Question

The paint in a container is sufficient to paint an area equal to 18.56cm how many bricks of dimensions 12cm*10cm*8cm can be painted by this paint

Answer 1

Therefore the number of brick can be painted by this paint container≈313.

Step-by-step explanation:

Given, the paint in a container is sufficient to paint an area equal to 18.56 m² = 185600 ²cm

The dimensions of one brick is 12 cm× 10 cm × 8 cm

Surface area of a cuboid=2(l×b+l×h+b×h)

The surface area of a brick is = 2(12×10+12×8+10×8) cm²

                                                =592 cm²

Therefore the number of brick can be painted by this paint container=(185600÷592) =313.51≈313.

Answer 2

L - 12 cm

L - 12 cmB - 10 cm

L - 12 cmB - 10 cmH - 8 cm

L - 12 cmB - 10 cmH - 8 cmTSA of bricks - 2( lb+ bh+hl)

L - 12 cmB - 10 cmH - 8 cmTSA of bricks - 2( lb+ bh+hl)2(12×10 + 10×8 + 8×12)

L - 12 cmB - 10 cmH - 8 cmTSA of bricks - 2( lb+ bh+hl)2(12×10 + 10×8 + 8×12)2(120+80+96)

L - 12 cmB - 10 cmH - 8 cmTSA of bricks - 2( lb+ bh+hl)2(12×10 + 10×8 + 8×12)2(120+80+96)2×296

L - 12 cmB - 10 cmH - 8 cmTSA of bricks - 2( lb+ bh+hl)2(12×10 + 10×8 + 8×12)2(120+80+96)2×296592 cm²

L - 12 cmB - 10 cmH - 8 cmTSA of bricks - 2( lb+ bh+hl)2(12×10 + 10×8 + 8×12)2(120+80+96)2×296592 cm² no. of bricks - TSA of bricks ÷ total paint

L - 12 cmB - 10 cmH - 8 cmTSA of bricks - 2( lb+ bh+hl)2(12×10 + 10×8 + 8×12)2(120+80+96)2×296592 cm² no. of bricks - TSA of bricks ÷ total paint 592÷1856×100

L - 12 cmB - 10 cmH - 8 cmTSA of bricks - 2( lb+ bh+hl)2(12×10 + 10×8 + 8×12)2(120+80+96)2×296592 cm² no. of bricks - TSA of bricks ÷ total paint 592÷1856×100592÷ 185600

L - 12 cmB - 10 cmH - 8 cmTSA of bricks - 2( lb+ bh+hl)2(12×10 + 10×8 + 8×12)2(120+80+96)2×296592 cm² no. of bricks - TSA of bricks ÷ total paint 592÷1856×100592÷ 185600313•51

L - 12 cmB - 10 cmH - 8 cmTSA of bricks - 2( lb+ bh+hl)2(12×10 + 10×8 + 8×12)2(120+80+96)2×296592 cm² no. of bricks - TSA of bricks ÷ total paint 592÷1856×100592÷ 185600313•51—> 313