Question 1. A park, in the shape of a quadrilateral ABCD, has C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?

Solution:

To start, make a quadrilateral ABCD and join BD. 

We know 

C = 90 °, AB = 9 metres, BC = 12 metres, CD = 5 metres, and AD = 8 metres 

The diagram is as follows:

Apply Pythagoras theorem to ΔBCD now. 

\( BC^{2} \)+ \( CD^{2} \) = \( BD^{2} \)

\( BD^{2} \) = 122+52 

\( BD^{2} \) = 169 

BD = 13 m 

Now, the area of ΔBCD is equal to (½ \( \times \)12 \( \times \) 5) = 30 \( m^{2} \). 

ΔABD's semi-perimeter 

(perimeter/2) = (s) 

= (8 + 9 + 13)/2 m 

= 30/2 m 

= 15 m 

Using Heron's formula as a guide, 

Area of ΔABD

= \( \sqrt{s (s-a) (s-b) (s-c)} \)

= \( \sqrt{15 (15-13) (15-9) (15-8)}~ m^{2} \)

= \( \sqrt{15 \times 2 \times 6 \times 7} \)

= \( \sqrt{1260} \)

= 6\( \sqrt{35} \) \( m^{2} \)

= 35.5 \( m^{2} \) (approximately) 

\( \therefore \) ABCD quadrilateral area = ΔBCD quadrilateral area + Area of ΔABD 

= 30 \( m^{2} \) + 35.5 \( m^{2} \)

= 65.5 \( m^{2} \)

Question 2.  Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

Solution:

Make a diagram with the parameters you've been given. 

Now, in ΔABC, apply the Pythagorean theorem. 

\( AC^{2} = AB^{2} + BC^{2} \)

32+42 = 52

 = 25

As a result, we can deduce that ΔABC is right-angled at B. 

As a result, BCD has a surface area of (1/2 \( \times \) 3 \( \times \) 4) = 6 \( cm^{2} \). 

Semi-perimeter ACD (s) = (perimeter/2) 

= (5+5+4)/2 cm 

= 14/2 cm 

= 7 m

Using Heron's formula as a guide, 

Area of ΔACD 

= \( \sqrt{s (s-a) (s-b) (s-c)} \)

= \( \sqrt{7 (7-5) (7-5) (7-4)} ~cm^{2} \)

= \( \sqrt{7 \times 2 \times 2 \times 3}~ cm^{2} \)

= \( \sqrt{ 84 }~cm^{2} \)

= 9.17 \( cm^{2} \)

= 221 \( cm^{2} \)  (approximately) 

ABCD quadrilateral area = ΔABC area + Δ ACD area = 6 \( cm^{2} \)  + 9.17 \( cm^{2} \)

= 15.17 \( cm^{2} \)

Question 3. Radha made a picture of an aeroplane with coloured paper as shown in Fig 12.15. Find the total area of the paper used.

Solution: 

For Triangle I Section,

The sides of the isosceles triangle are 5 cm, 1 cm, and 5 cm. 

Perimeter = 5+5+1 = 11 cm 

As a result, the semi-perimeter = 1 1/2 cm = 5.5 cm. 

Using Heron's formula as a guide, 

Area =  \( \sqrt{s (s-a) (s-b) (s-c)} \)

= \( \sqrt{5.5. (5.5-5) (5.5-5) (5.5-1)} cm^{2} \)

= \( \sqrt{5.5 \times 0.5 \times 0.5 \times 4.5}~ cm^{2} \)

= 6.1875 \( cm^{2} \)

= 0.75 \( \times \) 3.317 \( cm^{2} \) 

= 2.488 \( cm^{2} \) (approx) 

For the quadrilateral II section, 

use the following formula: 

The length and breadth of this quadrilateral are 6.5 cm and 1 cm, respectively. 

\( \therefore \) Area = 6.5\( \times \)1 \( cm^{2} \) = 6.5 \( cm^{2} \)

For the quadrilateral III section, 

use the following formula: 

It's a trapezoid with two 1 cm sides and a 2 cm third side.

Area of the trapezoid = Area of the parallelogram + Area of the equilateral triangle

The parallelogram's perpendicular height will be 

= \( \sqrt{1^{2} - 0.5^{2}} \)

= 0.86 cm. 

And the equilateral triangle's area will be (3/4) a2 = 0.43. 

The trapezoid's area is 0.86 + 0.43 = 1.3 cm2 (approximately). 

For triangles IV and V, use the following formula: 

These are two congruent right-angled triangles with a base of 6 cm and a height of 1.5 cm. 

Area of triangles IV and V = 2(1/2 x 6 x 1.5) \( cm^{2} \) = 9 \( cm^{2} \)

As a result, the total area of the paper used is 19.3 \( cm^{2} \) (2.488 + 6.5 + 1.3 + 9) \( cm^{2} \).

Question 4. A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.

Solution:

Now, 

It is assumed that the areas of the parallelogram and triangle are equal. 

The triangle's sides are 26 cm, 28 cm, and 30 cm long. 

As a result, the perimeter equals 26+28+30 = 84 cm. 

Its semi-perimeter is 84/2 cm, which equals 42 cm. 

Now, using Heron's formula, the triangle's area =

= \( \sqrt{s (s-a) (s-b) (s-c)} \)

= \( \sqrt{42 (42-26) (42-28) (42-30)}~ cm^{2} \)

=  \( \sqrt{42 \times 16 \times 14 \times 12}~cm^{2} \)

= \( \sqrt{112896}~cm^{2} \)

= 336 \( cm^{2} \)

Let h be the height of the parallelogram. 

Because the area of a parallelogram equals the area of a triangle, 

= 336 \( cm^{2} \times \) 28 cm h 

\( \therefore \) h = 336/28 cm 

Thus, the parallelogram's height is 12 cm.

Question 5. A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?

Solution: First, make a rhombus-shaped field with ABCD vertices. The rhombus is divided into two congruent triangles with equal areas by the diagonal AC. The following is a diagram.

Consider the BCD triangle. 

(48 + 30 + 30)/2 m = 54 m is its semi-perimeter. 

Using Heron's formula as a guide, 

Area of the ΔBCD = \( \sqrt{s (s-a) (s-b) (s-c)} \)

= \( \sqrt{54 (54-48) (54-48) (54-30)} \)

= \( \sqrt{54 \times 6 \times 24 \times 24} \)

= \( \sqrt{186624} \)

= 432 \( m^{2} \)

 Therefore, Field area = 2 \( \times \) ΔBCD area = \( (2 \times 432) m^{2} \) = 864 \( m^{2} \)

As a result, each cow will receive an area of the grass field equal to (864/18) \( m^{2} \) = 48 \( m^{2} \).

These were NCERT solutions for Class 9 Maths Chapter 12 Heron’s Formula. We hope you have understood all the problems and their solutions. If you need NCERT solutions of other maths and science topics, you may find them on the MSVGo app. 

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