Solution of NCERT Exercise 11.2-part2
Perimeter and Area 7th Math
Solution of NCERT Exercise 11.2-part2
Question (4) Find the missing values:
| Base | Height | Area of Triangle |
|---|---|---|
| 15 cm | – | 87 cm2 |
| – | 31.4 mm | 1256 mm2 |
| 22 cm | – | 170.5 cm2 |
Solution:
(a) Given,
Base of the triangle = 15 cm
Area of the triangle = 87 cm2
Thus, Height of the triangle = ?
Let height of the triangle = h
We know that,
Area of a triangle = `1/2` × Base × Height
⇒ 87 cm2 = `1/2` × 15 cm × h
⇒ `1/2` × 15 cm × h = 87 cm2
⇒ 15 cm × h = 87 cm2 × 2
⇒ h `=(87xx2 cm^2)/(15 cm)`
⇒ h = 11.6 cm
Thus, height of triangle = Missing value = 11.6 cm Answer
(b) Given,
Height of the triangle = 31.4 mm
Area of the triangle = 1256 mm2
Thus, Base of the triangle = ?
Let base of the triangle = b
We know that,
Area of a triangle = `1/2` × Base × Height
⇒ 1256 mm2 = `1/2` × b × 31.4 mm
⇒ `1/2` × b × 31.5 mm = 1256 mm2
⇒ b × 31.5 mm = 1256 mm2 × 2
⇒ b `=(1256 mm^2xx2)/(31.5 mm)`
⇒ b = 80 mm
Thus, base of triangle = Missing value = 80 mm Answer
(c) Given,
Base of the triangle = 22 cm
Area of the triangle = 1705 cm2
Thus, Height of the triangle = ?
Let height of the triangle = h
We know that,
Area of a triangle = `1/2` × Base × Height
⇒ 170.5 cm2 = `1/2` × 22 cm × h
⇒ 170.5 cm2 = 11 cm × h
⇒ 11 cm × h = 170.5 cm2
⇒ h `=(170.5 cm^2xx2)/(11 cm)`
⇒ h = 15.5 cm
Thus, height of triangle = Missing value = 15.5 cm Answer
Question (5) PQRS is a parallelogram (figure). 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
Solution:
(a) Given PQRS is a parallelogram, in which
SR = Base of parallelogram = 12 cm
And, QM = height of the parallelogram = 7.6 cm
Therefore, Area of the parallelogram = ?
We know that,
Area of a parallelogram = Base × Height
∴ Area of the given parallelogram, PQRS = SR × QM
= 12 cm × 7.6 cm
= 91.2 cm2
Thus, Area of the parallelogram, PQRS = 91.2 cm2 Answer
(b) Given,
PS = base of the parallelogram = 8 cm
∴ QN = height of the parallelogram = ?
Area of the parallelogram = 91.2 cm2 [As calculated above in (a)]
Now, We know that,
Area of a parallelogram = Base × Height
⇒ 91.2 cm2 = 8 cm × QN
⇒ 8 cm × QN = 91.2 cm2
⇒ QN `=(91.2 cm^2)/(8 cm)`
= 11.4 cm
Thus, QN = Height of the parallelogram = 11.4 cm Answer
Question (6) DL and BM are heights on sides AB and AD, respectively of parallelogram ABCD (figure). If the area of the parallelogram is 1470 cm2, AB = 35 cm and AD = 49 cm, find the length of BM and DL.
Solution:
Given, Area of parallelogram = 1470 cm2
Base, AB = 35 cm
& Base, AD = 49 cm
Therefore, Length of BM and DL =?
We know that,
Area of parallelogram = Base × Height
When, AB = 35 cm = Base, and DL = Height
Thus, Area of parallelogram, ABCD = Base (AB) × Height (DL)
⇒ 1470 cm2 = 35 cm × DL
⇒ 35 cm × DL = 1470 cm2
⇒ DL = `(1470 cm^2)/(35 cm)`
⇒ DL = 42 cm
Now, When, AD = 49 cm = Base, and BM = Height
Thus, Area of parallelogram, ABCD = Base (AD) × Height (BM)
⇒ 1470 cm2 = 49 cm × DM
⇒ 49 cm × DM = 1470 cm2
⇒ DM = `(1470 cm2)/(49 cm)`
⇒ DM = 30 cm
Thus, length of DL = 42 cm and DM = 30 cm Answer
Question (7) `triangle` ABC is right angled at A (Figure). Ad is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm. Find the area of `triangle`ABC. Also find the length of AD.
Solution:
Given,
ABC is a right angled triangle, in which
`/_A = 90^o`,
AB = Base = 5 cm,
AC = Height = 12 cm,
BC = Hypotenuse = 13 cm,
And, AD `_|_` BC
Therefore, Area of triangle ABC = ?
And length of AD = ?
We know that, Area of triangle ABC = `1/2` × Base × Height
= `1/2` × AB × AC
= `1/2` × 5 cm × 12 cm
= `1/2` × 60 cm2
⇒ Area of Triangle ABC = 30 cm2
Now, Let BC = Base = 13 cm
Therefore, Area of triangle ABC = `1/2` × Base (BC) × Height (AD)
⇒ 30 cm2= `1/2` × 13 cm × AD
⇒ `1/2` × 13 cm × AD = 30 cm2
⇒ 13 cm × AD = 30 cm2 × 2
⇒ AD = `(30 cm^2xx2)/(13 cm)`
⇒ AD = 4.615 cm
Thus, Area of ABC = 30 cm2 and Height of AD = 4.615 cm Answer
Question (8) `triangle` ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm (Figure). The height AD from A to BC, is 6 cm. Find the area of `triangle` ABC. What will be the height from C to AB i.e., CE?
Solution:
Given, ABC is an isosceles triangle, in which
AB = AC = 7.5 cm
BC = 9 cm
AD `_|_` BC = 6 cm
Therefore, Area of triangle ABC = ?
And, length of CE =?
We know that, Area of a triangle = `1/2` × Base × Height
Now, when BC = base = 9 cm and AD = Height = 6 cm
Thus, Area of triangle ABC = `1/2` × Base (BC) × Height (AD)
= `1/2` × 9 cm × 6 cm
= `1/2` × 54 cm2
= 27 cm2
Now, When AB = Base = 7.5 cm, and CE = Height
Thus, Area of triangle ABC = `1/2` × Base (AB) × Height (CE)
⇒ 27 cm2 = `1/2` × 7.5 cm × CE
⇒ `1/2` × 7.5 cm × CE = 27 cm2
⇒ 7.5 cm× CE = 27 cm2 × 2
⇒ CE = `(27 cm^2xx2)/(7.5 cm)`
⇒ CE = `(54 cm^2)/(7.5 cm)`
⇒ CE = 7.2 cm
Thus, Area of given triangle ABC = 27 cm2 and height of CE = 7.2 cm Answer
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