RBSE Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.2

Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.2 Textbook Exercise Questions and Answers.

RBSE Class 9 Maths Solutions Chapter 12 Heron’s Formula Exercise 12.2

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?
Answer:

Area of ∆BCD = \(\frac{1}{2}\) × BC × CD
= (\(\frac{1}{2}\) × 12 × 5)m²
= 30 m²
Using Pythagoras theorem, we have :
BD² = BC² + CD² ⇒ BD² = 12² + 5²
So, BD² = 144 + 25 ⇒ BD² = 169
∴ BD = √169 m = 13 m
For ∆ABD :
Let a = 13 m, b = 8 m and c = 9 m.
s = \(\frac{1}{2}\)(a + b + c) = \(\frac{1}{2}\)(13 + 8 + 9) m = \(\frac{1}{2}\) × 30 m = 15 m
s - a = (15 - 13) m = 2 m
s - b = (15 - 8) m = 7 m
and s - c = (15 - 9) m = 6 m
Area of ∆ABD
= \(\sqrt{15 \times 2 \times 7 \times 6}\) m²
= \(\sqrt{3 \times 5 \times 2 \times 7 \times 2 \times 3}\) m²
= \(\sqrt{2 \times 2 \times 3 \times 3 \times 5 \times 7}\) m²
= 2 × 3√35 m²
= 6 × 5.9 m² (approx.)
= 35.4 m² (approx.)
Required area = area of ∆ABD + area of ∆BCD = 35.4 m²+ 30 m² = 65.4 m².

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.
Answer:

Since AC² - AB² + BC²
(as 5² = 3² + 4², i.e. 25 = 9 + 16)
∴ ∠ABC = 90°
Area of right-angled ∠d ∆ABC = \(\frac{1}{2}\) × AB × BC
= (\(\frac{1}{2}\) × 3 × 4) cm² = 6 cm²
For ∆ ACD : Let a = 5 cm, b = 4 cm and c = 5 cm.
Then, s = \(\frac{1}{2}\)(a + b + c) = \(\frac{1}{2}\)(5 + 4 + 5) cm
= \(\frac{1}{2}\) × 14 cm = 7 cm
Now, s - a = (7 - 5) cm = 2 cm
s - b = (7 - 4) cm = 3 cm
and s - c = (7 - 5) cm = 2 cm
Area of ∆ ACD = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{7 \times 2 \times 3 \times 2}\) cm² = 2√21 cm²
= 2 × 4.6 cm² (approx.) = 9.2 cm² (approx.)
Area of quadrilateral ABCD = Area of ∆ ABC + Area of ∆ ACD
= (6 + 9.2) cm² = 15.2 cm² (approx.) 

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

For figure I:
Here, a = 5 cm, 6 = 5 cm and c = 1 cm.

For figure II:
Area (II) = 1 × 6.5 cm²
For figure III:

For ∆CEB
2s = 1 + 1 + 1 = 3 ⇒ s = \(\frac{3}{2}\) 

= (15.5 + 37.86) cm² = 19.286 cm² = 19.3 cm² (approx.)

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.
Answer:
For the triangle :
Let its sides be a, b and c such that a = 26 cm, b = 28 cm and c = 30.
Then, s = \(\frac{1}{2}\) (26 + 28 + 30) cm = \(\frac{1}{2}\) x 84 cm = 42 cm .
Now, s - a = (42 - 26) cm = 16 cm
s - b = (42 - 28) cm = 14 cm '
and s - c = (42 - 30) cm = 12 cm
Area of the triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{42 \times 16 \times 14 \times 12}\) cm²
= \(\sqrt{2 \times 3 \times 7 \times 4 \times 4 \times 2 \times 7 \times 2 \times 2 \times 3}\) cm²
= \(\sqrt{2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 4 \times 4 \times 7 \times 7}\) cm²
= (2 × 2 × 3 × 4 × 7) cm² = 336 cm²
For the parallelogram:
Area = Base × Height
[∵ Area of ||gm = Area of ∆ (given)
∴ Area of ||gm = 336 cm² 
and its base = 28 cm]
∴ Height = \(\frac{\text { Area }}{\text { Base }}\) 
= \(\left(\frac{336}{28}\right)\) cm = 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?
Answer:

We know that the diagonals of the rhombus bisect each other at right angles. Using Pythagoras theorem, we have :
OD = \(\sqrt{A D^{2}-A O^{2}}\) = \(\sqrt{30^{2}-24^{2}}\) m
= \(\sqrt{(30+24)(30-24)}\) m
= \(\sqrt{54 \times 6}\) m = \(\sqrt{9 \times 6 \times 6}\) m
= (3 × 6) m = 18 m
Area of one AAOD = (\(\frac{1}{2}\) × 24 × 18) m² = 216 m²
∴ Area of rhombus = 4 × ∆AOD = (4 × 216) m² = 864 m² 
∴ Grass area for 18 cows = 864 m²
Grass area for 1 cow = \(\left(\frac{864}{18}\right)\)m² = 48 m²

Question 6.
An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see figure), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?

Answer:
In one triangular piece, let a = 20 cm, b = 50 cm and c = 50 cm.
Now, s = \(\frac{1}{2}\)(a + b + c) = \(\frac{1}{2}\)(20 + 50 + 50) cm
= \(\frac{1}{2}\) × 120 cm = 60 cm
s - a = (60 - 20) cm = 40 cm
s - b - (60 - 50) cm = 10 cm
and s - c = (60 - 50) cm = 10 cm
∴ Area of one triangular piece = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{60 \times 40 \times 10 \times 10}\) cm²
= 200√6 cm²
∴ Area of 5 red triangles = 1000 √6 cm²
and area of 5 green triangles = 1000 √6 cm².

Question 7.
A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in figure. How much paper of each shade has been used in it?

Answer:
ABCD is a square such that AC = BD = 32 cm and AEF is an isosceles triangle in which AE = AF = 6 cm
For the area of shades I and II Clearly, from the figure
Area of shade I = Area of shade
II = Area of ∆CDB = \(\frac{1}{2}\) × DB × CM
= \(\frac{1}{2}\) × 32 × 16 sq cm = 256 sq cm
Since, areas of shades I and II are equal. Theresfore, area of shade II = 256 sq m.
For the area of shade III (i.e., ∆AEF)
EL = LF = \(\frac{1}{2}\)EL = \(\frac{1}{2}\) × 8 cm = 4 cm
and AE = 6 cm (Given)
∴ AL = \(\sqrt{A E^{2}-E L^{2}}\) = \(\sqrt{36-16}\) cm = √20 cm = 2√5 cm
Area of shade III = \(\frac{1}{2}\) × EF × AL = \(\frac{1}{2}\) × 8 × 2√5 sq cm = 8√5 sq cm
= 8 × 2.24 sq cm (approx.) = 17.92 sq cm (approx.)
Aliter : Let a = 6 cm, 6 = 6 cm and c = 8 cm
So, s = \(\left(\frac{a+b+c}{2}\right)\) = \(\frac{(6+6+8)}{2}\) cm = 10 cm
Area of shade III = \(\sqrt{s(s-a)(s-b)(s-c)}\)
Area of shade III = 7s(s - a)(s - b)(s - c)
= \(\sqrt{10 \times 4 \times 4 \times 2}\) cm²
= 8√5 cm² = 17.92 cm² 

Question 8.
A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see figure). Find the cost of polishing the tiles at the rate of 50 p per cm2.

Answer:
For one triangular tile :
Let a = 9 cm, 6 = 28 cm and c = 35 cm
Now, s = \(\frac{1}{2}\)(a + b + c) = \(\frac{1}{2}\)(9 + 28 + 35) cm
= \(\frac{1}{2}\) × 72 cm = 36 cm
s - a = (36 - 9) cm = 27 cm
s - b = (36 - 28) cm = 8 cm
and, s - c = (36 - 35) cm = 1 cm
∴  Area of one tile = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{36 \times 27 \times 8 \times 1}\) cm²
= \(\sqrt{9 \times 4 \times 9 \times 3 \times 4 \times 2}\) cm²
= \(\sqrt{2 \times 3 \times 4 \times 4 \times 9 \times 9}\) cm²
= 4 × 9√6 cm²
= 36√6 cm = 36 × 2.45 cm² (approx.)
= 88.2 cm² (approx)
∴ Area of 16 such tiles = (16 × 88.2) cm² = 1411.2 cm² (approx.)
Cost of polishing @ 50 p per cm² = ₹ (1411.2 × \(\frac{50}{100}\)) = ₹ 705.60 (approx.). 

Question 9.
A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.

Answer:
From C, draw CE || DA.
Clearly, ADCE is a parallelogram having AD || CE and CD || AE such that AD = 13 m and CD = 10 m.
AE = DC = 10 m and CE = AD = 13 m D 10 m C
Also, BE = AB - AE
= (25 - 10) m = 15 m

In ∆BCE, we have :
BC = 14 m, CE = 13 m and BE = 15 m. Then,
2s = (14 + 13 + 15) m = 42 m
So, s = 21 m
∴ Area (∆ BCE) = \(\sqrt{21 \times(21-14)(21-13)(21-15)}\) = \(\sqrt{21 \times 7 \times 8 \times 6}\)
= 7 × 3 × 4 m² = 84 m²
Area (∆ BCE) = \(\frac{1}{2}\) × BE × CL
So, 84 = \(\frac{1}{2}\) × 15 × CL
⇒ CL = \(\frac{168}{15}\) = \(\frac{56}{5}\)m
i.e. Height of the trapezium = \(\frac{56}{5}\) m
∴ Area of trapezium = \(\frac{1}{2}\)(AB + CD) × CL = \(\frac{1}{2}\) × (25 + 10) × \(\frac{56}{5}\)m²
= \(\frac{1}{2}\) × 35 × \(\frac{56}{5}\) m² = 7 × 28 m² = 196 m².

Aliter:

Clearly DM = CN
So,  DM² = CN²
⇒ 13² - x² = 14² - (15 - x)²
⇒ 169 - x² = 196 - 225 + 30x - x²
⇒ 30x = 169 - 196 + 225 = 198
So, x = 6.6
So, DM² = AD² = AM² = 13² - (6.6)² = 19.6 × 6.4
∴ DM = \(\sqrt{19.6 \times 6.4}\) = 11.2
∴ Area of the trapezium ABCD = \(\frac{1}{2}\) × (AB + CD) × DM
= \(\frac{1}{2}\) × (25 + 10) × 11.2 m² = 196 m²