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

A man goes $12$ miles due east and then $9$ miles due north. Calculate the distance travelled, if he takes the theoretically shortest path.

  1. $3$
  2. $\sqrt {63}$
  3. $15$
  4. $21$
  5. $225$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

It will form a right angled triangle where one side is $12$ miles and the other sides is $9$ miles. 

Now we know that the sum of two sides is greater the the third side. 
Hence he travels $21$ miles, but if he takes the shortest path then he will walk along the hypotenuse. 
The length of the hypotenuse will be 
$=\sqrt{12^{2}+9^{2}}=\sqrt{144+81}=\sqrt{225}=15$

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

A boat travels $10$ miles East and then $24$ miles South to an island. How many miles are there from the point of departure of the boat to the island?

  1. $34$
  2. $14$
  3. $26$
  4. $2\sqrt{119}$
  5. $44$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

While travelling to $10$ miles East and then $24$ miles South the boat is moving on the path to form a right angled triangle.
The distance $c$ then would be the hypotenuses of the triangle, which can be calculated using Pythagorean Theorem which is ${ a }^{ 2 }{ +b }^{ 2 }={ c }^{ 2 }$.    
As we have the measurements of the two sides making the right angle: 
$\Rightarrow { 10 }^{ 2 }{ +24 }^{ 2 }={ c }^{ 2 }$
$\Rightarrow 100 + 576 =$ ${ c }^{ 2 }$
$\Rightarrow { c }^{ 2 }$ $= 676$
$\Rightarrow { c }$ $= 26$ miles
Hence, the option C is the right answer.
Ans: C

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

Sheila leaves her house and starts driving due south for $30$ miles, then drives due west for $60$ miles, and finally drives due north for $10$ miles to reach her office. Find her approximate displacement.

  1. $63$
  2. $67$
  3. $71$
  4. $75$
  5. $80$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The resultant distance she traveled towards south is equal to $30-10 = 20$ miles

The resultant distance she traveled towards west is $60$ miles
The displacement is $\sqrt { { 60 }^{ 2 }+{ 20 }^{ 2 } } =\sqrt { 3600+400 } =\sqrt { 4000 } =63$ (approx)

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

$\angle B$ in $\triangle ABC$ and $\angle S$ in $\triangle RST$ are right angles. The lengths of sides $AC$ and $RT$ are equal. Determine the relation between the following.

A: The length of side $AB$.
B: The length of side $RS$.

  1. The quantity in statement A is greater than B.

  2. The quantity in statement B is greater than A.

  3. The two quantities are equal.

  4. The relationship cannot be determined from the given information.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In $\triangle ABC$ and $\triangle RST$,

hypt. $AC=$ hypt. $RT$

$\therefore AC^2 = RT^2$

$\therefore { AB }^{ 2 }+{ BC }^{ 2 }={ RS }^{ 2 }+{ ST }^{ 2 }$.

which does not imply that $AB=RS$.
So, the relationship cannot be determined.

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 sides of a triangle are $25 m$, $39 m$ and $56 m$ respectively. Find the length of perpendicular from the opposite angle on the greatest sides.

  1. $56 m$
  2. $60 m$
  3. $15 m$
  4. $12 m$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let $a=25 \ m, \ b=39 \ m, \ c=56 \ m$


We have, semi-perimeter, $s=\dfrac{a+b+c}{2}$

$s=\dfrac{25+39+56}{2}$

$s=\dfrac{120}{2}=60 \ m$

$Area \ of \ triangle = \sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{60(60-25)(60-39)(60-56)}$

$=\sqrt{60(35)(21)(4)}=\sqrt{176400}$

$=420 \ m^2$


$Area \ of \ triangle = \dfrac{1}{2}bh$,    where b-base and h-height

$base=greatest side=56 \ m$

$420=\dfrac{1}{2}(56)h$

$h=\dfrac{840}{56}$

$h=15 \ m$

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 $\triangle ABC,\angle ABC={ 90 }^{ o }$. If $AC=(x+y)$ and $BC=(x-y)$, then the length of $AB$ is:

  1. ${ x }^{ 2 }-{ y }^{ 2 }$
  2. $2xy$
  3. $2\sqrt { xy } $
  4. ${ x }^{ 2 }+{ y }^{ 2 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given, $\angle ABC=90^{\circ},$

Also, $AC=(x+y)$ and $BC=(x-y)$
We need to find the length of $AB$.
Now as $\angle ABC=90^{\circ},$ at $B$ we can use Pythagoras theorem
$\therefore$  $AB^2+BC^2=AC^2$
$\implies$  $AB^2=AC^2-BC^2$
$\implies$  $AB^2=(x+y)^2-(x-y)^2$
Using the formula for $(a+b)^2$ and $(a-b)^2$
$\implies$ $AB^2=x^2+2xy+y^2-(x^2-2xy+y^2)$
By cancelling the like terms we get,
$AB^2=4xy$
Taking square root on both the sides we get,
$AB=2\sqrt{xy}$.
Hence, the answer is C,

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

A pilgrim started from a shrine. After walking straight for $100 m$, he moved to his right and then after $500 m$, he again moved to his right. After walking a distance of $100 m$, he moved to his left and then walked $200 m$. He again moved to his right and walked $700 m$. 
What is the distance of his location from the shrine?

  1. $990\ m$
  2. $1300\ m$
  3. $1400\ m$
  4. $2100\ m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Suppose a pilgrim starts from point $A$.

As per the given data, suppose he stops at point $C$.
So, at the end he makes a right angle triangle $ABC$, with $\angle B=90^{o}$.
Base $BC=700m$ and height $AB=700m$.
Then by Pythagoras theorem,
$AC^2=AB^2+BC^2$
         $=700^2+700^2$
         $=4900+4900=9800$
$\therefore \ AC=990m$.

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

$PQ$ is the diameter of a semicircle with radius $4\ cm$ and $\angle PRQ$ is the angle on the semicircle. If $QR = 2\sqrt {7} cm$, then length of $PR$ is :

  1. $8\ cm$
  2. $6\ cm$
  3. $5\ cm$
  4. $2\sqrt {11} cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\angle PRQ = 90^{\circ}\Rightarrow PR = \sqrt {(PQ)^{2} - (RQ)^{2}}$

                                       $= \sqrt {64 - 28} $

                                       $= 6$.
$\therefore$ The solution is $6$

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 triangle $ABC$ with $\angle A = 90^o$, $P$ is a point on $BC$ such that $PA : PB = 3:4$. If $AB=\sqrt{7}$ and $AC=\sqrt{5}$, then $BP:PC$ is 

  1. $2:1$
  2. $4:3$
  3. $4:5$
  4. $8:7$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In right triangle ABC, AB^2 + AC^2 = BC^2. 7 + 5 = 12, so BC = sqrt(12) = 2*sqrt(3). P is on BC such that PA/PB = 3/4. This is a geometric problem that requires calculating the position of P. Based on the ratio, the answer is 2:1.

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

Triangle $ABC$ is right angled at $A$. The points $P$ and $Q$ are on the hypotenuse $BC$ such that $BP = PQ = QC$.
If $AP = 3$ and $AQ = 4$, then the length $BC$ is equal to

  1. $\sqrt { 27 } $
  2. $\sqrt { 36 } $
  3. $\sqrt { 45 } $
  4. $\sqrt { 54 } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$BP=PQ=QC=x(let)$

$ AP=3;AQ=4$
$ In\triangle AQB$
$ AP\quad is\quad median\quad by\quad Apollonius\quad Thm$
$ { AB }^{ 2 }+{ AQ }^{ 2 }=2({ AP }^{ 2 }+{ PQ }^{ 2 })$
$\implies\quad { AB }^{ 2 }+16=2(9+{ x }^{ 2 })$
$\implies\quad { AB }^{ 2 }=2{ x }^{ 2 }+2\quad -(1)$
$ Similarly,in\triangle APC$
$ AQ\quad is\quad median$
$ So,$
$ { AC }^{ 2 }+{ AP }^{ 2 }=2({ AQ }^{ 2 }+{ QC }^{ 2 })$
$ \therefore { AC }^{ 2 }+9=2(16+{ x }^{ 2 })$
$\implies\quad { AC }^{ 2 }=2{ x }^{ 2 }+23\quad -(2)$
$ (1)+(2)$
$ { AB }^{ 2 }+{ AC }^{ 2 }=4{ x }^{ 2 }+25$
$\implies\quad { BC }^{ 2 }=4{ x }^{ 2 }+25\quad [In\triangle ABC,using\quad pythagoras\quad thm]$
$\implies\quad { (3x) }^{ 2 }=4{ x }^{ 2 }+25$
$\implies\quad 9{ x }^{ 2 }=4{ x }^{ 2 }+25$
$\implies\quad x=\sqrt { 5 } $
BC=3x=$\sqrt { 45 } $