eSaral - JEE | NEET | Class 8-10 Exam Prep App

In the given figure, O is the centre of a circle and ∠AOC = 130°. Then, ∠ABC = ?

Question: In the given figure,Ois the centre of a circle andAOC= 130. Then, ABC= ?(a) 50(b) 65(c) 115(d) 130 Solution: (c) 115Take a point D on the remaining part of the circumference.Join AD and CD. Then $\angle \mathrm{ADC}=\frac{1}{2} \angle \mathrm{AOC}=\left(\frac{1}{2} \times 130^{\circ}\right)=65^{\circ}$ In cyclic quadrilateral ABCD, we have:ABC + ADC = 180 (Opposite angles of a cyclic quadrilateral)⇒ABC + 65 = 180⇒ABC = (180 - 65) = 115...

If three positive real numbers a, b, c are in A.P. and abc = 4,

Question: If three positive real numbersa,b,care in A.P. andabc= 4, then the minimum possible value ofbis_________. Solution: Let us suppose three positive real numbersa, bandcare in A.P andabc= 4 Since arithmetic mean geometric mean i. e $\frac{a+b+c}{3} \geq \sqrt[3]{a b c}$ i. e $\frac{a+b+c}{3} \geq \sqrt[3]{4}$ (given) Since $b=\frac{a+c}{2}$ $2 b=a+c$ i. e $\frac{2 b+b}{3} \geq(4)^{\frac{1}{3}}$ i.e $\frac{3 b}{3} \geq(4)^{\frac{1}{3}}$ i. e $b \geq(2)^{\frac{2}{3}}$ i.e minimum possible v...

If the coordinates of the mid-points of the

Question: If the coordinates of the mid-points of the sides of a triangle are (3, 4) (4, 6) and (5, 7), find its vertices. Solution: The co-ordinates of the midpoint $\left(x_{n}, y_{n}\right)$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by, $\left(x_{w}, y_{m}\right)=\left(\left(\frac{x_{1}+x_{2}}{2}\right),\left(\frac{y_{1}+y_{2}}{2}\right)\right)$ Let the three vertices of the triangle be $A\left(x_{A}, y_{A}\right), B\left(x_{B}, y_{n}\right)$ and ...

In the given figure ABCD is a cyclic quadrilateral in which AB || DC and ∠BAD = 100°.

Question: In the given figureABCDis a cyclic quadrilateral in whichAB||DCandBAD= 100. Then, ABC= ?(a) 80(b) 100(c) 50(d) 40 Solution: (b) 100Since ABCD is a cyclic quadrilateral, we have:BAD + BCD = 180 (Opposite angles of a cyclic quadrilateral)⇒ 100 + BCD = 180⇒BCD = (180 - 100) = 80Now,AB || DCand CB is the transversal.ABC + BCD = 180⇒ABC + 80 = 180⇒ABC = (180 - 80) = 100...

If A and G are the arithmetic and geometric means, respectively,

Question: IfAandGare the arithmetic and geometric means, respectively, of the roots of a quadratic equation. Then, the equation is _________. Solution: LetArepresent arithmetic mean andG represent geometric mean of the roots of a quadratic equation. Let us suppose the roots of the equation be aandb According t o given condition:- $A \cdot M=\frac{a+b}{2}=A$ i. e $a+b=2 A$ $G . M=\sqrt{a b}=G$ i. e $a b=G^{2}$ Hence, quadratic equation is given byx2 (a + b)x+ab= 0 [x2 (sum of roots)x+ (product of...

if

Question: If $A=\left[\begin{array}{r}-2 \\ 4 \\ 5\end{array}\right], B=\left[\begin{array}{lll}1 3 -6\end{array}\right]$, verify that $(A B)^{T}=B^{T} A^{T}$ Solution: Given : $A=\left[\begin{array}{c}-2 \\ 4 \\ 5\end{array}\right]$ $A^{T}=\left[\begin{array}{lll}-2 4 5\end{array}\right]$ $B=\left[\begin{array}{lll}1 3 -6\end{array}\right]$ $B^{T}=\left[\begin{array}{c}1 \\ 3 \\ -6\end{array}\right]$ Now, $A B=\left[\begin{array}{c}-2 \\ 4 \\ 5\end{array}\right]\left[\begin{array}{lll}1 3 -6\en...

In the given figure, AOB is a diameter and ABCD is a cyclic quadrilateral.

Question: In the given figure,AOBis a diameter andABCDis a cyclic quadrilateral. IfADC= 120, then BAC= ?(a) 60(b) 30(c) 20(d) 45 Solution: (b) 30We have:ABC + ADC = 180 (Opposite anglesof a cyclic quadrilateral)⇒ABC +120 = 180⇒ABC = (180 - 120) = 60Also, ACB = 90 (Angle in a semicircle)InΔABC, we have:BAC + ACB + ABC = 180 (Angle sum property of a triangle)⇒BAC + 90 + 60 = 180⇒BAC = (180 - 150) = 30...

Solve the following

Question: If $A_{1}, A_{2}$ are the two arithmetic means between two numbers a and $b$ and $G_{1}, G_{1}$ are two geometric means between same two numbers, then $\frac{A_{1}+A_{2}}{G_{1} G_{2}}=$ Solution: IfA1,A2are two arithmetic means between two numbersaandbandG1,G2are geometric means betweenaandb. Figure i. e $A_{1}=a+\frac{1}{3}(b-a)=\frac{3 a+b-a}{3}=\frac{2 a+b}{3}$ and $A_{2}=a+\frac{2}{3}(b-a)=\frac{3 a+2 b-2 a}{3}=\frac{a+2 b}{3}$ AlsoG1andG2are the geometric means betweenaandb. So, $...

In the given figure, O is the centre of a circle in which ∠OAB = 20° and ∠OCB = 50°.

Question: In the given figure,Ois the centre of a circle in whichOAB= 20 and OCB= 50. Then, AOC= ?(a) 50(b) 70(c) 20(d) 60 Solution: (d) 60We have:OA = OB (Radii of a circle)⇒OBA= OAB= 20InΔOAB, we have:OAB + OBA + AOB = 180 (Angle sum property of a triangle)⇒ 20 + 20 + AOB = 180⇒AOB = (180 - 40) = 140Again, we have:OB = OC(Radii of a circle)⇒OBC = OCB= 50InΔOCB, we have:OCB + OBC + COB = 180 (Angle sum property of a triangle)⇒ 50 + 50 + COB = 180⇒COB = (180 - 100) = 80SinceAOB = 140, we have:AO...

If two vertices of a parallelogram are (3, 2)

Question: If two vertices of a parallelogram are (3, 2) (1, 0) and the diagonals cut at (2, 5), find the other vertices of the parallelogram. Solution: We have a parallelogram ABCD in which A (3, 2) and B (1, 0) and the co-ordinate of the intersection of diagonals is M (2,5). We have to find the co-ordinates of vertices C and D. So let the co-ordinates be $\mathrm{C}\left(x_{1}, y_{1}\right)$ and $\mathrm{D}\left(x_{2}, y_{2}\right)$ In general to find the mid-point $\mathrm{P}(x, y)$ of two poi...

In a geometric progression consisting of positive terms, each term equals the sum of the next two terms.

Question: In a geometric progression consisting of positive terms, each term equals the sum of the next two terms. Then the common ratio of the progression is ____________. Solution: Leta, ar, ar2, .......arn1defines a G.P Sincea = ar + ar2 (given condition) i. e $1=r+r^{2} i. e $r^{2}+r-1=0$ i. e $r=\pm \frac{\sqrt{5}-1}{2} \quad D=1-4(-1)=5$ Since series is given to be positive $\Rightarrow r=\frac{\sqrt{5}-1}{2}$ i.e the common ratio is $\frac{\sqrt{5}-1}{2}$...

In the given figure, O is the centre of a circle and chords AC and BD intersect at E.

Question: In the given figure,Ois the centre of a circle and chordsACandBDintersect atE. IfAEB= 110 and CBE= 30, then ADB= ?(a) 70(b) 60(c) 80(d) 90 Solution: (c) 80We have:AEB + CEB= 180 (Linear pair angles)⇒ 110 +CEB= 180⇒CEB = (180 - 110) = 70InΔCEB, we have:CEB + EBC + ECB = 180 (Angle sum property of a triangle)⇒70 + 30 +ECB = 180⇒ECB = (180 - 100) = 80The angles in the same segment are equal.Thus,ADB=ECB =80...

Three vertices of a parallelogram are (a+b, a−b),

Question: Three vertices of a parallelogram are (a+b, ab), (2a+b, 2ab), (ab, a+b). Find the fourth vertex. Solution: Let $\mathrm{ABCD}$ be a parallelogram in which the co-ordinates of the vertices are $\mathrm{A}(a+b, a-b) ; \mathrm{B}(2 a+b, 2 a-b)$ and $\mathrm{C}(a-b, a+b)$. We have to find the co-ordinates of the forth vertex. Let the forth vertex be $\mathrm{D}(x, y)$ Since ABCD is a parallelogram, the diagonals bisect each other. Therefore the mid-point of the diagonals of the parallelogr...

The sum of first two terms of a G.P. is 1 and every term is twice the previous term.

Question: The sum of first two terms of a G.P. is 1 and every term is twice the previous term. The first term of the G.P. is ____________. Solution: Leta, ar, ar2, .......arn1defines a G.P According to given condition, ar= 2a i.er= 2 anda + ar= 1 a(1 +r) = 1 a(1 + 2) = 1 i. e $a=\frac{1}{3}$ i.e first term of g.p is $\frac{1}{3}$...

In the give figure, AB and CD are two intersecting chords of a circle. If ∠CAB = 40° and ∠BCD = 80°, then ∠CBD = ?

Question: In the give figure,ABandCDare two intersecting chords of a circle. IfCAB= 40 andBCD= 80, then CBD= ?(a) 80(b) 60(c) 50(d) 70 Figure Solution: (b) 60We have:CDB = CAB = 40 (Angles in the same segment of a circle)In Δ CBD, we have:CDB + BCD +CBD = 180 (Angle sum property of a triangle)⇒40 + 80+CBD = 180⇒CBD = (180 - 120) = 60...

The third term of a G.P. is the square of its first term. If its third term is 8,

Question: The third term of a G.P. is the square of its first term. If its third term is 8, then the common ratio is ____________. Solution: Let the G.P bea,ar,ar2 According to given condition, third term = (first term)2 i.ear2=a2 i.er2=a (1) givenar2= 8 (given) i.ea2= 8 i. e $a=\sqrt{8}$...

Let

Question: Let $A=\left[\begin{array}{rrr}1 -1 0 \\ 2 1 3 \\ 1 2 1\end{array}\right]$ and $B=\left[\begin{array}{lll}1 2 3 \\ 2 1 3 \\ 0 1 1\end{array}\right] .$ Find $A^{T}, B^{T}$ and verify that (i) $(A+B)^{T}=A^{\top}+B^{\top}$ (ii) $(A B)^{T}=B^{T} A^{T}$ (iii) $(2 A)^{\top}=2 A^{\top}$. Solution: Given : $A=\left[\begin{array}{ccc}1 -1 0 \\ 2 1 3 \\ 1 2 1\end{array}\right]$ and $B=\left[\begin{array}{lll}1 2 3 \\ 2 1 3 \\ 0 1 1\end{array}\right]$ $A^{T}=\left[\begin{array}{ccc}1 2 1 \\ -1 1...

If x, 2y, 3z are in A.P., where the distinct numbers x, y, z are in G.P.,

Question: Ifx, 2y, 3zare in A.P., where the distinct numbersx,y,zare in G.P., then the common ratio of the G.P. is (a) 3 (b) $\frac{1}{3}$ (c) 2 (d) $\frac{1}{2}$ Solution: Letx, 2y, 3zbe in A.P, for distinctx, yandz wherex, y, zare in G.P Sincex, 2yand 3zare in A.P $\therefore 2 y=\frac{x+3 y}{2} \quad(\because$ middle term $=$ arithmetic mean of adjacent terms) i. e $4 y=x+3 y \quad \ldots(1)$ and sincex, yandzare in G.P y2=xz...(2) Letrdenote the common ratio of G.P $\therefore \frac{y}{x}=r$...

In the given figure, O is the centre of a circle and ∠OAB = 50°.

Question: In the given figure,Ois the centre of a circle andOAB= 50. Then , CDA= ?(a) 40(b) 50(c) 75(d) 25 Solution: (b) 50We have:OA = OB (Radii of a circle)⇒OBA = OAB = 50CDA =OBA =50(Angles in thesame segment of a circle)...

Find the lengths of the medians of a triangle

Question: Find the lengths of the medians of a triangle whose vertices are A (1,3), B(1,1) and C(5, 1). Solution: We have to find the lengths of the medians of a triangle whose co-ordinates of the vertices are A (1, 3); B (1,1) and C (5, 1). So we should find the mid-points of the sides of the triangle. In general to find the mid-point $\mathrm{P}(x, y)$ of two points $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ we use section formula as, $\mathrm{P}(x, y)=\lef...

In the given figure, O is the centre of a circle and ∠AOC = 120°.

Question: In the given figure,Ois the centre of a circle andAOC= 120. Then, BDC= ?(a) 60(b) 45(c) 30(d) 15 Solution: (c) 30COB = 180 - 120 = 60 (Linear pair)Now, arc BC subtendsCOB at the centre and BDC at the point D of the remaining part of the circle.COB = 2BDC $\Rightarrow \angle \mathrm{BDC}=\frac{1}{2} \angle \mathrm{COB}=\left(\frac{1}{2} \times 60^{\circ}\right)=30^{\circ}$...

The minimum value of

Question: The minimum value of 4x+ 41 x,xR, is (a) 2 (b) 4 (c) 1 (d) 0 Solution: For $4^{x}+4^{1-x}: x \in R$ 4xand 41 xare 2 terms sayxandyrespectively Since arithmetic mean Geometric mean ofxandy i. e $\frac{x+4}{2} \geq \sqrt{x y}$ i. e $\frac{4^{x}+4^{1-x}}{2} \geq \sqrt{4^{x} 4^{1-x}}$ i. e $4^{x}+4^{1-x} \geq 2 \sqrt{4^{x} 4.4^{-x}}$ i. e $4^{x}+4^{1-x} \geq 2 \sqrt{4}$ i. e $4 x+4^{1-x} \geq 4$ minimum value of 4x+ 41 xis 4 Hence, the correct answer is option B....

In the given figure, O is the centre of a circle. Then, ∠OAB = ?

Question: In the given figure,Ois the centre of a circle. Then,OAB= ?(a) 50(b) 60(c) 55(d) 65 Solution: (d) 65We have:OA = OB (Radii of a circle)LetOAB= OBA =xIn Δ OAB, we have:x +x + 50 = 180 (Angle sum property of a triangle)⇒ 2x = (180 - 50) = 130 $\Rightarrow x=\left(\frac{130}{2}\right)^{\circ}=65^{\circ}$ Hence,OAB= 65...

In the given figure, O is the centre of a circle.

Question: In the given figure,Ois the centre of a circle. IfAOB= 100 and AOC= 90, then BAC= ?(a) 85(b) 80(c) 95(d) 75 Solution: (a) 85We have:BOC + BOA + AOC = 360⇒BOC + 100 + 90 = 360⇒BOC = (360 - 190) = 170 $\therefore \angle \mathrm{BAC}=\left(\frac{1}{2} \times \angle \mathrm{BOC}\right)=\left(\frac{1}{2} \times 170^{\circ}\right)=85^{\circ}$...

The lengths of three unequal edges of a rectangular solid block are in G.P.

Question: The lengths of three unequal edges of a rectangular solid block are in G.P. The volume of the block is 216 cm3and the total surface area is 252 cm2. The length of the longest edge is (a) 12 cm (b) 6 cm (c) 18 cm (d) 3 cm Solution: Since lengths of three unequal sides of a rectangular solid block are in G.P. Let the length, breadth and height of rectangular solid block be $\frac{a}{r}, a$ and ar respectively. Volume of rectangular block is $\frac{a}{r} \times a \times a r \quad[\because...

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