Haryana State Board HBSE 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.2 Textbook Exercise Questions and Answers.

## Haryana Board 7th Class Maths Solutions Chapter 11 Perimeter and Area Exercise 11.2

Question 1.

Find the area of each of the following parallelograms :

Solution:

(a) Length of base (6) = 7 cm

height (h) = 4 cm

∴ Area of the parallelogram

=6 x A = (7 x 4) cm^{2} = 28 cm^{2}.

(b) The length of base = 5 cm height (h) = 3 cm

∴ Area of the parallelogram

= b x h = (5 x 3) cm^{2} = 15 cm^{2}.

(c) The length of base = 2.5 cm

height (h) = 3.5 cm

∴ Area of the parallelogram

=b x h = (2.5 x 3.5) cm^{2} = 8.75 cm^{2}.

(d) The length of base (6) = 5 cm

height (h) = 4.8 cm

∴ Area of the parallelogram

=b x h = (5 x 4.8) cm^{2} = 24.0 cm^{2}.

(e) The length of base (b) = 2 cm

height (h) = 4.4 cm

∴ Area of the parallelogram

=b x h = (2 x 4.4) cm^{2} = 24.0 cm^{2}.

Question 2.

Find the area of each of the following triangles:

Solution:

(a) The length of base (b) = 4 cm

height (h) = 3

∴ Area of triangle = \(\frac{1}{2}\) x b x h = \(\frac{1}{2}\) x 3 x 4 = \(\frac{12}{2}\) = 6 cm^{2}.

(b) base (b) = 4 cm

height (h) = 3

∴ Area of triangle = \(\frac{1}{2}\) x b x h = \(\frac{1}{2}\) x 3 x 4 = \(\frac{16}{2}\) = 8 cm^{2}.

(c) base (b) = 3 cm

height (h) = 4

∴ Area of triangle = \(\frac{1}{2}\) x b x h = \(\frac{1}{2}\) x 3 x 4 = \(\frac{12}{2}\) = 6 cm^{2}.

(d) base (b) = 3 cm

height (h) = 2

∴ Area of triangle = \(\frac{1}{2}\) x b x h = \(\frac{1}{2}\) x 3 x 2 = \(\frac{6}{2}\) = 3 cm^{2}.

Question 3.

Find the missing values :

Base | Height | Area of the parallelogram |

(a) 20 cm (b) …………. (c) …………. (d) 15.6cm |
…………. 15 cm 8.4 cm …………. |
246 cm^{2
}154.5 cm^{2
}48.72 cm^{2
}16.38 cm^{2} |

Solution:

(a) Base = 20 cm,

Area of the parallellogram = 246 cm^{2}

Height = \(\frac{\text { Area }}{\text { Base }}=\frac{246}{20}\) cm

= 12.3 cm

∴ The missing value (height) = 12.3 cm^{2}

(b) Height = 15 cm,

∴ Area of the parallelogram = 154.5 cm^{2}

Area of the parallelogram = b x h

⇒ 154.5 = b x 15

⇒ b = \(\frac{154.5}{15}\) = 10.3 cm

∴ The missing value (base)

= 10.3 cm

(c) Height = 8.4 cm,

Area of the parallelogram

= 48.72 cm^{2}

Let the base of the parallelogram = b cm

∴ b x h = 48.72

⇒ b x 8.4 = 48.72

b = \(\frac{48.72}{8.4}\) = 5.8 cm

Thus, the missing value (base) = 5.8 cm

(d) Base (6) = 15.6 cm,

Area of the parallellogram

= 16.38 cm^{2}

Let the height be = h cm

b x h = 16.38 cm^{2}

⇒ 15.6 x h = 16.38 cm^{2}

⇒ h = \(\frac{16.38 \mathrm{~cm}^{2}}{15.6 \mathrm{~cm}}\) = 1.05 cm.

Thus the missing value (height) = 1.05 cm

Question 4.

Find the missing value.

Base | Height | Area of the triangle |

(a) 15 cm (b) …………. (b) 22 cm |
…………. 31.4 mm …………. |
87 cm^{2
}1256 mm^{2
}170.5 cm^{2} |

(i) Here,

Base(b) = 15 cm

Area of Triangle = 87cm^{2}

Let Height = h

∴ Area of Triangle = \(\frac{1}{2}\) x b x h

⇒ 87 = \(\frac{1}{2}\) x 15 x h

⇒ 87 x 2 = 15h

⇒ \(\frac{87 \times 2}{15}\) = h

⇒ h = \(\frac{174}{15}\)

= 11.6 cm.

The missing value (height) = 11.6 cm

(ii) Here,

Height (h) = 31.4 mm

Area of Triangle = 1256 mm^{2}

Let, Base = b mm

∴ \(\frac{1}{2}\) x b x h = 1256

⇒ \(\frac{1}{2}\) x 6 x 31.4 = 1256

⇒ b = \(\frac{2 \times 1256}{31.4}\)

∴ b = 80 mm

The missing value Base (b)

= 80 mm.

(iii) Here,

Base (b) = 22 cm

Area of Triangle = 170.5 cm2

Let, Height = h cm

∴ \(\frac{1}{2}\) x b x h = 170.5 cm^{2}

⇒ \(\frac{1}{2}\) x 22 x h = 170.5 cm^{2}

⇒ h = \(\frac{2 \times 170.5}{22}\) cm^{2}

= 15.5 cm.

Question 5.

PQRS is a parallelogram. 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) QM (height) = 7.6 cm

SR (base) = 12 cm

∴ Area of parallelogram PQRS

= b x h = (12 x 7.6) cm^{2} = 91.2 cm^{2}.

(b) QN (height) = ?

PS (base) = 8 cm.

and area of parallelogram PQRS = 91.2 cm^{2}

∴ Area of parallelogram PQRS = base x height

91.2 cm^{2} = 8 cm x h

or h = \(\frac{91.2 \mathrm{~cm}^{2}}{8 \mathrm{~cm}}\)

(QN)height = 11.4 cm.

Question 6.

DL and BM are the heights on sides AB & AD respectively of parallelogram ABCD. If the area of parallelogram is 1470

cm^{2} and AB = 35 cm and AD = 49 cm. Find the length of BM and DL.

Solution:

In parallelogram ABCD,

AB (base) = 35 cm

∴ Area of parallelogram = 1470 cm^{2}

DL (height) = ?

Area of parallelogram = b x h

h = \(\frac{1470}{35}\)

or, height (DL) = 42 cm

And AD(base) = 49 cm

BM (height) = ?

Area of parallelogram = 1470 cm^{2}

∴ Area of parallelogram = b x h

1470 = 49 x h

or, h = \(\frac{1470}{49}\)

or, height (BM) = 30 cm.

Question 7.

ΔABC is right angle at A. AD is per-pendicular BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm. Find the area of AABC. Also find the length of AD.

Solution:

In a right angled AABC,

base = 12 cm,

height = 5 cm

∴ Area of a right angled triangle = \(\frac{1}{2}\) x b x h

= \(\frac{1}{2}\) x 12 x 5

= 30 cm^{2}.

If base = 13 cm,

height (AD) = ?

Area = \(\frac{1}{2}\) xb x h = \(\frac{1}{2}\) x 13 x AD

or 30 cm^{2} = \(\frac{13 \mathrm{AD}}{2}\)

or 13 AD = 30 x 2 = 60 cm^{2}

∴ AD = \(\frac{60 \mathrm{~cm}^{2}}{13 \mathrm{~cm}}\) = 4.62cm

And, Second method : From Hero’s formula,

Question 8.

AABC is isosceles with AB = AC =

7,5 cm, and BC = 9 cm.

The height AD from A to BC is 6 cm. Find the area of ∆ABC. What will be the height from C to AB 8 i.e., CE.

Solution:

In a ∆ABC, base = 9 cm

height = 6 cm

∴ Area of AABC = \(\frac{1}{2}\) x b x h

= \(\frac{1}{2}\) x 9 x 6 = 27 cm^{2}.

Now, In a ∆ABC, if base = 7.5 cm and

height = ?

∴ Area of a ABC = \(\frac{1}{2}\) x base x height

27cm^{2} = \(\frac{1}{2}\) x 7.5 cm CE

or 7.5 cm x CE = 27 cm^{2} x 2

or CE = \(\frac{54 \mathrm{~cm}^{2}}{7.5 \mathrm{~cm}}\)

∴ height, CE = 7.2