Area and perimeter Questions and answers

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Elementary Mathematics

41). One side of a parallelogram is 8.06 cm and its perpendicular distance from opposite side is 2.08 cm. What is the approximate area of the parallelogram ?

A). \( \Large 12.56 cm^{2} \)

B). \( \Large 14.56 cm^{2} \)

C). \( \Large 16.76 cm^{2} \)

D). \( \Large 22.56 cm^{2} \)

View Answer

42). Floor of a square room of side 10 m is to be completely covered with square tiles, each having length 50 cm. The smallest number of tiles needed is

A). 200

B). 300

C). 400

D). 500

View Answer

Correct Answer: 400

Area of square room = \( \Large (10)^{2} \) = 100 sq

= \( \Large 100\times (100)^{2} \) sq cm

= \( \Large 100\times 100\times 100 \) sq cm

Now, area of tile = \( \Large (50)^{2}=50\times 50 \) sq cm

Number of tiles needed

= \( \Large \frac{Area \ of \ square \ room}{Area \ of \ tile} \)

= \( \Large \frac{100\times 100\times 100}{50\times 50} \) =400

Hence, 400 tiles will be needed.

43). The area of a trapezium is \( \Large \Large 384 cm^{2} \). If its parallel sides are in the ratio 3 : 5 and the perpendicular distance between them is 12 cm, the smaller of the parallel sides is

A). 20 cm

B). 24 cm

C). 30 cm

D). 36 cm

View Answer

Correct Answer: 24 cm

Let the sides of trapezium be 5x and respectively.

According to the question,

\( \Large \frac{1}{2}\times(5x+3x)\times 12 \)=384

=> 8x=\( \Large \frac{384\times 2}{12} \) => x=\( \Large \frac{64}{8} \)=8 cm

Length of smaller of the parallel sides

= \( \Large 8\times 3 \) = 24 cm

44). In a trapezium, the two non-parallel sides are equal in length, each being of 5 units. The parallel sides are at a distance of 3 units apart. If the smaller side of the parallel sides is of length 2 units, then the sum of the diagonals of the trapezium is

A). \( \Large 10\sqrt{5} \) units

B). \( \Large 6\sqrt{5} \) units

C). \( \Large 5\sqrt{5} \) units

D). \( \Large 3\sqrt{5} \) units

View Answer

Correct Answer: \( \Large 6\sqrt{5} \) units

In \( \Large \triangle BCF \), by Pythagoras theorem,

\( \Large (5)^{2}=(3)^{2}+(BF)^{2} \)

BF=4 cm

AB=2+4+4=10 cm

Now, in \( \Large \triangle ACF \), \( \Large AC^{2}=CF^{2}+FA^{2} \)

=> \( \Large AC^{2}=3^{2}+6^{2} \) => AC=\( \Large \sqrt{45} \)cm

Similarly, BD = \( \Large \sqrt{45} \)cm

Sum of diagonals

=AC+BD= \( \Large \sqrt{45}+\sqrt{45}=2\sqrt{45}=6\sqrt{5} \)cm

45). The ratio of the length of the parallel sides of a trapezium is 3 : 2. The shortest distance between them is 15 cm. If the area of the trapezium is 450 sq cm, find the sum of the lengths of the parallel sides.

A). 15 cm

B). 36 cm

C). 42 cm

D). 60 cm

View Answer

Correct Answer: 60 cm

Let the lengths of parallel sides an 3x and 2x.

We know that,

Area of trapezium = \( \Large \frac{1}{2} \) \( \Large (Sum \ of \ the \ parallel \ sides)\times Distance \ between \ them \)

\( \Large \frac{1}{2}(3x+2x)15=450 \)=>75x=900

x=\( \Large \frac{900}{75} \)=12 cm

Sum of the parallel sides = (3x + 2x)

= 5x = \( \Large 5\times 12 \)=60 cm

46). If area of a regular pentagon is \( \Large \Large 125\sqrt{3} \) units sq cm, how long is its each side

A). 10 cm

B). 15 cm

C). 16 cm

D). 25 cm

View Answer

Correct Answer: 10 cm

We know that,

Area of regular pentagon = \( \Large 5a^{2}\frac{\sqrt{3}}{4} \)

According to the question,

\( \Large \frac{5a^{2}\sqrt{3}}{4}=125\sqrt{3} \)=> \( \Large a^{2}=\frac{125\sqrt{3}\times 4}{5\sqrt{3}} \)=100

a = 10 cm

47). The sides of a parallelogram are 12 cm and 8 cm long and one of the diagonals is 10 cm long. If d is the length of other diagonal, then which one of the following is correct?

A). d < 8 cm

B). 8 cm < d < 10cm

C). 10 cm < d < 12 cm

D). d > 12 cm

View Answer

Correct Answer: d > 12 cm

In parallelogram.

\( \Large d_{1}^{2} + d_{2}^{2} = 2(l^{2} + b^{2})\)

\( \Large d^{2}+(10)^{2} \)=2(64+144)

=> \( \Large d^{2}=2\times 208-100 \)

=> \( \Large d^{2} \)=416-100=316

=> d=\( \Large \sqrt{316} \)

=> d=17.76 cm => d > 10

48). The radius of a circular field is 25 m. Find the area of the field.

A). \( \Large 125\pi \)sq m

B). \( \Large 625\pi \)sq m

C). \( \Large 25\pi \)sq m

D). \( \Large 135\pi \)sq m

View Answer

49). A railing of 288 m is required for fencing a semi-circular park. Find the area of the park. \( \Large ( \pi = 22/7) \)

A). \( \Large 4928 m^{2} \)

B). \( \Large 9865 m^{2} \)

C). \( \Large 8956 m^{2} \)

D). \( \Large 9856 m^{2} \)

View Answer

Correct Answer: \( \Large 4928 m^{2} \)

Let the radius of the park be r, then

\( \Large \pi r+2r \)=288

\( \Large (\pi+2)r \)=288

=> \( \Large \left(\frac{22}{7}+2 \right)r \)=288

=> r=\( \Large \frac{288\times 7}{36} \)=56

Area of the park = \( \Large \frac{1}{2}\pi r^{2} \)

= \( \Large \frac{1}{2}\times \frac{22}{7} \times 56\times 56 \)

= 4928

50). The area of a sector of a circle of radius 36 cm is \( \Large \Large 72\pi cm^{2} \). The length of the Corresponding are of the sector 1s

A). \( \Large \pi cm \)

B). \( \Large 2\pi cm \)

C). \( \Large 3\pi cm \)

D). \( \Large 4\pi cm \)

View Answer

Correct Answer: \( \Large 4\pi cm \)

Given, area of sector = \( \Large 72\pi cm^{2} \)

or \( \Large \frac{\pi r^{2}\theta}{360 degree}=72\pi \)

\( \Large \theta=\frac{72\times 360}{36\times 36} \)=> \( \Large \theta \)=20 degree

Now,length of arc

=\( \Large \frac{\pi r\theta}{180 degree}=\frac{\pi\times 36\times 20 degree}{180 degree}=4\pi cm \)

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