**Q1. The area of the floor of a rectangular hall of length 40m is 960 m**(Unsolved Exercise 16.1 from RD Sharma)

^{2}. Carpets of size 6m x 4m are available. Find how many carpets are required to cover the hall.Answer:

length = 6m breadth = 4m and Area = l x b

$\therefore$ Area of carpet is = 6 x 4 = 24 m

^{2.}

Floor area = 960 m

^{2}

Number of carpets =

**Q2: Find the area of a square the length of whose diagonal is 2.9 meters.**(Unsolved Ex 16.1 from RDS)

Answer: Diagonal (d) = 2.9 m. Let x be the length of each side of the square.

Applying Pythagoras theorem,

$ x^ + x^ = 2x^ = (2.9)^2$

$\Rightarrow x^2 = \text {area of square} = \frac{(2.9)^2}{2} = \frac{8.41}{2} = 4.205 m^2$

**Q3: The area of a square field is 0.5 hectares. Find the length of its diagonal in meters.**

Answer: 1 hect = 10000 m2

$\therefore$ 0.5 hec = 5000 m2 $

Applying Pythagoras theorem,

$ \frac{diagonal^2}{2} = 5000$

$\Rightarrow diagonal = \sqrt{10000} = 100 m$

**Q4: The diameter of a semi-circular field is 14 meters. What is the cost of fencing the plot at Rs. 10 per meter.**

Answer: Diameter of field (d) = 14m

$\Rightarrow r = 14 \div 2 = 7m$

$ \text{Perimeter of semicircle} = (\pi + 2) \times r$

$ \Rightarrow = (\frac{22}{7} + 2) \times 7 = 36m $

Cost of fencing per meter = Rs 10

$\therefore \text{cost of fencing for 36m} = 36 \times 10 = \text{Rs }360$

**Q5: Find the area of the quadrilateral ABCD as shown. The diagonal AC is 84 units and perpendiculars BE and FD are 21 and 28 units respectively**.

Answer: The area of a general quadrilateral = $\frac{1}{2}d \times (h_1 + h_2).$

Here d = 84, h1 = 21 units and h2 = 28 units.

Therefore area = 84 * (21+28)/2 = 2058 square units. (Answer).

thanks..............my papers are going good for you.

ReplyDeleteit gives us idea of questions. thank u so much!!!!!!

ReplyDeletevery nice...........

ReplyDeleteya tnx 2 u and yr ans is so gud.,.,.,

ReplyDeleteNakwas

ReplyDeletefantastic

ReplyDeletethank you for giving me a wonderful platform

ReplyDelete