Mensuration
- In an equilateral triangle ABC of side 10cm, the side BC is trisected at D. Then the length (in cm) of AD is
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AE ⊥ BC
∴ BE = EC = 5 cm
AC = 10 cm
AE = √10² - 5²
= √100 - 25 = √75
= 5√3 cm
DE = DC – EC= 2 × 10 - 5 = 5 cm. 3 3 ∴ AD = √ 
5 
² + (5√3)² 3 = √ 25 + 75 9 = √ 25 + 675 9 = √ 700 = 10√7 cm 9 3 Correct Option: C
AE ⊥ BC
∴ BE = EC = 5 cm
AC = 10 cm
AE = √10² - 5²
= √100 - 25 = √75
= 5√3 cm
DE = DC – EC= 2 × 10 - 5 = 5 cm. 3 3 ∴ AD = √ 5 ² + (5√3)² 3 = √ 25 + 75 9 = √ 25 + 675 9 = √ 700 = 10√7 cm 9 3
- The perimeter of a triangle is 40cm and its area is 60 cm². If the largest side measures 17cm, then the length (in cm) of the smallest side of the triangle is
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Smallest side of the triangle = x cm (let)
∴ Second side of triangle = 40 – 17 – x = 23 – xSemi-perimeter = s = 40 = 20 2
∴ √s (s - a)(s - b)(s - c)
⇒ √20(20 - 17)(20 - x)(20 - 23 - x) = 60
⇒ (20 - x)(x - 3) = 60
⇒ (20 – x) (x – 3) = 60
⇒ 20x – 60 –x² + 3x = 60
⇒ x² – 23x + 120 = 0
⇒ x² – 15x – 8x + 120 = 0
⇒ x (x – 15) – 8 (x – 15) = 0
⇒ (x – 8) (x – 15) = 0
⇒ x = 8 or 15
⇒ Smallest side = 8 cmCorrect Option: C
Smallest side of the triangle = x cm (let)
∴ Second side of triangle = 40 – 17 – x = 23 – xSemi-perimeter = s = 40 = 20 2
∴ √s (s - a)(s - b)(s - c)
⇒ √20(20 - 17)(20 - x)(20 - 23 - x) = 60
⇒ (20 - x)(x - 3) = 60
⇒ (20 – x) (x – 3) = 60
⇒ 20x – 60 –x² + 3x = 60
⇒ x² – 23x + 120 = 0
⇒ x² – 15x – 8x + 120 = 0
⇒ x (x – 15) – 8 (x – 15) = 0
⇒ (x – 8) (x – 15) = 0
⇒ x = 8 or 15
⇒ Smallest side = 8 cm
- The ratio of the area of two isosceles triangles having the same vertical angle (i.e. angle between equal sides) is 1 : 4. The ratio of their heights is
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Both the triangles are equiangular.
⇒ These are similar triangles.
∴ Ratio of their height = Square root of ratio of their area = 1 : 2Correct Option: C
Both the triangles are equiangular.
⇒ These are similar triangles.
∴ Ratio of their height = Square root of ratio of their area = 1 : 2
- The length of one side of a rhombus is 6.5 cm and its altitude is 10 cm. If the length of its diagonal be 26 cm, the length of the other diagonal will be :
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We know that if all the sides of a parallelogram are equal, it is called a rhombus.
Area = Base × Height = 6.5 × 10 = 65 cm²
Let the diagonals of the rhombus be d1 and d2.∴ Area = 1 d1d2 2 ⇒ 65 = 1 × 26 × d2 2 ⇒ d2 = 2 × 65 ⇒ d2 = 5 2
Hence, other diagonal of rhombus = 5 cm.Correct Option: A
We know that if all the sides of a parallelogram are equal, it is called a rhombus.
Area = Base × Height = 6.5 × 10 = 65 cm²
Let the diagonals of the rhombus be d1 and d2.∴ Area = 1 d1d2 2 ⇒ 65 = 1 × 26 × d2 2 ⇒ d2 = 2 × 65 ⇒ d2 = 5 2
Hence, other diagonal of rhombus = 5 cm.
- The outer and inner radii of a hollow metallic cylinder of height 24 cms. are respectively 6.75 cms. and 5.25 cms. It is melted to form a solid sphere. Find the surface area of the sphere.
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Volume of metallic cylinder
= π(r1² - r2²) × h
= π(6.752 – 5.252) × 24 = π × 12 × 1.5 × 24 cu.cm.
∴ Volume of sphere = π × 12 × 1.5 × 24⇒ 4 πr³ = π × 12 × 1.5 × 24 3 ⇒ r³ = 12 × 1.5 × 24 × 3 = 12 × 3 × 3 × 3 4
∴ r = ³√12 × 3 × 3 × 3
= 3³√12
∴ Surface area = 4πr²
= 4 × π × (3³√12)²
= 4 × π × 9 (12)²/3
= π × 36 (12)²/3 sq.cm.Correct Option: B
Volume of metallic cylinder
= π(r1² - r2²) × h
= π(6.752 – 5.252) × 24 = π × 12 × 1.5 × 24 cu.cm.
∴ Volume of sphere = π × 12 × 1.5 × 24⇒ 4 πr³ = π × 12 × 1.5 × 24 3 ⇒ r³ = 12 × 1.5 × 24 × 3 = 12 × 3 × 3 × 3 4
∴ r = ³√12 × 3 × 3 × 3
= 3³√12
∴ Surface area = 4πr²
= 4 × π × (3³√12)²
= 4 × π × 9 (12)²/3
= π × 36 (12)²/3 sq.cm.
