In the adjoining figure, BC is a diameter of a circle with centre O.
Question: In the adjoining figure,BCis a diameter of a circle with centreO. IfABandCDare two chords such thatAB||CD, prove thatAB=CD. Solution: Given:BCis a diameter of a circle with centreO. ABandCDare two chords such thatAB||CD.TO prove:AB = CDConstruction:DrawOLABandOMCD. Proof:In ΔOLBand ΔOMC, we have:OLB = OMC [90 each]OBL = OCD [Alternate angles asAB || CD]OB = OC [Radii of a circle] ΔOLB ΔOMC(AAS criterion)Thus,OL = OM (CPCT)We know that chords equidistant from the centre are equal.Hence,...
Read More →Which point on x-axis is equidistant from (5, 9) and (−4, 6) ?
Question: Which point on x-axis is equidistant from (5, 9) and (4, 6) ? 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 thexaxis which is equidistant from both the pointsA(5,9) andB(4,6) Let this point be denoted asC(x, y) Since the point lies on thex-axis the value of its ordinate will be 0. Or in other words we...
Read More →Find the matrix A such that
Question: Find the matrixA such that (i) $\left[\begin{array}{ll}1 1 \\ 0 1\end{array}\right] A=\left[\begin{array}{lll}3 3 5 \\ 1 0 1\end{array}\right]$ (ii) $A\left[\begin{array}{lll}1 2 3 \\ 4 5 6\end{array}\right]=\left[\begin{array}{rrr}-7 -8 -9 \\ 2 4 6\end{array}\right]$ (iii) $\left[\begin{array}{l}4 \\ 1 \\ 3\end{array}\right] A=\left[\begin{array}{lll}-4 8 4 \\ -1 2 1 \\ -3 6 3\end{array}\right]$ (iv) $\left[\begin{array}{lll}2 1 3\end{array}\right]\left[\begin{array}{ccc}-1 0 -1 \\ -1...
Read More →In the adjoining figure, O is the centre of a circle.
Question: In the adjoining figure,Ois the centre of a circle. IfABandACare chords of the circle such thatAB=AC,OPABandOQAC, prove thatPB=QC. Solution: Given:AB and AC are chords of the circle with centre O. AB = AC, OP AB and OQ AC To prove:PB = QCProof:AB = AC (Given) $\Rightarrow \frac{1}{2} \mathrm{AB}=\frac{1}{2} \mathrm{AC}$ The perpendicular from the centre of a circle to a chord bisects the chord. MB = NC ...(i)Also, OM = ON (Equal chords of a circle are equidistant from the centre)and OP...
Read More →Two vertices of an isosceles triangle are (2, 0) and (2, 5).
Question: Two vertices of an isosceles triangle are (2, 0) and (2, 5). Find the third vertex if the length of the equal sides is 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 $d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$ In an isosceles triangle two sides will be of equal length. Here two vertices of the triangle is given asA(2,0) andB(2,5). Let the third side of the triangle beC(x, y) I...
Read More →AB and AC are two chords of a circle of radius r such that AB = 2AC.
Question: $A B$ and $A C$ are two chords of a circle of radius $r$ such that $A B=2 A C$. If $p$ and $q$ are the distances of $A B$ and $A C$ from the centre then prove that $4 q^{2}=p^{2}+3 r^{2}$. Solution: LetAC=a.Since,AB= 2AC,AB= 2a. From centreO, perpendicular is drawn to the chordsABandACat pointsMandN, respectively.It is given thatOM=pandON=q.We know that perpendicular drawn from the centre to the chord, bisects the chord. AM=MB=a...(1) and $A N=N C=\frac{a}{2}$ ...(2) In ∆OAN, $(A N)^{2...
Read More →If a, b, c are in A.P. and a, b, d are in G.P.,
Question: Ifa,b,care in A.P. anda,b,dare in G.P., then prove thata,ab,dcare in G.P. Solution: $a, b$ and $c$ are in A.P. $\therefore 2 b=a+c \quad \ldots \ldots .(\mathrm{i})$ Also, $a, b$ and $d$ are in G.P. $\therefore b^{2}=a d \quad \ldots \ldots$ (ii) Now, $(a-b)^{2}$ $=a^{2}-2 a b+b^{2}$ $=a^{2}-a(a+c)+a d$ [ Using (1) and (2) ] $=a d-a c$ $=a(d-c)$ $\Rightarrow(a-b)^{2}=a(d-c)$ Therefore, $a,(a-b)$ and $(d-c)$ are in G.P....
Read More →Show that the quadrilateral whose vertices are (2, −1), (3, 4) (−2, 3)
Question: Show that the quadrilateral whose vertices are (2, 1), (3, 4) (2, 3) and (3,2) is a rhombus. 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 rhombus all the sides are equal in length. Here the four points areA(2,1), B(3,4),C(2,3) andD(3,2). First let us check if all the four sides are equal. $A B=\sqrt{(2-3)^{2}+(-1-4)^{2}}$ $=\sqrt...
Read More →Solve the following
Question: If the 4th, 10thand 16thterms of a G.P. arex,yandzrespectively. Prove thatx,y,zare in G.P. Solution: $a_{4}=x$ $\Rightarrow a r^{3}=x$ Also, $a_{10}=y$ $\Rightarrow a r^{9}=y$ And, $a_{16}=z$ $\Rightarrow a r^{15}=z$ $\because \quad \frac{y}{x}=\frac{a r^{9}}{a r^{3}}=r^{6}$ and $\frac{z}{y}=\frac{a r^{15}}{a r^{9}}=r^{6}$ $\therefore \frac{y}{x}=\frac{z}{y}$ Therefore, $x, y$ and $z$ are in G.P....
Read More →An equilateral triangle has two vertices at the points
Question: An equilateral triangle has two vertices at the points (3, 4) and (2, 3), find the coordinates of the third vertex. 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 equilateral triangle all the sides are of equal length. Here we are given thatA(3,4) andB(2,3) are two vertices of an equilateral triangle. LetC(x, y) be the third verte...
Read More →If a, b, c are in G.P., then prove that:
Question: If $a, b, c$ are in G.P., then prove that: $\frac{a^{2}+a b+b^{2}}{b c+c a+a b}=\frac{b+a}{c+b}$ Solution: $a, b$ and $c$ are in G. P. $\therefore b^{2}=a c \quad \ldots \ldots \ldots(\mathrm{i})$ Now, LHS $=\frac{a^{2}+a b+b^{2}}{b c+c a+a b}$ $=\frac{\mathrm{a}^{2}+\mathrm{ab}+\mathrm{ac}}{\mathrm{bc}+\mathrm{b}^{2}+\mathrm{ab}}$ [Using (i)] $=\frac{a(a+b+c)}{b(c+b+a)}$ $=\frac{a}{b}$ $=\frac{1}{r}$ Here, $r=$ common ratio $\mathrm{RHS}=\frac{b+a}{c+b}$ $=\frac{a r+a}{a r^{2}+a r}$ $...
Read More →In the given figure, AB is a chord of a circle with centre O and AB is produced to C such that BC = OB.
Question: In the given figure,ABis a chord of a circle with centreOandABis produced toCsuch thatBC=OB. Also,COis joined and produced to meet the circle inD. IfACD=y and AOD=x, prove thatx= 3y. Solution: We have:OB = OC, BOC = BCO =yExternal OBA = BOC + BCO = (2y)Again, OA = OB, OAB = OBA = (2y)External AOD = OAC + ACOOrx= OAB + BCOOrx= (2y) +y= 3yHence,x= 3y...
Read More →In the adjoining figure, two circles with centres at A and B, and of radii 5 cm and 3 cm touch each other internally.
Question: In the adjoining figure, two circles with centres atAandB, and of radii 5 cm and 3 cm touch each other internally. If the perpendicular bisector ofABmeets the bigger circle inPandQ, find the length ofPQ. Solution: Two circles with centresAandBof respective radii 5 cm and 3 cm touch each other internally.The perpendicular bisector ofABmeets the bigger circle atPandQ.JoinAP. LetPQintersectABat pointL.Here,AP= 5 cmThenAB= (5 - 3) cm = 2 cmSincePQis the perpendicular bisector ofAB,we have:...
Read More →If (a − b), (b − c), (c − a) are in G.P.,
Question: If (ab), (bc), (ca) are in G.P., then prove that (a+b+c)2= 3 (ab+bc+ca) Solution: $(a-b),(b-c)$ and $(c-a)$ are in G.P. $\therefore(b-c)^{2}=(a-b)(c-a)$ $\Rightarrow b^{2}-2 b c+c^{2}=a c-b c+a b-a^{2}$ $\Rightarrow a^{2}+b^{2}+c^{2}=a b+b c+c a$$\quad \ldots \ldots$ (i) Now, LHS $=(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a$ $=a b+b c+c a+2 a b+2 b c+2 c a \quad[$ Using (i) $]$ $=3 a b+3 b c+3 c a$ $=3(a b+b c+c a)$ $=\mathrm{RHS}$...
Read More →Prove that the points (2,3), (−4, −6) and (1, 3/2) do not form a triangle.
Question: Prove that the points (2,3), (4, 6) and (1, 3/2) do not form a 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 any triangle the sum of lengths of any two sides need to be greater than the third side. Here the three points are $A(2,3), B(-4,-6)$ and $C\left(1, \frac{3}{2}\right)$ Let us now find out the lengths of all the thr...
Read More →If a, b, c, d are in G.P., prove that:
Question: Ifa,b,c,d are in G.P., prove that: (i) (a2+b2), (b2+c2), (c2+d2) are in G.P. (ii) (a2b2), (b2c2), (c2d2) are in G.P. (iii) $\frac{1}{a^{2}+b^{2}}, \frac{1}{b^{2}-c^{2}}, \frac{1}{c^{2}+d^{2}}$ are in G.P. (iv) (a2+b2+c2), (ab+bc +cd), (b2+c2+d2) are in G.P. Solution: a, b, c and d are in G.P. $\therefore b^{2}=a c$ ...(1) $a d=b c$ $c^{2}=b d$ (i) $\left(b^{2}+c^{2}\right)^{2}=\left(b^{2}\right)^{2}+2 b^{2} c^{2}+\left(c^{2}\right)^{2}$ $\Rightarrow\left(b^{2}+c^{2}\right)^{2}=(a c)^{2...
Read More →If a diameter of a circle bisects each of the two chords of a circle then prove that the chords are parallel.
Question: If a diameter of a circle bisects each of the two chords of a circle then prove that the chords are parallel. Solution: Given:ABandCDare two chords of a circle with centreO. DiameterPOQbisects them at pointsLandM.To prove:AB||CDProof:ABandCDare two chords of a circle with centreO.DiameterPOQbisects them atLandM. ThenOLABAlso,OMCD ALM = LMD= 90oSince alternate angles are equal, we have:AB|| CD...
Read More →Without using the concept of inverse of a matrix, find the matrix
Question: Without using the concept of inverse of a matrix, find the matrix $\left[\begin{array}{ll}x y \\ z u\end{array}\right]$ such that $\left[\begin{array}{rr}5 -7 \\ -2 3\end{array}\right]\left[\begin{array}{ll}x y \\ z u\end{array}\right]=\left[\begin{array}{rr}-16 -6 \\ 7 2\end{array}\right]$ Solution: Given : $\left[\begin{array}{cc}5 -7 \\ -2 3\end{array}\right]\left[\begin{array}{ll}x y \\ z u\end{array}\right]=\left[\begin{array}{cc}-16 -6 \\ 7 2\end{array}\right]$ $\Rightarrow\left[...
Read More →Prove that the points (2a, 4a), (2a, 6a) and
Question: Prove that the points $(2 a, 4 a),(2 a, 6 a)$ and $(2 a+\sqrt{3} a, 5 a)$ are the vertices of an equilateral 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 equilateral triangle all the sides have equal length. Here the three points are $A(2 a, 4 a), B(2 a, 6 a)$ and $C(2 a+a \sqrt{3}, 5 a)$. Let us now find out the lengt...
Read More →Two equal circles intersect in P and Q. A straight line through P meets the circles in A and B.
Question: Two equal circles intersect inPandQ. A straight line throughPmeets the circles inAandB. Prove thatQA=QB. Solution: Given:Two equal circles intersect at pointPandQ.A straight line passes throughPand meets the circle at pointsAandB.To prove:QA = QBConstruction:JoinPQ. Proof:Two circles will be congruent if and only if they have equal radii.Here,PQis the common chord to both the circles.Thus, their corresponding arcs are equal (if two chords of a circle are equal, then their corresponding...
Read More →Prove that (2, −2) (−2, 1) and (5, 2) are the vertices of a right
Question: Prove that (2, 2) (2, 1) and (5, 2) are the vertices of a right angled triangle. Find the area of the triangle and the length of the hypotenuse. 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 right-angled triangle, by Pythagoras theorem, the square of the longest side is equal to the sum of squares of the other two sides in the tri...
Read More →Two circles of radii 10 cm and 8 cm intersect each other, and the length of the common chord is 12 cm.
Question: Two circles of radii 10 cm and 8 cm intersect each other, and the length of the common chord is 12 cm. Find the distance between their centres. Solution: Given:OA= 10 cm,O'A= 8 cm andAB= 12 cm $A D=\left(\frac{\mathrm{AB}}{2}\right)=\left(\frac{12}{2}\right)=6 \mathrm{~cm}$ Now, in right angled ΔADO, we have: $O A^{2}=A D^{2}+O D^{2}$ $\Rightarrow O D^{2}=O A^{2}-A D^{2}$ $=10^{2}-6^{2}$ = 100 - 36 = 64 OD= 8 cm Similarly, in right angled ΔADO', we have: $O^{\prime} A^{2}=A D^{2}+O^{\p...
Read More →If a, b, c are in G.P., prove that the following are also in G.P.:
Question: Ifa, b, care in G.P., prove that the following are also in G.P.: (i)a2,b2,c2 (ii)a3,b3,c3 (iii)a2+b2,ab+bc,b2+c2 Solution: a, b and c are in G.P. $\therefore b^{2}=a c \quad \ldots \ldots(1)$ (i) $\left(b^{2}\right)^{2}=(a c)^{2} \quad[$ Using $(1)]$ $\Rightarrow\left(b^{2}\right)^{2}=a^{2} c^{2}$ Therefore, $a^{2}, b^{2}$ and $c^{2}$ are also in $\mathrm{G} . \mathrm{P} .$ (ii) $\left(b^{3}\right)^{2}=\left(b^{2}\right)^{3}=(a c)^{3} \quad[$ Using $(1)]$ $\Rightarrow\left(b^{3}\right)...
Read More →If a, b, c are in G.P., prove that the following are also in G.P.:
Question: Ifa, b, care in G.P., prove that the following are also in G.P.: (i)a2,b2,c2 (ii)a3,b3,c3 (iii)a2+b2,ab+bc,b2+c2 Solution: a, b and c are in G.P. $\therefore b^{2}=a c \quad \ldots \ldots(1)$ (i) $\left(b^{2}\right)^{2}=(a c)^{2} \quad[$ Using $(1)]$ $\Rightarrow\left(b^{2}\right)^{2}=a^{2} c^{2}$ Therefore, $a^{2}, b^{2}$ and $c^{2}$ are also in $\mathrm{G} . \mathrm{P} .$ (ii) $\left(b^{3}\right)^{2}=\left(b^{2}\right)^{3}=(a c)^{3} \quad[$ Using $(1)]$ $\Rightarrow\left(b^{3}\right)...
Read More →Solve this
Question: If $A=\left[\begin{array}{lll}3 2 0 \\ 1 4 0 \\ 0 0 5\end{array}\right]$, show that $A^{2}-7 A+10 / 3=0$ Solution: Given : $A=\left[\begin{array}{lll}3 2 0 \\ 1 4 0 \\ 0 0 5\end{array}\right]$ Here, $A^{2}=A A$ $\Rightarrow A^{2}=\left[\begin{array}{lll}3 2 0 \\ 1 4 0 \\ 0 0 5\end{array}\right]\left[\begin{array}{lll}3 2 0 \\ 1 4 0 \\ 0 0 5\end{array}\right]$ $\Rightarrow A^{2}=\left[\begin{array}{ccc}9+2+0 6+8+0 0+0+0 \\ 3+4+0 2+16+0 0+0+0 \\ 0+0+0 0+0+0 0+0+25\end{array}\right]$ $\Ri...
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