## NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area (EX 11.2) Exercise 11.2

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## Access NCERT Solutions for Class 7 Chapter 11 – Perimeter and Area

Exercise 11.2

1. Find the area of each of the following parallelograms:

# (a)

Ans: Height of parallelogram is \[4cm\] .

Base of the parallelogram is \[7cm\] .

Therefore,

Area of the parallelogram is \[base \times height\] .

The area of parallelogram is \[7cm \times 4\;cm\].

So, the area of parallelogram is \[28c{m^2}\] .

# (b)

Ans: Height of parallelogram is \[3cm\] .

Base of the parallelogram is \[5cm\] .

Therefore,

Area of the parallelogram is\[base \times height\] .

Area of parallelogram is \[5cm \times 3cm\]

Area of the parallelogram is \[15c{m^2}\] .

# (c)

Ans: Height of parallelogram is \[3.5cm\] .

Base of parallelogram is \[2.5cm\] .

Therefore,

Area of parallelogram is \[base \times height\]

Area of parallelogram is \[2.5cm \times 3.5cm\]

Area of parallelogram is \[8.75c{m^2}\]

# (d)

Ans: Height of parallelogram is \[4.8cm\].

Base of the parallelogram is \[5cm\].

Therefore,

Area of the parallelogram is \[base \times height\] .

Area of the parallelogram is \[4.8cm \times 5cm\] .

Area of the parallelogram is \[24c{m^2}\] .

# (e)

Ans: Height of parallelogram is \[4.4cm\] .

Base of the parallelogram is \[2cm\] .

Therefore,

Area of the parallelogram is \[base \times height\] .

Area of the parallelogram is \[4.4cm \times 2cm\] .

Area of the parallelogram is \[8.8c{m^2}\] .

2. Find the area of each of the following triangles:

# (a)

Ans: Base of the triangle is \[4cm\] .

Height of height is \[3cm\] .

Therefore,

Area of the triangle is \[\dfrac{1}{2} \times base \times height\]..

Area of the triangle is \[\dfrac{1}{2} \times 4cm \times 3cm\] .

Area of the triangle is \[6c{m^2}\].

# (b)

Ans: Base of triangle is \[5cm\] .

Height of height is \[3.2cm\] .

Therefore,

Area of the triangle is \[\dfrac{1}{2} \times base \times height\].

Area of the triangle is \[\dfrac{1}{2} \times 5cm \times 3.2cm\] .

Area of the triangle is \[8c{m^2}\] .

# (c)

Ans: Base of triangle is \[4cm\] .

Height of height is \[3cm\] .

Therefore,

Area of the triangle is \[\dfrac{1}{2} \times base \times height\].

Area of the triangle is \[\dfrac{1}{2} \times 4cm \times 3cm\] .

Area of the triangle is \[6c{m^2}\] .

# (d)

Ans: Base of triangle is \[3cm\] .

Height of height is \[2cm\] .

Therefore,

Area of the triangle is \[\dfrac{1}{2} \times base \times height\].

Area of the triangle is \[\dfrac{1}{2} \times 3cm \times 2cm\] .

Area of the triangle is \[3c{m^2}\] .

3. Find the missing values:

S.No. | Base | Height | Area of the Parallelogram |

a. | \[20cm\] | \[246c{m^2}\] | |

b. | \[15cm\] | \[154.5c{m^2}\] | |

c. | \[8.4cm\] | \[48.72c{m^2}\] | |

d. | \[15.6cm\] | \[16.38c{m^2}\] |

Ans (a) : Base of parallelogram is \[20cm\] .

Height of the parallelogram is to be known.

Area of the parallelogram is \[246c{m^2}\] .

Therefore,

\[Area = base \times height\]

\[246c{m^2} = 20cm \times height\]

\[\dfrac{{246c{m^2}}}{{20cm}} = height\]

\[12.3cm = height\]

Therefore, the height of the parallelogram is \[12.3cm\] .

Ans (b) : Base of parallelogram is to be known .

Height of the parallelogram is \[15cm\] .

Area of the parallelogram is \[154.5c{m^2}\] .

Therefore,

\[Area = base \times height\]

\[154.5c{m^2} = base \times 15cm\]

\[\dfrac{{154.5c{m^2}}}{{15cm}} = base\]

\[10.3cm = base\]

Therefore, the base of the parallelogram is \[10.3cm\] .

Ans (c) : Base of parallelogram is to be known .

Height of the parallelogram is \[8.4cm\] .

Area of the parallelogram is \[48.72c{m^2}\] .

Therefore,

\[Area = base \times height\]

\[48.4c{m^2} = base \times 8.4cm\]

\[\dfrac{{48.4c{m^2}}}{{8.4cm}} = base\]

\[5.8cm = base\]

Therefore, the base of the parallelogram is \[5.8cm\] .

Ans (d) : Base of parallelogram is \[15.6cm\] .

Height of the parallelogram is to be known.

Area of the parallelogram is \[16.38c{m^2}\] .

Therefore,

\[Area = base \times height\]

\[16.38c{m^2} = 15.6cm \times height\]

\[\dfrac{{16.38c{m^2}}}{{15.6cm}} = height\]

\[1.05cm = height\]

Therefore, the height of the parallelogram is \[1.05cm\] .

So, the final table is

S.No. | Base | Height | Area of the Parallelogram |

a. | \[20cm\] | \[12.3cm\] | \[246c{m^2}\] |

b. | \[10.3cm\] | \[15cm\] | \[154.5c{m^2}\] |

c. | \[5.8cm\] | \[8.4cm\] | \[48.72c{m^2}\] |

d. | \[15.6cm\] | \[1.05cm\] | \[16.38c{m^2}\] |

4. Find the missing values:

S.No. | Base | Height | Area of Triangle |

a. | \[15cm\] | \[87c{m^2}\] | |

b. | \[31.4mm\] | \[1256m{m^2}\] | |

c. | \[22cm\] | \[170.5c{m^2}\] |

Ans (a) : Base of triangle is \[15cm\] .

Height of the triangle is to be known.

Area of the triangle is \[87c{m^2}\] .

Therefore,

\[Area = \dfrac{1}{2} \times base \times height\]

\[87c{m^2} = \dfrac{1}{2} \times 15cm \times height\]

\[\dfrac{{87c{m^2} \times 2}}{{15cm}} = height\]

\[11.6cm = height\]

Therefore, the height of the triangle is \[11.6cm\] .

Ans (b) : Base of triangle is to be known.

Height of the triangle is \[31.4mm\].

Area of the triangle is \[1256m{m^2}\] .

Therefore,

\[Area = \dfrac{1}{2} \times base \times height\]

\[1256m{m^2} = \dfrac{1}{2} \times base \times 31.4mm\]

\[\dfrac{{1256m{m^2} \times 2}}{{31.4m}} = base\]

\[80mm = base\]

Therefore, the base of the triangle is \[80mm\] .

Ans (c) : Base of triangle is \[22cm\] .

Height of the triangle is to be known.

Area of the triangle is \[170.5c{m^2}\] .

Therefore,

\[Area = \dfrac{1}{2} \times base \times height\]

\[170.5c{m^2} = \dfrac{1}{2} \times 22cm \times height\]

\[\dfrac{{170.5c{m^2} \times 2}}{{22cm}} = height\]

\[15.5cm = height\]

Therefore, the height of the triangle is \[15.5cm\] .

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.

Fig \[11.3\]

Find:

(a) The area of the parallelogram PQRS

Ans: Base of parallelogram is \[12cm\] .

Height of the parallelogram is \[7.6cm\] .

Therefore,

Area of parallelogram is \[base \times height\]

Area of the parallelogram is \[SR \times QM\] .

The area of parallelogram \[12cm \times 7.6cm\;\]

So, the area of parallelogram is \[91.2c{m^2}\].

(b) QN, if PS = 8 cm

Ans: The area of parallelogram is \[91.2c{m^2}\] .

Base of the parallelogram is \[8cm\] .

Height of the parallelogram is to be known.

Area of the parallelogram is \[base \times height\] .

\[Area = base \times height\]

\[Area = PS \times QN\]

\[91.2c{m^2} = 8cm \times QN\]

\[QN = \dfrac{{91.2c{m^2}}}{{8cm}}\]

\[QN = 11.4cm\].

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 \[1470c{m^2}\], \[AB = 35cm\]cm and \[AD = 49cm\], find the length of BM and DL.

Ans: Case 1 : Finding the value of DL.

Area of parallelogram is ABCD \[1470c{m^2}\] .

Base of parallelogram is ,

\[AB = 35cm\]

Height of parallelogram is,

\[DL = ?\]

\[Area = base \times height\]

\[Area = AB \times DL\]

\[1470c{m^2} = 35cm \times DL\]

\[DL = \dfrac{{1470c{m^2}}}{{35cm}}\]

\[DL = 42cm\]

Therefore, the value of DL is \[42cm\] .

Case 2: Finding the value BM.

Case 1 : Finding the value of DL.

Area of parallelogram is ABCD \[1470c{m^2}\] .

Base of parallelogram is ,

\[AD = 49cm\]

Height of parallelogram is,

\[BM = ?\]

\[Area = base \times height\]

\[Area = AD \times BM\]

\[1470c{m^2} = 49cm \times BM\]

\[BM = \dfrac{{1470c{m^2}}}{{49cm}}\]

\[BM = 30cm\]

Therefore, the value of DL is \[30cm\] .

7. \[\Delta ABC\] is right angled at A (Fig 11.25). AD is perpendicular to BC. If\[AB = 5cm\], \[BC = 13cm\] cm and \[AC = 12cm\], Find the area of \[\Delta ABC\]. Also find the length of AD.

Ans: Finding the area of \[\Delta ABC\] .

Base of \[\Delta ABC\] is \[5cm\] .

Height of \[\Delta ABC\] is \[12cm\] .

Therefore,

\[Area = \dfrac{1}{2} \times base \times height\]

\[Area = \dfrac{1}{2} \times AB \times AC\]

\[Area = \dfrac{1}{2} \times 5cm \times 12cm\]

\[Area = 30c{m^2}\]

Therefore, the area of \[\Delta ABC\] is \[30c{m^2}\] .

Finding the value of AD.

Area of \[\Delta ABC\] is \[30c{m^2}\]

Base of \[\Delta ABC\] is \[13cm\] .

Height of \[\Delta ABC\] is to be known.

Therefore,

\[Area = \dfrac{1}{2} \times base \times height\]

\[Area = \dfrac{1}{2} \times BC \times AD\]

\[30c{m^2} = \dfrac{1}{2} \times 13 \times AD\]

\[\dfrac{{30c{m^2} \times 2}}{{13cm}} = AD\]

\[AD = 4.6cm\]

Therefore, the value of AD is \[4.6cm\] .

8. \[\Delta ABC\] 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?

Ans: Finding the area of \[\Delta ABC\] .

Base of \[\Delta ABC\] is \[9cm\] .

Height of \[\Delta ABC\] is \[6cm\] .

Therefore,

\[Area = \dfrac{1}{2} \times base \times height\]

\[Area = \dfrac{1}{2} \times AD \times BC\]

\[Area = \dfrac{1}{2} \times 9cm \times 6cm\]

\[Area = 27c{m^2}\]

Therefore, the area of \[\Delta ABC\] is \[27c{m^2}\] .

Find the value of CE.

Area of \[\Delta ABC\] is \[27c{m^2}\]

Base of \[\Delta ABC\] is \[7.5cm\] .

Height of \[\Delta ABC\] is to be known.

Therefore,

\[Area = \dfrac{1}{2} \times base \times height\]

\[Area = \dfrac{1}{2} \times BA \times CE\]

\[27c{m^2} = \dfrac{1}{2} \times 7.5cm \times CE\]

\[\dfrac{{27c{m^2} \times 2}}{{7.5cm}} = CE\]

\[CE = 7.2cm\]

Therefore, the value of CE is \[7.2cm\].

## NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.2

Opting for the NCERT solutions for Ex 11.2 Class 7 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 11.2 Class 7 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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