Tag: diagonal of cube and cuboid

Questions Related to diagonal of cube and cuboid

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

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. $3$

  2. $\sqrt {63}$

  3. $15$

  4. $21$

  5. $225$

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Correct Option: C
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 
$=\sqrt{12^{2}+9^{2}}=\sqrt{144+81}=\sqrt{225}=15$

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?

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. $34$

  2. $14$

  3. $26$

  4. $2\sqrt{119}$

  5. $44$

Show answer
Correct Option: C
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

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.

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. $63$

  2. $67$

  3. $71$

  4. $75$

  5. $80$

Show answer
Correct Option: A
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)

$\angle B$ 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 $RS$.

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. 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.

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Correct Option: D
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.

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.

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. $56 m$

  2. $60 m$

  3. $15 m$

  4. $12 m$

Show answer
Correct Option: C
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

$base=greatest side=56 \ m$

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

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

$h=15 \ m$

In $\triangle ABC,\angle ABC={ 90 }^{ o }$. If $AC=(x+y)$ and $BC=(x-y)$, then the length of $AB$ 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. ${ x }^{ 2 }-{ y }^{ 2 }$

  2. $2xy$

  3. $2\sqrt { xy } $

  4. ${ x }^{ 2 }+{ y }^{ 2 }$

Show answer
Correct Option: C
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,

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?

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. $990\ m$

  2. $1300\ m$

  3. $1400\ m$

  4. $2100\ m$

Show answer
Correct Option: A
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,
$AC^2=AB^2+BC^2$
         $=700^2+700^2$
         $=4900+4900=9800$
$\therefore \ AC=990m$.

$PQ$ 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 :

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. $6\ cm$

  3. $5\ cm$

  4. $2\sqrt {11} cm$

Show answer
Correct Option: B
Explanation:

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

                                       $= \sqrt {64 - 28} $

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

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 

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. $2:1$

  2. $4:3$

  3. $4:5$

  4. $8:7$

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Correct Option: A

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

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 { 27 } $

  2. $\sqrt { 36 } $

  3. $\sqrt { 45 } $

  4. $\sqrt { 54 } $

Show answer
Correct Option: C
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 } $