MATHEMATICS IN EVERYDAY LIFE–8

MATHEMATICS IN EVERYDAY LIFE?8

Chapter 15 : Area of a Trapezium and a Polygon

ANSWER KEYS

EXERCISE 15.1

1. Side of the rhombus = 81 m Altitude = 25 m

81 m

altitude = 25 m

We know that,

Area of rhombus = side ? altitude

= (81 ? 25) m2

= 2025 m2

Area of square = Area of rhombus

Area of square = 2025 m2

(Side)2 = 2025 m2

[Given]

Side = 2025 m

Side = 45 m

Hence, side of the square is 45 m.

2. Let PQRS be the given rhombus whose diagonals intersect at O.

We have, PQ = 13 cm and QS = 10 cm

S

R

O

P

13 cm

Q

Since the diagonals of a rhombus bisect each other at right angles.

Therefore,

OQ =

1 QS 2

=

1 10 cm 2

= 5 cm

[ QS = 10 cm (given)]

Mathematics In Everyday Life-8

In right angled POQ,

By Pythagoras theorem,

PQ2 = PO2 + OQ2

132 = PO2 + 52

169 = PO2 + 25

PO2 = 169 ? 25

PO2 = 144

PO = 12

PR = 2 ? PO = 2 ? 12 cm = 24 cm

Now,

Area of rhombus = 1 ? product of diagonals 2

=

1 2

?

24

?

10

sq.

cm

= 120 sq.cm

3. Area of rhombus = 1 ? product of diagonals 2

=

1 2

?

12

?

8.5

sq.

cm

= 51 sq. cm

4. Area of rhombus = 1 ? product of diagonals 2

360 = 1 ? 40 ? other diagonal

2

Other diagonal = 360 ? 2 cm = 18 cm 40

Hence, the length of other diagonal is 18 cm.

5. Area of rhombus = 96 sq. m

Perimeter of rhombus = 32 m

4 ? Side = 32

Side = 8 m

Now,

Area of rhombus = side ? altitude

96 = 8 ? altitude

altitude =

96 8

= 12

The length of altitude is 12 m.

1

6. D

C

E

A

5 cm

5 cm

A 14 cm B

Area of square ABCD = (side)2 = (14)2 cm2 = 196 cm2

Now,

radius of semi-circle AED = 1 AD = 1 14 cm = 7 cm

2

2

Area of semi-circle AED = Area of semi circle BEC

= 1 (radius)2 2

= 1 22 72 cm2 = 77 cm2 27

Area of shaded region = Area of square ABCD ? Area of semi-circle AED ? Area of semi-circle BEC

= (196 ? 77 ? 77) cm2 = (196 ? 154) cm2 = 42 cm2

7. Let ABCD be a parallelogram having adjacent sides 12 cm and 8 cm respectively. Let AE and AF be the altitudes corresponding to the sides CD and BC.

A

B

6 cm

F 8 cm

B

D

C

6 cm

D = 90?, BD =

1 BC 2

=

1 6 2

cm = 3 cm

In right ADB,

AB2 = AD2 + BD2

[Pythagoras theorem]

AD2 = AB2 ? BD2

AD2 = 52 ? 32

AD2 = 25 ? 9 = 16

AD = 16 = 4

Area of triangle =

1 ? base ? height 2

= 1 ? BC ? AD 2

= 1 6 4 cm2 = 12 cm2. 2

9. Let ABCD be a quadrilateral. Its diagonal AC is 16 cm long and perpendiculars DE and BF drawn to it from the opposite vertices are 5.5 cm and 6.5 cm respectively.

D

D

E

12 cm

C

We have, BC = 8 cm, DC = 12 cm

Altitude AE = 6 cm

Area of parallelogram = base ? altitude

= 12 ? 6 cm2 = 72 cm2

Now, AF is an altitude along BC.

Area of parallelogram = base ? altitude

72 = 8 ? AF

AF = 9

Hence, length of the altitude corresponding is 9 cm.

8. Let ABC be an isosceles triangle in whic h AB = AC = 5 cm, BC = 6 cm. AD is the altitude of ABC.

Altitude of an isosceles triangle bisects its base.

A

E

F

C

B Area of quadrilateral = Area of ADC + Area of ABC

= 1 AC DE 1 AC BF

2

2

= 1 AC (DE BF) 2

= 1 ? 16 ? (5.5 6.5) 2

=

1 2

16

12

=

96

sq.

cm

Hence, the area of quadrilateral is 96 sq. cm.

2

Answer Keys

10. Length of rectangular wire = Perimeter of rectangle

= 2(length + breadth)

= 2(52 + 36) cm

= 2 ? 88 cm = 176 cm

It is bent in the form of a circle.

Circumference of circle = Perimeter of rectangle

2 ? radius = 176

2 ? 22 ? radius = 176

7

176 7 radius = 2 22 = 28 cm

Hence, the radius of the circle is 28 cm.

11. Let ABCD be a rectangular plot. AB = 110 m, BC = 70 m

It has a gravel path 3 m wide all around it on the side.

P

Q

A

B

D

C

S

R

It is shown by the shaded region.

PQRS is rectangular shape including rectangular plot ABCD and gravel path.

PQ = (110 + 3 + 3) m = 116 m,

QR = (70 + 3 + 3) m = 76 m

Area of gravel path = Area of PQRS ? Area of ABCD

= PQ ? QR ? AB ? BC

= [(116 ? 76) ? (110 ? 70)] m2

= (8816 ? 7700) m2 = 1116 m2

Cost of levelling 1 sq. m = `3.50

Cost of levelling 1116 sq. m = `(3.50 ? 1116)

= `3906

Hence, the total cost of levelling the path is `3906.

12. In the given figure, PQRS is a rectangle in which PQ = RS = 35 m and PS = QR = 80 m.

P

Q

80 m

R S 35 m The plot has a semi-circle on PQ. Mathematics In Everyday Life-8

Radius of semi-circle = 1 ? PQ 2

=

1 2

?

35

m

=

35 2

m

Area of semi-circle =

1 ? radius 2

=

1 22 35 35 27 2 2

sq. m

= 481.25 sq. m

Area of PQRS = PS ? SR

= 80 ? 35 sq. m = 2800 sq. m

Area of plot = Area of rectangle PQRS + Area of semi-circle

= (2800 + 481.25) sq. m

= 3281.25 sq. m

Hence, area of the plot is 3281.25 sq. m.

13. Let ABCD be a square plot having a 3 m wide path surrounding it. Now PQRS is a square including square plot ABCD and gravel path.

P

Q

A

B

D

C

S

R

Let AB = BC = CD = DA = x metre

Then, PQ = (x + 3 + 3) metre = (x + 6) metre

Now,

Area of square PQRS = Area of square ABCD + Area of gravel path

(PQ)2 = (AB)2 + 132 sq. m

(x + 6)2 = x2 + 132

x2 + 12x + 36 = x2 + 132

12x + 36 = 132

(Cancelling x2 from both sides)

12x = 96

x =8

Side of square plot ABCD = 8 m

Area of square plot ABCD = (8)2 m2

= 64 m2

14. Let ABCD be a rectangular piece of paper 25 cm long and 16 cm wide.

AB = CD = 25 cm

BC = DA = 16 cm

3

7 cm

A

B

16 cm

D

25 cm

C

From its four corners, quadrants of radii 7 cm have been cut.

Area of four quadrants = Area of circle of radius 7 cm

= ? (7)2

= ? 7 ? 7

= 22 ? 7 ? 7 = 154 sq. cm 7

Area of the remaining part = Area of rectangle ? Area of four quadrants

= (25 ? 16) ? 154 = 400 ? 154 = 246 sq. cm Hence, area of remaining part of paper is 246 sq. cm. 15. ABCD is a rectangle in which AB = CD = 18 cm BC = AD = 12 cm

A

E

B

12 cm

D

18 cm

C

Area of rectangle = length ? breadth = (18 ? 12) sq. cm = 216 sq. cm

Area of triangle DEC =

1 ? base ? height 2

= 1 ? CD ? BC 2

=

1 18 12 2

sq.

cm

= 108 sq. cm

Area of shaded region = Area of rectangle ABCD ? Area of triangle DEC

= (216 ? 108) sq. cm

= 108 sq. cm

16. ABCD is a quadrilateral in which AB CD and CD DA.

AB = 16 cm, CD = 10 cm, BC = 10 cm

We draw a line CE perpendicular to AB.

4

Now, ADCE is a rectangle, CD = AE = 10 cm BE = AB ? AE = (16 ? 10) = 6 cm

In CEB, E = 90?.

C

10 cm

D

10 cm

E

B

16 cm

A

Using Pythagoras theorem,

BC2 = CE2 + BE2

CE2 = BC2 ? BE2

CE2 = 102 ? 62

CE2 = 100 ? 36 = 64

CE = 64

CE = 8

Area of rectangle ADCE = EA ? CE = 10 ? 8 cm2 = 80 cm2

Area of CEB =

1 ? base ? height 2

=

1 BE CE 2

= 1 6 8 cm2 = 24 cm2 2

Area of quadrilateral ABCD = Area of rectangle

ADCE + Area of CEB = (80 + 24) cm2 = 104 cm2

EXERCISE 15.2

1.

Area of trapezium =

1 2

? sum of parallel sides ? height

= 1 50 11 sq. cm = 275 sq. cm 2

1 2. Area of trapezium = 2 ? sum of parallel sides ? height

= 1 (21 19) 13 sq. cm 2

= 1 40 13 sq. cm = 260 sq. cm 2

1 3. Area of trapezium = 2 ? sum of parallel sides ? height

450 =

1 (37 23) height 2

450 =

1 60 height 2

450 2

height =

= 15 cm

60

Hence, the distance between parallel sides is 15 cm.

Answer Keys

4.

Area of trapezium

=

1 2

? sum of parallel sides ? height

156

=

1 (9 x) 12 2

Let one of the parallel sides be x cm.

2 ? 156 = (9 + x) ? 12

312 = 108 + 12x

12x = 312 ? 108

12x = 204

x = 17

The length of other parallel side is 17 cm.

5. Area of trapezium = 1 ? sum of parallel sides ? height 2

= 1 (21.5 23.5) 14 cm2 2

= 1 45 14 cm2 = 315 cm2 2

6. Area of trapezium = 1 ? sum of parallel sides ? height 2

135 = 1 (19.8 7.2) height

2

270 = 27 ? height

height = 270

27

height = 10

Hence, the distance between the parallel sides is 10 cm.

7. Let the other side be x cm.

1 Area of trapezium = 2 ? sum of parallel sides ? height

441 =

1 (28 x) 18 2

882 = (28 + x) ? 18

28 + x =

882 18

= 49

x = 49 ? 28 = 21

Hence, the length of other parallel side is 21 cm.

8. Let one of the parallel sides be x cm.

Other parallel side = (x + 5) cm

1 Area of trapezium = 2 ? sum of parallel sides ? height

390 =

1 (x x 5) 12 2

390 = 6 ? (2x + 5)

390 = 12x + 30

12x = 390 ? 30 = 360

x = 30

x + 5 = 30 + 5 = 35

The length of two parallel sides are 30 cm and 35 cm respectively.

9. Let the parallel sides be 3x and 5x respectively.

1 Area of trapezium = 2 ? sum of parallel sides ? height

380 =

1 (3x 5x) 19 2

380 = 19 ? 8x

2

380 = 76x

x =

380 76

= 5

Two parallel sides are 3 ? 5 = 15 cm and 5 ? 5 = 25 cm respectively.

10. Let ABCD be a trapezium in which AB = 16 cm, CD = 40 cm. AD = BC = 13 cm

We draw BE CD and BF AD.

A

16 cm

B

13 cm

13 cm

D

F

E

C

40 cm

Now, ABFD is a parallelogram.

AB = DF = 16 cm

CF = CD ? DF = (40 ? 16) cm = 24 cm

Since, BE CD, and BFC being isosceles. (BF = BC)

FE

=

1 CF 2

=

1 24 2

cm = 12 cm

In right angled BEF,

BE2 = BF2 ? FE2

BE2 = 132 ? 122 = 169 ? 144 = 25

BE = 25

BE = 5 cm

1 Area of trapezium = 2 ? sum of parallel sides ? height

= 1 ( AB CD) BE 2

= 1 (16 40) 5 cm2 2

= 1 56 5 cm2 = 140 cm2 2

Hence, area of trapezium is 140 cm2.

Mathematics In Everyday Life-8

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