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Mensuration Exercise 2

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  1. Question 1 of 25
    1. Question

    In a parallelogram, the length of one diagonal and the perpendicular dropped on that diagonal are 30 and 20 metres respectively. Find its area.

    Hint

    In a parallelogram,

    Area = Diagonal × length of perpendicular on it

    = 30 × 20 = 600 m²

  2. Question 2 of 25
    2. Question

    ABCD is a parallelogram. P, Q, R and S are points on sides AB, BC, CD and DA, respectively such that AP = DR. If the area of the rectangle ABCD is 16 cm², then the area of the quadrilateral PQRS is:
    41995.png

    Hint

    Area of the quadrilateral PQRS

    = Area of ∆SPR + Area of ∆PQR

    mensuration-q-42182.png

    mensuration-q-42175.png

    mensuration-q-42169.png (Since PR = AD and AP + PB = AB)

    mensuration-q-42163.png

    mensuration-q-42157.png

  3. Question 3 of 25
    3. Question

    The cross section of a canal is a trapezium in shape. If the canal is 7 metres wide at the top and 9 metres at the bottom and the area of cross-section is 1280 square metres, find the length of the canal.

    Hint

    mensuration-q-42568.png

    Let the length of canal = h m. Then,

    area of canal

    mensuration-q-42562.png

    ⇒ mensuration-q-42556.png

    mensuration-q-42550.png

  4. Question 4 of 25
    4. Question

    A triangle and a parallelogram are constructed on the same base such that their areas are equal. If the altitude of the parallelogram is 100 m, then the altitude of the triangle is:

    Hint

    Let the common base be x m.

    Now, area of the triangle

    = area of the parallelogram

    ½ × x × altitude of the triangle = x × 100

    Altitude of the triangle = 200 m

  5. Question 5 of 25
    5. Question

    Two sides of a plot measure 32 metres and 24 metres and the angle between them is a perfect right angle. The other two sides measure 25 metres each and the other three angles
    43949.png What is the area of the plot?

    Hint

    mensuration-q-44097.png

    (32 – y)² + (24 -x)² = 625 ….(1)

    x² + y² = 625 …(2)

    ⇒ (24)² + (32)² – 64y – 48x = 0

    From (1) & (2)

    ⇒ 64y – 48x = 576 + 1024

    ⇒ 4y + 3x = 36 + 64 = 100

    ⇒ mensuration-q-44091.png

    ∴ mensuration-q-44085.png

    From (2)

    ⇒ -600x +16x² + 10000 + 9x² = 625 × 16

    ⇒ 25x² – 600x + 10000 – 625 × 16 = 0

    ⇒ x = 24 and y = 7

    ∴ Area = (24 × 25) + ½ 24 × 7 = 684

  6. Question 6 of 25
    6. Question

    One diagonal of a rhombus is 24cm side is 13 cm Find the area of the rhombus.

    Hint

    (side)² = (½ × one diagonal)² + (½ × other diagonal)²

    ⇒ 13² = (½ × one diagonal)² + (½ × 24)²

    ⇒ 169 – 144 = (½ × diagonal)²

    ⇒ 25 = (½ × diagonal)²

    ⇒ 5 = ½ × diagonal

    ∴ diagonal = 10

    ∴ Area = ½ × 10 x 24

    = 120 sq. cm.

  7. Question 7 of 25
    7. Question

    When the circumference and area of a circle are numerically equal, then the diameter is numerically equal to

    Hint

    According to question, circumference of circle = Area of circle

    ⇒ mensuration-q-41501.png [where d = diameter]

    ∴ d = 4

  8. Question 8 of 25
    8. Question

    The diameter of a garden roller is 1.4 m and it is 2 m long. How much area will it cover in 5 revolutions? [use π = ²²⁄₇]

    Hint

    Required area covered in 5 revolutions

    = 5 × 2πrh

    = 5 × 2 × ²²⁄₇ × 0.7 × 2 = 44 m²

  9. Question 9 of 25
    9. Question

    A cow is tethered in the middle of a field with a 14 feet long rope. If the cow grazes 100 sq. ft. per day, then approximately what time will be taken by the cow to graze the whole field?

    Hint

    Area of the field grazed

    = mensuration-q-41711.pngsq. ft.

    = 616 sq. ft.

    Number of days taken to graze the field

    = 616/100 days

    = 6 days (approx)

  10. Question 10 of 25
    10. Question

    A wire is looped in the form of a circle of radius 28 cm. It is re-bent into a square form. Determine the length of a side of the square.

    Hint

    Length of the wire = Perimeter of the circle

    = 2π × 28 = 176 cm²

    Side of the square

    = ¹⁷⁶⁄₄ = 4cm

  11. Question 11 of 25
    11. Question

    Semi-circular lawns are attached to both the edges of a rectangular field measuring 42 m × 35 m. The area of the total field is:

    Hint

    Area of the field

    mensuration-q-42192.png

    = 1470 + 1386 + 962.5

    = 3818.5 m²

  12. Question 12 of 25
    12. Question

    A circle and a rectangle have the same perimeter. The sides of the rectangle are 18 cm and 26 cm. What is the area of the circle?

    Hint

    Perimeter of the circle

    = 2πr = 2(18 + 26)

    ⇒ mensuration-q-42669.png

    ⇒ r = 14

    ∴ Area of the circle

    = mensuration-q-42663.png.

  13. Question 13 of 25
    13. Question

    A horse is tethered to one corner of a rectangular grassy field 40 m by 24 m with a rope 14 m long. Over how much area of the field can it graze?

    Hint

    mensuration-q-42657.png

    Area of the shaded portion

    mensuration-q-42650.png

    = 154 m²

  14. Question 14 of 25
    14. Question

    How many plants will be there in a circular bed whose outer edge measure 30 cms, allowing 4 cm² for each plant?

    Hint

    Circumference of circular bed = 30 cm

    Area of circular bed

    mensuration-q-42644.png

    Space for each plant = 4 cm²

    ∴ Required number of plants

    mensuration-q-42638.png

  15. Question 15 of 25
    15. Question

    From a square piece of a paper having each side equal to 10 cm, the largest possible circle is being cut out. The ratio of the area of the circle to the area of the original square is nearly :

    Hint

    Area of the square = (10)² = 100 cm²

    The largest possible circle would be as shown in the figure below :

    mensuration-q-42632.png

    Area of the circle

    mensuration-q-42626.png

    Required ratio

    mensuration-q-42620.png

    = 0.785 ≈ 0.8 = ⅘

  16. Question 16 of 25
    16. Question

    If the area of a circle decreases by 36%, then the radius of a circle decreases by

    Hint

    If area of a circle decreased by x % then the radius of a circle decreases by

    mensuration-q-42614.png

    = mensuration-q-42608.png

    mensuration-q-42602.png

    mensuration-q-42596.png

  17. Question 17 of 25
    17. Question

    A circular grass lawn of 35 metres in radius has a path 7 metres wide running around it on the outside. Find the area of path.

    Hint

    Radius of a circular grass lawn (without path) = 35 m

    ∴ Area = πr² = π (35)²

    Radius of a circular grass lawn ( with path)

    = 35 + 7 = 42 m

    ∴ Area = πr² = π(42)²

    ∴ Area of path = π(42)² – π(35)²

    = π(42² – 35²)

    = π( 42 + 35) (42 –35)

    = π × 77 × 7

    mensuration-q-42590.png

  18. Question 18 of 25
    18. Question

    Four equal circles are described about the four corners of a square so that each touches two of the others. If a side of the square is 14 cm, then the area enclosed between the circumferences of the circles is:

    Hint

    mensuration-q-43009.png

    The shaded area gives the required region.

    Area of the shaded region

    = Area of the square – area of four quadrants of the circles

    = (14)² – 4 × ¼ π (7)²

    mensuration-q-42997.png

  19. Question 19 of 25
    19. Question

    In a special racing event, the person who enclosed the maximum area would be the winner and would get Rs 100 every square metre of area covered by him/her. Jonnson, who successfully completed the race and was the eventual winner, enclosed the area shown in the figure below. What is the prize money won? (Note : The arc from C to D makes a complete semi-circle).
    AB = 3 m, BC = 10 m, CD = BE = 2 m
    44003.png

    Hint

    Area of the semicircle

    = π/2 = 1.571m²

    Area of ∆ ABE

    = ½ AB × BE

    = ½ 3 × 2 = 3m²

    Area of rectangle BCDE

    = 10 × 2 = 20 m²

    Total covered area

    = 1.571 + 3 + 20

    = 24.571 m²

    Prize money won

    = 24.571 × 100 = Rs 2457

  20. Question 20 of 25
    20. Question

    The figure shows a rectangle ABCD with a semi-circle and a circle inscribed inside it as shown. What is the ratio of the area of the circle to that of the semi-circle?
    43997.png

    Hint

    Let the radius of the semi- circle be R and that of the circle be r, then from the given data, it is not possible to express r in terms of R. Thus option (d) is the correct alternative.

  21. Question 21 of 25
    21. Question

    In ∆ACD, AD = AC and ∠C = 2∠E,. The distance between parallel lines AB and CD is h.
    44850.png
    Then
    I. Area of parallelogram ABCD
    II. Area of ∆ADE

    Hint

    mensuration-q-44826.png

    ∠A = ∠C 60° (alternative angles)

    ∠C = ∠D 60° ( since AC = AD and ∠A 60° )

    ∆ ACD is equilateral

    so its area mensuration-q-44819.png (where x is side)

    Area of parallelogram ABCD mensuration-q-44813.png

    Area of ∆ ADE = ½ × AD × AE

    = ½ × x × x tan 60°mensuration-q-44807.png

    Therefore we see,

    area of parallelogram ABCD = Area of ∆ADE

  22. Question 22 of 25
    22. Question

    Wheels of diameters 7 cm and 14 cm start rolling simultaneously from X and Y, which are 1980 cm apart, towards each other in opposite directions. Both of them make the same number of revolutions per second. If both of them meet after 10 seconds, the speed of the smaller wheel is:

    Hint

    Let each wheel make x revolutions per sec. Then,

    mensuration-q-44945.png

    mensuration-q-44939.png

    ⇒ 66x = 198

    ⇒ x = 3.

    Distance moved by smaller wheel in 3 revolutions

    mensuration-q-44933.png cm

    = 66 cm.

    ∴ Speed of smaller wheel

    = 66/3 cm/s = 22 cm/s.

  23. Question 23 of 25
    23. Question

    In the adjoining the figure, points A, B, C and D lie on the circle. AD = 24 and BC = 12. What is the ratio of the area of the triangle CBE to that of the triangle ADE
    44997.png

    Hint

    AD = 24, BC = 12

    In ∆BCE and ∆ADE

    since ∠CBA = ∠CDA (Angles by same arc)

    ∠BCE = ∠DAE (Angles by same arc)

    ∠BEC = ∠DEA (Opposite angles)

    BCE and DAE are similar ∆s

    with sides in the ratio 1 : 2

    Ratio of area = 1:4 ( i.e square of sides)

  24. Question 24 of 25
    24. Question

    The figure shows a circle of diameter AB and radius 6.5 cm. If chord CA is 5 cm long, find the area of triangle ABC
    44991.png

    Hint

    In the figure ∠ACB is 90°

    (angle subtended by diameter= 90°)

    AC = 5, AB = 13

    Using Pythagoras theorem,

    AB² = AC² + CB²

    mensuration-q-44920.png

    Area of ∆ ABC = ½ × 5 × 12 = 30

  25. Question 25 of 25
    25. Question

    There are two concentric circular tracks of radii 100 m and 102 m, respectively. A runs on the inner track and goes once round on the inner track in 1 min 30 sec, while B runs on the outer track in 1 min 32 sec. Who runs faster?

    Hint

    Radius of the inner track = 100 m

    and time = 1 min 30 sec = 90 sec.

    Also, Radius of the outer track = 102 m

    and time = 1 min 32 sec = 92 sec.

    Now, speed of A who runs on the inner track

    = mensuration-q-44914.png

    And speed of B who runs on the outer track

    = mensuration-q-44908.png

    Since, speed of A > speed of B

    ∴ A runs faster than B.

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