Tag: length of the diagonal

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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

One side other than the hypotenuse of a right-angled isosceles triangle is $4$ cm. The length of the perpendicular on the hypotenuse from the opposite vertex is:

  1. $8$cm
  2. $4\sqrt { 2 } $cm
  3. $4$ cm
  4. $2\sqrt { 2 } $cm
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Let assume A, B, C be the vertices of the given triangle and right-angled at A i.e, $\angle A = 90^{\circ}$, $AD \perp BC$.......(D is the point of intersecton of perpendicular from A on BC)
By Pythagoras theorem,
$BC^2 = AB^2 + AC^2$
$BC^2 = 4^2 + 4^2$
$BC = 4\sqrt{2}$ cm
Area of triangle = $\frac{1}{2} base \times height$
Thus, $\frac{1}{2} AB \times AC = \frac{1}{2} AD \times BC$
$4 \times 4 = 4\sqrt{2} \times AD$
$AD = 2 \sqrt{2}$ 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

The length of the hypotenuse of a right angled $\Delta$ whose two legs measure $12 \ cm$ and $0.35 \ m$ is:

  1. $37 \ cm$
  2. $3.72 \ cm$
  3. $0.372 \ cm$
  4. $37 \ m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$0.35 \ m = 0.35 \times 100 \ cm = 35 \ cm.$
We have,
$(hypotenuse)^2\, =\, (side)^2\, +\, (side)^2$
$=\, (12)^2\, +\, (35)^2$
$= 144 + 1225$
$= 1369$
Hypotenuse $=\sqrt{1369} = 37 \ cm.$

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

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 $122$.