anyone can give me the solutions of class 9th mathematics chapter 12 cbse
very very very urgent
Answers
Answer:
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Step-by-step explanation:
Answer:
lLet each side of the equilateral triangle be a.
Semi-perimeter of the triangle,
NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.1 Q1
Ex 12.1 Class 9 Maths Question 2.
The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see figure). The advertisements yield an earning of ₹5000 per m² per year. A company hired one of its walls for 3 months. How much rent did it pay?
NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.1 Q2
Solution:
Let the sides of the triangular will be
a = 122m, b = 12cm, c = 22m
Semi-perimeter, s = \frac { a+b+c }{ 2 }
(\frac { 122+120+22 }{ 4 })m = \frac { 264 }{ 2 } m = 132m
The area of the triangular side wall
NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.1 Q2a
Rent for 1 year (i.e. 12 months) per m2 = Rs. 5000
∴ Rent for 3 months per m2 = Rs. 5000 x \frac { 3 }{ 12 }
= Rent for 3 months for 1320 m2 = Rs. 5000 x \frac { 3 }{ 12 } x 1320 = Rs. 16,50,000.
Ex 12.1 Class 9 Maths Question 3.
There is a slide in a park. One of its side Company hired one of its walls for 3 months.walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN” (see figure). If the sides of the wall are 15 m, 11 m and 6m, find the area painted in colour.
NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.1 Q3
Solution:
Let the sides of the wall be
a = 15m, b = 11m, c = 6m
Semi-perimeter,
NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.1 Q3a
Thus, the required area painted in colour
= 20√2 m2
Ex 12.1 Class 9 Maths Question 4.
Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.
Solution:
Let the sides of the triangle be a =18 cm, b = 10 cm and c = x cm
Since, perimeter of the triangle = 42 cm
∴ 18cm + 10 cm + xcm = 42
x = [42 – (18 + 10)cm = 14cm
Now, semi-permimeter, s = \frac { 42 }{ 2 }cm = 21 cm
NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.1 Q4
Thus, the required area of the triangle = 21\sqrt { 11 } cm2
Ex 12.1 Class 9 Maths Question 5.
Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area.
Solution:
Let the sides of the triangle be
a = 12x cm, b = 17x cm, c = 25x cm
Perimeter of the triangle = 540 cm
Now, 12x + 17x + 25x = 540
⇒ 54x = 54 ⇒ x = 10
∴ a = (12 x10)cm = 120cm,
b = (17 x 10) cm = 170 cm
and c = (25 x 10)cm = 250 cm
Now, semi-perimeter, s = \frac { 540 }{ 2 }cm = 270 cm
NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.1 Q5
Ex 12.1 Class 9 Maths Question 6.
An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
Solution:
Let the sides of an isosceles triangle be
a = 12cm, b = 12cm,c = x cm
Since, perimeter of the triangle = 30 cm
∴ 12cm + 12cm + x cm = 30 cm
⇒ x = (30 – 24) = 6
Now, semi-perimeter, s = \frac { 30 }{ 2 }cm =15 cm
NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.1 Q6
Thus, the required area of the triangle = 9√15