If the A.M. of two positive numbers a and b (a > b) is twice their geometric mean. Prove that:
Question: If the A.M. of two positive numbersaandb(ab) is twice their geometric mean. Prove that: $a: b=(2+\sqrt{3}):(2-\sqrt{3})$ Solution: $\mathrm{AM}=2 \mathrm{GM}$ $\therefore \frac{a+b}{2}=2 \sqrt{a b}$ $\Rightarrow a+b=4 \sqrt{a b}$ Squaring both the sides: $\Rightarrow(a+b)^{2}=(4 \sqrt{a b})^{2}$ $\Rightarrow a^{2}+2 a b+b^{2}=16 a b$ $\Rightarrow a^{2}-14 a b+b^{2}=0$ Using the quadratic formula: $\Rightarrow a=\frac{-(-14 b)+\sqrt{(-14 b)^{2}-4 \times 1 \times b^{2}}}{2 \times 1} \qua...
Read More →In the given figure, ∠BAC = 30°.
Question: In the given figure,BAC= 30. Show thatBCis equal to the radius of the circumcircle of ∆ABCwhose centre isO. Solution: JoinOBandOC.BOC = 2BAC(As angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference) = 2 30 [∵BAC= 30] = 60...(i) Consider ΔBOC, we have:OB = OC [Radii of a circle]⇒OBC =OCB ...(ii)In ΔBOC, we have:BOC+OBC+OCB= 180 (Angle sum property of a triangle)⇒ 60+OCB+OCB= 180 [From (i) and (ii)]⇒ 2OCB= (180- 60)...
Read More →If B, C are n rowed square matrices and if A = B + C,
Question: If $B, C$ are $n$ rowed square matrices and if $A=B+C, B C=C B, C^{2}=0$, then show that for every $n \in N, A^{n+1}=B^{n}(B+(n+1) C)$. Solution: Let $P(n)$ be the statement given by $P(n): A^{n+1}=B^{n}(B+(n+1) C)$. For $n=1$, we have $P(1): A^{2}=B(B+2 C)$ Here, $\mathrm{LHS}=A^{2}$ $=(B+C)(B+C)$ $=B(B+C)+C(B+C)$ $=B^{2}+B C+C B+C^{2}$ $=B^{2}+2 B C$ $\left[\because B C=C B\right.$ and $\left.C^{2}=O\right]$ $=B(B+2 C)=\mathrm{RHS}$ Hence, the statement is true for $n=1$. If the stat...
Read More →In the figure given below, P and Q are centres of two circles, intersecting at B and C,
Question: In the figure given below,PandQare centres of two circles, intersecting atBandC,andACDis a straight line. If APB= 150 and BQD=x, find the value ofx. Solution: We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it on the remaining part of the circle.Here, arcAEBsubtends APBat the centre and ACBatCon the circle. APB= 2ACB $\Rightarrow \angle A C B=\frac{150^{\circ}}{2}=75^{\circ}$ ...(1) SinceACDis a straight line, ACB +BCD= 180∘⇒ BCD= 1...
Read More →Find the centre of the circle passing through
Question: Find the centre of the circle passing through (6, 6), (3, 7) and (3, 3). Solution: The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula $d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$ The centre of a circle is at equal distance from all the points on its circumference. Here it is given that the circle passes through the pointsA(6,6),B(3,7) andC(3,3). Let the centre of the circle be represented by th...
Read More →In the given figure, PQ is a diameter of a circle with centre O.
Question: In the given figure,PQis a diameter of a circle with centreO. IfPQR= 65, SPR= 40 and PQM= 50, find QPR, QPMand PRS. Solution: Here,PQis the diameter and the angle in a semicircle is a right angle.i.e.,PRQ= 90In ΔPRQ, we have:QPR+PRQ+PQR= 180 (Angle sum property of a triangle)⇒QPR+ 90 + 65 = 180⇒QPR= (180 155) = 25In ΔPQM, PQis the diameter.PMQ= 90In ΔPQM, we have:QPM+PMQ+PQM= 180 (Angle sum property of a triangle)⇒QPM+ 90 + 50 = 180⇒QPM= (180 140) = 40Now, in quadrilateralPQRS, we have...
Read More →In the given figure, O is the centre of the circle,
Question: In the given figure,Ois the centre of the circle,BD=ODandCDAB. Find CAB. Solution: In the given figure,BD=ODandCDAB. JoinACandOC.In ∆ODEand ∆DBE,DOE = DBE(given)DEO= DEB =90∘OD=DB(given) By AAS conguence rule, ∆ODE≌ ∆BDE,Thus,OE=EB ...(1)Now, in ∆COEand ∆CBE,CE =CE(common)CEO= CEB =90∘OE=EB(from (1)) By SAS conguence rule, ∆COE≌ ∆CBE,Thus,CO=CB ...(2)Also,CO=OB=OA(radius of the circle) ...(3)From (2) and (3),CO=CB=OB ∆COBis equilateral triangle. COB = 60∘ ...(4)We know that the angle s...
Read More →Solve this
Question: Let $A=\left[\begin{array}{lll}1 1 1 \\ 0 1 1 \\ 0 0 1\end{array}\right]$. Use the principle of mathematical induction to show that $A^{n}=\left[\begin{array}{ccc}1 n n(n+1) / 2 \\ 0 1 n \\ 0 0 1\end{array}\right]$ for every positive integer $n$. Solution: We shall prove the result by the principle of mathematical induction onn.Step 1: Ifn= 1, by definition of integral power of a matrix, we have $A^{1}=\left[\begin{array}{ccc}1 1 1(1+1) / 2 \\ 0 1 1 \\ 0 0 1\end{array}\right]=\left[\be...
Read More →Find the values of y for which the distance between
Question: Find the values ofyfor which the distance between the pointsP(2, 3) andQ(10,y) is 10 units. Solution: The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula $d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$ The distance between two pointsP(2,3) andQ(10,y) is given as 10 units. Substituting these values in the formula for distance between two points we have, $10=\sqrt{(2-10)^{2}+(-3-y)^{2}}$ Now, squarin...
Read More →Prove that the product of n geometric means between two quantities
Question: Prove that the product ofngeometric means between two quantities is equal to thenth power ofageometric mean of those two quantities. Solution: Let $G_{1}, G_{2}, G_{3}, G_{4}, \ldots, G_{n}$ be $n$ G. M.s between $a$ and $b$. Then, $a, G_{1}, G_{2}, G_{3}, G_{4}, \ldots, G_{n}, b$ is a G.P. Let $r$ be the common ratio. $\because b=a_{n+2}=a r^{(n+1)}$ $\Rightarrow r=\left(\frac{b}{a}\right)^{\frac{1}{(n+1)}}$ $\therefore G_{1}=a_{2}=a r$ $G_{2}=a_{3}=a r^{2}$ $G_{3}=a_{4}=a r^{3}$ $G_{...
Read More →If Q (0,1) is equidistant from P (5, −3) and R (x, 6),
Question: IfQ(0,1) is equidistant fromP(5, 3) andR(x, 6), find the values ofx. Also, find the distancesQRandPR. Solution: The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula $d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$ The three given points areQ(0,1), P(5,3)andR(x,6). Now let us find the distance between P and Q. $P Q=\sqrt{(5-0)^{2}+(-3-1)^{2}}$ $=\sqrt{(5)^{2}+(-4)^{2}}$ $=\sqrt{25+16}$ $P Q=\sqrt{41}$...
Read More →In the given figure, O is the centre of the circle and ∠BCO = 30°. Find x and y.
Question: In the given figure,Ois the centre of the circle and BCO= 30. Findxandy. Solution: In the given figure,ODis parallel toBC. BCO= COD(Alternate interior angles) $\Rightarrow \angle C O D=30^{\circ}$ We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it on the remaining part of the circle.Here, arcCDsubtends CODat the centre and CBDatBon the circle. COD= 2CBD $\Rightarrow \angle C B D=\frac{30^{\circ}}{2}=15^{\circ}$ (from (1)) $\therefor...
Read More →If the point P(x, y) is equidistant from the points
Question: If the pointP(x,y) is equidistant from the pointsA(5, 1) andB(1, 5), prove thatx=y. Solution: The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula $d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$ The three given points areP(x, y), A(5,1)andB(1,5). Now let us find the distance between P and A. $P A=\sqrt{(x-5)^{2}+(y-1)^{2}}$ Now, let us find the distance between P and B. $P B=\sqrt{(x-1)^{2}+(y-5)^{2...
Read More →Solve this
Question: If $A=\left[\begin{array}{cc}\cos \alpha+\sin \alpha \sqrt{2} \sin \alpha \\ -\sqrt{2} \sin \alpha \cos \alpha-\sin \alpha\end{array}\right]$, prove that $A^{n}=\left[\begin{array}{cc}\cos n \alpha+\sin n \alpha \sqrt{2} \sin n \alpha \\ -\sqrt{2} \sin n \alpha \cos n \alpha-\sin n \alpha\end{array}\right]$ for all $n \in N$ Solution: We shall prove the result by the principle of mathematical induction onn. Step 1: If $n=1$, by definition of integral power of a matrix, we have $A^{1}=\...
Read More →Prove that the points (0, 0), (5, 5) and (−5, 5)
Question: Prove that the points (0, 0), (5, 5) and (5, 5) are the vertices of a right isosceles triangle. Solution: The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula $d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$ In an isosceles triangle there are two sides which are equal in length. By Pythagoras Theorem in a right-angled triangle the square of the longest side will be equal to the sum of squares of the ...
Read More →If AM and GM of two positive numbers a and b are 10 and 8 respectively,
Question: If AM and GM of two positive numbersaandbare 10 and 8 respectively, find the numbers. Solution: $\mathrm{AM}=10$ $\therefore \frac{a+b}{2}=10$ $\Rightarrow a+b=20 \quad \ldots \ldots(\mathrm{i})$ Also, $G=8$ $\therefore \sqrt{a b}=8$ $\Rightarrow a b=8^{2}$ $\Rightarrow a b=64$ ....(ii) Using (i) and (ii): $\Rightarrow a(20-a)=64$ $\Rightarrow a^{2}-20 a+64=0$ $\Rightarrow a^{2}-16 a-4 a+64=0$ $\Rightarrow a(a-16)-4(a-16)=0$ $\Rightarrow(a-16)(a-4)=0$ $\Rightarrow a=4,16$ If $a=4$, the...
Read More →Find the value of x such that PQ = QR
Question: Find the value ofxsuch thatPQ=QRwhere the coordinates ofP,QandRare (6, 1) , (1, 3) and (x, 8) respectively. Solution: The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula $d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$ The three given points areP(6,1), Q(1,3)andR(x,8). Now let us find the distance between P and Q. $P Q=\sqrt{(6-1)^{2}+(-1-3)^{2}}$ $=\sqrt{(5)^{2}+(-4)^{2}}$ $=\sqrt{25+16}$ $P Q=\sqr...
Read More →If AM and GM of roots of a quadratic equation are 8 and 5 respectively,
Question: If AM and GM of roots of a quadratic equation are 8 and 5 respectively, then obtain the quadratic equation. Solution: Let the roots of the quadratic equation be $a$ and $b$. $\mathrm{AM}=8$ $\therefore \frac{a+b}{2}=8$ $\Rightarrow a+b=16 \quad \ldots \ldots \ldots$ (i) Also, $G=5$ $\Rightarrow \sqrt{a b}=5$ $\Rightarrow a b=5^{2}$ $\Rightarrow a b=25$ ....(ii) Now, the quadratic equation is given by $x^{2}-(a+b) x+a b=0$ $\Rightarrow x^{2}-16 x+25=0$...
Read More →In the given figure, O is the centre of a circle in which ∠OAB = 20° and
Question: In the given figure,Ois the centre of a circle in whichOAB= 20 and OCB= 55. Find ]\ (i) BOC, (ii) AOC Solution: (i)OB = OC(Radii of a circle)⇒OBC =OCB =55Considering ΔBOC, we have:BOC+OCB+OBC= 180(Angle sum property of a triangle)⇒BOC+ 55+ 55= 180⇒BOC= (180- 110) = 70(ii)OA = OB(Radii of a circle)⇒OBA=OAB= 20Considering ΔAOB, we have:AOB+OAB+OBA= 180 (Angle sum property of a triangle)⇒AOB+ 20+ 20= 180⇒AOB= (180- 40) = 140AOC =AOB -BOC = (140- 70) = 70 Hence,AOC= 70...
Read More →Find the point on x-axis which is equidistant
Question: Find the point on x-axis which is equidistant from the points (2, 5) and (2,3). Solution: The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula $d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$ Here we are to find out a point on thex-axis which is equidistant from both the pointsA(2,5) andB(2,3) Let this point be denoted asC(x, y). Since the point lies on thex-axis the value of its ordinate will be 0. ...
Read More →Solve this
Question: If $A=\left[\begin{array}{cc}\cos \theta i \sin \theta \\ i \sin \theta \cos \theta\end{array}\right]$, then prove by principle of mathematical induction that $A^{n}=\left[\begin{array}{cc}\cos n \theta i \sin \theta \\ i \sin n \theta \cos n \theta\end{array}\right]$ for all $n \in N$ Solution: We shall prove the result by the principle of mathematical induction onn. Step 1: If $n=1$, by definition of integral power of a matrix, we have $A^{1}=\left[\begin{array}{cc}\cos 1 \theta i \s...
Read More →The sum of two numbers is 6 times their geometric means,
Question: The sum of two numbers is 6 times their geometric means, show that the numbers are in the ratio $(3+2 \sqrt{2}):(3-2 \sqrt{2})$ Solution: Let the two numbers be $a$ and $b$. Let the geometric mean between them be $G$. We have: $a+b=6 G$ But, $G=\sqrt{a b}$ $\therefore a+b=6 \sqrt{a b}$ $\Rightarrow(a+b)^{2}=(6 \sqrt{a b})^{2}$ $\Rightarrow a^{2}+2 a b+b^{2}=36 a b$ $\Rightarrow a^{2}-34 a b+b^{2}=0$ Using the quadratic formula: $\Rightarrow a=\frac{-(-34 b) \pm \sqrt{(-34 b)^{2}-4 \tim...
Read More →In the adjoining figure, chords AC and BD of a circle with centre O,
Question: In the adjoining figure, chordsACandBDof a circle with centreO, intersect at right angles atE. IfOAB= 25, calculate EBC. Solution: OA = OB(Radii of a circle)Thus,OBA =OAB =25Join OB. Now in ΔOAB, we have:OAB+OBA+AOB= 180(Angle sum property of a triangle)⇒⇒25+ 25+AOB= 180⇒⇒50+AOB=180⇒⇒AOB= (180 50) = 130We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.i.e.,AOB = 2ACB $\Rightarrow \angle A C B=...
Read More →(i) Show that the points A(5, 6), B(1, 5), C(2, 1) and D(6,2) are the vertices of a square.
Question: (i) Show that the pointsA(5, 6),B(1, 5),C(2, 1) andD(6,2) are the vertices of a square.(ii) Prove hat the pointsA(2, 3)B(2,2)C(1,2), andD(3, 1) are the vertices of a squareABCD. Solution: The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula $d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$ In a square all the sides are equal to each other. And also the diagonals are also equal to each other. Here the ...
Read More →In the given figure, O is the centre of a circle,
Question: In the given figure,Ois the centre of a circle,AOB40 and BDC= 100, find OBC. Solution: We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.AOB = 2ACB = 2DCB [∵ACB=DCB] $\therefore \angle D C B=\frac{1}{2} \angle A O B$ $\Rightarrow \angle D C B=\left(\frac{1}{2} \times 40^{\circ}\right)=20^{\circ}$ Considering ΔDBC,we have:BDC+DCB+DBC= 180⇒ 100+ 20+DBC= 180⇒DBC=(180 120) = 60⇒OBC =DBC= 60Hence,O...
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