anyone can give me the solutions of class 9th mathematics chapter 12 cbse
very very very urgent
Answers
Answer:
= 21√11 cm2
5. Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540cm. Find its area.
Solution:
The ratio of the sides of the triangle are given as 12 : 17 : 25
Now, let the common ratio between the sides of the triangle be “x”
∴ The sides are 12x, 17x and 25x
It is also given that the perimeter of the triangle = 540 cm
12x+17x+25x = 540 cm
54x = 540cm
So, x = 10
Now, the sides of triangle are 120 cm, 170 cm, 250 cm.
So, the semi perimeter of the triangle (s) = 540/2 = 270 cm
Using Heron’s formula,
Area of the triangle
Ncert solutions class 9 chapter 12-7
= 9000 cm2
6. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
Solution:
First, let the third side be x.
It is given that the length of the equal sides is 12 cm and its perimeter is 30 cm.
So, 30 = 12+12+x
∴ The length of the third side = 6 cm
Thus, the semi perimeter of the isosceles triangle (s) = 30/2 cm = 15 cm
Using Heron’s formula,
Area of the triangle
=
Ncert solutions class 9 chapter 12-8
= √[15(15-12)(15-12)(15-6)] cm2
= √[15×3×3×9] cm2
= 9√15 cm2
Exercise: 12.2 (Page No: 206)
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:
First, construct a quadrilateral ABCD and join BD.
We know that
C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m
The diagram is:
Ncert solutions class 9 chapter 12-9
Now, apply Pythagoras theorem in ΔBCD
BD2 = BC2 +CD2
⇒ BD2 = 122+52
⇒ BD2 = 169
⇒ BD = 13 m
Now, the area of ΔBCD = (½ ×12×5) = 30 m2
The semi perimeter of ΔABD
(s) = (perimeter/2)
= (8+9+13)/2 m
= 30/2 m = 15 m
Using Heron’s formula,
Area of ΔABD
Ncert solutions class 9 chapter 12-10
= 6√35 m2 = 35.5 m2 (approximately)
∴ The area of quadrilateral ABCD = Area of ΔBCD+Area of ΔABD
= 30 m2+35.5m2 = 65.5 m2
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:
First, construct a diagram with the given parameter.
Ncert solutions class 9 chapter 12-11
Now, apply Pythagorean theorem in ΔABC,
AC2 = AB2+BC2
⇒ 52 = 32+42
⇒ 25 = 25
Thus, it can be concluded that ΔABC is a right angled at B.
So, area of ΔBCD = (½ ×3×4) = 6 cm2
The semi perimeter of ΔACD (s) = (perimeter/2) = (5+5+4)/2 cm = 14/2 cm = 7 m
Now, using Heron’s formula,
Area of ΔABD
Ncert solutions class 9 chapter 12-12
= 2√21 cm2 = 9.17 cm2 (approximately)
Area of quadrilateral ABCD = Area of ΔABC + Area of ΔABD = 6 cm2 +9.17 cm2 = 15.17 cm2
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.
Ncert solutions class 9 chapter 12-13
Solution:
For the triangle I section:
Ncert solutions class 9 chapter 12-14
It is an isosceles triangle and the sides are 5 cm, 1 cm and 5 cm
Perimeter = 5+5+1 = 11 cm
So, semi perimeter = 11/2 cm = 5.5 cm
Using Heron’s formula,
Area = √[s(s-a)(s-b)(s-c)]
= √[5.5(5.5- 5)(5.5-5)(5.5-1)] cm2
= √[5.5×0.5×0.5×4.5] cm2
= 0.75√11 cm2
= 0.75 × 3.317cm2
= 2.488cm2 (approx)
For the quadrilateral II section:
This quadrilateral is a rectangle with length and breadth as 6.5 cm and 1 cm respectively.
∴ Area = 6.5×1 cm2=6.5 cm2
For the quadrilateral III section:
It is a trapezoid with 2 sides as 1 cm each and the
A = (½)×(diagonal)2
Area of the kite = (½)×32×32 = 512 cm2.
The area of shade I = Area of shade II
512/2 cm2 = 256 cm2
BE parallel to the line AD. Then, from B, draw a perpendicular on the line segment CD.
Ncert solutions class 9 chapter 12-25
Now, it can be seen that the quadrilateral ABED is a parallelogram. So,
AB = ED = 10 m
AD = BE = 13 m
EC = 25-ED = 25-10 = 15 m
Now, consider the triangle BEC,
Its semi perimeter (s) = (13+14+15)/2 = 21 m
By using Heron’s formula,
Area of ΔBEC =
Ncert solutions class 9 chapter 12-26
= 84 m2
We also know that the area of ΔBEC = (½)×CE×BF
84 cm2 = (½)×15×BF
BF = (168/15) cm = 11.2 cm
So, the total area of ABED will be BF×DE i.e. 11.2×10 = 112 m2
∴ Area of the field = 84+112 = 196 m2
hope this helps u
