Geometry: Solved Examples

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
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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 m2
Example 4: A cube of 600cm2 surface area is melted to make x small cubes each of 96mm2 surface area. X is
Sol : Surface area of larger cube = 600cm2
6L2 = 600 L2 = 100 L = 10cm.
volume of larger cube = 10 10 10 = 1000 cm2 . Surface area of smaller cube = 100 mm2 = 100/100
cm2 . 6l2 = (100/100) ⇒ l2 = (1/6)cm
Volume of smaller cube = 1/6 1/6 1/6
= (1/216)cm2 .
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 b2 = a2 + a2
b2 = 2a2.
b = √a. So 2a + √2a = 40⇒ 3.144a = 40
a = 11.72m.
Required area = (1/2)a2 = (1/2)×(11.72)(11.72)=68.64m2
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 m2
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,4/3πR3 = πr2h 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 = a3.
Volume of sphere = 4π/3(a/2)3 = πa3/6 ∴ required ratio = a3/6 ⇒ 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.
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Sol : πrl.r =14/2 = 7
Now 1 = √242 + √72 = √625 = 25cm
Required area = 22/7 × 7 × 25 = 550cm2