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Ask QuestionPosted by Kumarimanja923@Gmail.Com Manthan 1 year, 10 months ago
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Preeti Dabral 1 year, 10 months ago
y-Coordinate. A y-coordinate is the second element in an ordered pair. When an ordered pair is graphed as the coordinates of a point in the coordinate plane, the y-coordinate represents the directed distance of the point from the x-axis. Another name for the y-coordinate is the ordinate.
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Preeti Dabral 1 year, 10 months ago
AB = 3 cm, AC = 4 cm
In {tex}\triangle{/tex}BAC, by pythagoras theorem
BC2 = AB2 + AC2
{tex}\Rightarrow{/tex}BC2 = 32 + 42
{tex}\Rightarrow{/tex}BC2 = 25
{tex}\Rightarrow{/tex}BC = {tex}\sqrt {25} {/tex} = 5 cm
In {tex}\triangle{/tex}AOB and {tex}\triangle{/tex}CAB
{tex}\angle{/tex}ABO = {tex}\angle{/tex}ABC [common]
{tex}\angle{/tex}AOB = {tex}\angle{/tex}BAC [each 90o]
Then, {tex}\triangle{/tex}AOB - {tex}\triangle{/tex}CAB [by AA similarity]
{tex}\therefore{/tex} {tex}\frac { A O } { C A } = \frac { O B } { A B } = \frac { A B } { C B }{/tex} [c.p.s.t]
{tex}\Rightarrow{/tex} {tex}\frac { A O } { 4 } = \frac { O B } { 3 } = \frac { 3 } { 5 }{/tex}
Then, AO = {tex}\frac{{4 \times 3}}{5}{/tex} and OB = {tex}\frac{{3 \times 3}}{5}{/tex}
{tex}\Rightarrow{/tex} AO = {tex}\frac{12}{5}{/tex} cm and OB = {tex}\frac{9}{5}{/tex} cm
{tex}\therefore{/tex}OC = 5 - {tex}\frac{9}{5}{/tex} = {tex}\frac{16}{5}{/tex}cm
{tex}\therefore{/tex} Volume of double cone thus generated = volume of first cone + volume of second cone
{tex}= \frac { 1 } { 3 } \pi ( A O ) ^ { 2 } \times B O + \frac { 1 } { 3 } \pi ( A O ) ^ { 2 } \times O C{/tex}
{tex}= \frac { 1 } { 3 } \times \frac { 22 } { 7 } \times \left( \frac { 12 } { 5 } \right) ^ { 2 } \times \frac { 9 } { 5 } + \frac { 1 } { 3 } \times \frac { 22 } { 7 } \times \left( \frac { 12 } { 5 } \right) ^ { 2 } \times \frac { 16 } { 5 }{/tex}
{tex}= \frac { 1 } { 3 } \times \frac { 22 } { 7 } \times \frac { 12 } { 5 } \times \frac { 12 } { 5 } \left[ \frac { 9 } { 5 } + \frac { 16 } { 5 } \right]{/tex}
{tex}= \frac { 1 } { 3 } \times \frac { 22 } { 7 } \times \frac { 12 } { 5 } \times \frac { 12 } { 5 } \times 5{/tex}
={tex}\frac{1056}{35}{/tex} = {tex}30 \frac { 6 } { 35 } \mathrm { cm } ^ { 3 }{/tex}.
Posted by Marh Power 1 year, 10 months ago
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Posted by Marh Power 1 year, 10 months ago
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Preeti Dabral 1 year, 10 months ago
ABC is a given triangle with, AB=AC.
To prove: Angle opposite to AB= Angle
opposite to AC (i.e) ∠C=∠B
Construction: Draw AD perpendicular to BC
∴∠ADB=∠ADC=90o
Proof:
Consider △ABD and △ACD
AD is common
AB=AC
∴∠ADB=∠ADC=90o
Hence ∠ABD=∠ACD
∠ABC=∠ACB
∠B=∠C. Hence the proof
This is known as Isosceles triangle theorem
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Prarthana P 1 year, 10 months ago
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