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Sia ? 6 years, 4 months ago
The required number when divides 615 and 963, leaves remainder 6, this means 615 - 6 = 609 and 963 - 6 = 957 are completely divisible by the number.
Therefore,
The required number = H.C.F. of 609 and 957.
By applying Euclid’s division lemma
957 = 609 {tex}\times{/tex} 1+ 348
609 = 348 {tex}\times{/tex} 1 + 261
348 = 216 {tex}\times{/tex} 1 + 87
261 = 87 {tex}\times{/tex} 3 + 0.
Therefore, H.C.F. of 957 and 609 is = 87.
Hence, the largest number which divides 615 and 963 leaving remainder 6 in each case is 87.
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Sia ? 6 years, 4 months ago
We have,
DE || BC
Now, In {tex}\triangle{/tex}ADE and {tex}\triangle {/tex}ABC
{tex}\angle A = \angle A{/tex} [common]
{tex}\angle A D E = \angle A B C{/tex} [{tex}\because{/tex} DE || BC {tex}\Rightarrow{/tex} Corresponding angles are equal]
{tex}\Rightarrow \triangle A D E= \triangle A B C{/tex} [By AA criteria]
{tex}\Rightarrow \frac { A B } { B C } = \frac { A D } { D E }{/tex} [{tex}\because{/tex} Corresponding sides of similar triangles are proportional]
{tex}\Rightarrow \frac { A B } { 5 } = \frac { 2.4 } { 2 }{/tex}
{tex}\Rightarrow A B = \frac { 2.4 \times 5 } { 2 }{/tex}
{tex}\Rightarrow{/tex} AB = 1.2 {tex}\times{/tex} 5
= 6.0 cm
{tex}\Rightarrow{/tex} AB = 6 cm
{tex}\therefore{/tex} BD = AB - AD
= 6 - 2.4
= 3.6 cm
{tex}\Rightarrow{/tex} DB = 3.6 cm
Now,
{tex}\frac { A C } { B C } = \frac { A E } { D E }{/tex} [{tex}\because{/tex} Corresponding sides of similar triangles are equal]
{tex}\Rightarrow \frac { A C } { 5 } = \frac { 3.2 } { 2 }{/tex}
{tex}\Rightarrow A C = \frac { 3.2 \times 5 } { 2 }{/tex}
= 1.6 {tex}\times{/tex} 5
= 8.0 cm
{tex}\Rightarrow{/tex} AC = 8 cm
{tex}\therefore{/tex} CE = AC - AE
= 8 - 3.2
= 4.8 cm
Hence, BD = 3.6 cm and CE = 4.8 cm
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Aayushvardhan Pandey 7 years, 4 months ago
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