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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