Tag: surface area of cubes and cuboids

Questions Related to 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 

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 } $

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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 } $

Given that in a right angled triangle the length of two sides are 11 and 60. Find the perimeter of the 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. $132$

  2. $145$

  3. $89$

  4. $200$

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Correct Option: A
Explanation:
$\begin{array}{l} { 11^{ 2 } }+{ 60^{ 2 } }={ 61^{ 2 } } \\ \Rightarrow Perimeter\, \, of\, \, \Delta =132\, \, cm \\ \left\{ { \because sides\, \, 11,60\, \, \& \, \, 61 } \right\}  \end{array}$

In a right angled triangle the hypotenuse is  $2\sqrt{2}$ times the length of the perpendicular drawn from the opposite vertex on the hypotenuse. The the other two angles are

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. $\left( \dfrac { \pi }{ 3 } ,\dfrac { \pi }{ 6 } \right)$

  2. $\left( \dfrac { \pi }{ 4 } ,\dfrac { \pi }{ 4 } \right)$

  3. $\left( \dfrac { \pi }{ 8} ,\dfrac { 3\pi }{ 8 } \right)$

  4. $\left( \dfrac { \pi }{ 12 } ,\dfrac { 5\pi }{ 12 } \right)$

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Correct Option: C

In a right angled triangle, the square of the hypotenuse is equal to twice the product of the other two sides. One of the acute angles of the triangle 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. $40^0$

  2. $42^0$

  3. $44^0$

  4. $45^0$

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Correct Option: D
Explanation:

Let $\triangle ABC$ be right angled at B.

Then, by Pythagoras theorem,

$AC^2 = AB^2+BC^2$

But $AC^2 = 2 \times AB . BC$     .....Given

Hence, $AB^2 + BC^2 = 2 \times AB. BC$

$AB^2 + BC^2 - 2 \times AB. BC = 0$

$(AB - BC)^2 = 0$

So, $AB = BC$

Therefore, $\triangle ABC$ is right isosceles triangle.

So, $\angle A = \angle C = 45^o$.

Evaluate cos$\begin{pmatrix}2csc^{-1}(\dfrac{x+4}{5})\end{pmatrix} = $

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. $\dfrac{x^2+8x-16}{x+4}$

  2. $\dfrac{x^2+8x-16}{(x+4)^2}$

  3. $\dfrac{x^2+8x-34}{x+4}$

  4. $\dfrac{x^2+8x-34}{(x+4)^2}$

  5. $\dfrac{-16-8x-x^2}{(x+4)^2}$

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Correct Option: D
Explanation:

Let $\theta =\csc ^{ -1 }{ \left( \dfrac { x+4 }{ 5 }  \right)  }$ implies that $\csc { (\theta )=\dfrac { x+4 }{ 5 }  }$ and therefore, $\sin { (\theta ) } =\dfrac { 5 }{ x+4 }$.


Use the pythagorean theorem to determine that the remaining leg of the right triangle has length:

$\sqrt { (x+4)^{ 2 }-5^{ 2 } } =\sqrt { x^{ 2 }+16+8x-25 } =\sqrt { x^{ 2 }+8x-9 }$

Therefore, $\cos { (\theta ) } =\dfrac { \sqrt { x^{ 2 }+8x-9 }  }{ x+4 }$  

Hence, $\cos { (2\theta ) } =\cos ^{ 2 }{ (\theta ) } -\sin ^{ 2 }{ (\theta ) } $
$=\left( \dfrac { \sqrt { x^{ 2 }+8x-9 }  }{ x+4 }  \right) ^{ 2 }-\left( \dfrac { 5 }{ x+4 }  \right) ^{ 2 }$
$=\dfrac { x^{ 2 }+8x-9 }{ \left( x+4 \right) ^{ 2 } } -\dfrac { 25 }{ \left( x+4 \right) ^{ 2 } } $
$=\dfrac { x^{ 2 }+8x-34 }{ \left( x+4 \right) ^{ 2 } }$ 

Diagonals $\overline{AC}$ and $\overline{BD}$ of quadrilateral $ABCD$ are perpendicular. $AD=DC=8, AC=BC=6, m\angle ADC = 60^o$. The area of $ABCD$ 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. $4\sqrt{5}+8\sqrt{3}$

  2. $16\sqrt{3}$

  3. $32\sqrt{3}$

  4. $8\sqrt{5}+16\sqrt{3}$

  5. $48$

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Correct Option: D
Explanation:

The altitude to the base of an isosceles triangle also bisects the vertex angle, so $m\angle ADE=30$.

With the hypotenuse of the triangle having a length of $8$, $AE=4$ and $DE=4\sqrt { 3 }$.
$\triangle AEC$ is a right angle with leg $4$ and hypotenuse $6$.
Use the pythagorean theorem to determine that 
$BE=\sqrt { 6^{ 2 }-4^{ 2 } } =\sqrt { 36-16 } =\sqrt { 20 } =2\sqrt { 5 }$
The area of the quadrilateral with perpendicular diagonals is equal to half the product of the diagonals, so the area of $ABCD$ is:
$A=\dfrac { 1 }{ 2 } \times 8\times (2\sqrt { 5 } +4\sqrt { 3 } )=4(2\sqrt { 5 } +4\sqrt { 3 } )=8\sqrt { 5 } +16\sqrt { 3 }$