Mensuration
- In a trapezium ABCD, AB and DC are parallel sides and ∠ADC = 90°. If AB = 15 cm, CD = 40 cm and diagonal AC = 41 cm. then the area of the trapezium ABCD is
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∠ADC = 90°
AC = 41 cm.
CD = 40 cm.
∴ AD = √AC² -CD²
= √41² - 40²
= √(41 + 40)(41 - 40)
= √81 = 9 cm.
∴ Area of trapezium ABCD= 1 (AB + CD) × AD 2 = 1 (15 + 40) × 9 2
== 1 × 55 × 9 2
= 247.5 sq. cm.Correct Option: C
∠ADC = 90°
AC = 41 cm.
CD = 40 cm.
∴ AD = √AC² -CD²
= √41² - 40²
= √(41 + 40)(41 - 40)
= √81 = 9 cm.
∴ Area of trapezium ABCD= 1 (AB + CD) × AD 2 = 1 (15 + 40) × 9 2
== 1 × 55 × 9 2
= 247.5 sq. cm.
- A circular wire of radius 42 cm is bent in the form of a rectangle whose sides are in the ratio of 6 : 5. The smaller side of the rectangle is (Take π = 22/7)
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Using Rule 7, Circumference of the circular wire = 2&pir
= 2 × 22 × 42 = 264 cm 7
⇒ Perimeter of rectangle = 264 cm
Let the sides of rectangle be 6x and 5x cm.
∴ 2 (6x + 5x) = 264
⇒ 2 × 11 x = 264⇒ x = 264 = 12 22
∴ The smaller side = 5x = 5 × 12 = 60 cm.Correct Option: A
Using Rule 7, Circumference of the circular wire = 2&pir
= 2 × 22 × 42 = 264 cm 7
⇒ Perimeter of rectangle = 264 cm
Let the sides of rectangle be 6x and 5x cm.
∴ 2 (6x + 5x) = 264
⇒ 2 × 11 x = 264⇒ x = 264 = 12 22
∴ The smaller side = 5x = 5 × 12 = 60 cm.
- The cost of levelling a circular field at 50 paise per square metre is Rs. 7700. The cost (in Rs.) of putting up a fence all round it at Rs. 1.20 per metre is (Use π = 22/7)
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Area of circular field
= Total Expenditure Rate per square metre = 
7700 
sq. metre 1 2
= (7700 × 2) sq. metre
= 15400 sq. metre
If radius of field = r metre then, pr² = 15400⇒ 22 r² = 15400 7 ⇒ r² = 15400 × 7 = 7 × 700 22
⇒ r = √7 × 7 × 100 = 70 metre
∴ Circumference of circular field = 2πr metre= 2 × 22 × 70 = 440 meter 7
∴ Total expenditure fencing = Rs. (440 × 1.2) = Rs. 528Correct Option: C
Area of circular field
= Total Expenditure Rate per square metre = 7700 sq. metre 1 2
= (7700 × 2) sq. metre
= 15400 sq. metre
If radius of field = r metre then, pr² = 15400⇒ 22 r² = 15400 7 ⇒ r² = 15400 × 7 = 7 × 700 22
⇒ r = √7 × 7 × 100 = 70 metre
∴ Circumference of circular field = 2πr metre= 2 × 22 × 70 = 440 meter 7
∴ Total expenditure fencing = Rs. (440 × 1.2) = Rs. 528
- The sum of the length and breadth of a rectangle is 6 cm. A square is constructed such that one of its sides is equal to a diagonal of the rectangle. If the ratio of areas of the square and rectangle is 5 : 2, the area of the square in cm² is
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Let, AB = a cm.
BC = b cm.
According to the question, a + b = 6 ..... (i)
and diagonal of rectangle
= √a² + b² = side of square∴ Area of square Area of rectangle = (√a² + b²)² ab ⇒ 5 = a² + b² 2 ab ⇒ 5 = a² + b² 4 2ab ⇒ 5 + 4 = a² + b² + 2ab 5 - 4 a² + b² - 2ab
[By componendo and dividendo]⇒ 9 = (a + b)² 1 (a - b)² ⇒ 9 = 6 × 6 1 (a - b)² ⇒ (a - b)² = 6 × 6 = 4 9
⇒ a – b = 2 ..... (ii)
∴ Area of square = a² + b²= 1 [(a + b)² + (a - b)²] 2 = 1 (36 + 4) 2 = 1 × 40 = 20 sq. cm. 2 Correct Option: A
Let, AB = a cm.
BC = b cm.
According to the question, a + b = 6 ..... (i)
and diagonal of rectangle
= √a² + b² = side of square∴ Area of square Area of rectangle = (√a² + b²)² ab ⇒ 5 = a² + b² 2 ab ⇒ 5 = a² + b² 4 2ab ⇒ 5 + 4 = a² + b² + 2ab 5 - 4 a² + b² - 2ab
[By componendo and dividendo]⇒ 9 = (a + b)² 1 (a - b)² ⇒ 9 = 6 × 6 1 (a - b)² ⇒ (a - b)² = 6 × 6 = 4 9
⇒ a – b = 2 ..... (ii)
∴ Area of square = a² + b²= 1 [(a + b)² + (a - b)²] 2 = 1 (36 + 4) 2 = 1 × 40 = 20 sq. cm. 2
- The length of a side of an equilateral triangle is 8 cm. The area of the region lying between the circum circle and the incircle of the triangle is (Use : π = 22/7)
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Radius of in circle = a cm. 2√3 Radius of circum-circle = a cm. √3
Where a = side of triangle
∴ Required area = Area of circum-circle – area of in–circle= π 
a 
² - a ² √3 2√3 = π a² - a² 3 12 = π 4a² - a² 12 = 3a²π = πa² 12 4 = 22 × 8 × 8 7 4 = 352 = 50 2 sq. cm. 7 7 Correct Option: B
Radius of in circle = a cm. 2√3 Radius of circum-circle = a cm. √3
Where a = side of triangle
∴ Required area = Area of circum-circle – area of in–circle= π a ² - a ² √3 2√3 = π a² - a² 3 12 = π 4a² - a² 12 = 3a²π = πa² 12 4 = 22 × 8 × 8 7 4 = 352 = 50 2 sq. cm. 7 7
