Tag: surface area and volume of cube and cuboid

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Questions Related to surface area and volume of cube and cuboid

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

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

  1. $1.5\;cm$
  2. $1\;cm$
  3. $2\;cm$
  4. $3\;cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
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$

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

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

  1. $S^2$
  2. $d^2$
  3. $S^2-d^2$
  4. $S^2+d^2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let dimensions be a, b, c. S = a+b+c. Diagonal d^2 = a^2+b^2+c^2. Surface area = 2(ab+bc+ca). Since (a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca), then S^2 = d^2 + Surface Area. Therefore, Surface Area = S^2 - d^2.

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

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

  1. $3 \ cm$
  2. $5 \ cm$
  3. $4 \ cm$
  4. $7 \ cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
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, $\displaystyle BM = \frac{1}{2} AC = \frac{10}{2} = 5 \ cm.$
Multiple choice maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

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

  1. $-1$
  2. $7$
  3. $6$

  4. Both A and B
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $.

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

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?

  1. $10$ km
  2. $8$ km
  3. $14$ km
  4. $13$ km
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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

  1. $24$ and $25$
  2. $24$ and $26$
  3. $22$ and $25$
  4. $23$ and $25$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $26$.

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

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

  1. $15$ and $17$
  2. $14$ and $17$
  3. $15$ and $16$
  4. $11$ and $17$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $17$.

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

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

  1. $124$ and $122$
  2. $123$ and $122$
  3. $121$ and $122$
  4. $120$ and $122$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $122$.

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

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

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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?

  1. $\sqrt{40}$
  2. $\sqrt{47}$
  3. $\sqrt{55}$
  4. $\sqrt{63}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.