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.

Solution:

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:

Data: ∠A = ∠C and ∠D = ∠B To Prove: ABCD is a parallelogram Proof: ∠A + ∠B + ∠C + ∠D = 360° x + y + x + y = 360° 2x + 2y = 360° 2(x + y) = 360° x + y = 360°/2 = 180° Adjacent angles are supplementary ∴ ABCD is a parallelogram. 8. In a quadrilateral ABCD, suppose AB = CD and AD = BC. Prove that ABCD is a parallelogram.

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:

ABCD is a rhombus

∠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:

PQRS is a rhombus

∠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:

ABCD is a square Bd is the diagonal The diagonal divides the square into two congruent triangles ∴ Δ ABD ≅ Δ BCD D C ∴ Area of Δ ABD =area of Δ BCD Area of Δ ABD= 36 cm² (given) ∴ Area of Δ BCD = 36cm² Area of the square ABCD = Area of Δ ABD + Area of ΔBCD =36cm² +36cm² Area of the squre ABCD =72cm²

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?

Solution:

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