Tag: surface area and volume of cube and cuboid

Questions Related to surface area and volume of cube and cuboid

The perpendicular from A on side BC of a $\Delta ABC$ intersects BC at D such that $DB = 3CD$, then $2A{B^2} = A{C^2} + B{C^2}$.

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: B

State true or false
In a triangle ABC, P and Q are the id-points of AC and AB respectively and angle $BAC=90^o$.
then $BP^2 + CQ^2=5PQ^2$

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
Explanation:

In $\triangle BAP$,
$BP^2 = AB^2+ AP^2$ (Pythagoras Theorem)

Since, Q is the mid-point of AB, then $AB = 2 AQ$
Hence, $BP^2 = (2 AQ)^2 + AP^2$
$BP^2 = 4 AQ^2 + AP^2$...(I)

In $triangle QAC$,
$QC^2 = AQ^2 + AC^2$ (Pythagoras theorem)
Since, P is the mid point of AC, $AC = 2 AP$
Hence, $QC^2 = AQ^2 + (2 AP)^2$ 
$QC^2 = AQ^2 + 4 AP^2$...(II)

By Adding I and II
$BP^2 + QC^2 = 4 AQ^2 + AP^2 + AQ^2 + 4 AP^2$
$BP^2 + QC^2 = 5 (AP^2 +AQ^2)$

In $\triangle QAP$,
$AQ^2 + AP^2 = PQ^2$
Hence, $BP^2 + QC^2 = 5 PQ^2$

In a quadrilateral ABCD, $\angle B\, =\, 90^{\circ}$ and $\angle D\, =\, 90^{o}$. Then:

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. $2{AC}^{2}\, -\, {AB}^{2}\, =\, {BC}^{2}\, +\, {CD}^{2}\, +\, {DA}^{2}$

  2. $2{AC}^{2}\, =\, {BC}^{2}\, +\, {CD}^{2}\, +\, {DA}^{2}$

  3. $2{AC}^{2}\, -\, 2 {AB}^{2}\, =\, {BC}^{2}\, +\, {CD}^{2}\, +\, {DA}^{2}$

  4. $2{AC}^{2}\, =\, 2{BC}^{2}\, +\, {CD}^{2}\, +\, {DA}^{2} + {BC}^2 $

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Correct Option: A
Explanation:

In quadrilateral $ABCD$, 

$\angle B = \angle D = 90^{\circ}$
Now, join $AC$
In $\triangle ABC$,
${AC}^2 = {AB}^2 + {BC}^2$
In $\triangle ADC$,
${AC}^2 = {AD}^2 + {CD}^2$
Add both the equations:
$2{AC}^2 = {AB}^2 + {BC}^2 + {AD}^2 + {CD}^2$
Hence, option A.

A grassy land in the shape of a right angled triangle has its hypotenuse $1$ metre more than twice the shortest side. If the third side is $7$ metres more than the shortest side. The sides of the grassy land 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. $8$m, $17$m, $15$m

  2. $2$m, $16$m, $13$m

  3. $10$m, $4$m, $5$m

  4. $7$m, $10$m, $14$m

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Correct Option: A
Explanation:

Let the length of the shortest side be $x$ meters. Then,
Hypotenuse = $(2x +1)$ meters, Third side = $(x + 7)$ meters
(Hyppotenuse)$^2$ = sum of the square of the remaining two sides    ....[By pythagorous theroem]
$\Rightarrow (2x + 1)^2 = x^2 + (x + 7)^2$
$\Rightarrow 4x^2 + 4x + 1 = 2x^2 + 14x + 49$
$\Rightarrow 2x^2 - 10x - 48 = 0$
$\Rightarrow x^2 - 5x - 24 = 0$
$\Rightarrow x^2 - 8x + 3x - 24 = 0$
$\Rightarrow x(x - 8) + 3 (x - 8) = 0$
$\Rightarrow (x - 8) (x + 3) = 0$
$\Rightarrow x = 8, -3$
$\Rightarrow x = 8$             [$\because x = -3$ is not possible]
Hence, the lengths of the sides of the grassy land are $8$ m., $17$ m. and $15$ m.

The hypotenuse of a right angled triangle is $25$cm. The other two sides are such that one is $5$cm longer than the other. Their lengths (in cm) 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. $10, 15$

  2. $20, 25$

  3. $15, 20$

  4. $25, 30$

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Correct Option: C
Explanation:

Given: Hypotenuse of right triangle $= 25$ cm
Let the other sides be $x$ and $x + 5$
Thus, By Pythagoras Theorem:
$25^2 = x^2 + (x + 5)^2$
$625 = x^2 + x^2 + 25 + 10x$
$2x^2 + 10x - 600 = 0 $
$x^2 + 5x - 300 = 0 $
$x^2 + 20x - 15x - 300 = 0$
$(x + 20) (x - 15) = 0 $
$x = 15, -20$
Thus, the other two sides are $15$ cm and $20$ cm.

Hypotenuse of a right triangle is $25cm$ and out of the remaining two sides, one is longer than the other by $5cm$. Find the lengths of the other two 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. $10$cm and $20$ cm

  2. $15$cm and $20$ cm

  3. $25$cm and $20$ cm

  4. $5$cm and $20$ cm

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Correct Option: B
Explanation:

Let one side be $xcm$. Then the other side will be $(x+5)cm$. Therefore, from Pythagoras theorem
${x}^{2}+{(x+5)}^{2}={25}^{2}$
$\Rightarrow { x }^{ 2 }+{ x }^{ 2 }+10x+25=625$
$\Rightarrow { x }^{ 2 }+5x-300=0\quad \quad \Rightarrow { x }^{ 2 }+20x-15x-300=0$
$\Rightarrow x(x+20)-15(x+20)=0$
$\Rightarrow (x-15)(x+20)=0\quad \Rightarrow \quad x=15\quad or\quad x=-20$
Rejecting $x=-20$, we have length of one side $=15cm$ and that of the other side $=(15+5)cm=20cm$

The lengths of the sides of a right-angled triangle are all given in natural numbers. If two of these numbers are odd and they differ by $50$, then the least possible value for the third 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. $61$

  2. $60$

  3. $51$

  4. $50$

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Correct Option: B
Explanation:

We can observe,considering general numbers for right angled triangle,
$5^2 = 3^2 + 4^2$ (5, 3 are odd and 4 lies between 5, 3)
and  $13^2 = 5^2 + 12^2$ (5, 13 are odd and 12 lies between 5, 13)
By trial and error
$61^2 = 60^2 + 11^2$
Since (61, 11 are odd and 61 hypotenuse)
$61 -11 = 50$
Hence, option 'B' is correct.

The distance between the top of two trees $20$m and $28$m high is $17$m. The horizontal distance between the trees 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. $11$m

  2. $31$m

  3. $15$m

  4. $9$m

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Correct Option: C
Explanation:

Consider the two trees as perpendiculars on the ground forming a right triangle, where the distance between their bases is the base $(B)$, the distance between their tops is the Hypotenuse $(H)$ and difference between their lengths is the perpendicular $(P)$.
Hypotenuse, $H = 17$ m
Perpendicular, $P=$ distance between their lengths $= 28 - 20 = 8$ m
Thus, By Pythagoras Theorem,
$H^2 = P^2 + B^2$
$17^2 = 8^2 + B^2$
$B^2 = 289 - 64$
$B = 15m$.

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:

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. $4\sqrt { 2 } $cm

  3. $4$ cm

  4. $2\sqrt { 2 } $cm

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Correct Option: D
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

The length of the hypotenuse of a right angled $\Delta$ whose two legs measure $12 \ cm$ and $0.35 \ m$ 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. $37 \ cm$

  2. $3.72 \ cm$

  3. $0.372 \ cm$

  4. $37 \ m$

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Correct Option: A
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 $=\sqrt{1369} = 37 \ cm.$