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3 years ago

2. IQS = 4 and QR is a perpendicular bisector of TS , determined which of the following claims is true

Mathematics

1 answer:

Pavel [41]3 years ago

3 0

Answer:

RS=5 TRUE

RS=4 FALSE (it’s 5)

ST=10 FALSE (it’s 8)

QR=4 FALSE (its 3)

Step-by-step explanation:

First, we have to determine the sides of the triangle.

4x-3=2x+1

4x-2x=1+3

2x=4

x=2

SO, RS=4x-3=4(2)-3=8-3=5

RT=2(x)+1=2(2)+1=5

QR=4²+b²=5². 16+b²=25. b²=25-16. b²=9. b=3

CLAIMS:

RS=5 TRUE

RS=4 FALSE (it’s 5)

ST=10 FALSE (it’s 8)

QR=4 FALSE (its 3)

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The box which measures 70cm X 36cm X 12cm is to be covered by a canvas. How many meters of canvas of width 80cm would be require

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Answer:

142.2 meters.

Step-by-step explanation:

We have been given that a box measures 70 cm X 36 cm X 12 cm is to be covered by a canvas.

Let us find total surface area of box using surface area formula of cuboid.

$\text{Total surface area of cuboid}=2(lb+bh+hl)$ , where,

$l$ = Length of cuboid,

$b$ = Breadth of cuboid,

$w$ = Width of cuboid.

$\text{Total surface area of box}=2(70\cdot36+36\cdot 12+12\cdot 70)$

$\text{Total surface area of box}=2(2520+432+840)$

$\text{Total surface area of box}=2(3792)$

$\text{Total surface area of box}=7584$

Therefore, the total surface area of box will be 7584 square cm.

To find the length of canvas that will cover 150 boxes, we will divide total surface area of 150 such boxes by width of canvass as total surface area of canvas will also be the same.

$\text{Width of canvas* Length of canvass}=\text{Total surface area of 150 boxes}$

$80\text{ cm}\times\text{ Length of canvass}=150\times 7584\text{cm}^2$

$\text{ Length of canvass}=\frac{150\times 7584\text{ cm}^2}{80\text{ cm}}$

$\text{ Length of canvass}=\frac{1137600\text{ cm}^2}{80\text{ cm}}$

$\text{ Length of canvass}=14220\text{ cm}$

Let us convert the length of canvas into meters by dividing 14220 by 100 as 1 meter equals to 100 cm.

$\text{ Length of canvass}=\frac{14220\text{ cm}}{100\frac{cm}{m}}$

$\text{ Length of canvass}=\frac{14220\text{ cm}}{100}\times\frac{m}{cm}$

$\text{ Length of canvass}=142.20\text{ m}$

Therefore, 142.2 meters of canvas of width 80 cm required to cover 150 such boxes.

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