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Ask QuestionPosted by Nishant Pal 7 years, 4 months ago
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Posted by Tanveer Arhaan 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Let the side of triangle = a
{tex}\therefore{/tex} BC = a {tex}\Rightarrow{/tex} BD = {tex}\frac{a}{2}{/tex}
Now area({tex}\triangle{/tex}BDE) = {tex}\frac{\sqrt{3}}{4}{/tex}({tex}\frac{a}{2}{/tex})2 = {tex}\frac{\sqrt{3}}{4}{/tex}{tex}\frac{a^2}{4}{/tex} = {tex}\frac{1}{4}{/tex}({tex}\frac{\sqrt{3}}{4}{/tex}a2) = {tex}\frac{1}{4}{/tex}area ({tex}\triangle{/tex}ABC).
or area({tex}\triangle{/tex}BDE) = {tex}\frac{1}{4}{/tex}area ({tex}\triangle{/tex}ABC).
Hence proved.
Posted by Diksha Dabas 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
PA = PB (Given)
{tex}\therefore {/tex} PA2 = PB2
{tex} \Rightarrow {/tex} (5 - x)2 + (1 - y)2 = (1 - x)2 + (5 - y)2
{tex} \Rightarrow {/tex} 25 + x2 - 10x + 1 + y2 - 2y = 1 + x2 - 2x + 25 + y2 - 10y
{tex} \Rightarrow {/tex} -8x = -10y + 2y
{tex} \Rightarrow {/tex} -8x = -8y
{tex} \Rightarrow {/tex} x = y
Posted by Tanya Tomar 7 years, 4 months ago
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Sanjivani Gawade 7 years, 4 months ago
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Omraje Deore 7 years, 4 months ago
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Posted by Akansha Mourya 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Let AB be the vertical stick and AC be its shadow. Also, let DE be the vertical tower and DF be its shadow. Join BC and EF. Let DE = x metres.
We have,
AB = 12 m, AC = 8m, and DF = 40m.
In {tex}\Delta{/tex} ABC and {tex}\Delta{/tex}DEF, we have,
{tex}\angle{/tex}A = {tex}\angle{/tex}D = 90° and {tex}\angle{/tex}C = {tex}\angle{/tex}F [Angular elevation of the sun]
Therefore, by AA- criterion of similarity, we obtain {tex}\Delta A B C \sim \Delta D E F{/tex}
{tex}\Rightarrow \quad \frac { A B } { D E } = \frac { A C } { D F }{/tex}
{tex}\Rightarrow \quad \frac { 12 } { x } = \frac { 8 } { 40 } \Rightarrow \frac { 12 } { x } = \frac { 1 } { 5 } \Rightarrow x {/tex} = 60 metres
Posted by Karan Preet 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Let perimeter of first square = x metres
Let perimeter of second square = (x +24) metres
Length of side of first square = {tex}\frac { x } { 4 }{/tex}metres {Perimeter of square = 4 × length of side}
Length of side of second square = {tex}\left( \frac { x + 24 } { 4 } \right){/tex}metres
Area of first square = side × side = {tex}\frac { x } { 4 } \times \frac { x } { 4 } = \frac { x ^ { 2 } } { 16 } m ^ { 2 }{/tex}
Area of second square = {tex}\left( \frac { x + 24 } { 4 } \right) ^ { 2 } m ^ { 2 }{/tex}
According to given condition:
{tex}\frac { x^{ 2 } } { 16 } + \left( \frac { x + 24 } { 4 } \right) ^ { 2 } = 468{/tex} {tex}\Rightarrow \frac { x ^ { 2 } } { 16 } + \frac { x ^ { 2 } + 576 + 48 x } { 16 } = 468{/tex}
{tex}\Rightarrow \frac { x ^ { 2 } + x ^ { 2 } + 576 + 48 x } { 16 } = 468{/tex} {tex}\Rightarrow{/tex} 2x2 + 576 + 48x = 468 × 16
{tex}\Rightarrow{/tex} 2x2 +48x + 576 = 7488 {tex}\Rightarrow{/tex} 2x2 + 48x - 6912 = 0
{tex}\Rightarrow{/tex} x2 + 24x - 3456 = 0
Comparing equation x2 + 24x - 3456 = 0 with standard form ax2 + bx + c = 0,
We get a = 1, b = 24 and c = -3456
Applying Quadratic Formula {tex}x = {-b \pm \sqrt{b^2-4ac} \over 2a}{/tex}
{tex}x = \frac { - 24 \pm \sqrt { ( 24 ) ^ { 2 } - 4 ( 1 ) ( - 3456 ) } } { 2 \times 1 }{/tex}
{tex}\Rightarrow x = \frac { - 24 \pm \sqrt { 576 + 13824 } } { 2 }{/tex}
{tex}\Rightarrow x = \frac { - 24 \pm \sqrt { 14400 } } { 2 } = \frac { - 24 \pm 120 } { 2 }{/tex}
{tex}\Rightarrow x = \frac { - 24 + 120 } { 2 } , \frac { - 24 - 120 } { 2 }{/tex}
{tex}\Rightarrow{/tex} x = 48, -72
Perimeter of square cannot be in negative. Therefore, we discard x = -72
Therefore, perimeter of first square = 48 metres
And, Perimeter of second square = x + 24 = 48 + 24 = 72 metres
{tex}\Rightarrow{/tex} Side of First square {tex}= \frac { \text { Perimeter } } { 4 } = \frac { 48 } { 4 } = 12 \mathrm { m }{/tex}
And, Side of second Square {tex}= \frac { \text { Permeter } } { 4 } = \frac { 72 } { 4 } = 18 \mathrm { m }{/tex}
Posted by Manshi Heda 7 years, 4 months ago
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Archit Raj 7 years, 4 months ago
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Posted by Muskan Shekhawat 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Thales' theorem states that if A, B, and C are distinct points on a circle where the line AC is a diameter, then the angle ∠ABC is a right angle.
Posted by Anas Ahamd 6 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
213 = 120 x 1 + 93
120 = 93 x 1 + 27
93 = 27 x 3 + 12
27 = 12 x 2 + 3
12 = 3 x 4 + 0
The HCF(213,120) = 3
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Sunil Kumar 7 years, 4 months ago
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