**Example 1:** The ratio of sides of two squares is 4:5. What is the ratio of their areas?

**Sol :** Ratio of areas of the squares = (Ratio of sides)2

(4:5)2 = 16 : 25

**Example 2:** If the area of a square is increases by 69%, then the side of the square was increased by ?

**Sol :** Now, area of the square has increased by 69%. Therefore, it has become 1.69 times of itself. Now, increase in the side = sq. rt (1.69) = 1.3. Therefore, the correct answer is 40%.

**Example 3:** A man cycling at the speed of 8km/hr crosses a square field diagonally in 6 minutes. The area of the field is

**Sol :** Speed of the man is given as 8km/hr = 20/9 m/sec.

Time taken to cover this distance is six minutes or 360 seconds.

Therefore, length of the diagonal = (20/9)(360) = 200m.

Now, area of square is ½ (d)^{2} = ½ (200)^{2} = 2(10)^{4} m^{2}

**Example 4:** A cube of 600cm^{2} surface area is melted to make x small cubes each of 96mm^{2} surface area. X is

**Sol :** Surface area of larger cube = 600cm^{2}

6L^{2} = 600 *L*^{2} = 100 *L* = 10cm.

volume of larger cube = 10 10 10 = 1000 cm^{2} . Surface area of smaller cube = 100 mm^{2} = 100/100

cm^{2} . 6l^{2} = (100/100) ⇒ l^{2} = (1/6)cm

Volume of smaller cube = 1/6 1/6 1/6

= (1/216)cm^{2} .

x = 1000/1/216 = 216000

**Example 5:** The perimeter of an isosceles right angled triangle is 40m, Find the area of this particular triangle.

**Sol :** In an isosceles right angled triangle, Height = Base. Let a be the base of this triangle and b be the hypotenuse of this triangle

a + a + b = 40 2a + b = 40.

Also b^{2} = a^{2} + a^{2}

b^{2} = 2a^{2}.

b = √a. So 2a + √2a = 40⇒ 3.144a = 40

a = 11.72m.

Required area = (1/2)a^{2} = (1/2)×(11.72)(11.72)=68.64m^{2}

Must Read Geometry Articles

- Geometry: Solved Examples

**Example 6:** Two cubes, each of edge 30 m, are joined to form a single cuboid. What is the surface area of the new cuboid so formed.

**Sol :** Here length of cuboid = 30 m.

Breadth and height of cuboid = 15 m.

surface area = 2(450 + 225 + 450) = 2250 m^{2}

**Example 7:** A copper sphere of diameter 36cm is drawn into a wire of diameter 8mm. Find the length of the wire.

**Sol :** In this case,4/3π*R*^{3} = *πr*^{2}h where R is the radius of sphere, r is the radius of the wire and h is the length of the wire.

Hence 4/3 (36/2)^{3} = (8/10.h/2)^{2}

So h =48600cm

**Example 8:** Find the ratio of the volumes of a cube to that of a sphere, which will fit inside the cube.

**Sol :** Let the edge of the cube = a

Radius of sphere = a/2.

Volume of cube = a^{3}.

Volume of sphere = 4π/3(a/2)^{3} = πa^{3}/6 ∴ required ratio = a^{3}/6 ⇒ 6:π

**Example 9:** The difference between the circumference & radius of a circle is 37 meters. Find its circumference.

**Sol :** 2πr - r =37 ⇒ r(2π - 1) = 37

(2× (22/7) - -) = 37 ⇒ r(37/7) =37 ⇒ r = 7

r ∴ 2πr-7 = 37 ⇒ 2πr = 44m.

**Example 10:** Find the area of the sheet metal required to make a hollow cone 24cm high whose base diameter is 14 cm.

**Sol :** πrl.r =14/2 = 7

Now 1 = √24^{2} + √7^{2} = √625 = 25cm

Required area = 22/7 × 7 × 25 = 550cm^{2}