Tag: surface area of cubes and cuboids

Questions Related to surface area of cubes and cuboids

If the measures of the sides of a triangle are ________, then it is not a right angled 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. $3,4,5$

  2. $5,12,13$

  3. $8,24,26$

  4. $7,24,25$

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Correct Option: C
Explanation:

$\begin{array}{l} 8,24,26 \ \sin  ce, \ { \left( 8 \right) ^{ 2 } }+{ \left( { 24 } \right) ^{ 2 } }\ne { \left( { 26 } \right) ^{ 2 } } \end{array}$

Instead of walking along two adjacent sides of a rectangular field, a boy took a short cut along the diagonal and saved the distance equal to half of the longer side. Then the ratio of the shorter side to the longer side 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. $\cfrac{2}{3}$

  2. $\cfrac{5}{3}$

  3. $\cfrac{4}{3}$

  4. $\cfrac{8}{3}$

Show answer
Correct Option: D
Explanation:

Atp,

$\sqrt{l^2+b^2} + \cfrac{l}{2} = l+b$
$\sqrt{l^2+b^2} = b+\cfrac{l}{2}$
$l^2+b^2 = b^2+2lb+\cfrac{l^2}{4}$
$\cfrac{3l^2}{4} = 2lb$
$\cfrac{l}{b} = \cfrac{8}{3}$

$\angle B$ is a right angle is in $\Delta ABC$v and P,Q are points of trisection of hypotenuse $\bar{AC}.$ then $BP^{2}+BQ^{2}=\frac{5}{9}AC^{2}.$

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. True

  2. False

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Correct Option: A

In $\triangle ABC$ right angled at $B, AB=5\ cm$ and $\angle ACB=30^{o}$ then the length of the sides $BC$ 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. $5\sqrt {3}$

  2. $2\sqrt {3}$

  3. $10\ cm$

  4. $none\ of\ these$

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Correct Option: A
Explanation:

Given:- $ABC$ is a right angled triangle in which $AB = 5 \; cm$ and 


$\angle{ACB} = 30°$

Using trigonometric ratio,

$\tan{C} = \cfrac{AB}{BC}$

$\Rightarrow \tan{30°} = \cfrac{5}{BC}$

$\Rightarrow BC = 5 \sqrt{3} \; cm$

ABC is a triangle, right-angled at B. M is a point on BC. Hence,
$AM^{2}\, +\, BC^{2}\, =\, AD^{2}\, +\, BM^{2}$
State true or false.

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. True

  2. False

Show answer
Correct Option: B
Explanation:

Given, $\angle ABC = 90$, M is a point of BC
In $\triangle ABM$, 
$AB^2 + BM^2 = AM^2$ (Pythagoras theorem)
$AB^2 = AM^2 - BM^2$ (I)

In $\triangle ABC$,
$AB^2 + BC^2 = AC^2$ (Pythagoras Theorem)
$AB^2 = AC^2 - BC^2$ (II)

Equating I and II,
$AM^2 - BM^2 = AC^2 - BC^2$
thus, $AM^2 + BC^2 = AC^2 + BM^2$

A guy wire attached to a vertical pole of height $18m$ is $24m$ long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?

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. $15.87m$

  2. $16.8m$

  3. $15$

  4. $15.67$

Show answer
Correct Option: A
Explanation:

Consider the wire and pole system as a right angled triangle such that length of pole is the perpendicular, length of wire the hypotenuse and distance between the base of pole and the stack is the base, such that, H = 24 m, P = 18 m
Now, from Pythagoras theorem,
$H^2 = P^2 +B^2$
$24^2 = 18^2 + B^2$
$576 = 324 + B^2$
$B^2 = 252$
$B = 15.87$ m
Thus, distance between the base of the pole and the stack is 15.87 m

The sides of a rectangular field are $80$ m and $18$ m respectively. The length of the diagonal 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. $84$ m

  2. $98$ m

  3. $82$ m

  4. $86$ m

Show answer
Correct Option: C
Explanation:

Given sides of rectangle are $80$ m and $18$ m.
Thus length of diagonal is $\sqrt{80^2\, +\, 18^2}\, =\, \sqrt{6 724}\, =\, 82\, $ m.

A person wishes to fit three rods together in the shape of a right-angled triangle so that the hypotenuse is to be $4:cm$ longer than the base and $8:cm$ longer than the altitude. The lengths of the rods 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. $3:cm$, $4:cm$, $5:cm$

  2. $1.5:cm$, $2:cm$, $2.5:cm$

  3. $6:cm$, $8:cm$, $10:cm$

  4. $12:cm$, $16:cm$, $20:cm$

Show answer
Correct Option: D
Explanation:

Let the altitude$=x:cm$
$\therefore$ The Base$=(x+4):cm$
and the Hypotenuse$=(x+8):cm$

Using Pythtegores Theorem
$(x+8)^2=(x+4)^2+x^2$
$x^2-8x-48=0$
$(x-12)(x+4)=0$
$x=12$
$\therefore$ The sides are $12:cm$, $16:cm$, $20:cm$ 

What is the value of the hypotenuse of a right triangle whose sides are $12$ and $18$?

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. $4.24$

  2. $3.46$

  3. $2.16$

  4. $21.63$

Show answer
Correct Option: D
Explanation:

Let the sides be $a=12$ and $b=18$

Le the hypotenuse be $c$
Using Pythagoras theorem
${ c }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }\ { c }^{ 2 }={ (12) }^{ 2 }+{ (18) }^{ 2 }\ { c }^{ 2 }=144+324\ { c }^{ 2 }=468\ c=\sqrt { 468 } \ c=21.63$

A triangle whose lengths of sides are $5$ cm, $12$ cm and $13$ cm. 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. Obtuse-angled triangle

  2. Acute-angled triangle

  3. Right-angled triangle

  4. Triangle is not formed

Show answer
Correct Option: C
Explanation:
Given sides of triangle are $5$ cm, $12$ cm, $13$ cm
Now applying pythagorus theorem:
$h^{2}= b^{2}+P^{2}$
$\Rightarrow 13^{2}= 12^{2}+5^{2}$
$ \Rightarrow 169 =144+25$
$ \Rightarrow 169 =169$
These three sides clearly satisfy ptyhagorous theorem.
Therefore, the triangle is RIGHT ANGLED triangle.