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

On a common hypotenuse AB, two right triangles ACB and ADB are situated on opposite sides.

Question: On a common hypotenuseAB, two right trianglesACBandADBare situated on opposite sides. Prove that BAC= BDC. Solution: Draw two right trianglesACBandADBin a circle with centreO, whereABis the diameter of the circle. JoinCO.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, arcCBsubtends COBat the centre and CABatAon the circle. COB= 2CAB...(1)Also, arcCBsubtends COBat the centre and CDBatDon t...

The diagonals of a cyclic quadrilateral are at right angles.

Question: The diagonals of a cyclic quadrilateral are at right angles. Prove that the perpendicular from the point of their intersection on any side when produced backwards, bisects the opposite side. Solution: Let ABCD be a cyclic quadrilateral whose diagonals AC and BD intersect at O at right angles.Let OL AB such that LO produced meets CD at M. Then we have to prove that CM = MDClearly, 1 = 2 [Angles in the same segment]2 + 3 = 90 [∵ OLB = 90]3 + 4= 90 [∵ LOM is a straight line and BOC = 90] ...

If A be one A.M. and p, q be two G.M.'s between two numbers,

Question: IfAbe one A.M. andp,qbe two G.M.'s between two numbers, then 2 A is equal to (a) $\frac{p^{3}+q^{3}}{p q}$ (b) $\frac{p^{3}-q^{3}}{p q}$ (c) $\frac{p^{2}+q^{2}}{2}$ (d) $\frac{p q}{2}$ Solution: (a) $\frac{p^{3}+q^{3}}{p q}$ Let the two positive numbers be $a$ and $b$. $a, A$ and $b$ are in A.P. $\therefore 2 A=a+b$ ...(i) Also, $a, p, q$ and $b$ are in G.P. $\therefore r=\left(\frac{b}{a}\right)^{\frac{1}{3}}$ Again, $p=a r$ and $q=a r^{2}$ ...(ii) Now, $2 A=a+b \quad[$ From (i) $]$ $...

In a cyclic quadrilateral ABCD, if (∠B − ∠D) = 60°

Question: In a cyclic quadrilateralABCD, if (B D) = 60, show that the smaller of the two is 60. Solution: In cyclic quadrilateral ABCD, we have:B +D = 180 ...(i) (Opposite angles of a cyclic quadrilateral )B -D = 60...(ii) (Given)From (i) and (ii), we get:2B = 240⇒B = 120D= 60Hence, the smaller of the two angles is 60....

If A (−1, 3), B (1, −1) and C (5, 1) are the vertices

Question: IfA(1, 3),B(1, 1) andC(5, 1) are the vertices of a triangleABC, find the length of the median throughA. 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 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_{m}, y_{m}\right)=\left(\...

Give a geometrical construction for finding the fourth point lying on a circle passing through three given points,

Question: Give a geometrical construction for finding the fourth point lying on a circle passing through three given points, without finding the centre of the circle. Justify the construction. Solution: Let A, B and C be the given points.With B as the centre and a radius equal to AC, draw an arc.With C as the centre and AB as radius, draw another arc intersecting the previous arc at D.Then D is the desired point.Proof:Join BD and CD. In ΔABC and ΔDCB, we have:AB = DCAC = DBBC = CBi.e., ΔABC ΔDCB...

If a, b, c are in G.P. and x, y are AM's between a, b and b,c respectively,

Question: Ifa,b,care in G.P. andx,yare AM's betweena,bandb,crespectively, then (a) $\frac{1}{x}+\frac{1}{y}=2$ (b) $\frac{1}{x}+\frac{1}{y}=\frac{1}{2}$ (c) $\frac{1}{x}+\frac{1}{y}=\frac{2}{a}$ (d) $\frac{1}{x}+\frac{1}{y}=\frac{2}{b}$ Solution: (d) $\frac{1}{x}+\frac{1}{y}=\frac{2}{b}$ $a, b$ and $c$ are in G.P. $\therefore b^{2}=a c \quad \ldots \ldots$ (i) $a, x$ and $b$ are in A.P. $\therefore 2 x=a+b \quad \ldots \ldots \ldots($ ii $)$ Also, $b, y$ and $c$ are in A.P. $\therefore 2 y=b+c$ ...

Find the ratio in which the point (2, y)

Question: Find the ratio in which the point (2,y) divides the line segment joining the pointsA(2, 2) andB(3, 7). Also, find the value ofy. Solution: The co-ordinates of a point which divided two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ internally in the ratio $m: n$ is given by the formula, $(x, y)=\left(\left(\frac{m x_{2}+n x_{1}}{m+n}\right),\left(\frac{m y_{2}+n y_{1}}{m+n}\right)\right)$ Here we are given that the pointP(2,y) divides the line joining the pointsA(2,...

A trust fund has Rs 30000 that must be invested in two different types of bonds.

Question: A trust fund has Rs 30000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs 30000 among the two types of bonds. If the trust fund must obtain an annual total interest of(i) Rs 1800 (ii) Rs 2000 Solution: If Rs $x$ are invested in the first type of bond and Rs ( $30000-x$ ) are invested in the second type of bond, then the matrix $A=[x \qu...

ABCD is a rectangle.

Question: ABCDis a rectangle. Prove that the centre of the circle thoughtA,B,C,Dis the point of intersection of its diagonals. Solution: Given:ABCD is a cyclic rectangle whose diagonals intersect at O.To prove:O is the centre of the circle.Proof: Here, BCD = 90 [Since it is a rectangle]So, BD is the diameter of the circle(if the angle made by the chord at the circle is right angle, then the chord is the diameter).Also, diagonals of a rectangle bisect each other and are equal. OA = OB = OC = ODBD...

Prove that the circles described with the four sides of a rhombus, as diameters,

Question: Prove that the circles described with the four sides of a rhombus, as diameters, pass through the point of intersection of its diagonals. Solution: Let ABCD be the rhombus with AC and BD as diagonals intersecting at point O. The diagonals of a rhombus bisect each other at right angles.i.e., AOB = BOC = COD = AOD = 90Now, circles with AB, BC, CD and DA as diameter passes through O (angle in a semi-circle is a right angle).Hence, the circle with four sides of a rhombus as diameter, pass ...

The points (3, −4) and (−6, 2) are the extremities of a diagonal

Question: The points (3, 4) and (6, 2) are the extremities of a diagonal of a parallelogram. If the third vertex is (1,3). Find the coordinates of the fourth vertex. Solution: Let ABCD be a parallelogram in which the co-ordinates of the vertices are A (3,4); B (1,3) and C (6, 2). We have to find the co-ordinates of the forth vertex. Let the forth vertex be Since ABCD is a parallelogram, the diagonals bisect each other. Therefore the mid-point of the diagonals of the parallelogram will coincide. ...

If second term of a G.P. is 2 and the sum of its infinite terms is 8,

Question: If second term of a G.P. is 2 and the sum of its infinite terms is 8, then its first term is (a) 1/4 (b) 1/2 (c) 2 (d) 4 Solution: (d) 4 $a_{2}=2$ $\therefore a r=2 \quad \ldots \ldots \ldots(\mathrm{i})$ Also, $S_{\infty}=8$ $\Rightarrow \frac{a}{(1-r)}=8$ $\Rightarrow \frac{a}{\left(1-\frac{2}{a}\right)}=8 \quad[\mathrm{U} \sin \mathrm{g}(\mathrm{i})]$ $\Rightarrow a^{2}=8(a-2)$ $\Rightarrow a^{2}-8 a+16=0$ $\Rightarrow(a-4)^{2}=0$ $\Rightarrow a=4$...

Prove that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.

Question: Prove that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent. Solution: Let ABCD be the cyclic quadrilateral and PO, QO, RO and SO be the perpendicular bisectors of sides AB, BC, CD and AD. We know that the perpendicular bisector of a chord passes through the centre of the circle.Since, AB, BC, CD and AD are the chords of a circle, PO, QO, RO and SO pass through the centre.i.e., PO, QO, RO and SO are concurrent.Hence, the perpendicular bisectors of the s...

Three consecutive vertices of a parallelogram are

Question: Three consecutive vertices of a parallelogram are (2,1), (1, 0) and (4, 3). Find the fourth vertex. Solution: Let ABCD be a parallelogram in which the co-ordinates of the vertices are A (2,1); B (1, 0) and C (4, 3). We have to find the co-ordinates of the forth vertex. Let the forth vertex be Since ABCD is a parallelogram, the diagonals bisect each other. Therefore the mid-point of the diagonals of the parallelogram will coincide. Now to find the mid-pointof two pointsandwe use section...

In a legislative assembly election,

Question: In a legislative assembly election, a political group hired a public relations firm to promote its candidates in three ways: telephone, house calls and letters. The cost per contact (in paise) is given matrixAas The number of contacts of each type made in two cities $X$ and $Y$ is given in matrix $B$ as Find the total amount spent by the group in the two citiesXandY. Solution: The cost per contact(inpaise)inpaiseis given by $A=\left[\begin{array}{c}40 \\ 100 \\ 50\end{array}\right] \be...

In the given figure, ABCD is a quadrilateral in which AD = BC and ∠ADC = ∠BCD.

Question: In the given figure,ABCDis a quadrilateral in whichAD=BCandADC= BCD. Show that the pointsA,B,C,Dlie on a circle. Solution: ABCD is a quadrilateral in which AD = BC andADC = BCD.Draw DE AB and CF AB.In ΔADE and ΔBCF, we have:ADE =ADC - 90 =BCD -90 =BCF (Given: ADC = BCD)AD = BC (Given)andAED = BCF =90 ΔADE ΔBCF (By AAS congruency)⇒A = BNow,A + B + C + D = 360⇒ 2B+2D= 360⇒B + D = 180Hence, ABCD is a cyclic quadrilateral....

The nth term of a G.P. is 128 and the sum of its n terms is 225.

Question: Thenth term of a G.P. is 128 and the sum of itsnterms is 225. If its common ratio is 2, then its first term is (a) 1 (b) 3 (c) 8 (d) none of these Solution: $a_{n}=128, S_{n}=225$ and $r=2$ $a_{n}=128$ $\therefore a r^{(n-1)}=128$ $\Rightarrow 2^{(n-1)} a=128$ $\Rightarrow \frac{2^{n} a}{2}=128$ $\Rightarrow 2^{n}=\frac{256}{a} \quad \ldots \ldots \ldots(\mathrm{i})$ Also, $S_{n}=225$ $\Rightarrow a\left(\frac{r^{n}-1}{r-1}\right)=225$ $\Rightarrow a\left(\frac{2^{n}-1}{2-1}\right)=225...

Prove that the points (3, −2), (4, 0), (6, −3)

Question: Prove that the points (3, 2), (4, 0), (6, 3) and (5, 5) are the vertices of a parallelogram. Solution: Let A (3,2); B (4, 0); C (6,3) and D (5,5) be the vertices of a quadrilateral. We have to prove that the quadrilateral ABCD is a parallelogram. We should proceed with the fact that if the diagonals of a quadrilateral bisect each other than the quadrilateral is a parallelogram. Now to find the mid-pointof two pointsandwe use section formula as, $\mathrm{P}(x, y)=\left(\frac{x_{1}+x_{2}...

In the given figure, ∠BAD = 75°, ∠DCF = x°

Question: In the given figure,BAD= 75, DCF=x and DEF=y. Find the values ofxandy. Solution: We know that if one side of a cyclic quadrilateral is produced, then the exterior angle is equal to the interior opposite angle.i.e.,BAD=DCF = 75⇒DCF =x= 75Again, the sum of opposite angles in a cyclic quadrilateral is 180.Thus,DCF + DEF = 180⇒75 +y= 180⇒y= (180 - 75) = 105Hence,x= 75 andy= 105...

Find the coordinates of the point where the diagonals

Question: Find the coordinates of the point where the diagonals of the parallelogram formed by joining the points (2, 1), (1, 0), (4, 3) and(1, 2) meet. Solution: The co-ordinates of the midpoint $\left(x_{n}, y_{E}\right)$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by, $\left(x_{w}, y_{m}\right)=\left(\left(\frac{x_{1}+x_{2}}{2}\right),\left(\frac{y_{1}+y_{2}}{2}\right)\right)$ In a parallelogram the diagonals bisect each other. That is the point of ...

If the sum of first two terms of an infinite GP is 1 every term is twice the sum of all the successive terms,

Question: If the sum of first two terms of an infinite GP is 1 every term is twice the sum of all the successive terms, then its first term is (a) 1/3 (b) 2/3 (c) 1/4 (d) 3/4 Solution: (d) $3 / 4$ Let the terms of the G.P. be $a, a_{2}, a_{3}, a_{4}, a_{5}, \ldots, \infty$. And, let the common ratio be $r$. Now, $a+a_{2}=1$ $\therefore a+a r=1 \quad \ldots \ldots$ (i) Also, $a=2\left(a_{2}+a_{3}+a_{4}+a_{5}+\ldots \infty\right)$ $\Rightarrow a=2\left(a r+a r^{2}+a r^{3}+a r^{4}+\ldots \infty\rig...

In the given figure, AB and CD are two parallel chords of a circle. If BDE and ACE are straight lines, intersecting at E,

Question: In the given figure,ABandCDare two parallel chords of a circle. IfBDEandACEare straight lines, intersecting atE, prove that∆AEBis isosceles. Solution: AB and CD are two parallel chords of a circle. BDE and ACE are two straight lines that intersect at E.If one side of a cyclic quadrilateral is produced, then the exterior angle is equal to the interior opposite angle. Exterior EDC = A ...(i) Exterior DCE = B ...(ii)Also, AB parallel to CD.Then, EDC = B (Corresponding angles)and DCE = A (...

Find the points of trisection of the line segment joining the points:

Question: Find the points of trisection of the line segment joining the points: (a) 5, 6 and (7, 5),(b) (3, 2) and (3, 4),(c) (2, 2) and (7, 4). Solution: The co-ordinates of a point which divided two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ internally in the ratio $m: n$ is given by the formula, $(x, y)=\left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\right)$ The points of trisection of a line are the points which divide the line into the ratio. (i) Here ...

The cooperative stores of a particular school has 10 dozen physics books,

Question: The cooperative stores of a particular school has 10 dozen physics books, 8 dozen chemistry books and 5 dozen mathematics books. Their selling prices are Rs. 8.30, Rs. 3.45 and Rs. 4.50 each respectively. Find the total amount the store will receive from selling all the items. Solution: Stock of various types of books in the store is given by Selling price of various types of books in the store is given by $Y=\left[\begin{array}{l}8.30 \\ 3.45 \\ 4.50\end{array}\right] \begin{gathered}...

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