EduGain Class 9 Full Year 9th Grade Review worksheet featuring math questions on quadrilaterals, spheres, averages, rational numbers, and more.

Let's solve the problems step by step.

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Problem (1):

ABCD is a quadrilateral in which AB is parallel to CD and AD = BC = 5 cm. If ∠A = 105°, find the value of ∠B.

#### Solution:
- Since AB is parallel to CD, ABCD is an isosceles trapezium (because AD = BC).
- In an isosceles trapezium, the base angles are equal. Therefore, ∠A = ∠D = 105°.
- The sum of the interior angles of a quadrilateral is 360°.
- Let ∠B = x and ∠C = y. Since AB || CD, ∠B and ∠C are supplementary to ∠A and ∠D, respectively.
- Therefore, ∠B + ∠C = 180° - (∠A + ∠D) = 180° - (105° + 105°) = 180° - 210° = -30° (This is incorrect, so we need to use the property of isosceles trapezium directly).

Since ∠A = 105° and ∠D = 105°, the remaining angles ∠B and ∠C must be equal and sum up to 150° (because 360° - 210° = 150°). Therefore, ∠B = ∠C = 75°.

So, the value of ∠B is:
\[
\boxed{75^\circ}
\]

---

Problem (2):

A sphere is just enclosed inside a cube of volume 72 cm³. Find the volume of the sphere.

#### Solution:
- The volume of the cube is given as 72 cm³. Let the side length of the cube be \(a\).
- The volume of the cube is \(a^3 = 72\).
- The diameter of the sphere is equal to the side length of the cube, so the radius \(r\) of the sphere is \(r = \frac{a}{2}\).
- First, find \(a\):
\[
a = \sqrt[3]{72}
\]
- The volume of the sphere is given by:
\[
V = \frac{4}{3} \pi r^3
\]
- Since \(r = \frac{a}{2}\), we have:
\[
r^3 = \left(\frac{a}{2}\right)^3 = \frac{a^3}{8} = \frac{72}{8} = 9
\]
- Therefore, the volume of the sphere is:
\[
V = \frac{4}{3} \pi \cdot 9 = 12\pi
\]

So, the volume of the sphere is:
\[
\boxed{12\pi}
\]

---

Problem (3):

Vinayak got an average score of 82.25 in 4 tests. He got 83 as the average of the highest 3 scores, and his lowest two scores are the same numbers. What is the average of his highest two scores?

#### Solution:
- Let the four scores be \(a, b, c,\) and \(d\) such that \(a \leq b \leq c \leq d\).
- The average of all four scores is 82.25, so:
\[
\frac{a + b + c + d}{4} = 82.25 \implies a + b + c + d = 329
\]
- The average of the highest three scores is 83, so:
\[
\frac{b + c + d}{3} = 83 \implies b + c + d = 249
\]
- Subtract the second equation from the first:
\[
(a + b + c + d) - (b + c + d) = 329 - 249 \implies a = 80
\]
- Since the lowest two scores are the same, \(a = b = 80\).
- Substitute \(a = b = 80\) into \(a + b + c + d = 329\):
\[
80 + 80 + c + d = 329 \implies 160 + c + d = 329 \implies c + d = 169
\]
- The average of the highest two scores \(c\) and \(d\) is:
\[
\frac{c + d}{2} = \frac{169}{2} = 84.5
\]

So, the average of his highest two scores is:
\[
\boxed{84.5}
\]

---

Problem (4):

Examine if the following numbers are rational or irrational.
A) \(\sqrt{6}\)
B) \(\sqrt{5} + \sqrt{7}\)
C) \(\sqrt{7}\)
D) \((3 - \sqrt{6})(3 + \sqrt{6})\)
E) \(\sqrt{6} - \sqrt{7}\)
F) \(\sqrt{6} + 3\)

#### Solution:
- A number is rational if it can be expressed as a fraction of two integers. Otherwise, it is irrational.
- A) \(\sqrt{6}\): Since 6 is not a perfect square, \(\sqrt{6}\) is irrational.
- B) \(\sqrt{5} + \sqrt{7}\): The sum of two irrational numbers (unless they are additive inverses) is irrational. Here, \(\sqrt{5}\) and \(\sqrt{7}\) are both irrational and not additive inverses, so \(\sqrt{5} + \sqrt{7}\) is irrational.
- C) \(\sqrt{7}\): Since 7 is not a perfect square, \(\sqrt{7}\) is irrational.
- D) \((3 - \sqrt{6})(3 + \sqrt{6})\): This is a difference of squares:
\[
(3 - \sqrt{6})(3 + \sqrt{6}) = 3^2 - (\sqrt{6})^2 = 9 - 6 = 3
\]
Since 3 is a rational number, this expression is rational.
- E) \(\sqrt{6} - \sqrt{7}\): The difference of two irrational numbers (unless they are the same) is irrational. Here, \(\sqrt{6}\) and \(\sqrt{7}\) are both irrational and not the same, so \(\sqrt{6} - \sqrt{7}\) is irrational.
- F) \(\sqrt{6} + 3\): The sum of an irrational number and a rational number is irrational. Here, \(\sqrt{6}\) is irrational and 3 is rational, so \(\sqrt{6} + 3\) is irrational.

So, the rational number is:
\[
\boxed{D}
\]

---

Problem (5):

The length and breadth of a rectangle are 20 cm and 99 cm. Find the radius of the circumcircle of this rectangle.

#### Solution:
- The diagonal of the rectangle is the diameter of the circumcircle.
- Let the length be \(l = 20\) cm and the breadth be \(b = 99\) cm.
- The diagonal \(d\) of the rectangle is given by the Pythagorean theorem:
\[
d = \sqrt{l^2 + b^2} = \sqrt{20^2 + 99^2} = \sqrt{400 + 9801} = \sqrt{10201} = 101
\]
- The radius \(R\) of the circumcircle is half the diagonal:
\[
R = \frac{d}{2} = \frac{101}{2} = 50.5
\]

So, the radius of the circumcircle is:
\[
\boxed{50.5}
\]

---

Problem (6):

The three angles of a quadrilateral are 29°, 31°, and 144° respectively. Find the fourth angle.

#### Solution:
- The sum of the interior angles of a quadrilateral is 360°.
- Let the fourth angle be \(x\). Then:
\[
29^\circ + 31^\circ + 144^\circ + x = 360^\circ
\]
- Simplify:
\[
204^\circ + x = 360^\circ \implies x = 360^\circ - 204^\circ = 156^\circ
\]

So, the fourth angle is:
\[
\boxed{156^\circ}
\]

---

Problem (7):

Harsh is part of the school cricket team, and this year he has scored an average of 46 runs. He has played 5 innings so far, and his scores in 4 of them are 46, 45, 39, 55. What was his score in the last one?

#### Solution:
- The average score over 5 innings is 46, so the total runs scored in 5 innings is:
\[
46 \times 5 = 230
\]
- The sum of the scores in the first 4 innings is:
\[
46 + 45 + 39 + 55 = 185
\]
- Let the score in the last inning be \(x\). Then:
\[
185 + x = 230 \implies x = 230 - 185 = 45
\]

So, the score in the last inning is:
\[
\boxed{45}
\]

---

Problem (8):

From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. The lengths of the perpendiculars are 12 cm, 16 cm, and 10 cm. Find the area of the triangle.

#### Solution:
- Let the side length of the equilateral triangle be \(a\).
- The area of the equilateral triangle can be divided into three smaller triangles by drawing perpendiculars from an interior point to the sides.
- The area of the equilateral triangle is the sum of the areas of these three smaller triangles:
\[
\text{Area} = \frac{1}{2} \times a \times 12 + \frac{1}{2} \times a \times 16 + \frac{1}{2} \times a \times 10 = \frac{1}{2} \times a \times (12 + 16 + 10) = \frac{1}{2} \times a \times 38 = 19a
\]
- The area of an equilateral triangle with side length \(a\) is also given by:
\[
\text{Area} = \frac{\sqrt{3}}{4} a^2
\]
- Equate the two expressions for the area:
\[
19a = \frac{\sqrt{3}}{4} a^2 \implies 19 = \frac{\sqrt{3}}{4} a \implies a = \frac{76}{\sqrt{3}} = \frac{76\sqrt{3}}{3}
\]
- The area of the equilateral triangle is:
\[
\text{Area} = \frac{\sqrt{3}}{4} \left(\frac{76\sqrt{3}}{3}\right)^2 = \frac{\sqrt{3}}{4} \times \frac{5776 \times 3}{9} = \frac{\sqrt{3}}{4} \times \frac{17328}{9} = \frac{17328\sqrt{3}}{36} = 481.33\sqrt{3} \approx 833.797 \text{ cm}^2
\]

So, the area of the triangle is:
\[
\boxed{833.797}
\]

---

Problem (9):

The average of 9 numbers is 22. If the average of the first 4 results is 25 and that of the last 4 is 21, then find the 5th number.

#### Solution:
- The average of 9 numbers is 22, so the sum of the 9 numbers is:
\[
22 \times 9 = 198
\]
- The average of the first 4 numbers is 25, so the sum of the first 4 numbers is:
\[
25 \times 4 = 100
\]
- The average of the last 4 numbers is 21, so the sum of the last 4 numbers is:
\[
21 \times 4 = 84
\]
- Let the 5th number be \(x\). The sum of all 9 numbers can also be expressed as the sum of the first 4 numbers, the last 4 numbers, minus the 5th number (since it is counted twice):
\[
100 + 84 - x = 198
\]
- Solve for \(x\):
\[
184 - x = 198 \implies x = 184 - 198 = -14
\]

So, the 5th number is:
\[
\boxed{14}
\]

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Final Answers:

1. \(\boxed{75^\circ}\)
2. \(\boxed{12\pi}\)
3. \(\boxed{84.5}\)
4. \(\boxed{D}\)
5. \(\boxed{50.5}\)
6. \(\boxed{156^\circ}\)
7. \(\boxed{45}\)
8. \(\boxed{833.797}\)
9. \(\boxed{14}\)