Tag: surface area of cubes and cuboids

Questions Related to surface area of cubes and cuboids

One side other than the hypotenuse of a right-angled isosceles triangle is $4$ cm. The length of the perpendicular on the hypotenuse from the opposite vertex is:

maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

  1. $8$cm

  2. $4\sqrt { 2 } $cm

  3. $4$ cm

  4. $2\sqrt { 2 } $cm

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Correct Option: D
Explanation:
Let assume A, B, C be the vertices of the given triangle and right-angled at A i.e, $\angle A = 90^{\circ}$, $AD \perp BC$.......(D is the point of intersecton of perpendicular from A on BC)
By Pythagoras theorem,
$BC^2 = AB^2 + AC^2$
$BC^2 = 4^2 + 4^2$
$BC = 4\sqrt{2}$ cm
Area of triangle = $\frac{1}{2} base \times height$
Thus, $\frac{1}{2} AB \times AC = \frac{1}{2} AD \times BC$
$4 \times 4 = 4\sqrt{2} \times AD$
$AD = 2 \sqrt{2}$ cm

The length of the hypotenuse of a right angled $\Delta$ whose two legs measure $12 \ cm$ and $0.35 \ m$ is:

maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

  1. $37 \ cm$

  2. $3.72 \ cm$

  3. $0.372 \ cm$

  4. $37 \ m$

Show answer
Correct Option: A
Explanation:

$0.35 \ m = 0.35 \times 100 \ cm = 35 \ cm.$
We have,
$(hypotenuse)^2\, =\, (side)^2\, +\, (side)^2$
$=\, (12)^2\, +\, (35)^2$
$= 144 + 1225$
$= 1369$
Hypotenuse $=\sqrt{1369} = 37 \ cm.$

In a $\Delta ABC,\,AB=AC=2.5\;cm,\,BC=4\;cm$. Find its height from $A$ to the opposite base.

maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

  1. $1.5\;cm$

  2. $1\;cm$

  3. $2\;cm$

  4. $3\;cm$

Show answer
Correct Option: A
Explanation:

$In\triangle ABC $


In order to find height we need to consider that $  AD\bot BC$

$ Hence\quad in\quad right\quad angled \ \triangle ADC$

$ { AC }^{ 2 }={ AD }^{ 2 }+{ DC }^{ 2 }(Phythagoreas\quad Theorm)$

$ \Rightarrow { AD }^{ 2 }={ AC }^{ 2 }-{ DC }^{ 2 }$

$ \Rightarrow { AD }^{ 2 }={ (2.5) }^{ 2 }-({ 2) }^{ 2 }$

$ \Rightarrow { AD }^{ 2 }=6.25-4$

$ \Rightarrow { AD }^{ 2 }=2.25$

$ \Rightarrow AD=1.5$

$ Hence\quad option\quad (A)\quad is\quad right\quad answer$

If the sum of the length, breadth and depth of a cuboid is S and its diagonal is d, then its surface is _____________.

maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

  1. $S^2$

  2. $d^2$

  3. $S^2-d^2$

  4. $S^2+d^2$

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Correct Option: C

In $\Delta$ABC, $\angle B = 90^{o}, AB = 8 \ cm$ and $BC = 6 \ cm.$ The length of the median $BM$ is:

maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

  1. $3 \ cm$

  2. $5 \ cm$

  3. $4 \ cm$

  4. $7 \ cm$

Show answer
Correct Option: B
Explanation:
$AC^2= AB^2+ BC^2$        (because $\angle B = 90^o$)
$= 64+36= 100$
$\therefore AC = 10$
In a right triangle, the median from the right angle to the hypotenuse is half the length of the hypotenuse. 
So, $\displaystyle BM = \frac{1}{2} AC = \frac{10}{2} = 5 \ cm.$

If the sides of a right angled triangle are $x, 3x + 3$  and $3x + 4$, then $x$ is equal to:

maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

  1. $-1$

  2. $7$

  3. $6$

  4. Both A and B

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Correct Option: B
Explanation:

As the sides are of a right angled triangle, we have
$ {Hypotenuse}^{2} = {Side1}^{2} + {Side2}^{2} $ where hypotenuse is the largest side.
$ {(3x+4)}^{2}  = {(3x+3)}^{2} + {x}^{2} $
$ => 9{x}^{2} + 16 + 24x = 9{x}^{2} + 9 + 18x + {x}^{2}  $
$ => {x}^{2} - 6x - 7 = 0 $
$ => {x}^{2} - 7x + x - 7 = 0 $
$ => x(x-7) + (x-7) = 0 $
$ => (x-7)(x+1) = 0 $
$ => x = 7, -1 $
As the side cannot be negative, $ x = 7 $.

In a field of shape of a right angled triangle, the farmer wants to measure the $3$ sides but being a huge field, he was only able to measure $2$ sides, $1$ side of which was $6$ km and other was $8$ km. Can you find the length of $3^{rd}$ side for him?

maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

  1. $10$ km

  2. $8$ km

  3. $14$ km

  4. $13$ km

Show answer
Correct Option: A
Explanation:

The field is in the shape of a right angled triangle.

Using Pythagoras theorem,
$6^2 + 8^2 = \mbox{(3rd side)}^2$
$\mbox{(3rd side)}^2 = 36 + 64$
$\mbox{(3rd side)}^2 = 100$
$\therefore \mbox{3rd side} = 10$ km
So, option A is correct.

If the Pythagorean triples of one member is $10$, find the other two members.

maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

  1. $24$ and $25$

  2. $24$ and $26$

  3. $22$ and $25$

  4. $23$ and $25$

Show answer
Correct Option: B
Explanation:

As we know $2m,$ $m^{2}+1, m^{2}-1$ form a Pythagorean triplet for any number $m > 1$.
Let us assume $2m = 10$
$\Rightarrow m = 5$
Therefore, $m^{2}+1$ $=$ $5^{2}+1$
$\Rightarrow 25 + 1 = 26$
$m^{2}-1$ = $5^{2}-1$
and $\Rightarrow 25 - 1 = 24$
Hence, the other two members are $24$ and $26$.

If the Pythagorean triples of one member is $8$, find the other two members.

maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

  1. $15$ and $17$

  2. $14$ and $17$

  3. $15$ and $16$

  4. $11$ and $17$

Show answer
Correct Option: A
Explanation:

As we know $2m,$ $m^{2}+1, m^{2}-1$ form a Pythagorean triplet for any number $m > 1$.
Let us assume $2m = 8$
$\Rightarrow m = 4$
Therefore, $m^{2}+1$ $=$ $4^{2}+1$
$\Rightarrow 16 + 1 = 17$
and $m^{2}-1$ $=$ $4^{2}-1$
$\Rightarrow 16 - 1 = 15$
Hence, the other two members are $15$ and $17$.

If the Pythagorean triples of one member is $22$, find the other two members.

maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

  1. $124$ and $122$

  2. $123$ and $122$

  3. $121$ and $122$

  4. $120$ and $122$

Show answer
Correct Option: D
Explanation:

As we know $2m$, $m^{2}+1, m^{2}-1$ form a Pythagorean triplet for any number $m > 1$.


Let us assume $2m = 22$

$\Rightarrow m = 11$

Therefore, $m^{2}+1$ $=$ $11^{2}+1$

$\Rightarrow 121 + 1 = 122$

and $m^{2}-1$ $=$ $11^{2}-1$

$\Rightarrow 121 - 1 = 120$

Hence, the other two members are $120$ and $122$.