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Sia ? 6 years, 6 months ago

from right angle triangle BOC,
BC2 = OB2 + OC2
{tex}\Rightarrow{/tex} BC2 = 102 + 102
{tex}\Rightarrow{/tex} BC2 = 100 + 100
{tex}\Rightarrow{/tex} BC2 = 200
{tex}\Rightarrow{/tex} BC = {tex}\sqrt{200}{/tex}
{tex}\Rightarrow{/tex} BC = {tex}10\sqrt{2}{/tex}
So, the length of the chord is {tex}10\sqrt{2}{/tex} cm.
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Sia ? 6 years, 6 months ago
Let A → (1, 2), B → (4, y), C→ (x, 6) and D→ (3, 5).
We know that the diagonals of parallelogram bisect each other.
So, Coordinates of the mid-point of diagonal AC
= Coordinates of the mid-point of diagonal BD
{tex}\Rightarrow \left( {\frac{{1 + x}}{2},\frac{{2 + 6}}{2}} \right) = \left( {\frac{{4 + 3}}{2},\frac{{y + 5}}{2}} \right){/tex}
{tex}\Rightarrow \left( {\frac{{1 + x}}{2},4} \right) = \left( {\frac{7}{2},\frac{{y + 5}}{2}} \right){/tex}
{tex}\Rightarrow \frac{{1 + x}}{2} = \frac{7}{2}{/tex}
{tex}\Rightarrow{/tex} 1 + x = 7
{tex}\Rightarrow{/tex} x = 6
and {tex}4 = \frac{{y + 5}}{2}{/tex}
{tex}\Rightarrow{/tex} y + 5 = 8
{tex}\Rightarrow{/tex} y = 3
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Sia ? 6 years, 6 months ago

Given, AB is the hill and P and Q are two consecutive km stones.
Let the height of the hill AB be h m and
{tex}BP = x \ m.{/tex}
{tex}PQ= 1 km = 1000m{/tex}
In {tex}\Delta \mathrm { ABP },{/tex}
{tex}\tan 45 ^ { \circ } = \frac { A B } { B P }{/tex}
{tex}\therefore 1 = \frac { h } { x }{/tex}
{tex}\Rightarrow{/tex} {tex}x = h{/tex} ...(i)
In {tex}\triangle \mathrm { ABQ },{/tex}
{tex}\tan 30 ^ { \circ } = \frac { \mathrm { AB } } { \mathrm { BQ } }{/tex}
{tex}\therefore \frac { 1 } { \sqrt { 3 } } = \frac { h \mathrm { m } } { ( x + 1000 ) \mathrm { m } }{/tex} {tex}[ \because \mathrm { BQ } = \mathrm { BP } + \mathrm { PQ } = x + 1000]{/tex}
{tex}\Rightarrow{/tex}{tex}x + 1000 ={/tex} {tex}\sqrt { 3 } h{/tex}
{tex}\Rightarrow \sqrt { 3 } h = h + 1000{/tex}[Using (i)]
{tex}\Rightarrow ( \sqrt { 3 } - 1 ) h = 1000{/tex}
{tex}\Rightarrow h = \frac { 1000 } { \sqrt { 3 } - 1 }{/tex}
{tex}\Rightarrow h = \frac { 1000 ( \sqrt { 3 } + 1 ) } { ( \sqrt { 3 } - 1 ) ( \sqrt { 3 } + 1 ) }\\ = \frac { 1000 ( \sqrt { 3 } + 1 ) } { ( 3 - 1 ) } \\= \frac { 1000 ( \sqrt { 3 } + 1 ) } { 2 }\\ = 500 ( \sqrt { 3 } + 1 ){/tex}
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Rohan Chavan 7 years, 1 month ago
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