Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 11 Perimeter and Area Ex 11.2 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2

1. Estimate the area of the following figures by counting unit squares.

Question (i).

Solution:

In the given figure, number of squares covered completely = 135

Area of a square = 1 sq. unit

Area of (135 square) figure = 135 sq. units, (approx.)

Question (ii).

Solution:

In the given figure number of square covered completely = 114

Area of one square = 1 unit

∴ Area of 114 squares = 114 sq units approx

Thus area of given figure = 114 sq units approx.

2. In the following figures find the area of

Question (i).

ΔABC

Solution:

Given length of rectangle = 15 cm

Breadth of rectangle = 8 cm

The diagonal AC divides the rectangle into two triangles ΔABC and ΔADC

So, area of ΔABC = \(\frac {1}{2}\) × Area of rectangle ABCD

= \(\frac {1}{2}\) × length × breadth

= \(\frac {1}{2}\) × 15 × 8

= 60 cm^{2}

Question (ii).

ΔCOD

Solution:

Given side of square = 6 cm

The diagonals AC and BD divides the square into four equal posses (triangles)

So, area of ΔCOD = \(\frac {1}{4}\) × Area of square

= \(\frac {1}{4}\) × 6 × 6

= 9 cm^{2}

3. Find the area of following parallelograms.

Question (i).

Solution:

Given base of parallelogram = 9 cm

Height of parallelogram = 6 cm

Area of parallelogram = Base × height

= 9 × 6

= 54 cm^{2}

Question (ii).

Solution:

Given base of parallelogram = 6.5 cm

Height of parallelogram= 8.4 cm

Area of parallelogram = Base × height

= 6.5 × 8.4

= 54.6 cm^{2}

4. Find the value of x in the following parallelograms.

Question (i).

Solution:

Given base (AD) of parallelogram = 5.6 cm

Corresponding height of parallelogram = 9 cm

Area of parallelogram = 5.6 × 9 cm^{2} ….(1)

Also in the paralleogram, base (AB) = x

Corresponding height of parallelogram = 7 cm

Area of parallelogram will be = x × 7 ….(2)

From (1) and (2), we get

x × 7 = 5.6 × 9

x = \(\frac{5.6 \times 9}{7}\)

= 7.2

Question (ii).

Solution:

Given base (AB) of parallelogram = 15 cm

Corresponding height = 6 cm

Area of parallelogram =15 × 6 cm^{2} ….(1)

Also Base (AD || BC) of parallelogram = 9 cm

Corresponding height = x

So area of parallelogram = 9 × x ….(2)

From (1) and (2)

9 × x = 15 × 6

x = \(\frac{15 \times 6}{9}\)

= 10 cm.

5. The adjacent sides of a parallelogram are 28 cm and 45 cm and the altitude on longer side is 18 cm. Find the area of parallelogram.

Solution:

Given base of the parallelogram = 45 cm

Corresponding height = 18 cm

Area of parallelogram = Base × Height

= 45 × 18

= 810 cm2

6. ABCD is a parallelogram given in figure. DN and DM are the altitudes on side AB and CB respectively. If area of the parallelogram is 1225 cm^{2}, AB = 35 cm and CB = 25 cm, find DN and DM.

Solution:

In the given parallelogram ABCD

Base (AB) = 35 cm

Let height (DN) = x cm

So area of parallelogram = 35 × x cm^{2}

But given area of parallelogram (ABCD) = 1225 cm^{2}

Therefore 35x = 1225

x = \(\frac {1225}{35}\)

= 35 cm

Similarly, for base (BC) and height (DM)

1225 = BC × DM

\(\frac{1225}{\mathrm{BC}}\) = DM

or DM = \(\frac {1225}{25}\)

= 49 cm.

7. Find the area of the following triangles.

Question (i).

Solution:

Given base of triangle = 7 cm

Height of triangle = 4.8 cm

Area of triangle = \(\frac {1}{2}\) × Base × Height

= \(\frac {1}{2}\) × 7 × 4.8

= 16.8 cm^{2}.

Question (ii).

Solution:

Given base of triangle =6 cm

Height of triangle = 9 cm

Area of triangle = \(\frac {1}{2}\) × Base × Height

= \(\frac {1}{2}\) × 6 × 9

= 27 cm^{2}

8. Find the value of x in the following triangles.

Question (i).

Solution:

In ΔABC, BC = 8 cm, AC = 15 cm

Area of triangle ABC = \(\frac {1}{2}\) × Base × height

= \(\frac {1}{2}\) × BC × AC

= \(\frac {1}{2}\) × 8 × 15

= 60 cm^{2} …(1)

Also, in ΔABC, AB = 20 cm

height = x

Area of triangle ABC = \(\frac {1}{2}\) × Base × Height

= \(\frac {1}{2}\) × 20 × x ….(2)

From (1) and (2)

\(\frac {1}{2}\) × 20 × x = 60

x = \(\frac{60 \times 2}{20}\)

x = 6 cm.

Question (ii).

Solution:

In ΔABC, base (AC) = 25 cm

height = 14 cm

Area of triangle ABC = \(\frac {1}{2}\) × Base × height

\(\frac {1}{2}\) × 14 × 25 ….(1)

Also, in ΔABC, base AB = x cm

height = 20 cm

So, area of ΔABC = \(\frac {1}{2}\) × Base × Height

= \(\frac {1}{2}\) × x × 20 ….(2)

From (1) and (2) we get

\(\frac {1}{2}\) × x × 20 = \(\frac {1}{2}\) × 14 × 25

x = 17.5 cm

9. ABCD is a square, M is a point on AB such that AM = 9 cm and area of ΔDAM is 171 cm^{2}. What is the area of the square ?

Solution:

Given area of ΔDAM = 171 cm^{2}

Base of triangle = 9 cm

As, area of triangle ΔDAM = \(\frac {1}{2}\) × base × height

171 = \(\frac {1}{2}\) × 9 × (DA)

Hence height (DA) = \(\frac{171 \times 2}{9}\)

= 18 cm

Hence side of square (DA) = 18 cm

Therefore area of square = (side)^{2}

= (18)^{2}

= 324 cm^{2}

10. ΔABC is right angled at A as shown in figure. AD is perpendicular to BC, if AB = 9 cm, BC = 15 cm and AC = 12 cm. Find the area of ΔABC, also find file length of AD.

Solution:

Given AB = 9 cm

BC = 15 cm

AC = 12 cm

Let AD = x cm

Area of triangle = \(\frac {1}{2}\) × Base × height

= \(\frac {1}{2}\) × 12 × 9 cm^{2}.

= 54 cm^{2} ….(1)

Since, AD is perpendicular to BC

So, area of triangle = \(\frac {1}{2}\) × BC × AD

= \(\frac {1}{2}\) × 15 × AD ….(2)

From (1) and (2) we get

\(\frac {1}{2}\) × 15 × AD = 54

AD = \(\frac{54 \times 2}{15}\)

AD = 7.2 cm

11. ΔABC is isosceles with AB = AC = 9 cm, BC = 12 cm and the height AD from A to BC is 4.5 cm. Find the area of ΔABC. What will be the height from B to AC i.e. BN ?

Solution:

In triangle ABC, Base (BC) = 12 cm

AD = 4.5 cm

AD is perpendicular to BC

So, Area of ΔABC = \(\frac {1}{2}\) × base × height

= \(\frac {1}{2}\) × 12 × 4.5 cm

= 27 cm ….(1)

Also, in ΔABC, Base (AC) = 9 cm

Let corresponding height (BN) = x

So area of ΔABC = \(\frac {1}{2}\) × base × height

= \(\frac {1}{2}\) × 9 × BN ….(2)

From (1) and (2)

\(\frac {1}{2}\) × 9 × BN = 27

BN = \(\frac{27 \times 2}{9}\)

= 6 cm.

12. Multiple choice questions :

Question (i).

Find the height of a parallelogram whose area is 246 cm^{2} and base is 20 cm.

(a) 1.23 cm^{2}

(b) 13.2 cm^{2}

(c) 12.3 cm^{2}

(d) 1.32 cm^{2}

Answer:

(c) 12.3 cm^{2}

Question (ii).

One of the side and the corresponding height of a parallelogram are 7 cm and 3.5 cm respectively. Find the area of the parallelogram.

(a) 21 cm^{2}

(b) 24.5 cm^{2}

(c) 21.5 cm^{2}

(d) 24 cm^{2}

Answer:

(b) 24.5 cm^{2}

Question (iii).

The height of a triangle whose base is 13 cm and area is 65 cm^{2} is :

(a) 12 cm

(b) 15 cm

(c) 10 cm

(d) 20 cm

Answer:

(c) 10 cm

Question (iv).

Find the area of an isosceles right angled triangle, whose equal sides are of length 40 cm each.

(a) 400 cm^{2}

(b) 200 cm^{2}

(c) 600 cm^{2}

(d) 800 cm^{2}

Answer:

(d) 800 cm^{2}

Question (v).

If the sides of a parallelogram are increased to twice of its original length, how much will be the perimeter of the new parallelogram ?

(a) 1.5 times

(b) 2 times

(c) 3 times

(d) 4 times

Answer:

(b) 2 times

Question (vi).

In a right angled triangle one leg is double the other and area is 64 cm^{2} find the smaller leg.

(a) 8 cm

(b) 16 cm

(c) 24 cm

(d) 32 cm.

Answer:

(a) 8 cm