Mensuration
- If the area of a circle is A, radius of the circle is r and circumference of it is C, then
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Using Rule 14,
A = πr²
C = 2πr∴ A = πr² = r C 2πr 2
⇒ rC = 2 ACorrect Option: B
Using Rule 14,
A = πr²
C = 2πr∴ A = πr² = r C 2πr 2
⇒ rC = 2 A
- If D, E and F are the mid–points of the sides of an equilateral triangle ABC, then the ratio of the area of triangle DEF and DCF is :
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BD = DC = CF = AF = AE = BE = DE = EF = DF
∆BDE ≅ ∆DCF ≅ ∆AEF ≅ ∆DEF
∴ Area of ∆DEF = Area of ∆DCF
∴ Required ratio = 1 : 1Correct Option: D
BD = DC = CF = AF = AE = BE = DE = EF = DF
∆BDE ≅ ∆DCF ≅ ∆AEF ≅ ∆DEF
∴ Area of ∆DEF = Area of ∆DCF
∴ Required ratio = 1 : 1
- ∆ABC is similar to ∆DEF. If the area of ∆ABC is 9 sq.cm. and the area of ∆DEF is 16 sq.cm. and BC = 2.1 cm, then the length of EF will be
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∆ABC ~ ∆DEF
∴ Area of ∆ABC = BC² Area of ∆DEF EF² ⇒ 9 = (2.1)² 16 (EF)² ⇒ 3 = 2.1 4 EF ⇒ EF = 4 × 2.1 2.8 cm 3 Correct Option: B
∆ABC ~ ∆DEF
∴ Area of ∆ABC = BC² Area of ∆DEF EF² ⇒ 9 = (2.1)² 16 (EF)² ⇒ 3 = 2.1 4 EF ⇒ EF = 4 × 2.1 2.8 cm 3
- D and E are points on the sides AB and AC respectively of ∆ABC such that DE is parallel to BC and AD : DB = 4 : 5, CD and BE intersect each other at F. Then find the ratio of the areas of ∆DEF and ∆CBF.
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DE || BC
∠ADE = ∠ABC
∠AED = ∠ACB
By AA–similarity. ∆ABC ~ ∆ADE∴ AD = DE AB BC ∴ AD = 4 AB 5 ⇒ DB = 5 AD 4 ⇒ DB + AD = 5 + 4 AD 4 ⇒ AB = 9 = BC AD 4 DE
∆DEF ~ ∆CBF∴ Area of ∆DEF = DE² Area of ∆CBF BC² = 16 = 16 : 81 81 Correct Option: B
DE || BC
∠ADE = ∠ABC
∠AED = ∠ACB
By AA–similarity. ∆ABC ~ ∆ADE∴ AD = DE AB BC ∴ AD = 4 AB 5 ⇒ DB = 5 AD 4 ⇒ DB + AD = 5 + 4 AD 4 ⇒ AB = 9 = BC AD 4 DE
∆DEF ~ ∆CBF∴ Area of ∆DEF = DE² Area of ∆CBF BC² = 16 = 16 : 81 81
- In ∆ABC, D and E are two points on the sides AB and AC respectively so that DE||BC and
AD = 2 BD 3 Then the area of trapezium DECB is equal to the area of ∆ABC
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DE || BC
∴ ∠ADE = ∠ABC
∠AED = ∠ACB
∴ ∆ADE ~ ∆ABC∴ AD = 2 BD 3 ⇒ BD = 3 AD 2 = BD + 1 = 3 + 1 AD 2 ⇒ BD + AD AD ⇒ 3 + 2 ⇒ AB = 5 2 AD 2 ∴ Area of ∆ADE = AD² = 
2 
² = 4 Area of ∆ABC AB² 5 25 ⇒ 1 - Area of ∆ADE = 1 - 4 Area of ∆ABC 25 ⇒ Area of ∆ABC - Area of ∆ADE = 25 - 4 Area of ∆ABC 25 ∴ Area of trapezium DECB = 21 Area of ∆ABC 25 Correct Option: B
DE || BC
∴ ∠ADE = ∠ABC
∠AED = ∠ACB
∴ ∆ADE ~ ∆ABC∴ AD = 2 BD 3 ⇒ BD = 3 AD 2 = BD + 1 = 3 + 1 AD 2 ⇒ BD + AD AD ⇒ 3 + 2 ⇒ AB = 5 2 AD 2 ∴ Area of ∆ADE = AD² = 2 ² = 4 Area of ∆ABC AB² 5 25 ⇒ 1 - Area of ∆ADE = 1 - 4 Area of ∆ABC 25 ⇒ Area of ∆ABC - Area of ∆ADE = 25 - 4 Area of ∆ABC 25 ∴ Area of trapezium DECB = 21 Area of ∆ABC 25