Question

PLEASE ANSWER THE ALL QUESTIONS​ ILL MARK U AS BRAINLIEST​​

Answer 1

Question no (1)

Given -

Length (l) = 1.5 m

Breadth (b ) = 1.25 m

Height (h) = 65 cm 0.65m

Solution

(1) We know that the plastic box is open at the top so we have to reduce the area of top from area of plastic box .

Hence

(1)

$= (2(l + b) \times h + lb \: {m}^{2} \\ = (2(1.5 + 1.25) \times .65 + 1.5 \times 1.25) {m}^{2} \\ = (2 \times 2.75 \times .65 + 1.875) {m}^{2} \\ = 5.45 {m}^{2}$

(2) cost of 1m^2 of sheet = Rs 20

$hence \: \: total \: cost \: of \: \: 5.45 {m}^{2} of \: sheet \: \\ = rs(5.45 \times 20) = rs109$

Question no (2)

Solution

Length = 5m

Breadth = 4 m

Height = 3 m

According to the question

Area of four wall and ceiling

= Lateral surface area + Area of ceiling

$= 2(l + b)h + lb \\ = 2(5 + 4)3 + 5 \times 4 \\ = 54 + 20 \\ = 74 {m}^{2}$

Cost of whitewashing at the rate of rs 7.50/ m^ 2

$= 74 \times 7.50 = rs \: 555$

Question no (3)

Solution

Cost of painting the four walls = RS 15000

Rate of painting is RS 10 per m^2

$area \: of \: four \: walls \: = ( \frac{1500}{10} ) {m}^{2} \\ = 1500 {m}^{2} \\ = 2(l + b)h = 1500 = > perimeter \: \times height \: = 1500 \\ = 250 \times height = 1500 \\ = > height \: = \frac{1500}{250} = 6$

Hence the height of the hall is => 6m

Question (4)

Surface area of 1 break = 2(lb+bh + hl )

$= ( \frac{22.5}{100} \times \frac{10}{100} + \frac{10}{100} \times \frac{7.5}{100} + \frac{7.5}{100} \times \frac{22.5}{100} ) {m}^{2}$

$= 2 \times \frac{1}{100} \times \frac{1}{100} (22.5 \times 10 + 10 \times 7.5 + 7.5 \times 22.5) {m}^{2}$

$= \frac{1}{5000} \times (225 + 75 + 168.75) {m}^{2} \\ = \frac{1}{5000} \times 468.75 {m }^{2} = 0.09375 {m}^{2}$

Area for which the paint is just sufficient is 9.375 m^ 2

Hence the no. of the bricks that can be painted with the available paint .

$= \frac{9.375}{0.09375} = 100$

Question no (5)

(1)

Lateral surface area of the cubical box of edge 10cm

$= 4 \times {10}^{2} {cm}^{2} = 400 {cm}^{2} \\ lateral \: surface \: area \: of \: cuboidl \: \: box \: \\ = 2(l + b) \times h \\ = 2 \times (12.5 + 10) \times 8 \\ = 2 \times 22.5 \times 8 {cm}^{2} \\ = 360 {cm}^{2}$

Thus , lateral surface area of cubical box is greater and is more by

$= (400 - 360) {cm}^{2} = 40 {cm}^{2}$

(2)

Total surface area of cubical box of edge 10cm

$= 6 \times {10}^{2} {cm }^{2} = 600 {cm}^{2}$

Total surface area of cubical box

$= 2(lb + bh + hl) \\ = 2(125 + 80 + 100) {vm}^{2} \\ = (2 \times 305) {cm}^{2} = 610 {cm \: }^{2}$

Thus total surface area of the cuboidal box is greater by 10 cm^2 .

Hope it is helpful for.