1 Marks Question

Question 11 Mark

The Surface area of a cuboid is $1300 \mathrm{~cm}^2$. If its breadth is 10 cm and height is $20 \mathrm{~cm}^2$, find its length.

Answer

Let, $l \rightarrow$ Length of the cuboid
Breadth of the cuboid (b) $=10 \mathrm{~cm}$
Height of the cuboid $(h)=20 \mathrm{~cm}$
Surface area of the cuboid $(A)=1300 \mathrm{~cm}^2$
We have to find the length of the cuboid
We know that,
$A=2(l b+b h+h l)$
$1300=2(10 l+10 \times 20+20 l)$
$1300=2(200+30 l)$
$1300=400+60 \mathrm{l}$
$1=\frac{1300-400}{600}$
$=\frac{900}{60}$
$=15 \mathrm{~cm}$
Length of the cuboid is $15 \ cm .$

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Question 21 Mark

Three cubes of metal whose edges are in the ratio $3 : 4 : 5$ are melted down into a single cube whose diagonal is $12\sqrt{3}\text{cm}.$ Find the edge of three cubes.

Answer

Let the edge of the three cubes be $3x, 4x$ and $5x$ respectively.
Volume of three cubes $= (3x)^3 + (4x)^3 + (5x)^3 = 216x^3 cm^3$
Let a be the edge of new cube so formed.
Now, Volume of the cube = Volume of three cubes $\Rightarrow\text{a}^3=216\text{x}^3$
$\Rightarrow\text{a}=6\text{x}$ The diagonal of new cube is $12\sqrt{3}\text{cm}$
$\Rightarrow\sqrt{\text{a}^2+\text{a}^2+\text{a}^2}=12\sqrt{3}$
$\Rightarrow\sqrt{3}\text{a}=12\sqrt{3}$
$\Rightarrow\text{a}=12$
$\Rightarrow\text{x}=2$
$\Rightarrow3\text{x}=6\text{cm},4\text{x}=8\text{cm}$ and $5\text{x}=10\text{cm}$

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Question 31 Mark

Find the edge of a cube whose surface area is $432 \mathrm{~m}^2$.

Answer

Let, $\mathrm{a} \rightarrow$ Edge of the cube Surface area of the cube $=6 \mathrm{a}^2$ So, $6 \mathrm{a}^2=432 \mathrm{a}^2=\frac{432}{6}=72 \mathrm{a}=6 \sqrt{2} \mathrm{~m}$ Side of the cube is $6 \sqrt{2} \mathrm{~m}$.

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Question 41 Mark

Three cubes of each sides $4\ cm$ are joined end. Find the surface area of the resulting cuboid.

Answer

Side of each cube $(a)=4 \mathrm{~cm}$
We need to find the surface area of the resulting cuboid
Dimensions of the resulting cuboid,
Length $(l)=3 a$
Breadth $(b) = a$
Height $(h) = a$
Surface area of the cuboid,
$=2(l b+b h+h l)$
$=2[(3 a) a+(a)(a)+a(3 a)]$
$=2\left(7 a^2\right)$
$=14 a^2$
$=14 \times 4^2$
$=224 \mathrm{~cm}^2$
Surface area of the cuboid is $224 \mathrm{~cm}^2$.

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Question 51 Mark

If two cubes each of sides $6\ cm$ are jioned face to face, then find the volume of the resulting cuboid.

Answer

We have,
Side of each cube $(a)=6 \mathrm{~cm}$
We need to find the volume of resulting cuboid
Hence, dimensions of the resulting cuboid are,
$\text { Length }(l)=2 \mathrm{a}$
$=2 \times 6$
$=12 \mathrm{~cm}$
Breadth (b) = a
$=6 \mathrm{~cm}$
Height (h) $=\mathrm{a}$
$=6 \mathrm{~cm}$
Hence, volume of the resulting cuboid,
$V=l b h$
$=12 \times 6 \times 6$
$=432 \mathrm{~cm}^3$
Hence, volume of the resulting cuboid is $432 \mathrm{~cm}^3$.

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Question 61 Mark

If the perimeter of each face of a cube is $32\ cm$, find its lateral surface area. Note that four faces which meet the base of a cube are called its lateral faces.

Answer

Let, $a \rightarrow$ Side of the cube Perimeter of each face is $32 \mathrm{~cm} .4 a=32 a=8 \mathrm{~cm}$ Lateral surface area, $=4 a^2=4 \times 8^2=$
$256 \mathrm{~cm}^2$
So the lateral surface area of the cube is $256 \mathrm{~cm}^2$.

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Question 71 Mark

A cuboid has total surface area of $372 \mathrm{~cm}^2$ and its lateral surface area is $180 \mathrm{~cm}^2$, find the area of its base.

Answer

We have, Total surface area of the cuboid $(A)=372 \mathrm{~cm}^2$
Lateral surface area of the cuboid $\left(A^{\prime}\right)=180 \mathrm{~cm}^2$
Let, a $\rightarrow$ Area of the base
We know that, $\mathrm{A}=\mathrm{A}^{\prime}+2 \mathrm{a} \mathrm{a}=\frac{\mathrm{A}-\mathrm{A}^{\prime}}{2}=\frac{372-180}{2}=\frac{192}{2}=96 \mathrm{~cm}^2$
Area of the base is $96 \mathrm{~cm}^2$.

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