Chapter 6 - Measuring Space: Perimeter and Area (Continued)

End-Of-Chapter Exercises

Q. Find the area of a triangle, given that its sides are 8 cm and 11 cm long, and its perimeter is 32 cm.

Solution: The perimeter of the triangle is 32 cm. Let the sides be a = 8 cm, b = 11 cm, and c.
Third side c = 32 - (8 + 11) = 32 - 19 = 13 cm.
The semi-perimeter s = 32 / 2 = 16 cm.
Using Heron's formula: Area = √[s(s-a)(s-b)(s-c)]

Area = √[16 × (16 - 8) × (16 - 11) × (16 - 13)]
Area = √[16 × 8 × 5 × 3]
Area = √1920 = 8√30 cm²

The area of the triangle is 8√30 cm².

Q. The sides of a triangular plot are in the ratio 3:5:7; its perimeter is 300 m. Find its area.

Solution: Let the sides of the triangle be 3x, 5x, and 7x.
Perimeter = 3x + 5x + 7x = 15x.
Given Perimeter = 300 m, so 15x = 300 ⇒ x = 20.
The sides are 60 m, 100 m, and 140 m.
Semi-perimeter s = 300 / 2 = 150 m.

Area = √[150 × (150 - 60) × (150 - 100) × (150 - 140)]
Area = √[150 × 90 × 50 × 10]
Area = √6750000 = 1500√3 m²

The area of the plot is 1500√3 m².

Q. The sides of a triangle have lengths 7 cm, 24 cm, 25 cm. Find the area of the triangle in two different ways.

Solution:
Method 1: Using Heron's Formula
Semi-perimeter s = (7 + 24 + 25) / 2 = 56 / 2 = 28 cm.
Area = √[28 × (28 - 7) × (28 - 24) × (28 - 25)]
Area = √[28 × 21 × 4 × 3] = √7056 = 84 cm².

Method 2: Using the Right-Angled Triangle Property
Check if the sides satisfy the Pythagoras Theorem: 7² + 24² = 49 + 576 = 625, which equals 25².
Since it is a right-angled triangle, we use Area = 1/2 × base × height.
Area = 1/2 × 7 × 24 = 84 cm².

Chapter 8 - Predicting What Comes Next: Exploring Sequences

End-Of-Chapter Exercises

1. Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.

Solution: The formula for the nth term of an AP is t_n = a + (n - 1)d.
Given: t₁₁ = a + 10d = 38    (Equation 1)
t₁₆ = a + 15d = 73    (Equation 2)
Subtracting Equation 1 from Equation 2:

(a + 15d) - (a + 10d) = 73 - 38
5d = 35 ⇒ d = 7

Substitute d = 7 into Equation 1:
a + 10(7) = 38 ⇒ a + 70 = 38 ⇒ a = -32.
Now, to find the 31st term (t₃₁):
t₃₁ = a + 30d = -32 + 30(7) = -32 + 210 = 178.
The 31st term is 178.

2. Determine the AP whose third term is 16 and whose 7th term exceeds the 5th term by 12.

Solution: Given t₃ = 16, so a + 2d = 16.
We are also given that the 7th term exceeds the 5th term by 12: t₇ - t₅ = 12.

(a + 6d) - (a + 4d) = 12
2d = 12 ⇒ d = 6

Substitute d = 6 back into the first equation:
a + 2(6) = 16 ⇒ a + 12 = 16 ⇒ a = 4.
The Arithmetic Progression is 4, 10, 16, 22, 28, ...

3. How many three-digit numbers are divisible by 7?

Solution: The smallest 3-digit number divisible by 7 is 105. The largest 3-digit number divisible by 7 is 994.
These numbers form an AP: 105, 112, 119, ..., 994.
Here, a = 105, d = 7, and t_n = 994.

t_n = a + (n - 1)d
994 = 105 + (n - 1)7
889 = (n - 1)7
n - 1 = 127 ⇒ n = 128

There are 128 three-digit numbers divisible by 7.

4. How many multiples of 4 lie between 10 and 250?

Solution: The first multiple of 4 after 10 is 12. The last multiple of 4 before 250 is 248.
This forms an AP: 12, 16, 20, ..., 248.
Here, a = 12, d = 4, and t_n = 248.

t_n = a + (n - 1)d
248 = 12 + (n - 1)4
236 = (n - 1)4
n - 1 = 59 ⇒ n = 60

There are 60 multiples of 4 between 10 and 250.

5. Find a GP for which the sum of the first two terms is −4 and the fifth term is 4 times the third term.

Solution: Let the GP be a, ar, ar², ar³, ....
Given a + ar = -4 ⇒ a(1 + r) = -4    (Equation 1).
Also given t₅ = 4 × t₃.
ar⁴ = 4ar²
Dividing by ar² (assuming a ≠ 0, r ≠ 0), we get r² = 4 ⇒ r = ±2.
Case 1: If r = 2
a(1 + 2) = -4 ⇒ 3a = -4 ⇒ a = -4/3.
The GP is: -4/3, -8/3, -16/3, -32/3, ...
Case 2: If r = -2
a(1 - 2) = -4 ⇒ -a = -4 ⇒ a = 4.
The GP is: 4, -8, 16, -32, ...

7. The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of the 2nd hour, 4th hour and nth hour?

Solution: This creates a Geometric Progression because the quantity doubles (ratio r = 2) every hour.
Original amount a = 30.

• End of 1st hour = 30 × 2 = 60
• End of 2nd hour = 60 × 2 = 120
• End of 3rd hour = 120 × 2 = 240
• End of 4th hour = 240 × 2 = 480

In general, at the end of the nth hour, the amount is multiplied by 2, n times.
Amount after nth hour = 30 × 2ⁿ.

8. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

Solution:
Given t₄ + t₈ = 24.
(a + 3d) + (a + 7d) = 24 ⇒ 2a + 10d = 24 ⇒ a + 5d = 12    (Equation 1).
Given t₆ + t₁₀ = 44.
(a + 5d) + (a + 9d) = 44 ⇒ 2a + 14d = 44 ⇒ a + 7d = 22    (Equation 2).
Subtracting Equation 1 from Equation 2:
(a + 7d) - (a + 5d) = 22 - 12 ⇒ 2d = 10 ⇒ d = 5.
Substitute d = 5 in Equation 1:
a + 5(5) = 12 ⇒ a + 25 = 12 ⇒ a = -13.
The first three terms of the AP are: -13, -8, -3.

10. Which term of the GP: 2, 8, 32, … is 131072? Write the explicit formula as well as the recursive formula for the nth term.

Solution: The sequence is 2, 8, 32, ….
First term a = 2, Common ratio r = 8 / 2 = 4.
We are looking for n when t_n = 131072.

t_n = a × r^n-1
131072 = 2 × 4^n-1
65536 = 4^n-1

Since 4⁸ = 65536, we have n - 1 = 8 ⇒ n = 9.
131072 is the 9th term.

Explicit Formula: t_n = 2 × 4^n-1.
Recursive Formula: t₁ = 2, and t_n = 4 × t_n-1 for n ≥ 2.

11. The sum of the first three terms of a GP is 13/12 and their product is −1. Find the common ratio and the terms.

Solution: Let the first three terms of the GP be a/r, a, and ar.
Product = (a/r) × a × (ar) = a³ = -1. Thus, a = -1.
Sum = a/r + a + ar = 13/12.
Substitute a = -1:

-1/r - 1 - r = 13/12
-(1 + r + r²) / r = 13/12
-12(1 + r + r²) = 13r
-12 - 12r - 12r² = 13r
12r² + 25r + 12 = 0

Factoring the quadratic equation:
12r² + 16r + 9r + 12 = 0
4r(3r + 4) + 3(3r + 4) = 0
(4r + 3)(3r + 4) = 0.
So, r = -3/4 or r = -4/3.

If r = -3/4, the terms are: 4/3, -1, 3/4.
If r = -4/3, the terms are: 3/4, -1, 4/3.