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Sia ? 6 years, 6 months ago
In {tex}\Delta{/tex}BMC and {tex}\Delta{/tex} EMD, we have
MC = MD [{tex}\because{/tex} M is the mid-point of CD]
{tex}\angle \mathrm { CMB } = \angle E M D{/tex} [Vertically opposite angles]
and, {tex}\angle M B C = \angle M E D{/tex} [Alternate angles]
So, by AAS-criterion of congruence, we have
{tex}\therefore \quad \Delta B M C \cong \Delta E M D{/tex}
{tex}\Rightarrow{/tex} BC = DE .......(i)
Also, AD = BC [{tex}\because{/tex} ABCD is a parallelogram] ...... (ii)
Adding (i) and (ii),we get,
AD + DE = BC + BC
{tex}\Rightarrow{/tex} AE = 2 BC ...(iii)
Now, in {tex}\Delta{/tex}AEL and {tex}\Delta{/tex}CBL, we have
{tex}\angle A L E = \angle C L B{/tex} [Vertically opposite angles]
{tex}\angle E A L = \angle B C L{/tex} [Alternate angles]
So, by AA-criterion of similarity of triangles, we have
{tex}\Delta A E L \sim \Delta C B L{/tex}
{tex}\Rightarrow \quad \frac { E L } { B L } = \frac { A E } { C B }{/tex}
{tex}\Rightarrow \quad \frac { E L } { B L } = \frac { 2 B C } { B C }{/tex} [Using equations (iii)]
{tex}\Rightarrow \quad \frac { E L } { B L } = 2{/tex}
{tex}\Rightarrow{/tex} EL = 2 BL
Posted by Samarthsunil3 Samarthsunil3 7 years, 3 months ago
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Sia ? 6 years, 6 months ago
Let AB be the mountain of height h kilometres. Let C be a point at a distance of x km. from the base of the mountain such that the angle of elevation of the top at C is 30o. Let D be a point at a distance of 10 km from C such that the angle of elevation at D is of 15o.
In {tex}\triangle C A B,{/tex} we have
{tex}\tan 30 ^ { \circ } = \frac { A B } { A C }{/tex}
{tex}\Rightarrow \quad \frac { 1 } { \sqrt { 3 } } = \frac { h } { x }{/tex}
{tex}\Rightarrow \quad x = \sqrt { 3 } h{/tex} ...(i)
In {tex}\triangle D A B,{/tex} we have
{tex}\tan 15 ^ { \circ } = \frac { A B } { A D }{/tex}
{tex}\Rightarrow \quad 0.27 = \frac { h } { x + 10 }{/tex}
{tex}\Rightarrow{/tex} (0.27) (x +1) = h ...(ii)
Substituting x = {tex}\sqrt { 3 }{/tex}h obtained from equation (i) in equation (ii), we get
{tex}0.27 ( \sqrt { 3 } h + 10 ) = h{/tex}
{tex}\Rightarrow \quad 0.27 \times 10 = h - 0.27 \times \sqrt { 3 } h{/tex}
{tex}\Rightarrow \quad h ( 1 - 0.27 \times \sqrt { 3 } ) = 2.7{/tex}
{tex}\Rightarrow{/tex} h (1 - 0.46) = 2.7
{tex}\Rightarrow \quad h = \frac { 2.7 } { 0.54 } = 5{/tex}
Hence, the height of the mountain is 5 km.
Posted by Shubh Walia 7 years, 3 months ago
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Sia ? 6 years, 6 months ago
Given,
{tex}\frac { 1 } { ( a + b + x ) } = \frac { 1 } { a } + \frac { 1 } { b } + \frac { 1 } { x }{/tex}
{tex}\Rightarrow \quad \frac { 1 } { ( a + b + x ) } - \frac { 1 } { x } = \frac { 1 } { a } + \frac { 1 } { b } \Rightarrow \frac { x - ( a + b + x ) } { x ( a + b + x ) } = \frac { b + a } { a b }{/tex}
{tex}\Rightarrow \quad \frac { - ( a + b ) } { x ( a + b + x ) } = \frac { ( a + b ) } { a b }{/tex}
On dividing both sides by (a+b)
{tex}\Rightarrow \quad \frac { - 1 } { x ( a + b + x ) } = \frac { 1 } { a b }{/tex}
Now cross multiply
{tex}\Rightarrow{/tex} x(a + b + x) = -ab
{tex}\Rightarrow{/tex} x2 + ax + bx + ab = 0
{tex}\Rightarrow{/tex} x(x +a) + b(x +a) = 0
{tex}\Rightarrow{/tex} (x + a) (x + b) = 0
{tex}\Rightarrow{/tex} x + a = 0 or x + b = 0
{tex}\Rightarrow{/tex} x = -a or x = -b.
Therefore, -a and -b are the roots of the equation.
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Yogita Ingle 7 years, 3 months ago
In a right-angled triangle, “The sum of squares of the lengths of the two sides is equal to the square of the length of the hypotenuse (or the longest side).”
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Sia ? 6 years, 6 months ago
Given polynomial is p(x) = 2x3 - x2- 5x - 2
and -1 and 2 are zeroes of polynomial.
{tex}\therefore{/tex} {x - (-1)} (x - 2)= ( x + 1) (x - 2) = x2 - 2x + x - 2 = x2- x - 2 is a factor of p(x)
For other zeroes, 2x + 1 = 0
{tex}\Rightarrow x = \frac { - 1 } { 2 }{/tex}
{tex}\therefore{/tex} Other zero = {tex}\frac { - 1 } { 2 }{/tex}
Posted by Ankit Rajput 7 years, 3 months ago
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Sia ? 6 years, 6 months ago
Given: A square ABCD and equilateral triangle BCE and ACF have been described on the side BC and diagonal AC respectively.
To Prove {tex}\operatorname { ar } ( \Delta B C E ) = \frac { 1 } { 2 } \operatorname { ar } ( \Delta A C F ){/tex}.
Proof: Since each of the {tex}\triangle BCE{/tex} and {tex}\triangle ACF{/tex} is an equilateral triangle, so each angle of each one of them is 60°.
So, the triangles are equiangular and hence similar.
{tex}\therefore \quad \triangle B C E \sim \triangle A C F{/tex}
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}\therefore \quad \frac { \operatorname { ar } ( \triangle B C E ) } { \operatorname { ar } ( \triangle A C F ) } = \frac { B C ^ { 2 } } { A C ^ { 2 } } = \frac { B C ^ { 2 } } { 2 ( B C ) ^ { 2 } }{/tex} {tex}[ \because A C = \sqrt { 2 } B C ]{/tex}
{tex}= \frac { 1 } { 2 }{/tex}
Hence, {tex}\operatorname { ar } ( \triangle B C E ) = \frac { 1 } { 2 } \times \operatorname { ar } ( \triangle A C F ){/tex} .
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