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Yogita Ingle 6 years, 3 months ago
The base of an equilateral triangle is 8cm and one of its equal sides is 5cm.
Please check question how is it possible equilateral triangle of 8cm base and one of its equal sides is 5cm?
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Yogita Ingle 6 years, 3 months ago
Since, 237 > 81
On applying Euclid’s division algorithm, we get
237 = 81 × 2 + 75 ...(i)
81 = 75 × 1 + 6 ...(ii)
75 = 6 × 12 + 3 ...(iii)
6 = 3 × 2 + 0 ...(iv)
Hence, and HCF (81, 237) = 3. 1 Write 3 in the form of 81x + 237y, move backwards
3 = 75 – 6 × 12 [From (iii)]
= 75 – (81 – 75 × 1) × 12 [Replace 6 from (ii)]
= 75 – (81 × 12 – 75 × 1 × 12)
= 75 – 81 × 12 + 75 × 12
= 75 + 75 × 12 – 81 × 12
= 75 ( 1 + 12) – 81 × 12
= 75 × 13 – 81 × 12
= 13(237 – 81 × 2) – 81 × 12 [Replace 75 from (i)]
= 13 × 237 – 81 × 2 × 13 – 81 × 12
= 237 × 13 – 81 (26 + 12)
= 237 × 13 – 81 × 38
= 81 × (– 38) + 237 × (13)
= 81x + 237y
Hence, x = – 38 and y = 13
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Yogita Ingle 6 years, 3 months ago
Given: A square ABCD an equilateral triangle ABC and ACF have been described on side BC and diagonal AC respectively.
To Prove :- {tex}ar( ∆BCE ) = \bf{ \frac{1}{2} ar( \triangle ACF ) . }{/tex}
Proof :- Since each of the ∆ABC and ∆ACF is an equilateral triangle, so each angle of his strength is one of them is 60°. So, the angles are equiangular, and hence similar.
=> ∆BCE ~ ∆ACF.
We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.
{tex}\bf{ \frac{ ar( \triangle BCE ) }{ ar( \triangle ACF ) } = \frac{{BC}^{2} }{ {AC}^{2}} = \frac{{BC}^{2}}{2{(BC)}^{2}}. }{/tex}
[ Because, AC is hypotenuse => AC = √2BC. ]
{tex}\bf{ \implies \frac{ ar( \triangle BCE ) }{ ar( \triangle ACF ) } = \frac{1}{2} . }{/tex}
Hence, {tex}\boxed{ \sf ar( \triangle BCE ) = \frac{1}{2} \times ar( \triangle ACF ) . }{/tex}
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Yogita Ingle 6 years, 3 months ago
Let x be the numerator .
Since we are given that denominator is 2 more than the numerator .
So, denominator = x+2
So Fraction = {tex}\frac{x}{x+2}{/tex}
Reciprocal of fraction = {tex}\frac{x+2}{x}{/tex}
Now we are given that the sum of fraction and its reciprocal is 34/15.
{tex}⇒\frac{x}{x+2}+\frac{x+2}{x} =\frac{34}{15}{/tex}
{tex}⇒\frac{(x*x)+[(x+2)*(x+2)]}{(x+2)(x)}=\frac{34}{15}{/tex}
{tex}⇒\frac{2x^{2}+4x+4}{x^{2}+2x}=\frac{34}{15}{/tex}
{tex}⇒15*(2x^{2}+4x+4)=34*(x^{2}+2x){/tex}
{tex}⇒30x^{2}+60x+60)=34x^{2}+68x{/tex}
{tex}⇒4x^{2}+8x-60=0{/tex}
{tex}⇒x^{2}+2x-15=0{/tex}
{tex}⇒x^{2}+5x-3x-15=0{/tex}
⇒x(x+5)-3(x+5)=0
⇒(x-3)(x+5)=0
⇒x-3=0, x+5=0
⇒ x = 3 , x= -5
So Fraction when x = 3
{tex}\frac{x}{x+2} =\frac{3}{3+2}=\frac{3}{5}{/tex}
Fraction when x = -5
{tex}\frac{x}{x+2} =\frac{-5}{-5+2}=\frac{5}{3}{/tex}
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