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Given:

The hypotenuse of the right-angled isosceles triangle = 6√2 cm

Concept used:

Isosceles right-angled triangle:

• The two sides on the right angle are equal (let a).
• One angle is 90°
• Hypotenuse = a√2

Area of right triangle = \(\frac{1}{{2}}\) × base × height

Calculation:

According to the above concept,

Hypotenuse of the triangle = 6√2 cm

⇒ a√2 = 6√2

⇒ a = 6 cm

Now, the area of a right triangle

⇒ \(\frac{1}{{2}}\) × base × height 

⇒ (\(\frac{1}{{2}}\) × 6 × 6) cm²  = 18 cm²

∴ The required area is 18 cm².

Given:

The diameter of a circle circumscribing a square = 15√2 cm

Concept used:

By Pythagoras theorem

AC² = AB² + BC²

Calculation:

According to the above figure,

AC = 15√2 cm

Let the side of a square be x.

⇒ x² + x² = (15√2)²

⇒  2x² = 450

⇒ x² = 225

⇒ x = 15 cm

∴ The length of the side of the square is 15 cm.

Concept used:

1 mm = 0.1 cm

1 m = 100 cm

Calculation:

According to the above concept,

1 mm = 0.1 cm

⇒ 200 mm = 20 cm

Now, 1 m = 100 cm

⇒ 10 m = 1000 cm

Now, according to the question

⇒ (1000/20) = 50

∴ The required answer is 50.

Given:

The ratio of outer and inner perimeters of a circular path = 23 ∶ 22

The path is 5 m wide.

Concept used:

Circumference of circle = 2πr

Diameter = 2 × Radius

Calculation:

Let the radius of the outer circle be R.

And, the radius of the inner circle is r.

Now, according to the question

⇒ \(\frac{2\pi R}{{2\pi r}}\) = \(\frac{23}{{22}}\)

⇒ \(\frac{R}{{r}}\) = \(\frac{23}{{22}}\)

So, R = 23x and r = 22x    (x is common ratio)

According to the question

R - r = 5 

⇒ 23x - 22x = 5

⇒ x = 5 m

Now, the inner radius of the circle = 22x = (22 × 5) = 110 m

And, the diameter of the inner circle = (2 × 110) m = 220 m

∴  The diameter of the inner circle is 220 m.

Given:

The product of the lengths of the diagonals of a square = 50 square units.

Concept used:

Diagonal of square = a√2

Calculation:

Let the side of the square be a.

Diagonal of square = √2 a

Both the diagonals have the same length,

Product of length of diagonals = (a√2)² = 2a²

⇒ 50 = 2 × a²

⇒ a = 5 units

∴ The length of a side of the square is 5 units.

Concept used:

Volume of cylinder = πr²h

Volume of sphere = \(\frac{4}{{3}}\)πr³

Calculation:

The volume of the cylindrical log = πr²h

But h = 2r

⇒ Volume of the cylindrical log = 2πr³

The radius of the greatest possible sphere = r

So, the volume of this sphere = \(\frac{4\pi r^3}{{3}}\)

⇒ The required fraction = (\(\frac{4\pi r^3}{{3}}\)/2πr³) = 2/3

∴ The fraction of the original log which is cut away is 2/3.

Given:

The diagonal of a cube = 4√3 cm

Formula used:

Diagonal of cube = √3a

Volume of cube = a³

Calculation:

According to the question

The diagonal of a cube = 4√3 cm

⇒ √3a = 4√3

⇒ a = 4 cm

Now, volume of cube = a³

⇒ (4)³ = 64 cm³

∴ The required volume is 64 cu cm.

Given:

The size of a cube is increased by 100%

Concept used:

Total surface area of cube = 6a²

Calculation:

Let edge of a smaller cube be 1 m.

Then, edge of larger cube be 2 m.

Surface area of smaller cube = 6 × (1)² = 6 m²

And, surface area of larger cube = 6 × (2)² = 24 m²

% increase in surface area = (\(\frac{24-6}{{6}}\) × 100) = 300%

∴ The surface area of the cube increased is 300%.

Concept used:

Bounded plane figure: Simple plane closed figure bounded by three line segments is called a triangle

Explanation:

According to the above concept,

If a plane figure is bounded by straight lines only, Here at least three lines are required because the triangle is only a closed figure.

The least number of straight lines for a bounded plane figure is 3.

∴ The required answer is 3.

Given:

There are five lines in a plane, no two of which are parallel.

Concept used:

Max number of points of intersection = Number of ways of selecting two lines out of the given line

Calculation:

By using the above concept, 

⇒ Number of points of intersection = 5C2

⇒ Number of points of intersection = 

⇒ Number of points of intersection =  = 10

∴ The required value is 10.