Perimeter and Area
Exercise 11.2
1). Find the area of each of the following parallelograms:
a) Given: base = 7 cm
height = 4 cm
To find: Area of parallelogram
Area of parallelogram = base X height
= 7 X 4
= 28 Sq cm
b) Given: base = 5 cm
height = 3 cm
To find: Area of parallelogram
Area of parallelogram = base X height
= 5 X 3
= 15 Sq cm
c) Given: base = 2.5 cm
height = 3.5 cm
To find: Area of parallelogram
Area of parallelogram = base X height
= 2.5 X 3.5
= 8.75 Sq cm
d) Given: base = 5 cm
height = 4.8 cm
To find: Area of parallelogram
Area of parallelogram = base X height
= 5 X 4.8
= 24 Sq cm
e) Given: base = 2 cm
height = 4.2 cm
To find: Area of parallelogram
Area of parallelogram = base X height
= 2 X 4.2
= 8.4 Sq cm
2). Find the area of each of the following triangles:
a) Given: base = 4 cm
height = 3 cm
To find: Area of triangle
Area of triangle = ½ X base X height
= ½ X 4 X 4
= 8 sq cm
b) Given: base = 5 cm
height = 3.2 cm
To find: Area of triangle
Area of triangle = ½ X base X height
= ½ X 5 X 3.2
= 8.0 sq cm
c) Given: base = 3 cm
height = 4 cm
To find: Area of triangle
Area of triangle = ½ X base X height
= ½ X 3 X 4
= 6 sq cm
d) Given: base = 3 cm
height = 2 cm
To find: Area of triangle
Area of triangle = ½ X base X height
= ½ X 3 X 2
= 3 sq cm
3). Find the missing values:
Sr | Base | Height | Area of the parallelogram |
a | 20 cm | 246 cm^{2} | |
b | 15 cm | 154.5 cm^{2} | |
c | 8.4 cm | 48.72 cm^{2} | |
d | 15.6 cm | 16.38 cm^{2} |
Answer:
Sr | Base | Height | Area of the parallelogram |
a | 20 cm | 12.3 cm | 246 cm^{2} |
b | 10.3 cm | 15 cm | 154.5 cm^{2} |
c | 5.8 cm | 8.4 cm | 48.72 cm^{2} |
d | 15.6 cm | 1.05 cm | 16.38 cm^{2} |
a) Given: base = 20 cm
Area of parallelogram = 246 cm^{2}
To find: height of parallelogram
Area of parallelogram = base X height
246 = 20 X height
20 X height = 246
height = 246 ÷ 20
height = 12.3 cm
b) Given: height = 15 cm
Area of parallelogram = 154.5 cm^{2}
To find: base of parallelogram
Area of parallelogram = base X height
154.5 = base X 15
base X 15 = 154.5
base = 154.5 ÷ 15
base = 10.3 cm
c) Given: height = 8.4 cm
Area of parallelogram = 48.72 cm^{2}
To find: base of parallelogram
Area of parallelogram = base X height
48.72 = base X 8.4
base X 8.4 = 48.72
base = 48.72 ÷ 8.4
base = 5.8 cm
d) Given: base = 15.6 cm
Area of parallelogram = 16.38 cm^{2}
To find: height of parallelogram
Area of parallelogram = base X height
16.38 = 15.6 X height
15.6 X height = 16.38
height = 16.38 ÷ 15.6
height = 1.05 cm
4). Find the missing values:
Sr | Base | Height | Area of triangle |
a | 15 cm | 87 cm^{2} | |
b | 31.4 mm | 1256 mm^{2} | |
c | 22 cm | 8.4 cm | 170.5 cm^{2} |
Answer:
Sr | Base | Height | Area of triangle |
a | 15 cm | 11.6 cm | 87 cm^{2} |
b | 80 cm | 31.4 mm | 1256 mm^{2} |
c | 22 cm | 15.5 cm | 170.5 cm^{2} |
a) Given: base = 15 cm
Area of triangle = 87 cm^{2}
To find: height of triangle
Area of triangle = ½ X base X height
87 = ½ X 15 X height
½ X 15 X height = 87
height = (87 X 2 )÷ 15 = 174 ÷ 15
height = 11.6 cm
b) Given: height = 31.4 mm
Area of triangle = 1256 mm^{2}
To find: base of triangle
Area of triangle = ½ X base X height
1256 = ½ X base X 31.4
½ X base X 31.4 = 1256
base = (1256 X 2) ÷ 31.4
base = 80 mm
c) Given: height = 22 cm
Area of triangle = 170.5 cm^{2}
To find: height of triangle
Area of triangle = ½ X base X height
170.5 = ½ X 22 X height
½ X height X 22 = 170.5
height = (170.5 X 2) ÷ 22
height = 15.5 cm
5). PQRS is a parallelogram (Fig 11.23). QM is the height from Q to SR and QN is the height from Q to PS. If SR = 12 cm and QM = 7.6 cm. Find:
(a) the area of the parallegram PQRS
(b) QN, if PS = 8 cm
i) Given: SR = 12 cm
QM = 7.6 cm
To find: Area of parallelogram
Area of parallelogram = base X height
= 12 X 7.6
= 91.2 sq cm
ii) PS = 7.6 cm
To find: QN
Area of parallelogram = base X height
= PS X QN
91.2 = 8 X QN
QN X 8 = 91.2
QN = 91.2 ÷ 8
= 11.4 cm
Ans: area of parallelogram = 91.2 sq cm
QN = 11.4 cm
6). DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD (Fig 11.24). If the area of the parallelogram is 1470 cm^{2}, AB = 35 cm and AD = 49 cm, find the length of BM and DL.
Given: □ABCD is a parallelogram
AB = 35 cm
AD = 49 cm
Area of □ABCD = 1470 cm^{2}
To find: BM =?
DL =?
Area of parallelogram = base X height
1470 = AB X DL
AB X DL = 1470
35 X DL = 1470
DL = 1470 ÷ 35
= 42 cm
Area of parallelogram = base X height
1470 = AD X BM
AD X BM = 1470
49 X BM = 1470
BM = 1470 ÷ 49
= 30 cm
Ans: BM = 30 cm
DL = 42 cm
7). DABC is right angled at A (Fig 11.25). AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, Find the area of DABC. Also find the length of AD.
Given: DABC is right angled at A
AB = 5 cm
BC = 13 cm
AC = 12 cm
To find: AD = ?
A(DABC) = ½ X base X height
= ½ X AB X AC
= ½ X 5 X 12
= 5 X 6
= 30 sq cm
A(DABC) = ½ X base X height
= ½ X BC X AD
30 = ½ X 13 X AD
½ X 13 X AD = 30
AD = (30 X 2) ÷ 13
= 60 ÷ 13
= 4.61 cm
8). DABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm (Fig 11.26). The height AD from A to BC, is 6 cm. Find the area of DABC. What will be the height from C to AB i.e., CE?
Given: DABC is isosceles triangle
AB = AC =7.5 cm
BC = 9 cm
AD = 6 cm
To find: A(DABC) = ?
CE = ?
A(DABC) = ½ X base X height
= ½ X AD X BC
= ½ X 6 X 9
= 3 X 6
= 27 sq cm
A(DABC) = ½ X base X height
= ½ X AB X CE
27 = ½ X 7.5 X CE
½ X 7.5 X CE = 27
CE = (27 X 2) ÷ 7.5
= 54 ÷ 7.5
= 7.2 cm
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