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Posted by Abhishek Kumar 7 years, 3 months ago
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Posted by Tanisha Singh 7 years, 3 months ago
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Gaurav Seth 7 years, 3 months ago
Let us assume that ΔABC is an isosceles triangle having CA and CB of equal lengths, base AB of 8 cm, and AD is the altitude of 4 cm.
</header>A ΔAB'C' whose sides are
times of ΔABC can be drawn as follows.
Step 1
Draw a line segment AB of 8 cm. Draw arcs of same radius on both sides of the line segment while taking point A and B as its centre. Let these arcs intersect each other at O and O'. Join OO'. Let OO' intersect AB at D.
Step 2
Taking D as centre, draw an arc of 4 cm radius which cuts the extended line segment OO' at point C. An isosceles ΔABC is formed, having CD (altitude) as 4 cm and AB (base) as 8 cm.
Step 3
Draw a ray AX making an acute angle with line segment AB on the opposite side of vertex C.
Step 4
Locate 3 points (as 3 is greater between 3 and 2) A1, A2, and A3 on AX such that AA1 = A1A2 = A2A3.
Step 5
Join BA2 and draw a line through A3 parallel to BA2 to intersect extended line segment AB at point B'.
Step 6
Draw a line through B' parallel to BC intersecting the extended line segment AC at C'. ΔAB'C' is the required triangle.

Justification
The construction can be justified by proving that

In ΔABC and ΔAB'C',
∠ABC = ∠AB'C' (Corresponding angles)
∠BAC = ∠B'AC' (Common)
∴ ΔABC ∼ ΔAB'C' (AA similarity criterion)
… (1)
In ΔAA2B and ΔAA3B',
∠A2AB = ∠A3AB' (Common)
∠AA2B = ∠AA3B' (Corresponding angles)
∴ ΔAA2B ∼ ΔAA3B' (AA similarity criterion)

On comparing equations (1) and (2), we obtain

⇒
This justifies the construction.
Ram Kushwah 7 years, 3 months ago
If base =2x=8
x=4
Side2 =x2+h2=16+16=32
side=4√2=4×1.41=5.64 cm
AB=AC=5.64 Cm BC=8 cm
Posted by Durga Appa 7 years, 3 months ago
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Gurjar Kabir (बैंसला) 7 years, 3 months ago
Gurjar Kabir (बैंसला) 7 years, 3 months ago
Posted by Gurjar Kabir (बैंसला) 7 years, 3 months ago
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Anushka Jugran ? 7 years, 3 months ago
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Jasline S.? 7 years, 3 months ago
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Posted by Deepika Pansotra 6 years, 9 months ago
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Sia ? 6 years, 9 months ago
The greatest number of cartons is the HCF of 144 and 90
Now the prime factorization of 144 and 90 are
{tex}144=16×9=2^4{/tex} {tex}\times{/tex} 32.{tex}{/tex}
{tex}90=2\times3\times3\times5=2\times3^2\times5{/tex}
HCF (144,90)= 2 {tex}\times{/tex} 32 = 18
{tex}\therefore{/tex} The greatest number of cartons each stack would have= 18.
Posted by Twinkle Star ( Atul )....... ?? 7 years, 3 months ago
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Puja Sahoo? 7 years, 3 months ago
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Yaar 7 years, 3 months ago
Posted by Fehmina Tabassum 6 years, 9 months ago
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Sia ? 6 years, 9 months ago
Given,

AB II PR
{tex}\angle{/tex}B = {tex}\angle{/tex}Q [Alternate interior angles]
{tex}\angle{/tex}B = 70o
{tex}\triangle{/tex}AQB is an Isosceles triangle,
AQ = BQ
So, {tex}\angle{/tex}BQR = {tex}\angle{/tex}BAQ
{tex}\angle{/tex}A = 70o
In {tex}\triangle{/tex}AQB,
{tex}\angle{/tex}QAB+ {tex}\angle{/tex}ABQ + {tex}\angle{/tex}BQA = 180o
{tex}\Rightarrow{/tex} {tex}70° + 70° +{/tex} {tex}\angle{/tex}{tex}Q = 180^o{/tex}
{tex}\Rightarrow{/tex} {tex}\angle{/tex}B{tex}QA = 180^o - 140^o = 40^o{/tex}
Posted by Yashika Goyal 7 years, 3 months ago
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Gaurav Seth 7 years, 3 months ago
Let a and d be the first term and common difference respectively of the given AP,then
mth term = a+(m-1)d ==> a+(m-1)d = 1/n -----(1)
and nth term=a+(n-1)d ==> a+(n-1)d = 1/m -----(2)
subtracting (2) from (1), we have
Let the first term of the A.P be 'a ' and its common difference be 'd '
Now, mth term of the A.P is 1/n
which means, a + (m - 1)d = 1/n --- (i)
Again,
its nth term is 1/m
which means, a + (n - 1)d = 1/m --- (ii)
on substracting eqn. (ii) from (i) we get,
a + (m - 1)d - a - (n - 1)d = 1/n - 1/m
=> d (m - 1 - n + 1) = (m - n)/mn
=> d (m - n) = (m - n) / mn
=> d = 1/mn
so, a = 1/m - (n - 1)(1/mn) = n - n + 1 / mn =1/mn
ie., a = d = 1/mn
Now, Smn = mn/2 [2a + (mn - 1)d]
= mn/2 [2/mn + (mn - 1)1/mn]
= mn/2 [2 + mn - 1]/mn
= (mn + 1)/2
Posted by Aa Bisht 6 years, 9 months ago
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Sia ? 6 years, 9 months ago
Radius of well {tex} = \frac{7}{4}m{/tex}
Depth of well = 20m
{tex}\therefore {/tex} Volume of earth taken out {tex} = \frac{{22}}{7} \times \frac{7}{4} \times \frac{7}{4} \times 20c{m^3} = \frac{{385}}{2}{m^3}{/tex}
Area of field {tex} = 20m \times 14m = 280{m^2}{/tex}
Area of field excluding well {tex} = \left( {280 - \frac{{22}}{7} \times \frac{7}{4} \times \frac{7}{4}} \right){m^2} = \frac{{2163}}{8}{m^2}{/tex}
{tex}\therefore {/tex} Level of earth raised {tex} = \frac{{volume\;of\;earth\;taken\;out}}{{Area\;of\;field}}{/tex}
{tex} = \frac{{385}}{2} \times \frac{8}{{2163}}m = 0.7119m{/tex}
= 71.19 cm
Posted by Aa Bisht 7 years, 3 months ago
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Chētnà Pandey✌️ 7 years, 3 months ago
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Chētnà Pandey✌️ 7 years, 3 months ago
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Vishakha Singh? 7 years, 3 months ago

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Harsh Pandey 7 years, 3 months ago
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