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# If three diagonals of the faces of a rectangular brick have their diag

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If three diagonals of the faces of a rectangular brick have their diag  [#permalink]

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03 Apr 2020, 00:55
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55% (hard)

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62% (02:44) correct 38% (03:16) wrong based on 51 sessions

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If three diagonals of the faces of a rectangular brick have their diagonals in the ratio $$3 : 2√3 : √15$$, then the ratio of the length of the shortest edge of the brick to that of its longest edge is

A. $$√3 : 2$$

B. $$2 : √5$$

C. $$1 : √3$$

D. $$√2 : √3$$

E. $$√2 : √5$$

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Re: If three diagonals of the faces of a rectangular brick have their diag  [#permalink]

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03 Apr 2020, 05:11
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Let a, b, and c be length of the three sides of the brick.

From the given ratio of the brick's face diagonals, we can calculate as follows:
(1) (a^2+b^2) = (3)^2 = 9
(2) (b^2+c^2) = (2√3)^2 = 12
(3) (a^2+c^2) = (√15)^2 = 15

(1)+(2)+(3)
2*(a^2+b^2+c^2) = 36
(a^2+b^2+c^2) = 18
(9+c^2) = 18
c^2 = 9

(2)
(b^2+c^2) = 12
(b^2+9) = 12
b^2 = 3

(1)
(a^2+b^2) = 9
(a^2+3) = 9
a^2 = 6

From values of a^2, b^2, and c^2, the shortest and the longest sides are obviously b and c, respectively.

b^2 : c^2 = 3 : 9 = 1 : 3
Hence, b : c = 1 : √3

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Re: If three diagonals of the faces of a rectangular brick have their diag  [#permalink]

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03 Apr 2020, 01:44
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Quote:
If three diagonals of the faces of a rectangular brick have their diagonals in the ratio 3:2√3:√15, then the ratio of the length of the shortest edge of the brick to that of its longest edge is

A. √3:2

B. 2:√5

C. 1:√3

D. √2:√3

E. √2:√5

Let sides are of length a < b < c

c = ?

3:2√3:√15

i.e. √(b^2+c^2) = √15
i.e. (b^2+c^2) = 15----------(1)

similarly
(a^2+c^2) = (2√3)^2 = 12----------(2)

and

(a^2+b^2) = (3)^2 = 9----------(3)

Subtracting eq(3) from eq (1)

c^2 - a^2 = 15-9 = 6
also (a^2+c^2) = 12

Adding them we get, 2c^2 = 18 i.e. c^2 = 9
i.e. a^2 = 3

Ratio of longest to shortest = c/a = √(c^2/a^2) = √(9/3) = √3

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Re: If three diagonals of the faces of a rectangular brick have their diag  [#permalink]

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03 Apr 2020, 01:52
1
a^2+c^2=9
a^2+b^2=12
b^2+c^2=15

a^2=9-c^2
b^2=c^2+3
b^2+c^2=15

a^2=9-c^2
b^2=c^2+3
c^2=6

c^2=6
b^2=9
a^2=3

a:b=sqrt(3):sqrt(9)
a:b=1:sqrt(3)

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Re: If three diagonals of the faces of a rectangular brick have their diag  [#permalink]

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03 Apr 2020, 04:07
Quote:
If three diagonals of the faces of a rectangular brick have their diagonals in the ratio 3:2√3:√15, then the ratio of the length of the shortest edge of the brick to that of its longest edge is

A. √3:2
B. 2:√5
C. 1:√3
D. √2:√3
E. √2:√5

a^2+b^2=3^2=9
a^2+c^2=(2√3)^2=12
b^2+c^2=(√15)^2=15
2(a^2+b^2+c^2)=36
a^2+b^2+c^2=18
a^2=18-15=3, a=√3
b^2=18-12=6, b=√6
c^2=18-9=9, c=√9
a/c=√3/√9=√(1/3)=1/√3

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Re: If three diagonals of the faces of a rectangular brick have their diag  [#permalink]

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04 Apr 2020, 00:01
ratios of diagonals=3:2√3:√15
√l^2+b^2:√h^2+b^2:√l^2+h^2 = 3:2√3:√15.....can be written in other combinations as well, but it doesnt matter since we need the highest and the lowest sides ratio
squaring on both sides
l^2+b^2:h^2+b^2:l^2+h^2 = = 9:12:15
after equating individual terms and solving them we get the ratio as √3:√2:1

ratio of the length of the shortest edge of the brick to that of its longest edge is 1:√3

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Re: If three diagonals of the faces of a rectangular brick have their diag  [#permalink]

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04 Apr 2020, 06:12
1
Option C
let l>b>h

l^2+b^2=15
l^2+h^2=12
b^2+h^2=9

solving above 3 equations, we get l=3 and h=root3
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Re: If three diagonals of the faces of a rectangular brick have their diag  [#permalink]

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04 Apr 2020, 20:30
let the common factor be k
and the sides be a, b and c
thus$$a^2+ b^2 = 9k^2$$
$$b^2+ c^2 = 12k^2$$
$$c^2+ a^2 = 15k^2$$

$$2(a^2+b^2+c^2) = 36 k^2$$
or
$$(a^2+b^2+c^2) = 18 k^2$$

subtract each of the above equations

so
$$c^2 =9k^2$$
$$b^2 =6k^2$$
$$a^2 =3k^2$$

thus ratio of smallest to longest edge will be$$a:c =\sqrt{ \frac{3k^2}{9k^2}}$$
=$$1:\sqrt{3}$$
thus C
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Re: If three diagonals of the faces of a rectangular brick have their diag  [#permalink]

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04 Apr 2020, 21:14
1
l > b> h
Diagonals = (l^2 + b^2 )^1/2 , (l^2 + h^2 )^1/2 , and (h^2 + b^2 )^1/2
Thus ratio of squares of diagonals = (l^2 + b^2 ) : (l^2 + h^2) : (h^2 + b^2 ) = (√15)^2 : (2√3)^2 : 3^2
Or (l^2 + b^2 ) : (l^2 + h^2) : (h^2 + b^2 ) = 15 : 12 : 9 = ( 9 + 6) : (9 + 3) : (3 + 6)
By comparing we can say l^2 = 9, h^2 = 3 and b^2 = 6
So l = 3 and h = √3
Required ratio of h/l = √3/3 = 1: √3
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Re: If three diagonals of the faces of a rectangular brick have their diag  [#permalink]

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05 Apr 2020, 20:08
If three diagonals of the faces of a rectangular brick have their diagonals in the ratio 3:2√3:√153:2√3:√15, then the ratio of the length of the shortest edge of the brick to that of its longest edge is

Solving :

a^2 + b^2 = 9k
b^2 + c ^2 = 12k
b^2 + c ^2 = 15k

a^2 + b^2 + c ^2 = 18k --------> a=\sqrt{6k} b=\sqrt{3k} & c = \sqrt{9k}

c/a = sqrt(3)/3 ANS C
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Re: If three diagonals of the faces of a rectangular brick have their diag  [#permalink]

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05 Apr 2020, 22:35
1
Let's say that the lengths of diagonals are 3, √12 and √15.
And we got :
--> $$a^{2}+ b^{2}= 9$$
--> $$a^{2}+ c^{2}= 12$$
--> $$b^{2}+ c^{2} = 15$$
----------------------
--> $$2 (a^{2}+b^{2} +c^{2}) =36$$
$$a^{2}+b^{2} +c^{2}= 18$$
$$a^{2}= 18- 15 = 3$$
$$a = √3$$
$$c= 3$$
$$b = √6$$

the ratio of the length of the shortest edge of the brick to that of its longest edge is

--> $$\frac{a}{ c}= \frac{√3}{3 }= \frac{1}{√3}$$

Re: If three diagonals of the faces of a rectangular brick have their diag   [#permalink] 05 Apr 2020, 22:35