Tag: length of the diagonal

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Questions Related to length of the diagonal

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 } $

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

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

  1. $132$
  2. $145$
  3. $89$
  4. $200$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\begin{array}{l} { 11^{ 2 } }+{ 60^{ 2 } }={ 61^{ 2 } } \\ \Rightarrow Perimeter\, \, of\, \, \Delta =132\, \, cm \\ \left\{ { \because sides\, \, 11,60\, \, \& \, \, 61 } \right\}  \end{array}$
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 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

  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. $\left( \dfrac { \pi }{ 12 } ,\dfrac { 5\pi }{ 12 } \right)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the altitude to the hypotenuse be h. Hypotenuse c = 2*sqrt(2)*h. In a right triangle, h = (a*b)/c. Also, h = c*sin(A)*cos(A). So 2*sqrt(2)*h = c, meaning sin(A)cos(A) = 1/(2*sqrt(2)). This leads to the angles pi/8 and 3pi/8.

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 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:

  1. $40^0$
  2. $42^0$
  3. $44^0$
  4. $45^0$
Reveal answer Fill a bubble to check yourself
D Correct answer
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, $\angle A = \angle C = 45^o$.

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

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

  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. $\dfrac{-16-8x-x^2}{(x+4)^2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 } } $
$=\dfrac { x^{ 2 }+8x-34 }{ \left( x+4 \right) ^{ 2 } }$ 

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

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

  1. $4\sqrt{5}+8\sqrt{3}$
  2. $16\sqrt{3}$
  3. $32\sqrt{3}$
  4. $8\sqrt{5}+16\sqrt{3}$
  5. $48$
Reveal answer Fill a bubble to check yourself
D Correct answer
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:
$A=\dfrac { 1 }{ 2 } \times 8\times (2\sqrt { 5 } +4\sqrt { 3 } )=4(2\sqrt { 5 } +4\sqrt { 3 } )=8\sqrt { 5 } +16\sqrt { 3 }$