KSEEB Solutions for Class 9 Maths Chapter 8 Heron’s Formula Ex 8.2

Students can Download Class 9 Maths Chapter 8 Heron’s Formula Ex 8.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 9 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

Karnataka State Syllabus Class 9 Maths Chapter 8 Heron’s Formula Ex 8.2

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?
Solution:
BD is joined.
In Right angled ABCD, ∠C = 90°

As per Pythagoras Theorem,
BD² = BC² + CD²
= (12)² + (5)²
= 144 + 25
BD² = 169
∴ BD = 13m.
Area of ∆BCD = \(\frac{1}{2}\) × BC × CD
= \(\frac{1}{2}\) × 12 × 5 = 30 m²
Now, in ∆ABD,

By Heron’s formula,

∴ Area of quadrilateral ABCD,
= Area(∆ABD) + Area (∆BCD)
= 35.46 + 30
= 65.46 sq.m.
Area of quadrilateral ABCD = 65.46 sq.m.

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

Solution:
ln ∆ABC
AC² = AB² + BC²
5² = 3² + 4²
∴ ∆^le ABC is a right angled ∆^le
Area of ∆ABC = \(\frac{1}{2}\) × AB × BC
= \(\frac{1}{2}\) × 3 × 4 = 6 cm²
In ∆ADC
Perimeter = 2s = AC + CD + DA
= 5 +4 + 5 = 14 cm
s = 7cm
By Heron’s formula,

= 2 × 4.583 = 9.166 cm²
Area of ∆ADC = 9.16 sq.cm.
∴ Area of quadrilateral ABCD,
= Area(∆ABC) + Area(∆ADC)
= 6 + 9.16
= 15.16 sq. cm.

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

Solution:
(I) ∆ABC is an isosceles triangle.
AB = AC = 5 cm, and BC = 1 cm.
∴ a = 5 cm, b = 5 cm, c = 1 cm.
∴ Perimeter = a + b + c = 5 + 5 + 1 = 11 cm.
s = \(\frac{11}{2}\) = 5.5 cm

∴ Area(∆ABC) = 24.85 sq. cm.

(II) PBCS is a rectangle.
l = 6.5 cm, b = 1 cm.
Area of rectangle PBCS = l × b
= 6.5 × 1 = 6.5 sq.cm.

(III) PQRS is a trapezium PT ⊥ QR
In Right angle ∆PTV,

Area = Area of parallelogram + Area of equilateral ∆^le
= (0.866) 1 + \(\frac{\sqrt{3}}{4}\) (1)²
= 0.866 + 0.433 = 1.299 cm²

Area of ∆^le IV = Area of ∆^le V
\(\frac{1}{2}\) = 1.5 × 6 – 4.5 cm²
Total area of paper used = 2.488 + 6.5 + 1.299 + 4.5 × 2
= 19.287 cm².

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:
Perimeter of ∆ABC = 26 + 28 + 30 = 84 cm.
2s = 84
s = 42 cm
By Heron’s formula,

Let the height of the parallelogram be h
Area of parallelogram Area of triangle
h × 28 = 336 cm²
h = 12 cm
∴ the height of the parallelogram 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:
ABCD is a rhombus.
Area(∆ABD) = Area(∆BDC).
In ∆ABD, a = 30 m, b = 30 m, c = 48 m.
∴ Perimeter= a + b + c = 30 + 30 + 48 = 108 m

= 3 × 6 × 24 = 432 m²
But, Area of whole field = 2 × area of ∆ABD
= 2 × 432
= 864 sq.m.
∴ Are required for growing grass for 18 cows is 864 sq.m.
∴ Area required for growing grass for 1 cow ………. ??
\(\frac{864}{18}\) = 48 sq.m.

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

Solution:
Each side of the triangular-shaped cloth
Let a = 50 cm, b = 50 cm, c = 20 cm.
∴ Perimeter = a + b + c = 50 + 50 + 20 = 120 cm.

∴ Total Area of the Umbrella mad up with 10 triangular shaped cloth is
= 5 × 200√6
= 100√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 Fig. How much paper of each shade has been used in it?

Solution:
We know that
Area of a square = \(\frac{1}{2}\) (diagonal)²
Area ofakite = \(\frac{1}{2}\) (32)2 = 512 cm²
Area of the 1^st shape = Area of 2^nd shape
= \(\frac{512}{2}\) = 256 cm²
∴ Area of paper required in each shape is 256 cm²
For IIIrd triangle,

= 8√5 cm² = 8 × 2.24 = 17.92 cm²
∴ Area of paper required for III^rd shape
= 17.92 cm²

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

Solution:
In the given figure there are 16 triangles.
Each sides of triangle are
a = 28 cm, b = 9 cm, c = 35 cm.
Perimeter of triangle = a + b + c
= 28 + 9 + 35 = 72 cm.

∴ Area of 16 triangles = 16 × 88.2 = 1411.2 sq.cm.
Polish expenditure for 1 sq.cm. = 50 p = Re. 0.5.
∴ Polish expenditure for 1411.2sq. cm. =?
= 1411.2 × 0.5
= Rs 705.6

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.

Solution:
ABCD is a trapezium.
AB || DC.
Draw AE || BC and AF ⊥ DC.
∴ ABCE is a parallelogram.
DE = DC – EC = 25 – 10
DE = 15 cm.
Sides of ∆ADE
a = 14 cm, b = 13 cm. c = 15 cm.
∴ Perimeter= a + b + c
= 14 + 13 + 15 = 42 cm.
∴ s = \(\frac{a+b+c}{2}=\frac{42}{2}=21 \mathrm{cm}\)
∴ Area of ∆ADE,

(ii) Area of quadrilateral ABCE = base × height
= EC × AF
= 10 × 11.2
= 112.0
= 112 sq.cm.
∴ Complete area of trapezium ABCD, A
A = Area of ∆ADE + Area of quadrilateral ABCE
= 84 + 112
= 196 sq.cm.