Maths

### In the following figure, XY is parallel to BC, AX = 9 cm, XB = 4.5 cm and BC = 18 cm. Find: XY

Solution: According to the given question, $XY\text{ }||\text{ }BC$ So, In $\Delta \text{ }AXY\text{ }and\text{ }\Delta \text{ }ABC$ $\angle AXY\text{ }=\angle ABC$ [Corresponding angles]...

### In the following figure, XY is parallel to BC, AX = 9 cm, XB = 4.5 cm and BC = 18 cm. Find: (i) AY/YC (ii) YC/AC

Solution: According to the given question, $XY\text{ }||\text{ }BC$ So, In $\Delta \text{ }AXY\text{ }and\text{ }\Delta \text{ }ABC$ $\angle AXY\text{ }=\angle ABC$[Corresponding angles]...

### In the following figure, point D divides AB in the ratio 3: 5. If BC = 4.8 cm, find the length of DE.

Solution: Because, $\vartriangle ADE\text{ }\sim\text{ }\vartriangle ABC\text{ }by\text{ }AA$ criterion for similarity So, we have $AD/AB\text{ }=\text{ }DE/BC$ $3/8\text{ }=\text{ }DE/4.8$...

### Describe: i) The locus of points at distances less than 3 cm from a given point. ii) The locus of points at distances greater than 4 cm from a given point.

i) The locus of the points will be the space inside of the circle whose radius is $3\text{ }cm$and centre as the given point. ii) The locus of the points will be the space outside of the circle...

### The speed of sound is 332 meters per second. A gun is fired. Describe the locus of all the people on the Earth’s surface, who hear the sound exactly one second later.

The locus of all the people on Earth’s surface is the circumference of a circle whose radius is $332\text{ }m$and centre is the point where the gun is fired.

### Describe the locus for the locus of a point in rhombus ABCD, so that it is equidistant from i) AB and BC; ii) B and D.

(i) The locus of the point in a rhombus $ABCD$which is equidistant from $AB\text{ }and\text{ }BC$ will be the diagonal $BD$of the rhombus.   (ii) The locus of the point in a rhombus...

### Describe the locus for the locus of a point P, so that: AB^2 = AP^2 + BP^2, where A and B are two fixed points.

The locus of the point $P$is the circumference of a circle with $AB$ as diameter and satisfies the condition $A{{B}^{2}}~=\text{ }A{{P}^{2~}}+\text{ }B{{P}^{2}}$

### Describe the locus for the locus of a point in space which is always at a distance of 4 cm from a fixed point.

The locus of a point in space is the surface of the sphere whose centre is the fixed point and radius equal to $4\text{ }cm.$

### Describe the locus for the locus of vertices of all isosceles triangles having a common base.

The locus of vertices of all isosceles triangles having a common base will be the perpendicular bisector of the common base of the triangles.

### Describe the locus for the locus of the centres of all circles passing through two fixed points.

The locus of the centres of all the circles passing through two fixed points will be the perpendicular bisector of the line segment joining the two given fixed points.

### The locus of a points inside a circle and equidistant from two fixed points on the circumference of the circle.

The locus of the points inside the circle which are equidistant from the fixed points on the circumference of a circle will be a diameter which is the perpendicular bisector of the line joining the...

### Describe the locus for The locus of the door-handle, as the door opens.

The locus of the door handle will be the circumference of a circle with centre at the axis of rotation of the door and the radius equal to distance between the door handle and the axis of rotation...

### The locus of a runner, running around a circular track and always keeping a distance of 1.5 m from the inner edge.

The locus of the runner, running around a circular track and always keeping a distance of $1.5\text{ }m$ from the inner edge will be the circumference of a circle where the radius is equal to the...

### The locus of a stone dropped from the top of a tower.

Solution: As per the given question, Locus of stone is dropped from the top of tower will be vertical line through the point from which the stone is dropped.

### The locus of the moving end of the minute hand of a clock.

The locus of the moving end of the minute hand of the clock will be a circle whose radius will be the length of the minute hand.

### The locus of the centre of a wheel of a bicycle going straight along a level road.

The locus of the centre of wheel, is going straight along the level road will be a straight line parallel to the road at a distance equal to the radius of the wheel.

### The locus of a points at a distance of 2 cm from a fixed line.

The locus of points which are at a distance of $2\text{ }cm$from a fixed line $AB$are a pair of straight lines $l\text{ }and\text{ }m$which are parallel to the given line at a distance of...

### The locus of a point at a distance of 3 cm from a fixed point.

The locus of a point is at a distance of $3\text{ }cm$from a fixed point is circumference of a circle whose radius is $3\text{ }cm$and the fixed point is the centre of the circle.

### In the given figure, AB is a side of a regular six-sided polygon and AC is a side of a regular eight-sided polygon inscribed in the circle with centre O. calculate the sizes of: ∠ABC.

Solution: According to the given question, $AC$ is the side of a regular octagon, $\angle AOC\text{ }=\text{ }{{360}^{o}}/\text{ }8\text{ }=\text{ }{{45}^{o}}$ Hence, $arc\text{ }AC$subtends...

### In the given figure, AB is a side of a regular six-sided polygon and AC is a side of a regular eight-sided polygon inscribed in the circle with centre O. calculate the sizes of: (i) ∠AOB, (ii) ∠ACB,

Solution: (i) $Arc\text{ }AB$ subtends$\angle AOB$at the centre and $\angle ACB$at the remaining part of the circle. $\angle ACB\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\angle AOB$...

### The given figure show a circle with centre O. Also, PQ = QR = RS and ∠PTS = 75°. Calculate: ∠PQR.

Join $OP,\text{ }OQ,\text{ }OR\text{ }and\text{ }OS$ Given, $PQ\text{ }=\text{ }QR\text{ }=\text{ }RS$ So, $\angle POQ\text{ }=\angle QOR\text{ }=\angle ROS$ [Equal chords subtends equal...

### The given figure show a circle with centre O. Also, PQ = QR = RS and ∠PTS = 75°. Calculate: (i) ∠POS, (ii) ∠QOR,

Solution: Join $OP,\text{ }OQ,\text{ }OR\text{ }and\text{ }OS$ Given, $PQ\text{ }=\text{ }QR\text{ }=\text{ }RS$ So, $\angle POQ\text{ }=\angle QOR\text{ }=\angle ROS$ [Equal chords subtends...

### Calculate BC.

SOLUTION: In ∆ADC, $\begin{array}{*{35}{l}} CD/AD\text{ }=\text{ }tan\text{ }{{42}^{o}} \\ CD\text{ }=\text{ }20\text{ x }0.9004\text{ }=\text{ }18.008\text{ }m \\ \end{array}$ In ∆ADB,...

### In the following diagram, AB is a floor-board; PQRS is a cubical box with each edge = 1 m and ∠B = 60o. Calculate the length of the board AB.

SOLUTION: In ∆PSB, $\begin{array}{*{35}{l}} PS/PB\text{ }=\text{ }sin\text{ }{{60}^{o}} \\ PB\text{ }=\text{ }2/\text{ }\surd 3\text{ }=\text{ }1.155\text{ }m \\ \end{array}$ In ∆APQ,...

### If p – 15 = 0 and 2×2 + px + 25 = 0: find the values of x.

According to the given question, $p\text{ }\text{ }15\text{ }=\text{ }0$ And $2{{x}^{2}}~+\text{ }px\text{ }+\text{ }25\text{ }=\text{ }0$ Thus, $p\text{ }=\text{ }15$ Using$p$in the...

### Solve:

According to the given question, $\left( x\text{ }+\text{ }5 \right)\text{ }\left( x\text{ }\text{ }5 \right)\text{ }=\text{ }24$ Or, ${{x}^{2}}~\text{ }25\text{ }=\text{ }24$ Or,...

### Solve: (x^2+1/x^2)-3(x-1/x)-2=0

Let $\left( x\text{ }\text{ }1/x \right)\text{ }=\text{ }y\text{ }\ldots .\text{ }\left( 1 \right)$ squaring on both sides ${{\left( x\text{ }\text{ }1/x \right)}^{2}}~=\text{ }{{y}^{2}}$ or,...

### Solve: 2(x^2+1/x^2)-(x+1/x)=11

let $\left( x\text{ }+\text{ }1/x \right)\text{ }=\text{ }y\text{ }\ldots .\text{ }\left( 1 \right)$ squaring on both sides ${{\left( x\text{ }+\text{ }1/x \right)}^{2}}~=\text{ }{{y}^{2}}$...

### Solve:9(x^2+1/x^2)-9(x+1/x)-52=0

Let, $~\left( x\text{ }+\text{ }1/x \right)\text{ }=\text{ }y\text{ }\ldots .\text{ }\left( 1 \right)$ squaring both sides ${{\left( x\text{ }+\text{ }1/x \right)}^{2}}~=\text{ }{{y}^{2}}$...

### If given matrix, find the matrix X such that: A + X = 2B + C

Solution: As per the given question,

### Solve: (i) A (BA) (ii) (AB) B.

Solution: $\left( i \right)\text{ }A\text{ }\left( BA \right)$ $\left( ii \right)\text{ }\left( AB \right)\text{ }B$

### 3A x M = 2B; find matrix M.

Solution: According to the given question, $3A\text{ }x\text{ }M\text{ }=\text{ }2B$ Suppose the order of the $matrix\text{ }M\text{ }be\text{ }\left( a\text{ }x\text{ }b \right)$ Now, we know...

### Two trains leave a railway station at the same time. The first train travels due west and the second train due north. The first train travels 5 km/hr faster than the second train. If after 2 hours, they are 50 km apart, find the speed of each train.

Let the speed of the second train be $~x\text{ }km/hr.$ Then, the speed of the first train is $\left( x\text{ }+\text{ }5 \right)\text{ }km/hr$ Let O be the position of the railway station,...

### Solve: (i) A – B (ii) A^2

Solution: $\left( \mathbf{i} \right)\text{ }\mathbf{A}\text{ }-\text{ }\mathbf{B}\text{ }$ $\text{ }\left( \mathbf{ii} \right)\text{ }{{\mathbf{A}}^{\mathbf{2}~}}$

### BA = M^2, find the values of a and b.

Solution: $BA$ ${{M}^{2}}$ So, $BA\text{ }={{M}^{2}}$ On comparison, we get $-2b\text{ }=\text{ }-2$ $b\text{ }=\text{ }1$ And, $a\text{ }=\text{ }2$

### If the given matrix and I is a unit matrix of the same order as that of M; show that: M2 = 2M + 3I

Solution: ${{M}^{2}}$ $2M\text{ }+\text{ }3I$ Hence, ${{M}^{2}}~=\text{ }2M\text{ }+\text{ }3I$

### Is the following possible: A^2

Solution: ${{A}^{2}}~=\text{ }A\text{ }x\text{ }A,\text{ }$isn’t possible because the number of columns isn’t equal to its number of rows in matrix A.

### Rs 6500 was divided equally among a certain number of persons. Had there been 15 persons more, each would have got Rs 30 less. Find the original number of persons.

let the original number of persons to be x. According to ques, Total money which was divided is $=\text{ }Rs\text{ }6500$ Each person’s share is $=\text{ }Rs\text{ }6500/x$ Then, as the question...

### Is the following possible: (i) AB (ii) BA

Solution: $\left( i \right)\text{ }AB$ $\left( ii \right)\text{ }BA$

### If find x and y when x and y when A2 = B.

Solution: ${{A}^{2}}~$ ${{A}^{2}}~=\text{ }B$ On comparison, we get $4x\text{ }=\text{ }16$ $x\text{ }=\text{ }4$ And, $1\text{ }=\text{ }-y$ $y\text{ }=\text{ }-1$

### If the given matrix and I is a unit matrix of order 2×2, find: (i) A^2 (ii) B^2A

Solution: (i) ${{A}^{2}}$   (ii) $~{{B}^{2}}$ ${{B}^{2}}A$

### A trader buys x articles for a total cost of Rs 600. (i) Write down the cost of one article in terms of x. If the cost per article were Rs 5 more, the number of articles that can be bought for Rs 600 would be four less. (ii) Write down the equation in x for the above situation and solve it for x.

According to ques, Number of articles $=\text{ }x$ And, the total cost of articles $=\text{ }Rs\text{ }600$ Again, (i) Cost of one article $=\text{ }Rs\text{ }600/x$ (ii) also,...

### If the given matrix and I is a unit matrix of order 2×2, find: (i) AI (ii) IB

Solution: (i) AI = (ii) IB=

### If given matrix and I is a unit matrix of order 2×2, find: (i) AB (ii) BA

Solution: According to the given question (i) (ii)

### The distance by road between two towns A and B is 216 km, and by rail it is 208 km. A car travels at a speed of x km/hr and the train travels at a speed which is 16 km/hr faster than the car. Calculate: (i) the time taken by the car to reach town B from A, in terms of x; (ii) the time taken by the train to reach town B from A, in terms of x.

According to ques, Speed of car = $x\text{ }km/hr$ Speed of train = $\left( x\text{ }+\text{ }16 \right)\text{ }km/hr$ Time = $Distance/\text{ }Speed$ (i)Time taken by the car to reach town B...

### (i) find the matrix 2A + B. (ii) find a matrix C such that:

(ii) Solution: (i) $2A\text{ }+\text{ }B$ (ii)

### Solve:

Solution: According to the given question, the matrix is

### From given data below find (i) 2A – 3B + C (ii) A + 2C – B

Solution: $\left( i \right)\text{ }2A\text{ }-\text{ }3B\text{ }+\text{ }C$ $\left( ii \right)\text{ }A\text{ }+\text{ }2C\text{ }-\text{ }B$

### Find x and y if: (i) 3[4 x] + 2[y -3] = [10 0]

(ii) Solution: From L.H.S, we have $3\left[ 4\text{ }x \right]\text{ }+\text{ }2\left[ y\text{ }-3 \right]$ $=\text{ }\left[ 12\text{ }3x \right]\text{ }+\text{ }\left[ 2y\text{ }-6 \right]~$...

### Evaluate:

(I) (ii) Solution: According to the given ques, (i) (ii)

### Evaluate: (i) 3[5 -2]

(ii)   Solution: (i) $3\left[ 5\text{ }-2 \right]\text{ }=\text{ }\left[ 3\times 5\text{ }3x-2 \right]\text{ }=\text{ }\left[ 15\text{ }-6 \right]$ (ii)

### In a G.P., the ratio between the sum of first three terms and that of the first six terms is 125: 152. Find its common ratio.

According to the given ques, Hence, the common ratio is $3/5.$

### How many terms of the series 2 + 6 + 18 + ….. must be taken to make the sum equal to 728?

. According to the given question, G.P: $2\text{ }+\text{ }6\text{ }+\text{ }18\text{ }+\text{ }\ldots ..$ Here, $a\text{ }=\text{ }2$ And $r\text{ }=\text{ }6/2\text{ }=\text{ }3$ Also given,...

### The first term of a G.P. is 27 and its 8th term is 1/81. Find the sum of its first 10 terms.

$First\text{ }term\text{ }\left( a \right)\text{ }of\text{ }a\text{ }G.P\text{ }=\text{ }27$ And, ${{8}^{th}}~term\text{ }=\text{ }{{t}_{8}}~=\text{ }a{{r}^{8\text{ }-\text{ }1}}~=\text{ }1/81$...

### Prove that : (iii) (iv)

(iii)  $\sin \left(28^{\circ}+\mathrm{A}\right)=\sin \left[90^{\circ}-\left(62^{\circ}-\mathrm{A}\right)\right]=\cos \left(62^{\circ}-\mathrm{A}\right)$ (iv)

### Find the seventh term from the end of the series: √2, 2, 2√2, …… , 32

Given series: $\surd 2,\text{ }2,\text{ }2\surd 2,\text{ }\ldots \ldots \text{ },\text{ }32$ Here, $a\text{ }=\text{ }\surd 2$ $r\text{ }=\text{ }2/\text{ }\surd 2\text{ }=\text{ }\surd 2$...

### The product of 3rd and 8th terms of a G.P. is 243. If its 4th term is 3, find its 7th term

According to the given question, Product of ${{3}^{rd}}~and\text{ }{{8}^{th}}$ terms of a G.P. is $243$ The general term of a G.P. First term $a$ And Common ratio $r$is given by,...

(i) (ii) 1+1=2

### (i) If , show that: (ii) If , show that:

(i) Since, $2 \sin \mathrm{A}-1=0$ Therefore, $\sin A=1 / 2$ since, $\sin 30^{\circ}=1 / 2$ Hence, $\mathrm{A}=30^{\circ}$ LHS = $\sin 3 A=\sin 3\left(30^{\circ}\right)=\sin 30^{\circ}=1$...

### If tan A = n tan B and sin A = m sin B, prove that: cos^2 A = m^2 – 1/ n^2 – 1

$\tan A=n \tan B$ $n=\tan A / \tan B$ And, $\sin A=m \sin B$ $\mathrm{m}=\sin \mathrm{A} / \sin \mathrm{B}$ Substituting RHS in $\mathrm{m}$ and $\mathrm{n}$ $m^{2}-1 / n^{2}-1$...

### Which term of the G.P. :

Solution: In the given G.P. First term, $a\text{ }=\text{ }-10$ Common ratio, $r\text{ }=\text{ }\left( 5/\surd 3 \right)/\text{ }\left( -10 \right)\text{ }=\text{ }1/\left( -2\surd 3 \right)$...

### Prove : (xiii) (xiv)

LHS = = RHS (xiv) LHS = = RHS

### The mid-point of the line segment joining (2a, 4) and (-2, 2b) is (1, 2a + 1). Find the value of a and b.

According to the given question, The mid-point of $\left( 2a,\text{ }4 \right)\text{ }and\text{ }\left( -2,\text{ }2b \right)\text{ }is\text{ }\left( 1,\text{ }2a\text{ }+\text{ }1 \right)$ By...

### Prove : (xi) (xii)

LHS = = RHS (xii) LHS = = RHS

### Prove (ix) (x)

LHS= = RHS (x)  LHS =

### Prove : (vii) (viii)

(vii) LHS =$=(\sin A /(1-\cos A))-\cot A$ Since, $\cot A=\cos A / \sin A$ $=\left(\sin ^{2} A-\cos A+\cos ^{2} A\right) /(1-\cos A) \sin A$ $=(1-\cos A) /(1-\cos A) \sin A$ $=1 / \sin \mathrm{A}$...

### Prove: (v) (vi)

(v)  LHS= cot A/ (1 – tan A) + tan A/ (1 – cot A) = RHS (vi)  LHS= cos A/ (1 + sin A) + tan A = RHS