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Sia ? 6 years, 6 months ago
Given: {tex}\triangle {/tex} ABC in which AB=BC=CA and AD {tex} \bot {/tex} BC
To prove: 3BC2 = 4AD2
Proof: {tex}\triangle {/tex} ABC is an equilateral triangle. AD {tex} \bot {/tex} BC
{tex}\Rightarrow {/tex} BD = DC [In an equilateral,{tex}\triangle {/tex} altitude bisect the opposite side]
Now in {tex}\triangle {/tex}ADB, {tex}\angle{/tex} D = 90o
{tex}\Rightarrow {/tex} AB2 = AD2 + BD2 [Pythagoras theorem]
{tex}\Rightarrow {/tex} BC2 = AD2+{tex}{\left( {\frac{{BC}}{2}} \right)^2}{/tex}
{tex}\Rightarrow {/tex} 4BC2 = 4AD2 + BC2
{tex}\Rightarrow {/tex} 3BC2 = 4AD2
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Yogita Ingle 6 years, 11 months ago
tan 45 = 1 cos30= √ 3/2 sin 60 = √ 3/2
2 tan245 + cos230 - sin2 60
= 2(1) + (√ 3/2)2 - (√ 3/2)2
= 2 + 3/4 - 3/4
= 2
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![](“https://images.app.goo.gl/Z6yv5GAmRxyqGCb27")
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Sia ? 6 years, 6 months ago
We have to find the HCF, by Euclid's division algorithm of the numbers 92690, 7378 and 7161.
Here, we will apply Euclid's Division Lemma, we have
92690 = 7378 × {tex}{/tex}12 + 4154
Again we apply Euclid's Division Lemma of divisor 7,378 and remainder 4154, thus we have
7378 = 4154 × {tex}{/tex}1 + 3,224
4154 = 3224 × {tex}{/tex}1 + 930
3224 = 930 × 3 {tex}{/tex} + 434
930 = 434 × 2 {tex}{/tex}+ 62
434 = 62 × {tex}{/tex}7 + 0
HCF of 92690 and 7378= 62
Now, using Euclid's Division Lemma on 7161 and 62, we have
7161 = 62 × {tex}{/tex}115 + 31
Again, applying Euclid's Division Lemma on divisor 62 and remainder 31, we have
62 = 31 × {tex}{/tex}2 + 0
Clearly, HCF of 7161 and 62 = 31
Hence, HCF of 92690, 7378 and 7161 is 31.
Posted by Anshika & Aliens???? 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
{tex}{/tex}According to the question,we have the following information,
Volume of water in 30 minutes {tex}= \pi \times ( 2 ) ^ { 2 } \times 700 \times 60 \times 30 \mathrm { cm } ^ { 3 }{/tex}
Let height of water in tank = h cm
and radius = 40 cm
Volume of water in the tank {tex}\pi ( 40 ) ^ { 2 } \times h = 700 \times 60 \times 30 \times 4 \times \pi{/tex}
or, {tex}h = \frac { 700 \times 60 \times 30 \times 4 } { 40 \times 40 }{/tex}
{tex}= \frac { 6300 } { 2 } \mathrm { cm } = \frac { 63 } { 2 } \mathrm { }{/tex}
=31.5 meters
Hence, water level increased = 31.5 metres
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Ram Kushwah 6 years, 11 months ago
PLEASE DO NOT CHAT AND DOWNLOAD SOLUTION FROM
https://www.vedantu.com/ncert-solutions/campve-ncert-solutions-class-10-maths-chapter-3-pair-of-linear-equations-in-two-variables
0Thank You