This page is prepared by The Brainbox Tutorials. It contains ICSE Class 10 Maths Ratio and Proportion PYQs with Solution. The PYQs are from the year 2010 to 2023. Ratio and proportion is an important chapter in ICSE Class 10 Mathematics. Solving previous years’ questions from this chapter will not only increase the understanding of the chapter but also boost the confidence of the aspiring students.
The PDF of the PYQs is also provided for download.
ICSE Class 10 Maths Ratio and Proportion PYQs PDF
ICSE Class 10 Maths Ratio and Proportion PYQs with Solution
Q1. Using Componendo and Dividendo solve for x: [2023]
\[\frac{\sqrt{2x+2}+\sqrt{2x-1}}{\sqrt{2x+2}-\sqrt{2x-1}}=3\]
Solution: x = 1
Step-by-step Explanation:
\[\frac{\sqrt{2x+2}+\sqrt{2x-1}}{\sqrt{2x+2}-\sqrt{2x-1}}=3\\u\sin g\;componendo\;and\;dividendo,\\\frac{\sqrt{2x+2}+\sqrt{2x-1}+\sqrt{2x+2}-\sqrt{2x-1}}{\sqrt{2x+2}+\sqrt{2x-1}-\sqrt{2x+2}+\sqrt{2x-1}}=\frac{3+1}{3-1}\\\frac{2\sqrt{2x+2}}{2\sqrt{2x-1}}=\frac42\\\frac{\sqrt{2x+2}}{\sqrt{2x-1}}=2\\squaring\;both\;sides,\;we\;get\\\frac{2x+2}{2x-1}=4\\8x-4=2x+2\\8x-2x=2+4\\6x=6\\x=1\]
You can also watch the video solution of ICSE Class 10 Maths Ratio and Proportion PYQs here.
Q2. What number must be added to each of the numbers 4, 6, 8, 11 in order to get the four numbers in proportion? [2023]
Solution: 4
Step-by-step Explanation:
\[Let\;x\;be\;added\;to\;each\;of\;the\;numbers.\\According\;to\;the\;problem,\\\frac{4+x}{6+x}=\frac{8+x}{11+x}\\(4+x)(11+x)=(8+x)(6+x)\\44+11x+4x+x^2=48+6x+8x+x^2\\15x+44=14x+48\\15x-14x=48-44\\x=4\]
Q3. The mean proportional between 4 and 9 is
(a) 4
(b) 6
(c) 9
(d) 36 [2023]
Solution: (b)
Step-by-step Explanation:
Let the mean proportional be x.
Therefore, 4/x = x/9
x2 = 36
x = 6
Q4. If x, y, z are in continued proportion then (y2 + z 2): (x2 + y2 ) is equal to: [2]
(a) z: x
(b) x : z
(c) zx
(d) (y + z):(x + y) [2021 Semester-1]
Solution: (a)
Step-by-step Explanation:
\[Given,\;x,y,z\;are\;in\;continued\;proportion.\\Therefore,\;y^2=zx\\Now,\;\frac{y^2+z^2}{x^2+y^2}\\=\frac{zx+z^2}{x^2+zx}\\=\frac{z(x+z)}{x(x+z)}\\=\frac zx\]
Q5. If a, b, c, and d are proportional then (a+b)/(a-b) is equal to:
(a) c/d
(b) (c-d)/(c+d)
(c) d/c
(d) (c+d)/( c-d) [2021 Semester-1]
Solution: (d)
Step-by-step Explanation:
\[Given,\;\;a,b,c,d\;are\;in\;proportion.\\Therefore,\;\frac ab=\frac cd=k(say)\\so,\;a=bk\;and\;c=dk\\Now,\;\frac{a+b}{a-b}\\=\frac{b\left({\displaystyle\frac ab}+1\right)}{b\left(\frac ab-1\right)}\\=\frac{\left({\displaystyle\frac cd}+1\right)}{\left({\displaystyle\frac cd}-1\right)}\\=\frac{\left({\displaystyle\frac{c+d}d}\right)}{\left({\displaystyle\frac{c-d}d}\right)}\\=\frac{c+d}{c-d}\]
ICSE Class 10 Maths PYQs with Solution
Q6. If x, 5.4, 5, 9 are in proportion then x is: [1]
(a) 3
(b) 9.72
(c) 25
(d) 25/3 [2021 Semester-1]
Solution: (a)
Step-by-step Explanation:
x/5.4 = 5/9
9x = 27
x = 3
\[Q7.\;If\;x=\frac{\sqrt{2a+1}+\sqrt{2a-1}}{\sqrt{2a+1}-\sqrt{2a-1}},\;prove\;that\;x^2-4ax+1=0\;\;\;\;\;\;\;\lbrack2020\rbrack\]
Step-by-step Explanation:
\[\;\;x=\frac{\sqrt{2a+1}+\sqrt{2a-1}}{\sqrt{2a+1}-\sqrt{2a-1}}\\u\sin g\;componendo\;and\;dividendo,\\\frac{x+1}{x-1}=\frac{\sqrt{2a+1}+\sqrt{2a-1}+\sqrt{2a+1}-\sqrt{2a-1}}{\sqrt{2a+1}+\sqrt{2a-1}-\sqrt{2a+1}+\sqrt{2a-1}}\\\frac{x+1}{x-1}=\frac{2\sqrt{2a+1}}{2\sqrt{2a-1}}\\\frac{x+1}{x-1}=\frac{\sqrt{2a+1}}{\sqrt{2a-1}}\\squaring\;both\;sides\\\frac{{(x+1)}^2}{{(x-1)}^2}=\frac{2a+1}{2a-1}\\\frac{x^2+2x+1}{x^2-2x+1}=\frac{2a+1}{2a-1}\\u\sin g\;componendo\;and\;dividendo,\\\frac{x^2+2x+1+x^2-2x+1}{x^2+2x+1-x^2+2x-1}=\frac{2a+1+2a-1}{2a+1-2a+1}\\\frac{2x^2+2}{4x}=\frac{4a}2\\\frac{2(x^2+1)}{4x}=2a\\x^2+1=4ax\\x^2-4ax+1=0\\Proved.\]
Q8. Using properties of proportion find 𝑥 ∶ 𝑦 , given: [2020]
\[\frac{x^2+2x}{2x+4}=\frac{y^2+3y}{3y+9}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\]
Solution: x : y= 2 : 3
Step-by-step Explanation:
\[\frac{x^2+2x}{2x+4}=\frac{y^2+3y}{3y+9}\;\\U\sin g\;componendo\;and\;dividendo,\\\frac{x^2+2x+2x+4}{x^2+2x-2x-4}\;=\frac{y^2+3y+3y+9}{y^2+3y-3y-9}\;\;\;\;\\\frac{x^2+4x+4}{x^2-4}=\frac{y^2+6y+9}{y^2-9}\;\;\;\;\;\\\frac{{(x+2)}^2}{(x+2)(x-2)}=\frac{({y+3)}^2}{(y+3)(y-3)}\;\;\;\;\;\;\;\;\;\;\\\frac{x+2}{x-2}\;=\frac{y+3}{y-3}\;\;\;\\u\sin g\;componendo\;and\;dividendo,\\\frac{x+2+x-2}{x+2-x+2}=\frac{y+3+y-3}{y+3-y+3}\\\frac{2x}4=\frac{2y}6\\By\;alternendo,\\\frac{2x}{2y}=\frac46\\\frac xy=\frac23\]
Q9. The following numbers, K + 3, K + 2, 3K – 7 and 2K – 3 are in proportion. Find K. [2019]
Solution: k= -1 or 5
Step-by-step Explanation:
\[\frac{k+3}{k+2}=\frac{3k-7}{2k-3}\;\\\Rightarrow(k+3)(2k-3)=(3k-7)(k+2)\\\Rightarrow2k^2+6k-3k-9=3k^2-7k+6k-14\\\Rightarrow-k^2+4k+5=0\\\Rightarrow k^2-4k-5=0\\\Rightarrow k^2-5k+k-5=0\\\Rightarrow k(k-5)+1(k-5)=0\\\Rightarrow(k+1)(k-5)=0\\Either\;(k+1)=0\;or\;(k-5)=0\\\Rightarrow k=-1\;or\;k=5\\\]
Q10. Using properties of proportion solve for x, given [2019]
\[\frac{\sqrt{5x}+\sqrt{2x-6}}{\sqrt{5x}-\sqrt{2x-6}}=4\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\]
Solution: x = 30
Step-by-step Explanation:
\[\frac{\sqrt{5x}+\sqrt{2x-6}}{\sqrt{5x}-\sqrt{2x-6}}=4\;\\By\;componendo\;and\;dividendo,\\\frac{\sqrt{5x}+\sqrt{2x-6}+\sqrt{5x}-\sqrt{2x-6}}{\sqrt{5x}+\sqrt{2x-6}-\sqrt{5x}+\sqrt{2x-6}}\;\;=\frac{4+1}{4-1}\\\Rightarrow\frac{2\sqrt{5x}}{2\sqrt{2x-6}}=\frac53\;\;\;\;\\\Rightarrow\frac{\sqrt{5x}}{\sqrt{2x-6}}=\frac53\;\;\;\\squaring\;both\;sides,\\\frac{5x}{2x-6}\;=\frac{25}9\;\;\\\Rightarrow50x-150=45x\\\Rightarrow5x=150\\\Rightarrow x=30\;\;\;\;\;\;\;\;\;\;\;\;\;\]
Q11. Using properties of proportion, solve for x. Given that x is positive : [3] [2018]
\[\frac{2x+\sqrt{4x^2-1}}{2x-\sqrt{4x^2-1}}=4\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\]
Solution: x = 5/8
Step-by-step Explanation:
\[\frac{2x+\sqrt{4x^2-1}}{2x-\sqrt{4x^2-1}}=4\;\\by\;componendo\;and\;dividendo,\\\frac{2x+\sqrt{4x^2-1}+2x-\sqrt{4x^2-1}}{2x+\sqrt{4x^2-1}-2x+\sqrt{4x^2-1}}=\frac{4+1}{4-1}\\\frac{4x}{2\sqrt{4x^2-1}}=\frac53\\\frac{2x}{\sqrt{4x^2-1}}\;=\frac53\;\;\\squaring\;both\;sides,\\\frac{4x^2}{4x^2-1}\;\;=\frac{25}9\\100x^2\;-25=36x^2\;\;\\64x^2\;\;=25\\x^2=\frac{25}{64}\\x=\pm\sqrt{\frac{25}{64}}\\x=\frac58\;\;\;\;\;\;\;\;\;\;\;\;\]
Q12. If b is the mean proportion between a and c, show that: [3] [2017]
\[\frac{a^4+a^2b^2+b^4}{b^4+b^2c^2+c^4}=\frac{a^2}{c^2}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\]
Step-by-step Explanation:
\[Given,\;b\;is\;the\;mean\;proportional\;between\;a\;and\;c.\\Therefore,\;b^2=ac\\Now,\;\\L.H.S.=\frac{a^4+a^2b^2+b^4}{b^4+b^2c^2+c^4}\;\;\;\;\;\;\;\\=\frac{a^4+a^2.ac+{(ac)}^2}{{(ac)}^2+ac.c^2+c^4}\\=\frac{a^4+a^3c+a^2c^2}{a^2c^2+ac^3+c^4}\;\\=\;\frac{a^2(a^2+ac+c^2)}{c^2(a^2+ac+c^2)}\;\\=\frac{a^2}{c^2}\;\;\;\\=R.H.S.\\proved.\;\;\;\;\;\;\;\;\;\]
Q13. If (7m+2n)/(7m-2n)=5/3 , use properties of proportion to find :
(i) m : n
(ii) (m2 + n2)/(m2 – n2)[3] [2017]
Solution: (i) 8:7 (ii) 113/15
Step-by-step Explanation:
\[(i)\;\frac{7m+2n}{7m-2n}=\frac53\\by\;componendo\;and\;dividendo,\\\frac{7m+2n+7m-2n}{7m+2n-7m+2n}=\frac{5+3}{5-3}\\\Rightarrow\frac{14m}{4n}=\frac82\\\Rightarrow\frac{7m}{2n}=4\\\Rightarrow7m=8n\\\Rightarrow\frac mn=\frac87\\(ii)\;\frac mn=\frac87\\squaring\;both\;sides,\\\Rightarrow\frac{m^2}{n^2}=\frac{64}{49}\\by\;componendo\;and\;dividendo,\\\Rightarrow\frac{m^2+n^2}{m-n^2}=\frac{64+49}{64-49}\\\Rightarrow\frac{m^2+n^2}{m-n^2}=\frac{113}{15}\\\]
Q14. If (3a + 2b) : (5a + 3b) = 18 : 29. Find a : b. [3] [2016]
Solution: a : b= 4 : 3
Step-by-step Explanation:
\[\frac{3a+2b}{5a+3b}=\frac{18}{29}\\\Rightarrow90a+54b=87a+58b\\\Rightarrow90a-87a=58b-54b\\\Rightarrow3a=4b\\\Rightarrow\frac ab=\frac43\\\\\]
Q15. If x/a = y/b = z/c show that x3/a3 + y3/b3 + z/c3 = 3xyz/abc [3] [2016]
Step-by-step Explanation:
\[\frac xa=\frac yb=\frac zc=k\;(say)\\x=ak,\;y=bk,\;z=ck\\L.H.S.\\=\frac{x^3}{a^3}+\frac{y^3}{b^3}+\frac{z^3}{c^3}\\=\frac{a^3k^3}{a^3}+\frac{b^3k^3}{b^3}+\frac{c^3k^3}{c^3}\\=k^3+k^3+k^3\\=3k^3\\R.H.S.\\=\frac{3xyz}{abc}\\=\frac{3.ak.bk.ck}{abc}\\=3k^3\\L.H.S.=R.H.S.\\Proved.\]
Q16. If a, b, c are in continued proportion, prove that (a + b + c) (a – b + c) = a2+b2+c2. [2015]
Step-by-step Explanation:
\[a,\;b,\;c\;are\;in\;continued\;proportion.\\\frac ab=\frac bc=k\;(say)\\a=bk=ck.k=ck^2\\b=ck\\L.H.S.\\(a+b+c)(a-b+c)\\=(ck^2+ck+c)(ck^2-ck+c)\\=c(k^2+k+1)c(k^2-k+1)\\=c^2\lbrack{(k^2+1)}^2-k^2\rbrack\\=\;c^2\;(k^4+2k^2+1-k^2)\\=\;c^2(k^4+k^2+1)\\R.H.S.\\a^2+b^2+c^2\\=c^2k^4+c^2k^2+c^2\\=c^2(k^4+k^2+1)\\L.H.S.=R.H.S.\\Proved.\]
Q17. Given (x3+12x)/(6x2+8)=(y3+27y)/(9y2+27). Using Componendo and Dividendo find x : y. [2015]
Solution: x : y= 2 : 3
Step-by-step Explanation:
\[\frac{x^3+12x}{6x^2+8}=\frac{y^3+27y}{9y^2+27}\\By\;componendo\;and\;dividendo,\\\frac{x^3+12x+6x^2+8}{x^3+12x-6x^2-8}=\frac{y^3+27y+9y^2+27}{y^3+27y-9y^2-27}\\\Rightarrow\frac{{(x+2)}^3}{{(x-2)}^3}=\frac{{(y+3)}^3}{{(y-3)}^3}\\\Rightarrow\frac{x+2}{x-2}=\frac{y+3}{y-3}\\by\;componendo\;and\;dividendo,\\\Rightarrow\frac{x+2+x-2}{x+2-x+2}=\frac{y+3+y-3}{y+3-y+3}\\\Rightarrow\frac{2x}4=\frac{2y}6\\\Rightarrow\frac x2=\frac y3\\by\;alternendo\\\Rightarrow\frac xy=\frac23\]
Q18. If (x2+y2)/(x2-y2)=17/8,then find the value of: [3]
(i) x : y
(ii) (x3 + y3)/(x3 – y3) [2014]
Solution: (i) 5 : 3 (ii) 76/49
Step-by-step Explanation:
\[(i)\;\frac{x^2+y^2}{x^2-y^2}=\frac{17}8\\by\;componendo\;and\;dividendo,\\\frac{x^2+y^2+x^2-y^2}{x^2+y^2-x^2+y^2}=\frac{17+8}{17-8}\\\Rightarrow\frac{2x^2}{2y^2}=\frac{25}9\\\Rightarrow\frac{x^2}{y^2}=\frac{25}9\\\Rightarrow\frac xy=\frac53\\(ii)\;\frac xy=\frac53\\cubing\;both\;sides,\\\frac{x^3}{y^3}=\frac{125}{27}\\by\;componendo\;and\;dividendo\\\Rightarrow\frac{x^3+y^3}{x^3-y^3}=\frac{125+27}{125-27}\\\Rightarrow\frac{x^3+y^3}{x^3-y^3}=\frac{152}{98}\\\Rightarrow\frac{x^3+y^3}{x^3-y^3}=\frac{76}{49}\]
Q19. If (x – 9) : (3x + 6) is the duplicate ratio of 4 : 9, find the value of x. [3] [2014]
Solution: x= 25
Step-by-step Explanation:
\[\frac{x-9}{3x+6}=\left(\frac49\right)^2\\\Rightarrow\frac{x-9}{3x+6}=\frac{16}{81}\\\Rightarrow81x-729=48x+96\\\Rightarrow33x=825\\\Rightarrow x=\frac{825}{33}\\\Rightarrow x=25\]
Q20. What number must be added to each of the number 6, 15, 20 and 43 to make them proportional? [3] [2013]
Solution: 3
Step-by-step Explanation:
\[let\;x\;be\;added\;to\;the\;numbers.\\Therefore,\;\\\frac{6+x}{15+x}=\frac{20+x}{43+x}\\\Rightarrow258+43x+6x+x^2=300+15x+20x+x^2\\\Rightarrow49x-35x=300-258\\\Rightarrow14x=42\\\Rightarrow x=3\]
Q21. Using the properties of proportion, solve for x, given [3]
(x4+1)/(2x2 )=17/8 [2013]
Solution: x= ±2
Step-by-step Explanation:
\[\frac{x^4+1}{2x^2}=\frac{17}8\\by\;componendo\;and\;dividendo,\\\Rightarrow\frac{x^4+1+2x^2}{x^4+1-2x^2}=\frac{17+8}{17-8}\\\Rightarrow\frac{{(x^2+1)}^2}{{(x^2-1)}^2}=\frac{25}9\\\Rightarrow\frac{x^2+1}{x^2-1}=\frac53\\\Rightarrow5x^2-5=3x^2+3\\\Rightarrow2x^2=8\\\Rightarrow x^2=4\\\Rightarrow x=\pm2\\\]
Q22. The monthly pocket money of Ravi and Sanjeev are in the ratio 5 : 7. Their expenditures are in the ratio 3 : 5. If each saves Rs.80 every month, find their monthly pocket money. [2012]
Solution: Ravi- Rs 200, Sanjeev- Rs 280
Step-by-step Explanation:
\[Let\;monthly\;pocket\;money\;of\;Ravi\;and\;Sanjeev\;be\;5x\;and\;7x\;respectively.\\They\;save\;Rs.\;80\;per\;month.\\Therefore,\;by\;the\;problem\\\frac{5x-80}{7x-80}=\frac35\\\Rightarrow25x-400=21x-240\\\Rightarrow4x=160\\\Rightarrow x=40\\Therefore,\;Ravi’s\;pocket\;money=5\times40=Rs.200\\and\;Sanjeev’s\;pocket\;money=7\times40=Rs.\;280\]
\[Q23.\;If\;x\;=\;\frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a+1}-\sqrt{a-1}},\;u\sin g\;properties\;of\;proportion\;show\;that\;\;x^2-2ax+1=0\;\;\;\;\;\;\lbrack2012\rbrack\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\]
Step-by-step Explanation:
\[x\;=\;\frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a+1}-\sqrt{a-1}}\;\\By\;componendo\;and\;dividendo,\\\;\;\frac{x+1}{x-1}=\frac{\sqrt{a+1}+\sqrt{a-1}+\sqrt{a+1}-\sqrt{a-1}}{\sqrt{a+1}+\sqrt{a-1}-\sqrt{a+1}+\sqrt{a-1}}\;\;\\\;\Rightarrow\frac{x+1}{x-1}=\frac{2\sqrt{a+1}}{2\sqrt{a-1}}\\squaring\;both\;sides,\;\\\frac{\left(x+1\right)^2}{\left(x-1\right)^2}\;=\frac{a+1}{a-1}\\\Rightarrow\frac{x^2+2x+1}{x^2-2x+1}=\frac{a+1}{a-1}\\By\;componendo\;and\;dividendo,\\\frac{x^2+2x+1+x^2-2x+1}{x^2+2x+1-x^2+2x-1}=\frac{a+1+a-1}{a+1-a+1}\\\Rightarrow\frac{2x^2+2}{4x}=\frac{2a}2\\\Rightarrow\frac{2\left(x^2+1\right)}{4x}=a\\\Rightarrow x^2+1=2ax\\\Rightarrow x^2-2ax+1=0\\Proved.\]
Q24. Using componendo and dividendo, find the value of x. [3] [2011]
\[\frac{\sqrt{3x+4}+\sqrt{3x-5}}{\sqrt{3x+4}-\sqrt{3x-5}}=9\;\;\;\;\;\;\;\;\;\;\;\;\;\]
Solution: x= 7
Step-by-step Explanation:
\[\frac{\sqrt{3x+4}+\sqrt{3x-5}}{\sqrt{3x+4}-\sqrt{3x-5}}=9\\u\sin g\;componendo\;and\;dividendo,\\\frac{\sqrt{3x+4}+\sqrt{3x-5}+\sqrt{3x+4}-\sqrt{3x-5}}{\sqrt{3x+4}+\sqrt{3x-5}-\sqrt{3x+4}+\sqrt{3x-5}}=\frac{9+1}{9-1}\;\;\;\;\;\;\;\;\\\Rightarrow\frac{2\sqrt{3x+4}}{2\sqrt{3x-5}}=\frac{10}8\;\;\\\Rightarrow\frac{\sqrt{3x+4}}{\sqrt{3x-5}}=\frac54\;\\squaring\;both\;sides\\\Rightarrow\frac{3x+4}{3x-5}=\frac{25}{16}\\\Rightarrow75x-125=48x+64\\\Rightarrow27x=189\\\Rightarrow x=\frac{189}{27}\\\Rightarrow x=7\;\;\;\]
Q25. 6 is the mean proportion between two numbers x and y and 48 is the third proportional of x and y. Find the numbers. [3] [2011]
Solution: x = 3, y = 12
Step-by-step Explanation:
\[6\;is\;the\;mean\;proportional\;between\;x\;and\;y.\\Therefore,\;xy=36\;\\\Rightarrow x=\frac{36}y\\48\;is\;the\;third\;proportional\;of\;x\;and\;y.\\Therefore,\;y^2=48x\\putting\;the\;value\;of\;x,\;we\;get\\y^2=\frac{48\times36}y\\y^3=48\times36\\y=\sqrt[3]{48\times36}\\y=2\times6\\y=12\\Hence,\;x=\frac{36}{12}=3\]
Q26. If x, y, z are in continued proportion, prove that
\[\;\frac{{(x+y)}^2}{{(y+z)}^2}=\frac xz\;.\;\lbrack2010\rbrack\;\;\;\;\;\;\;\;\;\]
Step-by-step Explanation:
\[\;x,\;y,\;z\;are\;in\;continued\;proportion.\\Therefore,\;y^2=xz\\L.H.S.\\=\frac{\left(x+y\right)^2}{\left(y+z\right)^2}\\=\frac{x^2+2xy+y^2}{y^2+2yz+z^2}\\=\frac{x^2+2xy+xz}{xz+2yz+z^2}\\=\frac{x(x+2y+z)}{z(x+2y+z)}\\=\frac xz\\=R.H.S.\;\\Proved.\\\]
\[Q27.\;Given,\;x=\frac{\sqrt{a^2+b^2}+\sqrt{a^2-b^2}}{\sqrt{a^2+b^2}-\sqrt{a^2-b^2}},\;use\;componendo\;and\;dividendo\;to\;prove\;that\;b^2=\frac{2a^2x}{x^2+1}\;.\;\lbrack2010\rbrack\;\;\;\;\;\;\;\;\;\]
Step-by-step Explanation:
\[x=\frac{\sqrt{a^2+b^2}+\sqrt{a^2-b^2}}{\sqrt{a^2+b^2}-\sqrt{a^2-b^2}}\;\;\\U\sin g\;componendo\;and\;dividendo,\\\frac{x+1}{x-1}=\frac{\sqrt{a^2+b^2}+\sqrt{a^2-b^2}+\sqrt{a^2+b^2}-\sqrt{a^2-b^2}}{\sqrt{a^2+b^2}+\sqrt{a^2-b^2}-\sqrt{a^2+b^2}+\sqrt{a^2-b^2}}\\\frac{x+1}{x-1}=\frac{2\sqrt{a^2+b^2}}{2\sqrt{a^2-b^2}}\;\\\frac{x+1}{x-1}=\frac{\sqrt{a^2+b^2}}{\sqrt{a^2-b^2}}\\squaring\;both\;sides,\\\frac{{(x+1)}^2}{{(x-1)}^2}=\frac{a^2+b^2}{a^2-b^2}\;\\\frac{x^2+2x+1}{x^2-2x+1}=\frac{a^2+b^2}{a^2-b^2}\;\;\;\\u\sin g\;componendo\;and\;dividendo,\\\frac{x^2+2x+1+x^2-2x+1}{x^2+2x+1-x^2+2x-1}=\frac{a^2+b^2+a^2-b^2}{a^2+b^2-a^2+b^2}\\\frac{2x^2+2}{4x}=\frac{2a^2}{2b^2}\\\frac{x^2+1}{2x}=\frac{a^2}{b^2}\\b^2(x^2+1)=2a^2x\\b^2=\frac{2a^2x}{x^2+1}\\Proved.\]
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