Mensuration
- A wire of length 44 cm is first bent to form a circle and then rebent to form a square. The difference of the two enclosed areas is
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Using Rule 14,
Circumference of circle
= 2πr = 44⇒ 2 × 22 × r = 44 7 ⇒ r = 44 × 7 = 7cm. 2 × 22
Area of circle = πr²= 22 × 7 × 7 = 154 sq.cm. 7
Perimeter of square = 44 cm.Side of square = 44 = 11 cm. 4
Area of square = 11 × 11 = 121 sq. cm.
Difference = 154 – 121 = 33 sq. cm.Correct Option: B
Using Rule 14,
Circumference of circle
= 2πr = 44⇒ 2 × 22 × r = 44 7 ⇒ r = 44 × 7 = 7cm. 2 × 22
Area of circle = πr²= 22 × 7 × 7 = 154 sq.cm. 7
Perimeter of square = 44 cm.Side of square = 44 = 11 cm. 4
Area of square = 11 × 11 = 121 sq. cm.
Difference = 154 – 121 = 33 sq. cm.
- A parallelogram has sides 60 m and 40m and one of its diagonals is 80 m long. Its area is
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Using Rule 1,
Semiperimeter of ∆ABC(s) = a + b + c 2 = 60 + 40 + 80 = 90 metre 2
∴ Area of ∆ABC = √s(s - a)(s - b)(s - c)
= √90 (90 – 60) (90 – 40) (90 – 80)
= √90 × 30 × 50 × 10
= √3 × 30 × 30 × 5 × 10 × 10
= 30 × 10√15
= 300√15 sq. metre
∴ Area ∎ of ABCD
= 2 × Area of ∆ABC
= 2 × 300√15
= 600√15 sq. metreCorrect Option: B
Using Rule 1,
Semiperimeter of ∆ABC(s) = a + b + c 2 = 60 + 40 + 80 = 90 metre 2
∴ Area of ∆ABC = √s(s - a)(s - b)(s - c)
= √90 (90 – 60) (90 – 40) (90 – 80)
= √90 × 30 × 50 × 10
= √3 × 30 × 30 × 5 × 10 × 10
= 30 × 10√15
= 300√15 sq. metre
∴ Area ∎ of ABCD
= 2 × Area of ∆ABC
= 2 × 300√15
= 600√15 sq. metre
- ∠ACB is an angle in the semicircle of diameter AB = 5 and AC : BC = 3 : 4. The area of the triangle ABC is
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Angle at the semi-circle is a right angle.
∴ ∠ACB = 90°
AB = 5 cm.
AC = 3x cm. BC = 4x cm.
∴ (3x)² + (4x)² = (5)²
⇒ 9x² + 16x² = 25
⇒ 25x² = 25
⇒ x² = 1
⇒ x = 1∴ Area of ∆ABC = 1 × BC × AC 2 = 1 × 4 × 3 = 6 sq.cm. 2 Correct Option: D
Angle at the semi-circle is a right angle.
∴ ∠ACB = 90°
AB = 5 cm.
AC = 3x cm. BC = 4x cm.
∴ (3x)² + (4x)² = (5)²
⇒ 9x² + 16x² = 25
⇒ 25x² = 25
⇒ x² = 1
⇒ x = 1∴ Area of ∆ABC = 1 × BC × AC 2 = 1 × 4 × 3 = 6 sq.cm. 2
- If the perimeter of a right-angled triangle is 56 cm and area of the triangle is 84 sq. cm, then the length of the hypotenuse is (in cm)
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a + b + c = 56 ...(i)1 ac = 84 2
⇒ ac = 168 sq.cm.
∴ b² = a² + c²
⇒ b² = (a + c)² – 2ac
⇒ b² = (56 – b)² – 2 × 168 [By (i)]
⇒ b² = 3136 – 112 b + b² – 336
⇒ 112b = 2800⇒ b = 2800 = 25cm 112 Correct Option: A
a + b + c = 56 ...(i)1 ac = 84 2
⇒ ac = 168 sq.cm.
∴ b² = a² + c²
⇒ b² = (a + c)² – 2ac
⇒ b² = (56 – b)² – 2 × 168 [By (i)]
⇒ b² = 3136 – 112 b + b² – 336
⇒ 112b = 2800⇒ b = 2800 = 25cm 112
- The length and perimeter of a rectangle are in the ratio 5:18. Then length and breadth will be in the ratio
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l = 5 ⇒ l = 5 2(l + b) 18 l + b 9 ⇒ l + b = 9 ⇒ l + b - 1 = 9 - 1 l 5 l 5 ⇒ b = 4 l 5
⇒ l : b = 5 : 4Correct Option: C
l = 5 ⇒ l = 5 2(l + b) 18 l + b 9 ⇒ l + b = 9 ⇒ l + b - 1 = 9 - 1 l 5 l 5 ⇒ b = 4 l 5
⇒ l : b = 5 : 4
