Tag: diagonal of cube and cuboid

Questions Related to diagonal of cube and cuboid

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. $3\;cm$

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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. $7 \ cm$

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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

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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 $25$

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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 $17$

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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 $122$

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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 $122$.

Which of the following can't be  the lengths of the sides of a right-angled triangle?

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. $5$ inches, $12$ inches, $13$ inches

  2. $\displaystyle\frac{1}{3}$ of a foot, $\displaystyle\frac{1}{4}$ of a foot, $\displaystyle\frac{1}{5}$ of a foot

  3. $9$cm, $40$cm, $41$cm

  4. $\displaystyle\frac{3}{4}$ of a foot, $1$ foot, $15$ inches

Show answer
Correct Option: B
Explanation:

For a set of numbers to be sides of a right angled triangle, the square of the longest side must equal the sum of the squares of the other two sides.

In option A, $13^2 = 169, 5^2 + 12^2 = 169$ so possible.
In option B, $\left (\dfrac{1}{3}\right)^2 = \dfrac{1}{9}, \left (\dfrac{1}{4}\right)^2 + \left (\dfrac{1}{5}\right)^2 = \dfrac{41}{400}$
Since, these two values are not equal, this set cannot form a right angled triangle.

The ratio of the two legs of a right-angled triangle is $3:1$. If the lengths of the legs are whole numbers, what can be the possible value of the hypotenuse?

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. $\sqrt{40}$

  2. $\sqrt{47}$

  3. $\sqrt{55}$

  4. $\sqrt{63}$

Show answer
Correct Option: A
Explanation:

The legs of a right angled triangle are in the ratio $3:1$ and they are whole numbers.

The possibilities are as listed below:
$3,1$ - The hypotenuse becomes $\sqrt{(3)^2 + (1)^2} = \sqrt{10}$
$6,2$ - The hypotenuse becomes $\sqrt{(6)^2 + (2)^2} = \sqrt{40}$
$9,3$ - The hypotenuse becomes $\sqrt{(9)^2 + (3)^2} = \sqrt{90}$
Thus, the hypotenuse will always be square root of a multiple of $10$, which is in option A.