admissing values:
Base
Height
Area of Triangle
87 cm
15 cm
P
31.4 mm
1256 mm
7.6 cm
170.5 cm
A
22 cm
R
S
C С
&
B
M
PQRS is a parallelogram (Fig 11.23). QM is the height from Q
- SR and QN is the height from Q to PS. IF SR = 12 cm and
Fig 11.23
D
QM = 7.6 cm. Find:
(a) the area of the parallegram PQRS (b) QN, if PS = 8 cm M
DL and BM are the heights on sides AB and AD respectively of
parallelogram ABCD (Fig 11.24). If the area of the parallelogram
А
is 1470 cm. AB = 35 cm and AD= 49 cm, find the length of BM
L
and DL.
Fig 11.24
7. AABC 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 AABC. Also find the length of
AD.
A
7.5 cm
7.5 cm
12 cm
6 cm
5 cm
C С
B
D 9 cm
C С
B В
D 13 cm
Fig 11.26
Fig 11.25
som and BC = 9 cm (Fig 11.26). The height
from
Answers
Answer:
(a)
From the table,
Height of triangle =?
Base of triangle = 15 cm
Area of the triangle = 16.38 cm2
Then,
Area of triangle = ½ × base × height
87 = ½ × 15 × height
Height = (87 × 2)/15
Height = 174/15
Height = 11.6 cm
∴Height of the triangle is 11.6 cm.
(b)
From the table,
Height of triangle =31.4 mm
Base of triangle =?
Area of the triangle = 1256 mm2
Then,
Area of triangle = ½ × base × height
1256 = ½ × base × 31.4
Base = (1256 × 2)/31.4
Base = 2512/31.4
Base = 80 mm = 8 cm
∴Base of the triangle is 80 mm or 8 cm.
(c)
From the table,
Height of triangle =?
Base of triangle = 22 cm
Area of the triangle = 170.5 cm2
Then,
Area of triangle = ½ × base × height
170.5 = ½ × 22 × height
170.5 = 1 × 11 × height
Height = 170.5/11
Height = 15.5 cm
∴Height of the triangle is 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 parallelogram PQRS (b) QN, if PS = 8 cm
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Image 11
Fig 11.23
Solution:-
From the question it is given that,
SR = 12 cm, QM = 7.6 cm
(a) We know that,
Area of the parallelogram = base × height
= SR × QM
= 12 × 7.6
= 91.2 cm2
(b) Area of the parallelogram = base × height
91.2 = PS × QN
91.2 = 8 × QN
QN = 91.2/8
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 cm2, AB = 35 cm and AD = 49 cm, find the length of BM and DL.
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Image 12
Fig 11.24
Solution:-
From the question it is given that,
Area of the parallelogram = 1470 cm2
AB = 35 cm
AD = 49 cm
Then,
We know that,
Area of the parallelogram = base × height
1470 = AB × BM
1470 = 35 × DL
DL = 1470/35
DL = 42 cm
And,
Area of the parallelogram = base × height
1470 = AD × BM
1470 = 49 × BM
BM = 1470/49
BM = 30 cm
7. ΔABC 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 ΔABC. Also find the length of AD.
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Image 13
Fig 11.25
Solution:-
From the question it is given that,
AB = 5 cm, BC = 13 cm, AC = 12 cm
Then,
We know that,
Area of the ΔABC = ½ × base × height
= ½ × AB × AC
= ½ × 5 × 12
= 1 × 5 × 6
= 30 cm2
Now,
Area of ΔABC = ½ × base × height
30 = ½ × AD × BC
30 = ½ × AD × 13
(30 × 2)/13 = AD
AD = 60/13
AD = 4.6 cm
8. ΔABC 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 ΔABC. What will be the height from C to AB i.e., CE?
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Image 14
Solution:-
From the question it is given that,
AB = AC = 7.5 cm, BC = 9 cm, AD = 6cm
Then,
Area of ΔABC = ½ × base × height
= ½ × BC × AD
= ½ × 9 × 6
= 1 × 9 × 3
= 27 cm2
Now,
Area of ΔABC = ½ × base × height
27 = ½ × AB × CE
27 = ½ × 7.5 × CE
(27 × 2)/7.5 = CE
CE = 54/7.5
CE = 7.2 cm
Step-by-step explanation:
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