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

In the given figure ∆ABC is an isosceles triangle in which AB = AC and a circle passing through B and C intersects AB and AC at D and E respectively.

Question: In the given figure∆ABCis an isosceles triangle in whichAB=ACand a circle passing throughBandCintersectsABandACatDandErespectively. Prove thatDE||BC. Solution: ABC is an isosceles triangle.Here, AB = AC ACB = ABC ...(i)So, exterior ADE = ACB= ABC [from(i)] ADE = ABC (Corresponding angles)Hence,DE || BC...

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

Question: In the given figure,Ois the centre of a circle. IfAOD= 140 and CAB= 50, calculate(i) EDB,(ii) EBD. Solution: O is the centre of the circle whereAOD= 140andCAB= 50.(i)BOD= 180AOD= (180 140)= 40We have the following:OB = OD(Radii of a circle)OBD =ODBIn ΔOBD, we have:BOD +OBD +ODB= 180⇒BOD +OBD +OBD =180 [∵OBD =ODB]⇒ 40+2OBD= 180⇒ 2OBD= (180 40) = 140⇒OBD= 70SinceABCDis a cyclic quadrilateral, we have:CAB+BDC= 180⇒CAB+ODB+ODC= 180⇒ 50+ 70+ODC= 180⇒ODC= (180 120) = 60ODC= 60EDB= (180 (ODC+...

Three shopkeepers A, B and C go to a store to buy stationary. A purchases 12 dozen notebooks,

Question: Three shopkeepersA,BandCgo to a store to buy stationary.Apurchases 12 dozen notebooks, 5 dozen pens and 6 dozen pencils.Bpurchases 10 dozen notebooks, 6 dozen pens and 7 dozen pencils.Cpurchases 11 dozen notebooks, 13 dozen pens and 8 dozen pencils. A notebook costs 40 paise, a pen costs Rs. 1.25 and a pencil costs 35 paise. Use matrix multiplication to calculate each individual's bill. Solution: Here, Cost of notebooks per dozen $=(12 \times 40)$ paise $=$ Rs $4.80$ Cost of pens per d...

Find the coordinates of the point which divides

Question: Find the coordinates of the point which divides the line segment joining (1,3) and (4, 7) internally in the ratio 3 : 4 Solution: We have A (1, 3) and B (4,7) be two points. Let a pointdivide the line segment joining the points A and B in the ratio 3:4 internally. Now according to the section formula if point a point P divides a line segment joiningandin the ratio m: n internally than, $\mathrm{P}(x, y)=\left(\frac{m x_{1}+m x_{2}}{m+n}, \frac{m y_{1}+m y_{2}}{m+n}\right)$ Now we will ...

The sum of an infinite G.P. is 4 and the sum of the cubes of its terms is 92.

Question: The sum of an infinite G.P. is 4 and the sum of the cubes of its terms is 92. The common ratio of the original G.P. is (a) 1/2 (b) 2/3 (c) 1/3 (d) 1/2 Solution: (a) 1/2 Let the G.P. be $a, a r, a r^{2}, a r^{3}, \ldots, \infty$. $S_{\infty}=4$ $\Rightarrow \frac{a}{1-r}=4$ ...(i) Also, sum of the cubes, $S_{1}=92$ $\Rightarrow \frac{a^{3}}{\left(1-r^{3}\right)}=92$ ...(ii) Putting the value of $a$ from $(\mathrm{i})$ to $(\mathrm{ii})$ : $\Rightarrow \frac{(4(1-r))^{3}}{\left(1-r^{3}\r...

Two chords AB and CD of a circle intersect each other at P outside the circle.

Question: Two chordsABandCDof a circle intersect each other atPoutside the circle. IfAB= 6 cm,BP= 2 cm andPD= 25 cm, findCD. Solution: ABandCDare two chords of a circle which intersect each other atPoutside the circle.AB= 6 cm,BP= 2 cm andPD= 2.5 cmAP BP = CP DP⇒ 8 2 = (CD + 2.5) 2.5 [∵CP = CD + DP]LetCD=xcmThus, 8 2 = (CD+ 2.5) 2.5⇒ 16 = 2.5x+ 6.25⇒ 2.5x= (16 - 6.25) = 9.75 $\Rightarrow x=\frac{9.75}{2.5}=3.9$ Hence,CD= 3.9 cm...

Find a relation between x and y such that the point (x, y)

Question: Find a relation betweenxandysuch that the point (x,y) is equidistant from the points (3, 6) and (3, 4). 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 $d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$ Let the three given points beP(x, y), A(3,6)andB(3,4). Now let us find the distance between P and A. $P A=\sqrt{(x-3)^{2}+(y-6)^{2}}$ Now, let us find the distance between P and B. $P B=\...

In the given figure, AB is a diameter of a circle with centre O and DO || CB.

Question: In the given figure,ABis a diameter of a circle with centreOandDO||CB.IfBCD= 120, calculate(i) BAD(ii) ABD(iii) CBD(iv) ADC.Also, show that ∆OADis an equilateral triangle. Solution: We have,ABis a diameter of the circle whereOis the centre,DO || BC andBCD= 120.(i)SinceABCDis a cyclic quadrilateral, we have:BCD +BAD= 180⇒ 120+BAD= 180⇒BAD= (180 120) = 60BAD= 60(ii)BDA= 90(Angle in a semicircle)In ΔABD,we have:BDA +BAD +ABD= 180⇒ 90+ 60+ABD= 180⇒ABD= (180 150)= 30ABD= 30(iii)OD = OA(Radi...

Let

Question: Let $A=\left[\begin{array}{lll}1 1 1 \\ 3 3 3\end{array}\right], B=\left[\begin{array}{rr}3 1 \\ 5 2 \\ -2 4\end{array}\right]$ and $C=\left[\begin{array}{rr}4 2 \\ -3 5 \\ 5 0\end{array}\right]$. Verify that $A B=A C$ though $B \neq C, A \neq O$. Solution: Here,$A=\left[\begin{array}{lll}1 1 1 \\ 3 3 3\end{array}\right]$ $B=\left[\begin{array}{cc}3 1 \\ 5 2 \\ -2 4\end{array}\right]$ and $C=\left[\begin{array}{cc}4 2 \\ -3 5 \\ 5 0\end{array}\right]$ Now, $A B=\left[\begin{array}{lll}...

Find a point on y-axis which is equidistant

Question: Find a point ony-axis which is equidistant form the points (5, 2) and (3, 2). 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 the yaxis which is equidistant from both the pointsA(5,2) andB(3,2). Let this point be denoted asC(x, y). Since the point lies on they-axis the value of its ordinate will be 0. O...

In the given figure, sides AD and AB of cyclic quadrilateral ABCD are produced to E and F respectively.

Question: In the given figure, sidesADandABof cyclic quadrilateralABCDare produced toEandFrespectively. IfCBF= 130 and CDE=x, find the value ofx. Solution: ABCDis a cyclic quadrilateral.We know that in a cyclic quadrilateral, the exterior angle is equal to the interior opposite angle.CBF=CDA⇒CBF= (180x)⇒ 130= 180x [∵CBF= 130]⇒x= (180 130) = 50Hence,x= 50...

The value of

Question: The value of 91/3. 91/9. 91/27... upto inf, is (a) 1 (b) 3 (c) 9 (d) none of these Solution: $(\mathrm{b}) 3$ $9^{\frac{1}{3}} \times 9^{\frac{1}{9}} \times 9^{\frac{1}{27}} \times \ldots \infty$ $=9\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\ldots \infty\right)$ Here, it is a G. P. with $a=\frac{1}{3}$ and $r=\frac{1}{3}$. $\therefore 9^{\left(\frac{\frac{1}{3}}{1-\frac{1}{3}}\right)}$ $=9^{\left(\frac{1}{2}\right)}=3$...

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

Question: In the given figure,Ois the centre of the circle andDAB= 50. Calculate the values ofxandy Solution: Ois the centre of the circle andDAB= 50.OA = OB(Radii of a circle)⇒OBA =OAB= 50In ΔOAB, we have:OAB+OBA+AOB= 180⇒ 50+ 50+AOB= 180⇒AOB= (180 100)= 80SinceAODis a straight line, we have:x= 180AOB= (180 80)= 100i.e.,x= 100The opposite angles of a cyclic quadrilateral are supplementary.ABCDis a cyclic quadrilateral.Thus,DAB+BCD= 180BCD= (180 50) = 130y= 130Hence,x= 100andy= 130...

If pth, qth and rth terms of an A.P. are in G.P., then the common ratio of this G.P. is

Question: Ifpth,qth andrth terms of an A.P. are in G.P., then the common ratio of this G.P. is (a) $\frac{p-q}{q-r}$ (b) $\frac{q-r}{p-q}$ (c)pqr (d) none of these Solution: (b) $\frac{q-r}{p-q}$ Letabe the first term anddbe the common difference of the given A.P. Then, we have: $\mathrm{p}^{\text {th }}$ term, $a_{p}=a+(p-1) d$ $q^{\text {th }}$ term, $a_{q}=a+(q-1) d$ $r^{\text {th }}$ term, $a_{r}=a+(r-1) d$ Now, according to the question the $p^{\text {th }}$, the $q^{\text {th }}$ and the $...

If pth, qth and rth terms of an A.P. are in G.P., then the common ratio of this G.P. is

Question: Ifpth,qth andrth terms of an A.P. are in G.P., then the common ratio of this G.P. is (a) $\frac{p-q}{q-r}$ (b) $\frac{q-r}{p-q}$ (c)pqr (d) none of these Solution: (b) $\frac{q-r}{p-q}$ Letabe the first term anddbe the common difference of the given A.P. Then, we have: $\mathrm{p}^{\text {th }}$ term, $a_{p}=a+(p-1) d$ $q^{\text {th }}$ term, $a_{q}=a+(q-1) d$ $r^{\text {th }}$ term, $a_{r}=a+(r-1) d$ Now, according to the question the $p^{\text {th }}$, the $q^{\text {th }}$ and the $...

In the seating arrangement of desks in a classroom three students Rohini,

Question: In the seating arrangement of desks in a classroom three students Rohini, Sandhya and Bina are seated at A(3, 1), B(6, 4), and C(8, 6). Do you think they are seated in a line? Solution: The distancedbetween two pointsandis given by the formula $d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$ For three points to be collinear the sum of distances between any two pairs of points should be equal to the third pair of points. The given points areA(3,1),B(6,4) andC(8,6). L...

In the given figure, O is the centre of a circle and ∠BOD = 150°.

Question: In the given figure,Ois the centre of a circle andBOD= 150. Find the values ofxandy. Solution: Ois the centre of the circle andBOD= 150.Thus, reflex angleBOD= (360 150) = 210 Now, $x=\frac{1}{2}(\operatorname{reflex} \angle B O D)=\left(\frac{1}{2} \times 210^{\circ}\right)=105^{\circ}$ x= 105Again,x+y= 180(Opposite angles of a cyclic quadrilateral)⇒ 105+y= 180⇒y= (108- 105)= 75y= 75Hence,x= 105 andy= 75...

If A and B are square matrices of the same order such that AB = BA, then show that

Question: If $A$ and $B$ are square matrices of the same order such that $A B=B A$, then show that $(A+B)^{2}=A^{2}+2 A B+B^{2}$. Solution: $(A+B)^{2}=(A+B)(A+B)$ $=A^{2}+A B+B A+B^{2}$ $=A^{2}+2 A B+B^{2}$$(\because A B=B A)$ Hence, $(A+B)^{2}=A^{2}+2 A B+B^{2}$....

In the adjoining figure, ABCD is a cyclic quadrilateral in which ∠BCD = 100°

Question: In the adjoining figure,ABCDis a cyclic quadrilateral in whichBCD= 100 and ABD= 50. Find ADB. Solution: Given:ABCDis a cyclic quadrilateral.DAB+DCB= 180 ( Opposite angles of a cyclic quadrilateral are supplementary)⇒DAB+ 100= 180⇒DAB= (180 100) = 80Now, in ΔABD,we have:⇒DAB +ABD +ADB= 180⇒ 80+ 50+ADB= 180⇒ADB= (180 130) = 50Hence,ADB= 50...

Let A and B be square matrices of the order 3 × 3.

Question: Let $A$ and $B$ be square matrices of the order $3 \times 3$. Is $(A B)^{2}=A^{2} B^{2} ?$ Give reasons. Solution: Yes, $(A B)^{2}=A^{2} B^{2}$ if $A B=B A$ If $A B=B A$, then $(A B)^{2}=(A B)(A B)$ $=A(B A) B$ (associative law) $=A(A B) B$ $=A^{2} B^{2}$...

In the given figure, ∆ABC is equilateral.

Question: In the given figure,∆ABCis equilateral. Find (i) BDC, (ii) BEC. Solution: (i)Given: ΔABCis an equilateral triangle.i.e., each of its angle = 60⇒BAC =ABC =ACB= 60Angles in the same segment of a circle are equal.i.e.,BDC =BAC= 60BDC= 60(ii)The opposite angles of a cyclic quadrilateral are supplementary.Then in cyclic quadrilateralABEC,we have:BAC +BEC= 180⇒ 60 +BEC= 180⇒BEC= (180 60) = 120BDC= 60 andBEC= 120...

The fractional value of 2.357 is

Question: The fractional value of 2.357 is (a) 2355/1001 (b) 2379/997 (c) 2355/999 (d) none of these Solution: (C) $\frac{2355}{999}$ $2 . \overline{357}=2.0+0.357+0.000357+0.000000357+\ldots \infty$ $\Rightarrow 2 . \overline{357}=2+\left[\frac{357}{10^{3}}+\frac{357}{10^{6}}+\frac{357}{10^{9}}+\ldots \infty\right]$ $\Rightarrow 2 . \overline{357}=2+\frac{\frac{357}{10^{3}}}{1-\frac{1}{10^{2}}}$ $\Rightarrow 2 . \overline{357}=2+\frac{357}{999}$ $\Rightarrow 2 . \overline{357}=\frac{2355}{999}$...

In the given figure, O is the centre of the given circle and measure of arc ABC is 100°.

Question: In the given figure,Ois the centre of the given circle and measure of arcABCis 100. Determine ADCand ABC. Solution: We know that the angle subtended by an arc is twice the angle subtended by it on the circumference in the alternate segment.Thus,AOC = 2ADC⇒ 100= 2ADCADC =50The opposite angles of a cyclic quadrilateral are supplementary and ABCD is a cyclic quadrilateral.Thus,ADC +ABC= 180⇒ 50+ABC= 180⇒ABC= (180 50) = 130ADC= 50andABC= 130...

If A and B are square matrices of the same order, explain, why in general

Question: IfAandBare square matrices of the same order, explain, why in general (i) $(A+B)^{2} \neq A^{2}+2 A B+B^{2}$ (ii) $(A-B)^{2} \neq A^{2}-2 A B+B^{2}$ (iii) $(A+B)(A-B) \neq A^{2}-B^{2}$. Solution: (i) $\mathrm{LHS}=(A+B)^{2}$ $=(A+B)(A+B)$ $=A(A+B)+B(A+B)$ $=A^{2}+A B+B A+B^{2}$ We know that a matrix does not have commutative property. So, ABBAThus, $(A+B)^{2} \neq A^{2}+2 A B+B^{2}$ (ii) LHS $=(A-B)^{2}$ $=(A-B)(A-B)$ $=A(A-B)-B(A-B)$ $=A^{2}-A B-B A+B^{2}$ We know that a matrix does n...

If A and B are square matrices of the same order, explain, why in general

Question: IfAandBare square matrices of the same order, explain, why in general (i) $(A+B)^{2} \neq A^{2}+2 A B+B^{2}$ (ii) $(A-B)^{2} \neq A^{2}-2 A B+B^{2}$ (iii) $(A+B)(A-B) \neq A^{2}-B^{2}$. Solution: (i) $\mathrm{LHS}=(A+B)^{2}$ $=(A+B)(A+B)$ $=A(A+B)+B(A+B)$ $=A^{2}+A B+B A+B^{2}$ We know that a matrix does not have commutative property. So, ABBAThus, $(A+B)^{2} \neq A^{2}+2 A B+B^{2}$ (ii) LHS $=(A-B)^{2}$ $=(A-B)(A-B)$ $=A(A-B)-B(A-B)$ $=A^{2}-A B-B A+B^{2}$ We know that a matrix does n...

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