Solution: Real numbers are simply the combination of rational and irrational numbers, in the number system. In general, all the arithmetic operations can be performed on these numbers and they...
Solution: Real numbers are simply the combination of rational and irrational numbers, in the number system. In general, all the arithmetic operations can be performed on these numbers and they...
Solution: Real numbers are simply the combination of rational and irrational numbers, in the number system. In general, all the arithmetic operations can be performed on these numbers and they...
Solution: Real numbers are simply the combination of rational and irrational numbers, in the number system. In general, all the arithmetic operations can be performed on these numbers and they...
Solution: Real numbers are simply the combination of rational and irrational numbers, in the number system. In general, all the arithmetic operations can be performed on these numbers and they...
Solution: Real numbers are simply the combination of rational and irrational numbers, in the number system. In general, all the arithmetic operations can be performed on these numbers and they...
Solution: Real numbers are simply the combination of rational and irrational numbers, in the number system. In general, all the arithmetic operations can be performed on these numbers and...
Solution: Real numbers are simply the combination of rational and irrational numbers, in the number system. In general, all the arithmetic operations can be performed on these numbers and they...
Solution: Real numbers are simply the combination of rational and irrational numbers, in the number system. In general, all the arithmetic operations can be performed on these numbers and they...
marbles. Both of them lost 
marbles each, and the product of the number of marbles they now have is 
. Form the quadratic equation to find how many marbles they to start with, if John had 
marbles.Solution: Quadratic equations are the polynomial equations of degree $2$ in one variable of type $f(x) = ax2 + bx + c$ where a, b, c, ∈ R and a ≠ 0. Given, John and Jilani...
Solution: “In a pair of similar triangles, the corresponding sides are proportional. When the sides are corresponding it means to go from one triangle to another you can multiply each side by the...
long casts a shadow 
long. At the same time, a tower casts a shadow 
long. Determine the height of the tower.Solution: Given: Length of stick = $10cm$ Length of the stick’s shadow = $8cm$ Length of the tower’s shadow = $30m$ = $3000cm$ To find: The height of the tower = PQ. In ΔABC ∼ ΔPQR...
Solution: (i) Given: Δ ABC and Δ AMP are the two right triangles. To prove : ΔABC ∼ ΔAMP Proof : ∠AMP = ∠B = 90o ∠MAP = ∠BAC (Vertically Opposite...
Solution: Given: OC is the point of intersection of AC and BD in the trapezium ABCD, with AB ∥ DC. To prove: OA/ OC = OB/ OD Proof : In ΔAOB and ΔCOD ∠AOB = ∠COD...
Solution: Given: AB ⊥ BC, DC ⊥ BC, DE ⊥ AC To prove: ΔCED∼ΔABC From ΔABC and ΔCED ∠B = ∠E = 90o [given] ∠BAC = ∠ECD ...
. If 
, find AD.Solution: Given: DE∥BC $AE = (1/4)AC$ $AB = 6 cm$. To find: AD. In ΔADE and ΔABC ∠A = ∠A (Common) (Corresponding angles as AB||QR with PQ as the transversal) ∠ADE = ∠ABC (By AA...
and BD ⊥ AC. If 
, 
and 
, Find BC.Solution: Given: BD ⊥ AC $AC = 5.7 cm$, $BD = 3.8 cm$ and $CD = 5.4 cm$ \[\angle ABC~=\text{ }{{90}^{o}}\] To find: BC ΔABC ∼ ΔBDC (By AA similarity) \[\angle BCA~=\angle DCA\text{ }=\text{...
and BD⊥AC. If 
, and 
, find CD.Solution: Given : $∠ABC = 90o$ and BD⊥AC $BD = 8 cm$ $AD = 4 cm$ To find: CD. ABC is a right angled triangle and BD⊥AC. (By AA similarity) ΔDBA∼ΔDCB BD/ CD = AD/ BD...
Given : Consider ΔABC to be a right angle triangle having sides a, b and hypotenuse c. Let BD be the altitude drawn on the hypotenuse AC. To prove: ab = cx Prove : In ΔACB and ΔCDB...
Solution: Given : XY∥BC $AX = 1 cm$, $XB = 3 cm$ and $BC = 6 cm$ To find: XY In ΔAXY and ΔABC ∠A = ∠A (Common) (Corresponding angles as AB||QR with PQ as the transversal) ∠AXY = ∠ABC...
electrical components:Lifetimes (in hours):0 – 2020 – 4040 – 6060 – 8080 – 100100 – 120No. of components: 103552613829 Solution: Statistics is the discipline...
Given : ΔPQR, AB ∥ QR $AB = 3 cm$, $QR = 9 cm$ and $PR = 6 cm$ To find: PB In ΔPAB and ΔPQR ∠P = ∠P (Common) (Corresponding angles as AB||QR with PQ as the transversal) ∠PAB = ∠PQR...
, 
, 
and 
, find CA and AQ.Solution: Given, ΔACB ∼ ΔAPQ $BC = 8 cm$, $PQ = 4 cm$, $BA = 6.5 cm$ and $AP = 2.8 cm$ To find: CA and AQ ΔACB ∼ ΔAPQ [given] BA/ AQ = CA/ AP = BC/ PQ (Corresponding Parts of Similar Triangles)...
Solution: Given, \[cot\text{ }\theta \text{ }=\text{ }1/\surd 3\ldots \ldots .\text{ }\left( 1 \right)\] By definition we know that, \[cot\text{ }\theta \text{ }=\text{ }1/\text{ }tan\text{ }\theta...
![\[\mathbf{cos}\text{ }\mathbf{\theta }\text{ }=\text{ }\mathbf{12}/\mathbf{13},\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-eb80fa11d8d93920874f6d261c69280c_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{sin}\text{ }\mathbf{\theta }\left( \mathbf{1}\text{ }\text{ }\mathbf{tan}\text{ }\mathbf{\theta } \right)\text{ }=\text{ }\mathbf{35}/\mathbf{156}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-bd70331e2de174a9b75008059c8fe4ea_l3.png "Rendered by QuickLaTeX.com")
Solution: Given, \[cos\text{ }\theta \text{ }=\text{ }12/13\ldots \ldots \text{ }\left( 1 \right)\]definition we know that, \[cos\text{ }\theta \text{ }=\]Base side adjacent to ∠θ /...
![\[\mathbf{sec}\text{ }\mathbf{\theta }\text{ }=\text{ }\mathbf{13}/\mathbf{5},\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9a2ce3144478cc2b7f9167d9e02c0f84_l3.png "Rendered by QuickLaTeX.com")
Solution: Given,\[sec\text{ }\theta \text{ }=\text{ }13/5\] We know that, \[sec\text{ }\theta \text{ }=\text{ }1/\text{ }cos\text{ }\theta \] \[\Rightarrowcos\text{ }\theta \text{ }=\text{ }1/\text{...
![\[\mathbf{tan}\text{ }\mathbf{\theta }\text{ }=\text{ }\mathbf{a}/\mathbf{b},\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-cd749b91ff9a4cbeb368c727034d5ea1_l3.png "Rendered by QuickLaTeX.com")
Solution: Given, \[tan\text{ }\theta \text{ }=\text{ }a/b\] From LHS, let’s divide the numerator and denominator by \[cos\text{ }\theta...
![\[\mathbf{3}\text{ }\mathbf{tan}\text{ }\mathbf{\theta }\text{ }=\text{ }\mathbf{4},\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-eb5fe060663e488e5fa796b41cf0b313_l3.png "Rendered by QuickLaTeX.com")
Solution: Given, \[3\text{ }tan\text{ }\theta \text{ }=\text{ }4\] \[\Rightarrow tan\text{ }\theta \text{ }=\text{ }4/3\] From, let’s divide the numerator and denominator by \[cos\text{ }\theta .\]...
![\[\left( \mathbf{cos}\text{ }\mathbf{\theta }\text{ }+\text{ }\mathbf{sin}\text{ }\mathbf{\theta } \right)/\text{ }\left( \mathbf{cos}\text{ }\mathbf{\theta }\text{ }\text{ }\mathbf{sin}\text{ }\mathbf{\theta } \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6a0ae0a08f140e73ca3eea0759231288_l3.png "Rendered by QuickLaTeX.com")
Solution: Given, \[tan\text{ }\theta \text{ }=\text{ }a/b\] And, we know by definition that \[tan\text{ }\theta \text{ }=\]opposite side/ adjacent side Thus, by comparison Opposite side \[=\text{...
![\[\mathbf{3cot}\text{ }\mathbf{A}\text{ }=\text{ }\mathbf{4},\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a4ab277eb5555e0440bfa2833c83f679_l3.png "Rendered by QuickLaTeX.com")
![\[(\mathbf{1}\mathbf{ta}{{\mathbf{n}}^{\mathbf{2}}}\mathbf{A})/(\mathbf{1}+\mathbf{ta}{{\mathbf{n}}^{\mathbf{2}}}\mathbf{A})\text{ }=\text{ }(\mathbf{co}{{\mathbf{s}}^{\mathbf{2}}}\mathbf{A}\text{ }\text{ }\mathbf{si}{{\mathbf{n}}^{\mathbf{2}}}\mathbf{A})~\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2de1029dce4a0ac05f4b54e69b56733e_l3.png "Rendered by QuickLaTeX.com")
Solution: Given, \[3cot\text{ }A\text{ }=\text{ }4\] \[\Rightarrow cot\text{ }A\text{ }=\text{ }4/3~~\] By definition, \[tan\text{ }A\text{ }=\text{ }1/\text{ }Cot\text{ }A\text{ }=\text{ }1/\text{...
![\[.~\mathbf{If}~\mathbf{cot}\text{ }\mathbf{\theta }\text{ }=\text{ }\mathbf{7}/\mathbf{8},\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b22d83fb3247f489ec296f6c18a0a470_l3.png "Rendered by QuickLaTeX.com")
\[~~~~\left( \mathbf{i} \right)~~~\left( \mathbf{1}+\mathbf{sin}\text{ }\mathbf{\theta } \right)\left( \mathbf{1}\mathbf{sin}\text{ }\mathbf{\theta } \right)/\text{ }\left(...
![\[\mathbf{PQ}\text{ }=\text{ }\mathbf{4cm}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9d842a14208db3cbfeb06d5b4edb129f_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{RQ}\text{ }=\text{ }\mathbf{3}\text{ }\mathbf{cm}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e604195fecf3fc272a1179372de4ad94_l3.png "Rendered by QuickLaTeX.com")
\[\mathbf{sin}\text{ }\mathbf{P},~\mathbf{sin}\text{ }\mathbf{R},\text{ }\mathbf{sec}\text{ }\mathbf{P}\text{ }\mathbf{and}\text{ }\mathbf{sec}\text{ }\mathbf{R}.\] Solution: Given: \[\vartriangle...
![\[\mathbf{15cot}\text{ }\mathbf{A}=\text{ }\mathbf{8}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-776813bfd1ebfb22e8e1cdc48e3713c2_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{sin}\text{ }\mathbf{A}\text{ }\mathbf{and}\text{ }\mathbf{sec}\text{ }\mathbf{A}.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ae90f950247a1680b7897bbf39cc3f45_l3.png "Rendered by QuickLaTeX.com")
Solution We have, \[15cot\text{ }A\text{ }=\text{ }8\] Required to find: \[sin\text{ }A\text{ }and\text{ }sec\text{ }A\] As, \[15cot\text{ }A\text{ }=\text{ }8\] \[\Rightarrow cot\text{ }A\text{...
![\[\mathbf{12}/\mathbf{13}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0dda7c2e841b9572a9700f0411221855_l3.png "Rendered by QuickLaTeX.com")
Solution: Given, tan θ = \[12/13\] …….. \[\left( 1 \right)\] We know that by definition, tan θ = Perpendicular side opposite to ∠θ / Base side adjacent to ∠θ …… \[\left( 2 \right)\] On comparing...
![\[\mathbf{OA}\text{ }=\text{ }\mathbf{3x}\text{ }\text{ }\mathbf{19},\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-83033fa8761a3156e55c3abe3214ef6e_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{OB}\text{ }=\text{ }\mathbf{x}\text{ }\text{ }\mathbf{4},\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-fcb15b4a8c4daf26a5c3871f58dff365_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{OC}\text{ }=\text{ }\mathbf{x}-\text{ }\mathbf{3}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d97e82e9a1787b66d655c5d17e868326_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{OD}\text{ }=\text{ }\mathbf{4},\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7c5571c8b093b249eb94ff6a912f1975_l3.png "Rendered by QuickLaTeX.com")
Figure 4.72 Solution: Given : AB∥CD. Required to find the value of x. Diagonals of a parallelogram bisect each other AO/ CO = BO/ DO \[\left( 3x\text{ }\text{ }19 \right)/\text{ }\left( x\text{...
Figure -4.70 Figure 4.71 Solution: Given: AB∥CD. Required to find the value of x. Diagonals of a parallelogram bisect each other. \[AO/\text{ }CO\text{ }=\text{ }BO/\text{ }DO\] \[\Rightarrow...
![\[\angle \mathbf{B}=\mathbf{7}{{\mathbf{0}}^{\mathbf{o}}},\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b9131325619eb2fd477f0dd672942dc6_l3.png "Rendered by QuickLaTeX.com")
![\[\angle \mathbf{C}\text{ }=\text{ }\mathbf{5}{{\mathbf{0}}^{\mathbf{o}}},\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4d07c7a5c98b9efb07ed4d1f2f8f07b7_l3.png "Rendered by QuickLaTeX.com")
Solution: Given: Δ ABCsuch that AB/AC = BD/DC, \[\angle B\text{ }=\text{ }70\] o \[\angle C\text{ }=\text{ }50\] o Required to find: ∠BAD In ΔABC, \[\angle...
![\[\mathbf{AB}\text{ }=\text{ }\mathbf{10}\text{ }\mathbf{cm},\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-49cecb59e896652d21a82c8db7df1c60_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{AC}\text{ }=\text{ }\mathbf{6}\text{ }\mathbf{cm},\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-8fcf95066978bb3a5c81e31da13743b6_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{BC}\text{ }=\text{ }\mathbf{12}\text{ }\mathbf{cm},\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2442b612d2d4da073e2ebd36b6adccf8_l3.png "Rendered by QuickLaTeX.com")
Solution: Given: AE is the bisector of the exterior ∠CAD \[AB\text{ }=\text{ }10\text{ }cm,\] \[AC\text{ }=\text{ }6\text{ }cm,\] and \[BC\text{ }=\text{ }12\text{ }cm.\] Required to find: CE...
![\[\mathbf{AB}\text{ }=\text{ }\mathbf{5}.\mathbf{6}\text{ }\mathbf{cm},\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5bda87937897bc7402395a0e7497aa6b_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{BC}\text{ }=\text{ }\mathbf{6}\text{ }\mathbf{cm},\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-cc05f2b9f3ba514ca9c9b35898d5010a_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{BD}\text{ }=\text{ }\mathbf{3}.\mathbf{2}\text{ }\mathbf{cm},\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2cb291882d238e8010337f40ada99f74_l3.png "Rendered by QuickLaTeX.com")
Solution: Given: Δ ABC and AD bisects ∠A, meeting side BC at D. \[AB\text{ }=\text{ }5.6\text{ }cm,\] \[BC\text{ }=\text{ }6\text{ }cm,\] and \[BD\text{ }=\text{ }3.2\text{ }cm\]....
![\[\mathbf{5}/\mathbf{4}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a3636d015b854f3d808bb36663cd1dcf_l3.png "Rendered by QuickLaTeX.com")
Solution: Given, sec θ = \[5/4\] We know that, sec θ = \[1/\] cos θ ⇒ cos θ = \[1/\text{ }\left( 5/4 \right)\text{ }=\text{ }4/5\text{ }\ldots \ldots \text{ }\left( 1 \right)\] By definition, cos θ...
Solution: Given, tan θ = \[1/\text{ }\surd 7\text{ }\ldots ..\left( 1 \right)\] By definition, we know that tan θ = Perpendicular side opposite to ∠θ / Base side adjacent to ∠θ \[\ldots \ldots...
![\[\mathbf{AC}\text{ }=\text{ }\mathbf{4}.\mathbf{2}\text{ }\mathbf{cm},\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-45f16cbaa06cd9242cc91c64e1a2f9e9_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{DC}\text{ }=\text{ }\mathbf{6}\text{ }\mathbf{cm},\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e63dbecea189eee6b029fe954d57a01b_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{BC}\text{ }=\text{ }\mathbf{10}\text{ }\mathbf{cm},\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b04409e4addd60622f51dff4d1577ee3_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{DC}\text{ }=\text{ }\mathbf{3}\text{ }\mathbf{cm},\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-359b596c36ee16c046abc40f5e03a7d8_l3.png "Rendered by QuickLaTeX.com")
Solution: Given: Δ ABC and AD bisects ∠A, meeting side BC at D. \[~AC\text{ }=\text{ }4.2\text{ }cm\],\[~DC\text{ }=\text{ }6\text{ }cm,\] and \[BC\text{ }=\text{ }10\text{ }cm.\] Required...
![\[~\mathbf{30}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-27c0d25af5903e046d1ab004ac0ffca2_l3.png "Rendered by QuickLaTeX.com")
Solution: It’s seen that, the cumulative frequency just greater than n/\[2\text{ }\left( i.e.\text{ }30/\text{ }2\text{ }=\text{ }15 \right)\text{ }is\text{ }19\], belongs to class interval...
Solution: Here, we have \[N\text{ }=\text{ }250,\] So, \[N/2\text{ }=\text{ }250/\text{ }2\text{ }=\text{ }125\] The cumulative frequency just greater than \[N/\text{ }2\text{ }is\text{ }127\]then...
Solution: Here, we have \[N\text{ }=\text{ }140,\] So, \[N/2\text{ }=\text{ }140/\text{ }2\text{ }=\text{ }70\] The cumulative frequency just greater than \[N/\text{ }2\text{ }is\text{ }98\]then...
![\[\mathbf{15}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c554aaf0e077efa7da9b881aa7445bc4_l3.png "Rendered by QuickLaTeX.com")
\[\mathbf{715},\text{ }\mathbf{724},\text{ }\mathbf{725},\text{ }\mathbf{710},\text{ }\mathbf{729},\text{ }\mathbf{745},\text{ }\mathbf{694},\text{ }\mathbf{699},\text{ }\mathbf{696},\text{...
![\[\mathbf{sin}\text{ }\mathbf{A}\text{ }=~\mathbf{9}/\mathbf{41}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6063fa81e0b4df08fc4ffa0227c86ce9_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{cos}\text{ }\mathbf{A}\text{ }\mathbf{and}\text{ }\mathbf{tan}\text{ }\mathbf{A}.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-16379522aa97447bd74fe8c6550eec80_l3.png "Rendered by QuickLaTeX.com")
Solution: Given that, \[sin\text{ }A\text{ }=\text{ }9/41~\ldots \ldots \ldots \ldots .\text{ }\left( 1 \right)\] Required to find: \[cos\text{ }A,\text{ }tan\text{ }A\] By definition, we...
![\[\mathbf{tan}\text{ }\mathbf{P}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-75c64146364f060928b9fe4a96ee1ab8_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{cot}\text{ }\mathbf{R}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-99124b15748c04a1575b60ef0b6c9bd9_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{tan}\text{ }\mathbf{P}\text{ }=\text{ }\mathbf{cot}\text{ }\mathbf{R}?~\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3ce0b48333f90b7ed7fcf5bb7884881b_l3.png "Rendered by QuickLaTeX.com")
Solution: By using Pythagoras theorem in \[\vartriangle PQR\], we have \[P{{R}^{2}}~=\text{ }P{{Q}^{2}}~+\text{ }Q{{R}^{2}}\] Putting the length of given side PR and PQ in the above equation...
![\[\mathbf{AB}\text{ }=\text{ }\mathbf{3}.\mathbf{5}\text{ }\mathbf{cm},\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-45dd439365476597b2fa8b5cef250c70_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{DC}\text{ }=\text{ }\mathbf{2}.\mathbf{8}\text{ }\mathbf{cm},\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-49cbe9239520f170aa18fbd2a6895b8d_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{AC}\text{ }=\text{ }\mathbf{14}\text{ }\mathbf{cm},\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c8e04fe6419b7ac7bf51ffff487aac9a_l3.png "Rendered by QuickLaTeX.com")
Given: Δ ABC and AD bisects ∠A, meeting side BC at D. \[~AB\text{ }=\text{ }3.5\text{ }cm\], \[AC\text{ }=\text{ }4.2\text{ }cm,\] and \[DC\text{ }=\text{ }2.8\text{ }cm\]. Required...
for the given distribution whose mean is 
$x:$$5$$8$$10$$12$$p$$20$$25$$f:$$2$$5$$8$$22$$7$$4$$2$ Solution: $x$$f$$fx$$5$$2$$10$$8$$5$$40$$10$$8$$80$$12$$22$$264$$P$$7$$7p$$20$$4$$80$$25$$2$$50$$N=50$$\sum{fx=524+7p}$ We know that, Mean...
Find the median height. Solution: Here, we have N = \[420,\] So\[,\text{ }N/2\text{ }=\text{ }420/\text{ }2\text{ }=\text{ }210\] The cumulative frequency just greater than \[N/2\text{ }is\text{...
, find 
$x:$$5$10$15$$20$$25$$f:$$6$$p$$6$$10$$5$ Solution: $x$$f$$fx$$5$$6$$30$$10$$p$$10p$$15$$6$$90$$20$$10$$200$$25$$5$$125$$N=p+27$$\sum{fx=445+10p}$ We know that, Mean $=\sum{fx/N=\left( 445+10p...
is the mean of given data. Find the value of $x:$$10$$15$$p$$25$$35$$f:$$3$$10$$25$$7$$5$ Solution: $x$$f$$fx$$10$$3$$30$$15$$10$$150$$p$$25$$25p$$25$$7$$175$$35$$5$$175$$N=50$$\sum{fx=530+25p}$ We know that, Mean $=\sum{fx/N=\left( 2620+25p...
![\[\mathbf{24cm}\text{ }\mathbf{and}\text{ }\mathbf{10cm}.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c85c0e710149fb9633b94c8c2abffd5a_l3.png "Rendered by QuickLaTeX.com")
Solution: Let ABCD be a rhombus and AC and BD be the diagonals of ABCD. So, AC = \[24cm\text{ }and\text{ }BD\text{ }=\text{ }10cm\] \[\] We know that diagonals of a rhombus bisect each other at...
Given: Δ ABC and AD bisects ∠A, meeting side BC at D. \[BD\text{ }=\text{ }2.5cm\], \[AB\text{ }=\text{ }5\text{ }cm,\] and\[AC\text{ }=\text{ }4.2\text{ }cm\]. Required to...
$x:$$19$$21$$23$$25$$27$$29$$31$$f:$$13$$15$$16$$18$$16$$15$$13$ Solution: $x$$f$$fx$$19$$13$$247$$21$$15$$315$$23$$16$$368$$25$$18$$450$$27$$16$$432$$29$$15$$435$$31$$13$$403$$N=106$$\sum{fx}=2620$...
![\[\mathbf{2a}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-dc21b199c4adb0b51f7839a0d5967962_l3.png "Rendered by QuickLaTeX.com")
(i) AD = a\[\surd \mathbf{3}\] (ii) Area (∆ABC) = \[\surd \mathbf{3}\text{ }{{\mathbf{a}}^{\mathbf{2}}}\] Solution: (i) In ∆ABD and ∆ACD, we have∠ADB = ∠ADC = \[{{90}^{o}}\]AB = AC [Given]AD =...
![\[\mathbf{13cm}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0a3890f5045c3fc67ed8d0dc3fee80ac_l3.png "Rendered by QuickLaTeX.com")
\[\mathbf{5cm},\]find BC. Solution: Given, An isosceles triangle ABC, AB = AC = \[13cm,\text{ }AD\text{ }=\text{ }5cm\] Required to find: BC In ∆ ADB, by using Pythagoras theorem, we have...
![\[\mathbf{108c}{{\mathbf{m}}^{\mathbf{2}}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-aea3dbba23caac08c8a64b556bd738bb_l3.png "Rendered by QuickLaTeX.com")
Solution: Given, ABCD is a square. And, F is the mid-point of AB. BE is one third of BC. Area of ∆ FBE = \[108c{{m}^{2}}\] Required to find: length of AC Let’s assume the sides of the square to be...
![\[\mathbf{5}\text{ }\mathbf{cm},\text{ }\mathbf{12}\text{ }\mathbf{cm}\text{ }\mathbf{and}\text{ }\mathbf{13}\text{ }\mathbf{cm}.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-55595a46a936f27b67a56d78fd6d9f6b_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{13}\text{ }\mathbf{cm}.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-dd9682929f90290113c48dab79d05fcc_l3.png "Rendered by QuickLaTeX.com")
Solution: From the fig. \[AB\text{ }=\text{ }5cm,\text{ }BC\text{ }=\text{ }12\text{ }cm\text{ }and\text{ }AC\text{ }=\text{ }13\text{ }cm.\] Then, \[A{{C}^{2}}~=\text{ }A{{B}^{2}}~+\text{...
![\[.\text{ }\mathbf{4}.\mathbf{219}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e0e6c8ae5a8e273c9a6be45b04989009_l3.png "Rendered by QuickLaTeX.com")
Solution: We have, In ∆BAC, by Pythagoras theorem, we have \[\begin{array}{*{35}{l}} B{{C}^{2}}~=\text{ }A{{B}^{2}}~+\text{ }A{{C}^{2}} \\ \Rightarrow...
![\[\mathbf{9}\text{ }\mathbf{in}\text{ }\mathbf{and}\text{ }\mathbf{14}\text{ }\mathbf{m}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a5efcdbe8ed132a615044b2b9e1efbd4_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{12}\text{ }\mathbf{m},\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-18666dfb8dc6e2f5280fce139ea621e6_l3.png "Rendered by QuickLaTeX.com")
Solution: Comparing with the figure, it’s given that AC = \[14\text{ }m,\text{ }DC\text{ }=\text{ }12m\text{ }and\text{ }ED\text{ }=\text{ }BC\text{ }=\text{ }9\text{ }m\] Construction: Draw...
![\[\mathbf{6}\text{ }\mathbf{m}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0a07f6483f265e6e47c316121ac8cca7_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{8}\text{ }\mathbf{m}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f5dd6c4c521acc9e7e7137b3036d18fb_l3.png "Rendered by QuickLaTeX.com")
Solution: Let’s assume the length of ladder to be, AD = BE = x m So, in ∆ACD, by Pythagoras theorem We have, \[\begin{array}{*{35}{l}} ...
![\[\mathbf{AC}\text{ }=\text{ }\mathbf{25}\text{ }\mathbf{cm},\text{ }\mathbf{BC}\text{ }=\text{ }\mathbf{14}\text{ }\mathbf{cm}.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-49a79c647484ca1f07eb3412c6395529_l3.png "Rendered by QuickLaTeX.com")
Solution: Given, ∆ABC, AB = AC = \[25\text{ }cm\text{ }and\text{ }BC\text{ }=\text{ }14.\] \[\] In ∆ABD and ∆ACD, we see...
![\[\mathbf{6}\text{ }\mathbf{m}\text{ }\mathbf{and}\text{ }\mathbf{11}\text{ }\mathbf{m}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ce51d1c0d04bc918a23933dbe2116e9a_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{12}\text{ }\mathbf{m}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5b1a778c62eb81bbb683b987a69b68d9_l3.png "Rendered by QuickLaTeX.com")
Solution: Let CD and AB be the poles of height \[11m\text{ }and\text{ }6m.\] Then, its seen that CP = \[11\text{ }\text{ }6\text{ }=\text{ }5m.\] From the figure, AP should be \[12m\] (given) In...
![\[\mathbf{17}\text{ }\mathbf{m}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a8857d36c65c9db76f01d8b00e922493_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{15}\text{ }\mathbf{m}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f363f724150e591edf9182f4311ab3e8_l3.png "Rendered by QuickLaTeX.com")
Solution: In ∆ABC, by Pythagoras theorem \[\begin{array}{*{35}{l}} A{{B}^{2}}~+\text{ }B{{C}^{2}}~=\text{ }A{{C}^{2}} \\ \Rightarrow {{15}^{2}}~+\text{...
![\[\mathbf{15}\text{ }\mathbf{metres}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-712e26540491c89b51fc3bb96dfabd2f_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{8}\text{ }\mathbf{metres}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9d6db488112b88ddeb15d838748fedad_l3.png "Rendered by QuickLaTeX.com")
Solution: ...
\[\begin{array}{*{35}{l}} \left( \mathbf{i} \right)\text{ }\mathbf{a}\text{ }=\text{ }\mathbf{7}\text{ }\mathbf{cm},\text{ }\mathbf{b}\text{ }=\text{ }\mathbf{24}\text{...
![\[\mathbf{3}\text{ }\mathbf{cm},\text{ }\mathbf{4}\text{ }\mathbf{cm},\text{ }\mathbf{and}\text{ }\mathbf{6}\text{ }\mathbf{cm}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4e7f08d95ac5afd7a7bf17c308701538_l3.png "Rendered by QuickLaTeX.com")
Solution: We have, Sides of triangle as \[\begin{array}{*{35}{l}} AB\text{ }=\text{ }3\text{ }cm \\ BC\text{ }=\text{ }4\text{ }cm \\ AC\text{...
$x:$$5$$6$$7$$8$$9$$f:$$4$$8$$14$$11$$3$ Solution: $x$$f$$fx$$5$$4$$20$$6$$8$$48$$7$$14$$98$$8$$11$$88$$9$$3$$27$$N=40$$\sum{fx=281}$ Mean $=\sum{fx/n=281/40}$ $\therefore $ Mean...
(i) All circles are ____________ (congruent, similar). (ii) All squares are ___________ (similar, congruent). (iii) All ____________ triangles are similar (isosceles, equilaterals). (iv) Two...
Solution: Let the fixed charge of the car be ₹ x and, Let the variable charges of the car be ₹ y per km. So according to the question, we get \[2\] equations \[x\text{ }+\text{ }12y\text{ }=\text{...
Solution: We need to prove that ΔABC is right angled. \[\angle A\text{ }=\text{ }{{x}^{o}},\angle B\text{ }=\text{ }3{{x}^{o}}\]and \[\angle C\text{ }=\text{ }{{y}^{o}}\] Sum of the three angles in...
Solution: Let the total number of correct answers be x and the total number of incorrect answers be y. Their sum will give the total number of questions in the test = x + y According to the question...
Solution: The sum of the opposite angles of cyclic quadrilateral should be 180o. And, in the cyclic quadrilateral ABCD, Angles \[\angle A\] and \[\angle C\] & angles \[\angle B\] and \[\angle...
Solution: Given, \[\angle A\text{ }=\text{ }{{x}^{o}}\] \[\angle B\text{ }=\text{ }{{\left( 3x\text{ }\text{ }2 \right)}^{o}},\] \[\angle C\text{ }=\text{ }{{y}^{o}}\]Also given, \[\angle C\text{...
Solution: Let the time required for a man alone to finish the work be x days The time required for a boy alone to finish the work be y days. Then, The work done by a man in one day \[=\text{ }1/x\]...
Solution: Let he money with A be ₹ x and the money with B be ₹ y. According to the question, Case 1: If A gives ₹ \[30\]to B, then B will have twice the money left with A. Equation, \[y\text{...
Solution: Leave the pay alone signified by x and the consumption be meant by y. Then, at that point, from the inquiry we have The pay of X is ₹ 8x and the use of X is 19y. The pay of Y is ₹ 7x and...
Solution: Let the length and breadth of the rectangle be x units and y units. The area of rectangle = xy sq.units Length is increased by 3 m ⇒ The new length is \[x+3\] Breadth is reduced by 4 m...
Solution: Let the length and breadth of the rectangle be x units and y units respectively. Hence, the area of rectangle = xy sq.units Length is increased by 7 m ⇒ now, the new length is...
Solution: Let the length and breadth of the rectangle be x units and y units respectively. The area of rectangle = xy sq.units Case 1: Length is increased by 2 units ⇒ The new length is \[x+2\]units...
Solution: Let the positive integer = a According to Euclid’s division algorithm, a = 6q + r, where 0 ≤ r < 6 \[{{a}^{2}}={{\left( 6q+r \right)}^{2}}=36{{q}^{2}}+{{r}^{2}}+12qr\],\[\because...
Let ‘a’ be any positive integer. Then from Euclid’s division lemma, a = bq+r; where 0 < r < b Putting b=6 we get, ⇒ a = 6q + r, 0 < r < 6 For r = 0, we get\[a=6q=2\left( 3q...
Solution: Let ‘a’ be any positive integer. Then, According to Euclid’s division lemma, a=bq+r According to the question, when b = 5. \[a=5k+r,n<r<5\] When r = 0, we get, a = 5k ...
Let ‘a’ be any positive integer. Then, According to Euclid’s division lemma, a=bq+r According to the question, when b = 4. \[a=4k+r,n<r<4\] When r=0, we get a=4k \[\to...
Let n= 6q+5 be a positive integer for some integer q. An integer is positive if it is greater than zero. We know that any positive integer can be of the form 3k, or 3k+1, or 3k+2. ∴ q can be 3k or,...
![\[\mathbf{16}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ec8d9d2d731b4ca460e57ad46a988282_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{380}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-cb340828e6c6101fbde249e54a9cd1cb_l3.png "Rendered by QuickLaTeX.com")
Solution: On dividing \[380\]by \[16\]we get,\[23\]as the quotient and\[12\]as the remainder. Now, since the LCM is not exactly divisible by the HCF its can be said that two numbers cannot have...
![\[\mathbf{84},\text{ }\mathbf{90}\text{ }\mathbf{and}\text{ }\mathbf{120}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ccfecca6ebaa7e44c50bda4ca8c75939_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{24},\text{ }\mathbf{15}\text{ }\mathbf{and}\text{ }\mathbf{36}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-94f859fd8a71046e251dab20e2d08dcb_l3.png "Rendered by QuickLaTeX.com")
Solution: First, Find the prime factors of the given integers: \[84,\text{ }90\text{ }and\text{ }120\] For, \[\begin{array}{*{35}{l}} ~84\text{...
![\[\mathbf{8},\text{ }\mathbf{9}\text{ }\mathbf{and}\text{ }\mathbf{25}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-78297b7f285a41777161d421fedfd300_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{40},\text{ }\mathbf{36}\text{ }\mathbf{and}\text{ }\mathbf{126}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-17883cf51f0a9421203bd7e5576dc901_l3.png "Rendered by QuickLaTeX.com")
Solution: First, find the prime factors of the given integers: 8, 9 and 25 For, \[8\text{ }=\text{ }2\text{ }\times...
Let, n be any positive integer. And since any positive integer can be of the form 6q, or 6q+1, or 6q+2, or 6q+3, or 6q+4, or 6q+5. (From Euclid’s division lemma for b= 6) We have n3 – n =...
(i) 12, 15 and 21(ii) \[\mathbf{17},\text{ }\mathbf{23}\text{ }\mathbf{and}\text{ }\mathbf{29}\] Solution: First, find the prime factors of the given integers: \[12,\text{ }15\text{ }and\text{...
(i) \[\mathbf{26}\text{ and }\mathbf{91}\](ii) \[\mathbf{510}\text{ and }\mathbf{92}\] Solution: Given integers are\[\mathbf{26}\text{ and...
![\[\mathbf{p},~\surd p\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e6eb15f7e07221ee9035e70e0a293ac9_l3.png "Rendered by QuickLaTeX.com")
Solution: Consider \[\surd p\] as a rational number Assume\[\surd p\text{ }=\text{ }a/b\] where a and b are integers and \[b\text{ }\ne \text{ }0\] By squaring on both sides \[p\text{ }=\text{...
Solution: Let’s assume on the contrary that \[\surd 2\text{ }+\text{ }\surd 3\]is a rational number. Then, there exist co prime positive integers a and b such that \[\surd 2\text{ }+\text{ }\surd...
![\[\surd 5\text{ }+\text{ }\surd 3~\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0efb8e4ff13d751cc9c194f8bc9b1b82_l3.png "Rendered by QuickLaTeX.com")
Solution: Let’s assume on the contrary that \[\surd 5\text{ }+\text{ }\surd 3\] is a rational number. Then, there exist co prime positive integers a and b such that\[\surd 5\text{ }+\text{ }\surd...
![\[\mathbf{2}\text{ }-\text{ }\mathbf{3}\surd \mathbf{5}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-492fc7220996be5495f8ab5970d52502_l3.png "Rendered by QuickLaTeX.com")
Solution: Let’s assume on the contrary that \[2\text{ }\text{ }3\surd 5\]is a rational number. Then, there exist co prime positive integers a and b such that \[2\text{ }\text{ }3\surd 5\text{...
![\[\mathbf{2}\surd \mathbf{3}\text{ }-\text{ }\mathbf{1}~\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f25784e4cbfcc2b6f77b98331c7909c1_l3.png "Rendered by QuickLaTeX.com")
Solution: Let’s assume on the contrary that \[2\surd 3\text{ }\text{ }1\] is a rational number. Then, there exist co prime positive integers a and b such that \[2\surd 3\text{ }\text{ }1\text{...
![\[\mathbf{5}\text{ }-\text{ }\mathbf{2}\surd \mathbf{3}~\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4fe99f6574a18dae472d801d66331d39_l3.png "Rendered by QuickLaTeX.com")
Solution: Let’s assume on the contrary that \[5\text{ }\text{ }2\surd 3\] is a rational number. Then, there exist co prime positive integers a and b such that \[5\text{ }\text{ }2\surd 3\text{...
![\[\mathbf{4}\text{ }-\text{ }\mathbf{5}\surd \mathbf{2}~\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-728fc3aae21f5413024bcf441fffe566_l3.png "Rendered by QuickLaTeX.com")
Solution: Let’s assume on the contrary that \[4\text{ }\text{ }5\surd 2\] is a rational number. Then, there exist co prime positive integers a and b such that \[4\text{ }\text{ }5\surd 2\text{...
![\[\mathbf{3}\text{ }+\text{ }\surd \mathbf{2}~\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-fc1c82b2f160b7a1dc310519dbde254a_l3.png "Rendered by QuickLaTeX.com")
Solution: Let’s assume on the contrary that \[3\text{ }+\text{ }\surd 2\] is a rational number. Then, there exist co prime positive integers a and b such that \[3\text{ }+\text{ }\surd 2=\text{...
![\[\mathbf{2}\text{ }-\text{ }\surd \mathbf{3}~\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d0a39d7c5e26f9ab889ba6084fa2705f_l3.png "Rendered by QuickLaTeX.com")
Solution: Let’s assume on the contrary that \[2\text{ }\text{ }\surd 3\] is a rational number. Then, there exist co prime positive integers a and b such that\[2\text{ }\text{ }\surd 3=\text{ }a/b\]...
Let n be any positive integer. Thus, the three consecutive positive integers are n, n+1 and n+2. We know that any positive integer can be of the form 6q, or...
![\[\mathbf{4}\text{ }+\text{ }\surd \mathbf{2}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e72e18bc9159e65a933a5182ce232b1e_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{5}\surd \mathbf{2}~~~~~~~~~\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f66067211a2d0fa3d76358415a69dc68_l3.png "Rendered by QuickLaTeX.com")
Solution: Let’s assume on the contrary that \[4\text{ }+\text{ }\surd 2\] is a rational number. Then, there exist co prime positive integers a and b such that \[4\text{ }+\text{ }\surd 2\text{...
\[\mathbf{13}.\]\[~\mathbf{x}/\mathbf{a}\text{ }+\text{ }\mathbf{y}/\mathbf{b}\text{ }=\text{ }\mathbf{a}\text{ }+\text{ }\mathbf{b}\] \[\mathbf{x}/{{\mathbf{a}}^{\mathbf{2}~}}+\text{...
57/(x + y) + 6/(x – y) = 5 38/(x + y) + 21/(x – y) = 9 Solution: Let substitute $\frac{1}{\left( x+y \right)}=u$ and $\frac{1}{\left( x-y \right)}=v$, \[57u\text{ }+\text{ }6v\text{ }=\text{ }5\]...
\[\mathbf{9}.\] \[\mathbf{5}/\left( \mathbf{x}\text{ }+\text{ }\mathbf{y} \right)\text{ }\text{ }\mathbf{2}/\left( \mathbf{x}\text{ }-\mathbf{y} \right)\text{ }=\text{ }-\mathbf{1}\]...
\[\mathbf{7}.\]\[~\mathbf{x}\text{ }+\text{ }\mathbf{ay}\text{ }=\text{ }\mathbf{b}\] \[\mathbf{ax}\text{ }+\text{ }\mathbf{by}\text{ }=\text{ }\mathbf{c}\] \[\mathbf{8}.\] \[\mathbf{ax}\text{...
\[\mathbf{5}.\] \[\left( \mathbf{x}\text{ }+\text{ }\mathbf{y} \right)/\text{ }\mathbf{xy}\text{ }=\text{ }\mathbf{2}\] \[\left( \mathbf{x}\text{ }\text{ }\mathbf{y} \right)/\text{...
2x + y = 35, 3x + 4y = 65 Solution: Given \[\mathbf{2x}\text{ }+\text{ }\mathbf{y}\text{ }\text{ }-\mathbf{35}\text{ }=\text{ }\mathbf{0}\] \[\mathbf{3x}\text{ }+\text{ }\mathbf{4y}\text{ }\text{...
(i) \[\mathbf{2}/\surd \mathbf{7}\] ii) \[\mathbf{3}/\left( \mathbf{2}\surd \mathbf{5} \right)\] Solution: Let’s assume on the contrary that \[2/\surd 7\] is a rational number. Then, there exist...
\[\mathbf{x}\text{ }+\text{ }\mathbf{2y}\text{ }+\text{ }\mathbf{1}\text{ }=\text{ }\mathbf{0}\] \[\mathbf{2x}\text{ }\text{ }\mathbf{3y}\text{ }\text{ }\mathbf{12}\text{ }=\text{...
![\[\mathbf{6}\text{ }+\text{ }\surd \mathbf{2}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3751ac6d34393366c041894ce911cbaa_l3.png "Rendered by QuickLaTeX.com")
Solution: Let’s assume on the contrary that \[6+\surd 2\] is a rational number. Then, there exist co prime positive integers a and b such that \[6\text{ }+\text{ }\surd 2\text{ }=\text{ }a/b\]...
![\[\mathbf{1}/\surd\mathbf{2}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7ec751ccbc3a9989b1ffe04529c04049_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{7}\surd\mathbf{5}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ee3f5e43889f09d58c678f460bdeeff3_l3.png "Rendered by QuickLaTeX.com")
Solution: Consider \[1/\surd 2\] is a rational number Let us assume \[1/\surd 2\] = r where r is a rational number On further calculation we get \[1/r\text{ }=\text{ }\surd 2\] Since r is a rational...