Maths

### Use trigonometrical tables to find tangent of: (i) 37° (ii) 42° 18′

(i) $\tan 37^{\circ}=0.7536$ (ii) $\tan 42^{\circ} 18^{\prime}=0.9099$

### Evaluate: (vii) (viii)

(vii) = 1 – 2 = -1 (viii)

### In the given figure, P (4, 2) is mid-point of line segment AB. Find the co-ordinates of A and B.

Solution: Suppose point A lies on x-axis, hence its co-ordinates can be $\left( x,\text{ }0 \right)$ And, Point B lies on y-axis, hence its co-ordinates can be $\left( 0,\text{ }y \right)$...

### Use ruler and compasses only for this question. (i) Construct ∆ABC, where AB = 3.5 cm, BC = 6 cm and ∠ABC = 60degree (ii) Construct the locus of points inside the triangle which are equidistant from BA and BC.

Steps of construction: (i) Draw line $BC\text{ }=\text{ }6\text{ }cm$ and construct angle $CBX\text{ }=\text{ }{{60}^{o}}$. Cut off $AB\text{ }=\text{ }3.5$. Join $AC$, triangle  $ABC$is...

### In triangle LMN, bisectors of interior angles at L and N intersect each other at point A. Prove that: i) point A is equidistant from all the three sides of the triangle. ii) AM bisects angle LMN.

Construction: $Join\text{ }AM$ Proof: (i) Since,$A$ lies on bisector of $\angle N$ So, $A$is equidistant from $MN\text{ }and\text{ }LN.$ Now again, as $A$ lies on the bisector...

### In parallelogram ABCD, side AB is greater than side BC and P is a point in AC such that PB bisects angle B. Prove that P is equidistant from AB and BC.

Construction: From $P$, draw $PL\bot AB~and~PM\bot BC$ Proof: In $\vartriangle PLB\text{ }and\text{ }\vartriangle PMB$, $\angle PLB\text{ }=\angle PMB\text{ }[Each\text{ }{{90}^{o}}]$...

### Construct a triangle ABC, in which AB = 4.2 cm, BC = 6.3 cm and AC = 5 cm. Draw perpendicular bisector of BC which meets AC at point D. Prove that D is equidistant from B and C.

Given: In triangle ABC, $AB\text{ }=\text{ }4.2\text{ }cm,\text{ }BC\text{ }=\text{ }6.3\text{ }cm\text{ }and\text{ }AC\text{ }=\text{ }5\text{ }cm$ Steps of Construction: i) Draw a line segment...

### ABCD is a cyclic quadrilateral in which AB and DC on being produced, meet at P such that PA = PD. Prove that AD is parallel to BC.

We assume that $ABCD$be the given cyclic quadrilateral. $PA\text{ }=\text{ }PD\text{ }\left[ Given \right]$ So, $\angle PAD\text{ }=\angle PDA\text{ }\ldots \ldots \text{ }\left( 1 \right)$...

### Two circles intersect at P and Q. Through P, a straight line APB is drawn to meet the circles in A and B. Through Q, a straight line is drawn to meet the circles at C and D. Prove that AC is parallel to BD.

Solution: Join $AC,\text{ }PQ\text{ }and\text{ }BD.$ As $ACQP$is a cyclic quadrilateral $\angle CAP\text{ }+\angle PQC\text{ }=\text{ }{{180}^{o}}~\ldots \ldots .\text{ }\left( i \right)$...

### Prove that: (i) the parallelogram, inscribed in a circle, is a rectangle. (ii) the rhombus, inscribed in a circle, is a square.

Solution: (i) Let’s expect that $ABCD$is a parallelogram which is inscribed in a circle. Hence, we know that $\angle BAD\text{ }=\angle BCD$  [Opposite angles of a parallelogram are equal] And...

### In the figure given below, ABCD is a cyclic quadrilateral in which ∠BAD = 75Degree; ∠ABD = 58degree and ∠ADC = 77degree. Find: (i) ∠BDC, (ii) ∠BCD,

Solution: (i) According to the given question, By angle sum property of triangle ABD, $\angle ADB\text{ }=\text{ }{{180}^{o}}-\text{ }{{75}^{o}}-\text{ }{{58}^{o}}~=\text{ }{{47}^{o}}$ Hence,...

### Calculate: ∠ACB.

Solution: According to the given question By angle sum property of a triangle we have $\angle ACB\text{ }=\text{ }{{180}^{o}}-\text{ }{{49}^{o}}-\text{ }{{43}^{o}}~=\text{ }{{88}^{o}}$

### Calculate: (i) ∠CDB, (ii) ∠ABC,

Solution: According to the given question We get, (i) $\angle CDB\text{ }=\angle BAC\text{ }=\text{ }{{49}^{o}}$ (ii) $\angle ABC\text{ }=\angle ADC\text{ }=\text{ }{{43}^{o}}$ [Angles subtended...

### In the figure, given below, O is the centre of the circle. If ∠AOB = 140degree and ∠OAC = 50degree; Find (i) ∠OAB, (ii) ∠CBA.

Solution: According to the given question, (i) In $\vartriangle AOB,$we have $OA\text{ }=\text{ }OB\text{ }\left( radii \right)$ Thus, $\angle OBA\text{ }=\angle OAB$ By angle sum property of...

### In each of the following figures, O is the center of the circle. Find the values of a, b, c and d.

SOLUTION: (i) In the following figure, $\angle BAD\text{ }=\text{ }{{90}^{o}}~$ [Angle in a semi-circle] Therefore, $\angle BDA\text{ }=\text{ }{{90}^{o}}-\text{ }{{35}^{o}}~=\text{ }{{55}^{o}}$...

### In each of the following figures, O is the centre of the circle. Find the values of a, b and c.

(i) (ii) Solution: (i)In the following figure, $b\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }x\text{ }{{130}^{o}}$ [Angle at the center is double the angle at the circumference...

### In the given figure, O is the center of the circle. ∠OAB and ∠OCB are 30Degree and 40Degree respectively. Find ∠AOC Show your steps of working.

Solution: Let’s join $AC.$ And, let $\angle OAC\text{ }=\angle OCA\text{ }=\text{ }x$ [Angles opposite to equal sides are equal] Thus, $\angle AOC\text{ }=\text{ }{{180}^{o}}-\text{ }2x$ Also,...

### If the sides of a parallelogram touch a circle, prove that the parallelogram is a rhombus.

Solution: Suppose a circle touch the sides $AB,\text{ }BC,\text{ }CD\text{ }and\text{ }DA$of parallelogram $ABCD\text{ }at\text{ }P,\text{ }Q,\text{ }R\text{ }and\text{ }S$respectively. Now,...

### If the sides of a quadrilateral ABCD touch a circle, prove that AB + CD = BC + AD.

Solution: Suppose, a circle touch the sides $AB,\text{ }BC,\text{ }CD\text{ }and\text{ }DA$ of quadrilateral $ABCD\text{ }at\text{ }P,\text{ }Q,\text{ }R\text{ }and\text{ }S$respectively. As,...

### Using the Remainder Theorem, factorise each of the following completely.

Let f(x) = ${{\mathbf{x}}^{\mathbf{3}}}~+\text{ }{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{4x}\text{ }\text{ }\mathbf{4}$ For $x\text{ }=\text{ }-1$, the value of f(x) will be...

### Using the Remainder Theorem, factorise each of the following completely.

Let f(x) = $\mathbf{4}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{7}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{36x}\text{ }\text{ }\mathbf{63}$ For $x\text{ }=\text{ }3$, the value of f(x)...

### Using the Remainder Theorem, factorise each of the following completely.

Given: f(x) = 2x3 + x2 – 13x + 6 By hit and trial we put x=-2 For x = -2, the value of f(x) will be f(-2) = 3(-2)3 + 2(-2)2 – 23(-2) – 30 = -24 + 8 + 46 – 30 = -54 + 54 = 0 As f(-2) = 0, so (x + 2)...

### Using the Remainder Theorem, factorise each of the following completely.

Let f(x) = $\mathbf{2}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{13x}\text{ }+\text{ }\mathbf{6}$ For $x\text{ }=\text{ }2$, the value of f(x) will...

### Using the Remainder Theorem, factorise each of the following completely.

Let f(x) = $\mathbf{3}{{\mathbf{x}}^{\mathbf{3}~}}+\text{ }\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}~-\text{ }\mathbf{19x}\text{ }+\text{ }\mathbf{6}$ For x=2, the value of f(x) will be...

### One year ago, a man was 8 times as old as his son. Now his age is equal to the square of his son’s age. Find their present ages.

ACCORDING TO QUES, How about we think about the current age of the child to be x years. Along these lines, the current age of the man $=\text{ }x2\text{ }years$ One year prior, Child's age...

### A car covers a distance of 400 km at a certain speed. Had the speed been 12 km/h more, the time taken for the journey would have been 1 hour 40 minutes less. Find the original speed of the car.

We should accept x km/h to be the first speed of the vehicle. We realize that, Time = Distance/Speed From the inquiry, The time taken by the vehicle to finish 400 km = $400/x\text{ }hrs$...

### If the speed of an aeroplane is reduced by 40 km/hr, it takes 20 minutes more to cover 1200 km. Find the speed of the aeroplane.

How about we think about the first speed of the plane to be x km/hr. Presently, the time taken to cover a distance of $1200\text{ }km\text{ }=\text{ }1200/x\text{ }hrs$ [Since, Time =...

### The speed of an ordinary train is x km per hr and that of an express train is (x + 25) km per hr. (i) Find the time taken by each train to cover 300 km. (ii) If the ordinary train takes 2 hrs more than the express train; calculate speed of the express train.

(i) Given, Speed of the conventional train $=\text{ }x\text{ }km/hr$ Speed of the express train $=\text{ }\left( x\text{ }+\text{ }25 \right)\text{ }km/hr$ Distance $=\text{ }300\text{ }km$ We...

### If x= (fig 1) ,find the value of:

fig 1: SOLUTION: $\begin{array}{*{35}{l}} x\text{ }=\text{ }2ab/\text{ }\left( a\text{ }+\text{ }b \right) \\ x/a\text{ }=\text{ }2b/\left( a\text{ }+\text{ }b \right) \\ \end{array}$ Applying...

### If (3x – 4y): (2x – 3y) = (5x – 6y): (4x – 5y), find x: y.

Since, $\left( 3x\text{ }-\text{ }4y \right):\text{ }\left( 2x\text{ }-\text{ }3y \right)\text{ }=\text{ }\left( 5x\text{ }-\text{ }6y \right):\text{ }\left( 4x\text{ }-\text{ }5y \right)$ This...

### Points A and B have co-ordinates (3, 4) and (0, 2) respectively. Find the image: (a)A” of A under reflection in the y-axis. (b) B” of B under reflection in the line AA”.

(a) $A''\text{ }=\text{ }Image\text{ }of\text{ }A\text{ }under\text{ }appearance$in the y-hub = (- 3, 4) (b) $B''\text{ }=\text{ }Image\text{ }of\text{ }B\text{ }under\text{ }appearance$in the...

Given, On cross multiplying, we get $\begin{array}{*{35}{l}} 45x\text{ }+\text{ }54y\text{ }=\text{ }64x\text{ }+\text{ }40y \\ 14y\text{ }=\text{ }19x \\ {} \\ x/y\text{ }=\text{ }14/19 \\... read more ### If a: b = 3: 5, find: (10a + 3b): (5a + 2b) \[\begin{array}{*{35}{l}} a/b\text{ }=\text{ }3/5 \\ \left( 10a\text{ }+\text{ }3b \right)/\text{ }\left( 5a\text{ }+\text{ }2b \right) \\ ~ \\ \end{array}$

(a) $A'\text{ }=\text{ }Image\text{ }of\text{ }A\text{ }under\text{ }appearance$ in the $x-hub\text{ }=\text{ }\left( 3,\text{ }-\text{ }4 \right)$ (b) \[B'\text{ }=\text{ }Image\text{ }of\text{...