of a quadrilateral are said to be opposite angles, if they do not contain a common side. Here ∠DAB and ∠BCD are opposite angles. Identify other pair of opposite angles.

Adjacent angles and opposite angles:

1. The two angles of a quadrilateral are adjacent angles or consecutive angles. If they have a common side to them. Thus ∠DAB and ∠ABC are adjacent angles
2. Two angles of a quadrilateral are said to be opposite angles, if they do not contain a common side. Here ∠DAB and ∠BCD are opposite angles. Identify other pair of opposite angles.

Diagonal Property:

The diagonal AC divides the quadrilateral into two triangles, namely, triangles ABC and triangles ADC. Name the triangles formed when the diagonal BD is drawn.

Angle Sum Property:

Theorem 1: The sum of the angles of quadrilateral is 360˚

Given:  ABCD is a quadrilateral

To prove: ∠A + ∠B + ∠C + ∠D = 360˚

Construction: Draw the diagonal AC

Proof: In triangle ADC, ∠1 + ∠2 + ∠3 = 180˚(angle sum property)

In triangle ABC, ∠4 + ∠5 + ∠6 = 180˚

Adding these two,

∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 = 360˚

But , ∠1 + ∠4 = ∠A and ∠3 + ∠6 = ∠C. Therefore, ∠A + ∠D + ∠B + ∠C = 360˚

Thus, the sum of the angles of quadrilateral is 360˚

Example 1: The four angle of a quadrilateral are in the ratio 2:3:4:6. Find the measures of each angle.

Solution:

Given: The ratio of the angles is 2:3:4:6

To find: The measures of each angle.

Observe that 2 + 3 + 4 + 6 = 15. Thus 15 parts accounts for 360˚. Hence,

15 parts à 360˚

2 parts à (360˚/15) * 2 = 48˚

3 parts à (360˚/15) * 3 = 72˚

4 parts à (360˚/15) * 4 = 96˚

6 parts à (360˚/15) * 6 = 144˚

Thus the angles are 48˚, 72˚, 96˚, 144˚

Example 2: In a quadrilateral ABCD, ∠A and ∠C are of equal measure; ∠B is supplementary to ∠D. Find the measure of ∠A and ∠C.

Solution:

We are given ∠B + ∠D = 180˚. Using angle sum property of a quadrilateral, we get,

∠A + ∠C = 360˚- 180˚ = 180˚

Since, ∠A + ∠C are of equal measure, we obtain ∠A = ∠C = 180˚/2 = 90˚

Example 3: Find all the angles in the given quadrilateral below:

Solution:

We know that, ∠P + ∠Q + ∠R + ∠S = 360˚. Hence,

x + 2x + 3 + x + 3x – 7 = 360˚

This gives, 7x – 4 = 360˚ or 7x = 364˚

Therefore, x = 364˚/7 = 52˚

We obtain ∠P = 52˚; ∠R = 52˚; ∠Q = 2x + 3 = 107˚; ∠S = 149˚. Check that, ∠P + ∠Q + ∠R + ∠S = 360˚

EXERCISE 3.5.2

1. Two angles of a quadrilateral are 70° and 130° and the other two angles are equal. Find the measure of these two angles.

Solution:

 Let ∠A and ∠B be x, ∠C=70°

and ∠D = 130°

∠A + ∠B + ∠C + ∠D = 360° (Theorem 7)

x + x + 70° + 130° = 360°

2x = 360° − 200°

X = 160°/2 = 80°

∴ ∠A = 80° and ∠B = 80°

1. In the figure suppose ∠P and ∠Q are supplementary angles and ∠R = 125°. Find the measures of ∠S.

Solution:

∠P + ∠Q = 180° (Supplementary angles)

∠P + ∠Q + ∠R + ∠S = 360° (Theorem 7)

180° + 125° + ∠S = 360°

∠S = 360°−(180° + 125°)

∠S = 360°−305°

∠S=55°

1. Three angles of a quadrilateral are in the ratio 2:3:5 and the fourth angle is

90°. Find the measures of the other three angles.

Solution:

 Let the angles be 2x, 3x, 5x.

∠A + ∠B + ∠C + ∠D = 360°

2x + 3x + 5x + 90° = 360°

10x = 360°−90°

x = 270°/10

x = 27°

∴ The angles are , ∠A = 2x = 2×27 = 54°

∠B = 3x = 3×27 = 81

∠C = 5x = 5×27 = 135°

4.In the adjoining figure, ABCD is a quadrilateral such that ∠D + ∠C =100°. The bisectors of ∠A and ∠B meet at ∠P. Determine ∠APB.

Solution:

AP and BP are angular bisectors

∠D + ∠C = 100°

To find: ∠APB

Proof : ∠A + ∠B + ∠C + ∠D = 360° (Theorem 7)

∠A + ∠B + 100° = 360°

∠A + ∠B = 360° − 100°

Multipyling by 12 ,

12 × ∠A + 12 × ∠B = 12 × 260

a + b =130°

In Δ APB,

a + b + ∠P = 180°

∠P = 180° − 130°

∠P = 50°

3.5.3 Trapezium

The set of fundamental have a pair of opposite sides which are parallel. Such a quadrilateral is known as trapezium.

For example:

Example 4: In the figure ABCD, suppose AB||CD; ∠A = 65° and ∠B = 75°. What is the measure of ∠C and ∠D?

Solution: 

Observe that, ∠A + ∠D = 180° (a pair of adjacent angles of a trapezium is supplementary.)

Thus, 65° + ∠D = 180°

This gives, ∠D = 180° – 65° = 115°

Similarly, ∠B + ∠C = 180°, which gives 75° + ∠C = 180°. Hence,

∠C = 180° – 75° = 105°

Example 5: In an isosceles trapezium PQRS, ∠P and ∠S are in the ratio 1:2. Find the measure of all angles.

Solution:

In an isosceles trapezium base angles are equal; ∠P = ∠Q. Let ∠P = xº and ∠S =  2xº. Since, ∠P + ∠S = 180º (one pair of adjacent angles of a trapezium is supplementary). We get,

x + 2x = 180º

x = 180º/3 = 60º

Hence, ∠P = 60º and ∠S = 2 x 60º = 120º. Since, ∠P = ∠Q, we get, ∠Q = 60º. But, ∠Q + ∠R = 180º. Hence, we get,

∠R = 180º – 60º = 120º

EXERCISE 3.5.3
1. In a trapezium PQRS, PQ || RS, and ∠P = 70° and ∠Q = 80° . Calculate the measure of ∠S and ∠R.

Solution:

∠P + ∠S = 180° (Supplementary angles) ∠S = 180° − 70° = 110° ∠Q + ∠R = 180° (Supplementary angles)

∠R = 180° −80° = 100°

2. In a trapezium ABCD with AB || CD. It is given that AD is not parallel to BC. Is ΔABC ≅ Δ ADC? Give reasons.

Solution:

In Δ ABC and Δ ADC,

(1) AC = AC (Common side)

(2) ∠BAC = ∠ACD (Alternate angles, AB || CD)
∴ Δ ABC ≅ Δ ADC It cannot be proved with the help of any beam postulated.

3. In the figure, PQRS is an isosceles trapezium; ∠SRP = 30°, and ∠PQS=40° Calculate the angles ∠RPQ and ∠RSQ.

Solution:

Data:PQRS is an isosceles trapezium, PS = RQ and ∠P = ∠Q ∠SRP = 30° and ∠PQS = 40° To find : ∠RPQ and ∠RSQ Proof: ∠RPQ = ∠SRP = 30° (Alternate angles, PQ || SR) ∠RSQ = ∠PRS = 40°

(Alternate angles, PQ || SR)

4. Proof that the base angles of an isosceles trapezium are equal.

Solution:

Data: ABCD is an isosceles trapezium. To Prove: ∠A = ∠B Construction: Join the diagonal BD and AC. Proof : In Δ ACB and ΔBDA (1) BC = AD (Isosceles trapezium) (2)AC = BD (ISosceles trapezium) (3) AB = AB (Common side) ∴ ACB ≅ Δ BDA (SSS postulate) ∴ ∠A = ∠B (Congruency property) 5. Suppose in a quadrilateral ABCD, AC= BD and AD = BC. Prove that ABCD is a trapezium.

Solution:

Data: ABCD is a quadrilateral, AC = BD and AD = BC To prove: ABCD is trapezium Proof: In Δ ADB and Δ BCA (1) AD = BC (Data)

(2) AC = BD (Data)

(3) AB = AB (Common side) ∴ Δ ADB ≅ Δ BCA (SSS postulate) ∴ ∠A = ∠B (Congruency property) AC = BD (Data), AD = BC (Data)

∴ ABCD is an isosceles trapezium

3.5.4 Parallelograms

Look at the following sets of quadrilaterals:

A quadrilateral in which both the pairs of opposite sides are parallel is called a parallelogram.

Proposition 1: In parallelogram, opposite sides are equal and opposite angles are equal.

Proof:

Let ABCD is a parallelogram. Join BD. Then ∠1 = ∠2 and ∠3 = ∠4. In triangles ABD and CBD, we observe that,

∠1 = ∠2 ; ∠4 = ∠3 , BD(common)

Hence,  ΔABD ≅ Δ CDB (ASA postulate). It follows that,

AB = DC, AD = BC and ∠A = ∠C.

Similarly join AC, and we can prove that, ΔADC ≅ Δ CBA. Hence, ∠D =  ∠B

Example 6: The ratio of two sides of parallelogram is 3:$ and its perimeter is 42cm. Find the measures of all sides of the parallelogram.

Solution:

Let the sides be 3x  and 4x. Then the perimeter of the parallelogram is 2(3x + 4x) = 2 x 7x = 4x.

The given data implies that, 42 = 14x, so, that, x = 42/14 = 3. Hence the sides of the parallelogram are 3 x 3 = 9cm and 3 x4 = 12cm

Example 7: In the adjoining figure, PQRS is a parallelogram. Find x and y in cm.

Solution:

In a parallelogram, we know, that the diagonals bisect each other. Therefore SO = OQ. This gives 16 = x + y. Similarly, PO = OR, so that 20 = y + 7. We obtain y = 20 – 7 = 13cm. Substituting the value of y in the first relation, we get 16 = x + 13. Hence x = 3cm

EXERCISE 3.5.4 1.The adjacent angles of a parallelogram are in the ratio 2:1 Find the measures of all the angles.

Solution:

Let the angles be 2x and x ∠A + ∠B = 2x + x = 180° (Adjacent angles of parallelograms are supplementary) 2x + x = 180° 3x = 180° X = 180°/3 = 60° ∴ ∠A = 2x = 60×2 = 120° ∴ ∠B = 60° ∴ ∠C = ∠A = 120° (Opposite angles of parallelogram) ∠D = ∠B = 60°

(Opposite angles of parallelogram)

2. A field is in the form of a parallelogram, whose perimeter is 450 m and one of its sides is larger than the other by 75m. Find the lengths of all sides
Solution:

Perimeter = AD + DC + CB + BA 450 = x + x + 75 + x + x + 75 450 = 4x + 150 450 – 150 = 4x 300𝑥 = x X = 75 ∴ Side = 75m(x)

∴ Opposite side = x + 75 = 75 + 75 = 150m

3. In the figure, ABCD is a parallelogram. The diagonals AC and BD intersect at O; and ∠DAC = 40°, ∠CAB = 35°, and ∠DOC = 10°. ∠ADC, ∠ACB, and ∠CBD.

Data: ABCD is a parallelogram AD || BC and DC || AB. The diagonals AC and BD intersect at O ∠DAC = 40°, ∠CAB = 35° and ∠DOC = 110° To find : (1) ∠ABO (2)∠ADC (3)∠ACB (4) ∠CBD Proof: ∠DAC + ∠CAB = ∠A 40° + 35 = ∠A

∠A = 75°

∠C = ∠A = 75° (Opposite angles of parallelogram are equal) ∠D + ∠A = 180° (Supplementary angles) ∠D = 180°−75° = 105° ∠B = ∠D = 105° (Opposite angles of parallelogram are equal) ∠DOC = ∠AOB = 110° (Vertically opposite angles) In Δ AOB, ∠A + ∠O + ∠B = 180° (Sum of all angles of a Δ is 180°) 35° + 110° + ∠B = 180° ∠B = 180°-145° (1) ∠ABO = 35° (2)∠ADC = 105° (Proved) (3)∠ACB = ∠CBD = 40° (Alternate angles, AD || BC) ∠CBD = 105°-35°

(4) ∠CBD = 70°

(1)∠ABO = 35°

(2) ∠ADC = 105°

(3)∠ACB = 40°

(4) ∠CBD = 70° 4. In a parallelogram ABCD, the side DC is produced to E and ∠BCE = 105°.

Calculate and ∠A, ∠B, ∠C and ∠D.

Solution:

∠BCD + ∠BCE = 180° (Linear pair)
∠BCD = 180° − 105° = 75°

In ABCD, ∠A = ∠C = 75° (Opposite angles of parallelogram) ∠ABC = ∠BCE = 105° (Alternate angles, AB || DE) ∠D = ∠B = 105° (Opposite angles of parallelogram)

5. 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:

Data: ABCD is a parallelogram ( ║ln) To Prove : AO = OC and DO = OB Proof: In ΔABO and DOC (1)AB = DC (Opposite sides of angles) (2) ∠AOB = ∠DOC (V.O.A) (3) ∠ABD = ∠BDC (Alternate angles AB ||CD) ∴ ΔABO ≅ ΔDOC (ASA portulate) ∴ AO = OC & DO = BO (Congruence properties) 6. In a parallelogram KLMN, ∟K = 60°. Find the measure of all the angles.

Solution:

∠M = ∠K = 180° (Supplementary angles) ∠N = 180°−60° = 120° ∠M = ∠K = 60° (Opposite angles of parallelogram) ∠L = ∠N = 120° 7. Let ABCD be a quadrilateral in which ∠A = ∠C and ∠B = ∠D. Prove that ABCD is a parallelogram.

Solution:

Solution:

Data: ABCD is a quadrilateral, AB = DC, AD = BC To Prove : ABCD is a parallelogram Construction: Join DB

Proof: In Δ ABD and Δ DBC (1) DB = DB (Common side) (2) AD = BC (Data) (3) AB = DC (Data) ∴ ABD ≅ Δ DBC (SSS postulate) ∴ ∠DBA = ∠BDC (Congruency property) But they are alternate angles These we have AB ║ DC ∴ ∠ADB = ∠DBC (C.P) But they are alternate angles These we have DA ║ BC

∴ ABCD is a parallelogram.

3.5.5 Special kinds of paralellograms

Rectangle:

A rectangle is a parallelogram whose all angles are right angles.

Diagonal properties of a rectangle:

(i) The diagonals of a rectangle are equal

(ii) The diagonals of a rectangle bisect each other.

Example 8: In a rectangle XYWZ, suppose O is the point of intersection of its diagonals. If ∠ZOW = 110º. Calculate the measure of ∠OYW.

Solution:

We know, ∠ZOW = 110º. Hence, ∠WOY = 180º – 110º = 70º (supplementary angles)

Now, OYW is an isosceles triangle, as OY = OW. Hence,

∠OYW = ∠OWY = (180º –  70º)/2  = 110º/2 = 55º

Rhombus:

A rhombus is a parallelogram in which all the four sides are equal.

Example 10: The diagonals of a rhombus are 24cm and 10cm. Calculate the area of the rhombus.

Solution: 

We are given that AC = 24cm; BD = 10cm. We know that the diagonals of a rhombus bisect each other at right angles. Let O be the point of intersection of these diagonals. Then we have AO = CO = 12cm and BO = DO = 5cm. We also know that AOD is right angled triangle. Hence the area of triangle AOD is

1/2 x OA x OD = 1/2 x 12 x 5 = 30 cm2

Since a rhombus has four congruent right triangles, is area is 4 x 30 = 120 cm2

Square

A square is a parallelogram, in which

(i) in which all the sides are equal

(ii) each angle is right angle

(iii) diagonals are equal

(iv) diagonals bisect at right angles.

Example 12: A field is in the shape of a sqaure with side 20m. A pathway of 2m width is surrounding it. Find the outer perimeter of the pathway.

Solution:

Width of the pathway is 2m. Length of the side of the outer sqaure = (20 + 2 + 2) = 24m

Hence, perimeter 4 x 24 = 96m

Kite:

Kite is a quadrilateral in which two of isosceles triangles are joined along the common base.

Example 14: In the figure PQRS is a kite; PQ = 3cm and QR = 6cm. Find the perimeter of PQRS.

Solution:

We have PQ = PS = 3cm, QR = SR = 6cm.

Hence, the perimeter = PQ + QR + RS + PS

3 + 6 + 6 + 3 = 18cm

EXERCISE 3.5.5 1. The sides of the rectangle are in the ratio 2:1. The perimeter is 30 cm. Calculate the measure of all the sides.

Solution:

ABCD is a rectangle Let AB:BC = 2:1 ∴ AB = 2x and BC = x Perimeter of the rectangle = 2(1+b) 2(1+b) = 30 2(2x+x) = 30 2×3x = 30 6x = 30 x = 306 = 5 2x = 2×x = 2×5 = 10 1x = 1×4 = 1×5 = 5 AB = 10m; BC = 5cm CD = 10cm; DA = 5cm

2. In the adjacent rectangle ABCD, ∠OCD = 30°. Calculate ∠BOC. What type of triangle is BOC?

Solution:

ABCD is a rectangle

Diagonals AC and BD bisect each other at O and AC = BD.

AO = OC = BO = OD

OCD = ODC = 30°

On ΔCOD, ∠ODC + ∠OCD + ∠COD = 180˚

30° + 30° + ∠COD = 180°

60° + ∠COD = 180°

∴ ∠COD = 180˚ – 60˚ =120°

∠COD + ∠COB = 180˚

∠COB = 180˚ – 120˚

∠COB = 60°

∴ Δ BOC is an isosceles triangle

1. All rectangle are parallelograms, but all parallelograms are not rectangles. Justify the statements.

Solution:

All rectangles have all the properties of parallelograms but a parallelogram

May not have all the properties of a rectangle.

1. In a rectangle all the angles are right angle but in a parallelogram only opposite angles are equal.
2. In a rectangle the diagonals are equal but in a parallelogram diagonals are not equal.

1. Prove logically that the diagonals of a rectangle are equal.

Solution:

Data : ABCD is a rectangle.

AC and BD are the diagonals.

To Prove: AC = BD

Proof : In ΔABC and Δ ABD

∠ABC = ∠BAD [90°]

BC = AD [Opp. Sides]

AB = AB [Common side]

∴ ΔABC ≅ ΔABD [SAS]

∴ AC = BD [CPCT]

1. The sides of a rectangular park are in the ratio 4:3. If the area is 1728m2 , find the cost of fencing it at the rate of Rs.2.50/m.

Solution:

 ABCD is a rectangular park let AB : BC = 4 : 3

∴ AB = 4x and BC = 3x

Area of rectangle = 1×b

4x × 3x = 1728

12x2 = 1728

X2 = 144

X = √144 = 12

Length = 4x = 4×12 = 48m

Breadth = 3x = 3×12 = 36m

Perimeter of a rectangle = 2(l + b)

2(48 + 36) = 2×84 = 168m

Cost of fencing = perimeter ×Rate

= 168 ×2.50

= Rs.420

1. A rectangular yard contains two flower beds in the shape of congruent isosceles right triangle. The remaining portion is a yard of trapezoidal.

Shape (see fig) whose parallel sides have lengths 15 m and 25 m. What fraction of the yard is occupied by the flower bed?

Solution:

ABCD is a rectangular yard

AEFB is a trapezium

AB ║ EF AB = 25 m and EF = 15m

AD = BC = 5m

Area of rectangle = ½ × b = 25×5 = 125m² Area of each flower bed = ½ ×b×h = ½ ×5×5 = (25/2) m² Area of both flower beds = 2× (25/2) = 25m²

Fraction of flower beds to yard = 25/125 = 1/5

7. In a rhombus ABCD ∠C = 70°. Find the other angle of the rhombus.

solution:

∠C = 70°
But ∠A = ∠C = 70 ∴ ∠A = 70°

∠A + ∠B =180° (adjacent angles) 70 + ∠B = 180° ∠B = 180 − 70 = 110° ∠B = ∠D = 110° ∴ ∠A = 70°; ∠B = 110° ; ∠D = 110°

8. In a rhombus PQRS, ∠SQR = 40° and PQ = 3 cm. Find ∠SPQ, ∠QSR and the perimeter of the rhombus.

Solution:

∠SQR = 40° and PQ = 3cm

∠SQR = 40° and PQ = 3cm ∠PQS = ∠SQR = 40° ∠PQS = 40° (diagonal bisector angles ) 3 cm But ∠PQS = ∠QSP [PQ = PS] ∴ ∠QSP = 40° In Δ PQS, ∠PQS + ∠QSP + ∠SPQ = 180° 40° + 40° + ∠SPQ = 180 ∠SPQ = 180 −80 = 100° ∴ ∠SPQ = 100° In a rhombus PQ = QR = RS = SP = 3cm Perimeter of the Rhombus PQRS = 3×4 = 12cm

9. In a rhombus PQRS, if PQ = 3x -7 and QR = x + 3, find PS.

Solution: In a rhombus all sides are equal PQ = QR = RS = SP ∴ PQ = QR 3x -7 = x + 3 3x -x = 3 + 7 2x = 10 X = 5 PS = x+3 = 5+3 = 8 cm 10. Let ABCD be a rhombus and ∟ABC = 124°. Calculate ∟A, ∟D and ∟C.

In a rhombus, opposite angles are equal

Solution: ∠B = ∠D = 124° ∠A + ∠B =180°

Consecutive angles

∠A + 124° = 180°

∠A + 124° = 180° ∠A = 180-124=56° ∴ ∠A = 56° ; ∠B = 124° and ∠C = 56°

11. Rhombus is a parallelogram: justify.

Solution: Rhombus has all the properties of parallelogram i.e a) Opposite sides are equal and parallel. b) Opposite angles are equal. c) Diagonals bisect each other ∴ Rhombus is a parallelogram. 12. In a given square ABCD, if the area of triangle ABD is 36 cm2 , find (i) the area of triangle BCD;

(ii) the area of the square ABCD.

Solution:

13. The side of a square ABCD is 5 cm and another square PQRS has perimeter equal to 40cm. Find the ratio of the area ABCD to the area of PQRS.

Solution:
Perimeter of the square ABCD = 4 ×side

= 4×5 = 20 Ratio of (perimeter of ABCD)/(Perimeter of PQRS) = 20/40 = 1/2 or 1:2 Area of ABCD = (side)² = (5)² = 25cm Side of PQRS = perimeter/4 = 40/4 = 10cm Area of PQRS = (side)² = (10)² = 100cm²

Ratio of (Area of ABCD)/(Area of PQRS) = 25/100 = 1/4 or 1:4

14. A square field has side 20m. Find the length of the wire required to fence it four times.

Solution: Length of one side of the square = 20 Rs Length of wire required to fence one round = 4×20 Length of wire required to fence four rounds = 80m = 4×80m = 320m

15. List out the differences between square and rhombus.

Solution: Square 1. All the angles are equal 2. Diagonals are equal

3. Area = side × side=(s)²

Rhombus

1. Opposite angles are equal

2. Diagonals are unequal

3. Area= 1/2 × Product of diagonals = 1/2 x d­1 d2
16. Four congruent rectangles are placed as shown in the figure. Area of the outer square is 4 times that of the inner square. What is the ratio os length to breadth of the congruent rectangles?

Let the length of rectangles be ‘a’ and breadth be ‘b’ Side of outer square –(a+b) units Side of outer square (a-b) units Area of outer square = 4 times area of inner square Area of ABCD = 4(Area of PQRS)

(a + b)² = 4(a-b)

(a + b) = 2(a – b)

2a – 2b = (a + b)

2a – a = b + 2b

a = 3b

a/b = 1/3 or  1 : 3