Question 1

The area of a square is equal to the area of a rectangle. The length of the rectangle is 5 cm more than a side of the square and its breadth is 3 cm less than the side of the square. What is the perimeter of the rectangle ?

Accepted Answer

Let the side of the square be 's' cm length of rectangle = (s+5) cm breadth of rectangle = (s-3)cm (s+5) (s-3) = s 2 - 5s - 3s - 15 = s 2 2s = 15 Perimeter of rectangle = 2(L+B) = 2(s+5 + s–3) = 2(2s + 2) = 2(15 + 2) = 34 cm

Question 2

The question given below contain two statements giving certain data. You have to decide whether the data given in the statements are sufficient for answering the question ? Question : What is the height of a right-angled triangle ? Statements : A. T…

Accepted Answer

From statement (A), 20b = (1/2) × b × h h = 40 cm.

Question 3

A room is half as long again as it is broad. The cost of carpeting the at Rs. 5 per sq.m is Rs. 270 and the cost of papering the four walls at Rs. 10 per sq.m is Rs. 1720. If a door and 2 windows occupy 8 sq. m, find the dimensions of the room.

Accepted Answer

Let breadth = x metres, length = 3x metres, height = H metres. Area of the floor=(Total cost of carpeting)/(Rate) = (270/5) sq.m = 54 sq.m x × 3 x 2 = 54 ⇒ x 2 = 54 × 2 ⇒ x = 6 So, breadth = 6 m and length = 3 6 2 = 9 m. Now, papered area = (1720/10) = 172 sq.m Area of 1 door and 2 windows = 8 sq.m Total area of 4 walls = (172 + 8) sq.m = 180 sq.m 2 × 9 + 6 H = 180 ⇒ H = 6

Question 4

A man runs round a circular field of radius 50m at the speed of 12 km/hr. What is the time taken by the man to take twenty rounds of the field?

Accepted Answer

speed = 12 km/h = 12 × 5 18 = 10 3 m / s distance covered = 20 × 2 × 22 7 × 50 = 44000 7 m time taken = distance /speed = 44000 7 × 3 10 s e c = 4400 × 3 7 × 1 60 m i n = 220 7 m i n

Question 5

A rectangular lawn 55m by 35m has two roads each 4m wide running in the middle of it. One parallel to the length and the other parallel to breadth. The cost of graveling the roads at 75 paise per sq meter is

Accepted Answer

area of cross roads = (55 x 4) + (35 x 4)- (4 x 4) = 344sq m cost of graveling = 344 x (75/100) = Rs. 258

Question 6

The cost of fencing a square field @ Rs. 20 per metre is Rs.10.080.How much will it cost to lay a three meter wide pavement along the fencing inside the field @ Rs. 50 per sq m

Accepted Answer

perimeter = total cost / cost per m = 10080 /20 = 504m side of the square = 504/4 = 126m breadth of the pavement = 3m side of inner square = 126 - 6 = 120m area of the pavement = (126 x126) - (120 x 120) = 246 x 6 sq m cost of pavement = 246*6*50 = Rs. 73800

Question 7

The length of a room is 5.5 m and width is 3.75 m. Find the cost of paying the floor by slabs at the rate of Rs. 800 per sq.metre.

Accepted Answer

Area = 5.5 × 3.75 sq. metre. Cost for 1 sq. metre. = Rs. 800 Hence total cost = 5.5 × 3.75 × 800 = 5.5 × 3000 = Rs. 16500

Question 8

The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is:

Accepted Answer

We have: (l - b) = 23 and 2(l + b) = 206 or (l + b) = 103. Solving the two equations, we get: l = 63 and b = 40. Area = (l x b) = (63 x 40) = 2520 sq.m

Question 9

The inner circumference of a circular race track, 14 m wide, is 440 m. Find radius of the outer circle

Accepted Answer

Let inner radius be r metres. Then, 2 π r = 440 ; r = 440 × 7 44 = 70 m. Radius of outer circle = (70 + 14) m = 84 m.

Question 10

The length of a rectangle is twice its breadth. If its length is decreased by 5 cm and breadth is increased by 5 cm, the area of the rectangle is increased by 75 sq.cm. Find the length of the rectangle?

Accepted Answer

Let the breadth of the given rectangle be x then length is 2x. thus area of the given rect is 2 x 2 after dec 5cm from length and inc 5cm breadth , new lenght becomes 2x-5 and breadth is x+5.thus new area =(2x-5)(x+5)= 2 x 2 + 5 x - 25 since new area is 75 units greater than original area thus 2 x 2 + 75 = 2 x + 5 x - 25 5x=75+25 5x=100 therefore x=20 hence length of the rectangle is 40 cm.

Question 11

The area of a rectangle is 460 square metres. If the length is 15% more than the breadth, what is the breadth of the rectangular field?

Accepted Answer

Let breadth=x meters. Then, Length = 115 x 100 meters Given that, x × 115 x 100 = 460 => x = 20 Breadth = 20

Question 12

A circular piece of thin wire is converted into a rhombus of side 11 cm. Find the diameter of the circular piece?

Accepted Answer

Circular piece is 4 x 11 = 44 cm long, Then Circumference of circle is given by, 44 = pi x D, where D is the diameter D = 44 / pi Take pi = 22 / 7, then D = 44 / (22/7) = (44 x 7) / 22 D = 14 cm.

Question 13

The area of a circle of radius 5 is numerically what percent its circumference?

Accepted Answer

required percentage = πr 2 2 πr × 100 = 250 %

Question 14

Find the length of a rope by which a cow must be tethered in order that it may be able to graze an area of 9856 sq meters.

Accepted Answer

clearly the cow will graze a circular field of area 9856 sq m and radius equal to the length of the rope. Let the length of the rope be r mts then, πr 2 = 9856 ⇒ r 2 = 9856 × 7 22 = 3136 ⇒ r = 56 m

Question 15

Square units 13 by 9 of an office area is

Accepted Answer

Square units 13 by 9 of an office means office of length 13 units and breadth 9 units. Now its area is 13x 9 = 117 square units or units square.

Question 16

Find the area of the trapezium, whose parallel sides are 12 cm and 10 cm, and distance between the parallel sides is 14 cm?

Accepted Answer

We know that, Area of trapezium = 1/2 x (Sum of parallel sides) x (Distance between Parallel sides) = 1/2 x (12 + 10) x 14 = 22 x 14/2 = 22 x 7 = 154 sq. cm

Question 17

The area of the circle is 220 sq.m, the area of the square inscribed in a circle will be

Accepted Answer

area of a square = 1 2 d i a g o n a l 2

Question 18

The sides of a right-angled triangle are 12 cm, 16 cm, 20 cm respectively. A new right angle Δ is made by joining the midpoints of all the sides. This process continues for infinite then calculate the sum of the areas of all the triangles so made.

Accepted Answer

128 sq.cm

Question 19

The difference between two parallel sides of a trapezium is 4 cm. perpendicular distance between them is 19 cm. If the area of the trapezium is 475 find the lengths of the parallel sides.

Accepted Answer

Let the two parallel sides of the trapezium be a cm and b cm. Then, a - b = 4 And, 1 2 × a + b × 19 = 475 = > a + b = 50 Solving (i) and (ii), we get: a = 27, b = 23. So, the two parallel sides are 27 cm and 23 cm.

Question 20

The length of a rectangular plot is 20 metres more than its breadth. If the cost of fencing the plot @ Rs. 26.50 per metre is Rs. 5300, what is the length of the plot in metres?

Accepted Answer

Let length of plot = L meters, then breadth = L - 20 meters and perimeter = 2[L + L - 20] = [4L - 40] meters [4L - 40] * 26.50 = 5300 [4L - 40] = 5300 / 26.50 = 200 4L = 240 L = 240/4= 60 meters.

Area Questions