Chapter 9: Perimeter and Area

Solution:-

(a) 2 : 1 : 3

We know that, area of triangle = ½ × base × height

So, area of triangle MNO = ½ × NO × MO

= ½ × 5 × 4

= ½ × 20

= 10 cm²

Area of triangle MOP = ½ × MO × OP

= ½ × 5 × 2

= ½ × 10

= 5 cm²

Area of triangle MPQ = ½ × MO × PQ [MP = MO]

= ½ × 5 × 6

= ½ × 30

= 15 cm²

So, the ratios of area = 10: 5: 15 … [divide each by 5]

Then we get, 2: 1: 3

Solution:-

(c) 40 cm²

From the figure,

Consider the parallelogram EFGH,

EF = HG = 10 cm … [from the question]

We know that, area of parallelogram EFGH = Base × corresponding height

= 10 × 4

= 40 cm²

Solution:-

(c) circumference/diameter

We know that, circumference of circle = 2πr

Then, π = circumference/2r

Therefore, π = circumference/diameter … [2r = diameter]

Solution:-

(a) more than three times of its diameter

Solution:-

(a) 20 cm

We know that, area of triangle = ½ × base × height

From the question, it is given that, area of triangle PQR = 100 cm²

Height of the triangle = 10 cm = altitude

Therefore, area of triangle = ½ × base × height

100 = ½ × PR × 10

PR = (100 × 2)/10

PR = 200/10

PR = 20 cm

Solution:-

(c) 4 cm

From the question it is given that, PR = 12 cm, QR = 6 cm, PL = 8 cm

We know that, by the Pythagoras theorem

Hypotenuse² = perpendicular² + base²

In the given figure,

PR² = PL² + LR²

LR² = PR² – PL²

LR² = 12² – 8²

LR² = 144 – 64

LR² = 80

By transferring the square it becomes,

LR = √80

Therefore, LR = 4√5

Then,

LR = LQ + QR

LQ = LR – QR

LQ = 4√5 – 6

We know that, area of triangle PLR = ½ × LR × PL

= ½ × 4√5 × 8

= 1× 4√5 × 4

= 16√5

Now, area of triangle PLQ = ½ × LQ × PL

= ½ × (4√5 – 6) × 8

= 4(4√5 – 6)

= 16√5 – 24 cm²

So, Area of PLR = Area of PLQ + Area of PQR

16√5 = (16√5 – 24) + Area of PQR

Therefore, Area of PQR = 24 cm²

½ × PR × QM = 24

½ × 12 × QM = 24

6 × QM = 24

QM = 24/6

QM = 4 cm

Solution:-

(a) 4.8 cm

From the question it is given that,

∆ MNO is a right-angled triangle, its legs are 6 cm and 8 cm long.

By the rule of Pythagoras theorem,

Hypotenuse² = perpendicular² + base²

In the given figure,

MO² = MN² + NO²

MO² = 6² + 8²

MO² = 36 + 64

MO² = 100

MO = √100

MO = 10 cm

Then, consider the triangle MNO,

Area of triangle MNO = ½ × MN × NO = ½ × MO × NP

= ½ × 6 × 8 = ½ × 10 × NP

Therefore, NP = 24/5

NP = 4.8 cm

Solution:-

(b) 13 cm

We know that, area of triangle = ½ × base (small side) × height

30 = ½ × 5 cm × height

Height = (30 × 2)/5

Height = 60/5

Height = 12 cm

By the rule of the Pythagoras theorem,

Hypotenuse² = height² + base²

Hypotenuse² = 12² + 5²

Hypotenuse² = 144 + 25

Hypotenuse² = 169

Hypotenuse = √169

Hypotenuse = 13 cm

Solution:-

(c) 15.7 cm

From the question, it is given that, diameter = 5 cm

Radius = diameter/2

Radius = 5/2 = 2.5 cm

Circumference of a circle = 2πr

= 2 × (22/7) × 2.5

= 15.7 cm

Solution:-

(c) 14 cm

Circumference of a circle = 2πr

88 = 2 × (22/7) × r

(88 × 7)/(2 × 22) = r

r = 14 cm