Find the area bounded by curves
Question: Find the area bounded by curves $(x-1)^{2}+y^{2}=1$ and $x^{2}+y^{2}=1$ Solution: The area bounded by the curves, $(x-1)^{2}+y^{2}=1$ and $x^{2}+y^{2}=1$, is represented by the shaded area as On solving the equations, $(x-1)^{2}+y^{2}=1$ and $x^{2}+y^{2}=1$, we obtain the point of intersection as $\mathrm{A}\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ and $\mathrm{B}\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$. It can be observed that the required area is symmetrical about $x$-axis. ...
Read More →Draw a pair of vertically opposite angles.
Question: Draw a pair of vertically opposite angles. Bisect each of the two angles. Verify that the bisecting rays are in the same line. Solution: Steps of Construction: 1. Draw a pair of vertically opposite angleAOCandDOB. 2. Keeping O as the center and any radius draw two arcs which intersect OA at P, OC at Q, OB at S and OD at R. 3. Keeping P and Q as center and radius more than half of PQ draw two arcs which intersect each other at T. 4. Join TO. 5. Keeping R and S as center and radius more ...
Read More →Draw a linear pair of angles. Bisect each of the two angles.
Question: Draw a linear pair of angles. Bisect each of the two angles. Verify that the two bisecting rays are perpendicular to each other. Solution: Steps of construction: 1. Draw two angles DCA and DCB forming linear pair 2. With center C and any radius draw an arc which intersects AC at P and CD at Q and CB at R 3. With center P and Q and any radius draw two arcs which intersect each other at S 4. Join SC 5. With Q and R as center and any radius draw two arcs which intersect each other at T 6....
Read More →Using the protractor, draw a right angle. Bisect it to get an angle of measure 45°.
Question: Using the protractor, draw a right angle. Bisect it to get an angle of measure 45. Solution: Steps of construction: 1. Draw an angle ABC of90. 2. With B as the centre and any radius draw an arc which intersects AB at P and BC at Q. 3. With P as center and radius more than half of PQ draw an arc. 4. With Q as center and same radius draw an arc which intersects the previous arc at R. 5. Join RB. Therefore RBC = 45...
Read More →If tan x
Question: If $\tan x=\frac{a}{b}$, show that $\frac{a \sin x-b \cos \mathrm{x}}{a \sin x+b \cos x}=\frac{a^{2}-b^{2}}{a^{2}+b^{2}}$. Solution: LHS: $\frac{a \sin x-b \cos x}{a \sin x+b \cos x}$ Dividing by $b \cos x:$ $=\frac{\frac{a \tan x}{b}-1}{\frac{a \tan x}{b}+1}$ Substituting the value of $\tan x$ $=\frac{a^{2}-b^{2}}{a^{2}+b^{2}}$ = RHS Hence proved....
Read More →Find the area of the circle
Question: Find the area of the circle $4 x^{2}+4 y^{2}=9$ which is interior to the parabola $x^{2}=4 y$ Solution: The required area is represented by the shaded area OBCDO. Solving the given equation of circle, $4 x^{2}+4 y^{2}=9$, and parabola, $x^{2}=4 y$, we obtain the point of intersection as $\mathrm{B}\left(\sqrt{2}, \frac{1}{2}\right)$ and $\mathrm{D}\left(-\sqrt{2}, \frac{1}{2}\right)$. It can be observed that the required area is symmetrical about $y$-axis. $\therefore$ Area $\mathrm{OB...
Read More →Using your protractor, draw an angle of 108°.
Question: Using your protractor, draw an angle of 108. With this given angle as given, draw an angle of 54. Solution: Steps of construction: 1. Draw an angle ABC of108. 2. With B as the center and any radius draw an arc which intersects AB at P and BC at Q. 3. With P as center and radius more than half of PQ draw an arc. 4. With Q as the centre and same radius draw an arc which intersects the previous arc at R. 5. Join BR. Therefore RBC = 54...
Read More →Determine whether the triangle having sides (a − 1) cm,
Question: Determine whether the triangle having sides $(a-1) \mathrm{cm}, 2 \sqrt{a} \mathrm{~cm}$ and $(a+1) \mathrm{cm}$ is a right angled triangle. Solution: Let $A=(a-1)$ $B=\sqrt{2} a$ $C=(a+1)$ Larger side is $C=(a+1)$ We know that any number plus 1 is always greater than that number minus 1 and product of 2 and its square root. For example : Ifa= 36 $a-1=35$ $a+1=37$ $\sqrt{2} a=12$ If $a=5$ $a-1=4$ $a+1=6$ $\sqrt{2} a=4.47$ In order to prove that the given sides forms a right angled tria...
Read More →If tan x
Question: If $\tan x=\frac{b}{a}$, then find the values of $\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}$. Solution: $\tan x=\frac{b}{a}$ Now, $\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}$ $=\sqrt{\frac{1+\frac{b}{a}}{1-\frac{b}{a}}}+\sqrt{\frac{1-\frac{b}{a}}{1+\frac{b}{a}}}$ $=\sqrt{\frac{1+\tan x}{1-\tan x}}+\sqrt{\frac{1-\tan x}{1+\tan x}}$ $=\frac{\tan x+1+1-\tan x}{\sqrt{1-\tan ^{2} x}}$ $=\frac{2}{\sqrt{1-\tan ^{2} x}}$ $=\frac{2 \cos x}{\sqrt{\cos ^{2} x-\sin ^{2} x}}$...
Read More →Draw an obtuse angle.
Question: Draw an obtuse angle. Bisect it. Measure each of the angles so formed. Solution: Steps of construction: 1. Draw an angleABCof120. 2. With B as a centre and any radius, draw an arc which intersects AB at P and BC at Q. 3. With P as center and radius more than half of PQ draw an arc. 4. With Q as a center and same radius draw an arc which cuts the previous arc at R. 5. Join BR. Therefore ABR = RBC = 60...
Read More →An aeroplane leaves an airport and flies due north at a speed of 1000 km/hr.
Question: An aeroplane leaves an airport and flies due north at a speed of $1000 \mathrm{~km} / \mathrm{hr}$. At the same time, another aeroplane leaves the same airport and flies due west at a speed of $1200 \mathrm{~km} / \mathrm{hr}$. How far apart will be the two planes after $1 \frac{1}{2}$ hours? Solution: Let us draw the figure first. An aeroplane which flies due north at a speed of $1000 \mathrm{~km} / \mathrm{hr}$ covers the distance $\mathrm{AB}$ after $1 \frac{1}{2} \mathrm{hr}$ and a...
Read More →Area of the region bounded by the curve
Question: Area of the region bounded by the curve $y^{2}=4 x, y$-axis and the line $y=3$ is A. 2 B. $\frac{9}{4}$ C. $\frac{9}{3}$ D. $\frac{9}{2}$ Solution: The area bounded by the curve, $y^{2}=4 x, y$-axis, and $y=3$ is represented as $\therefore$ Area $\mathrm{OAB}=\int_{0}^{3} x d y$ $=\int_{0}^{3} \frac{y^{2}}{4} d y$ $=\frac{1}{4}\left[\frac{y^{3}}{3}\right]_{0}^{3}$ $=\frac{1}{12}(27)$ $=\frac{9}{4}$ units Thus, the correct answer is B....
Read More →If sin x
Question: If $\sin x=\frac{a^{2}-b^{2}}{a^{2}+b^{2}}$, then the values of $\tan x, \sec x$ and $\operatorname{cosec} x$ Solution: $\sin x=\frac{a^{2}-b^{2}}{a^{2}+b^{2}}$' We know, $\sin ^{2} x+\cos ^{2} x=1$ $\cos ^{2} x=1-\sin ^{2} x$ $=1-\left(\frac{a^{2}-b^{2}}{a^{2}+b^{2}}\right)^{2}$ $=\frac{\left(a^{4}+b^{4}+2 a^{2} b^{2}\right)-\left(a^{4}+b^{4}-2 a^{2} b^{2}\right)}{\left(a^{2}+b^{2}\right)^{2}}$ $=\frac{4 a^{2} b^{2}}{\left(a^{2}+b^{2}\right)^{2}}$ $\Rightarrow \cos x=\frac{2 a b}{\lef...
Read More →Draw an angle and label it as ∠BAC.
Question: Draw an angle and label it as BAC. Construct another angle, equal to BAC Solution: Steps of construction: 1. Draw an angle ABC and a line segment QR. 2. With center A and any radius, draw an arc which intersectsBAC at E and D. 3. With Q as a centre and same radius draw an arc which intersects QR at S. 4. With S as center and radius equal to DE, draw an arc which intersects the previous arc at T. 5. Draw a line segment joining Q and T. Therefore PQR = BAC...
Read More →A guy wire attached to a vertical pole of height 18 m is 24 m long has a stake
Question: A guy wire attached to a vertical pole of height 18 m is 24 m long has a stake attached to the other end. How far from the base of pole should the stake be driven so that the wire will be taut? Solution: We will draw the figure from the given information as below, Let AB be the vertical pole of length 18 m and let the stake be at the point C so the wire will be taut. Therefore, we have $A B=18 \mathrm{~m}, A C=24 \mathrm{~m}$ and we have to find $\mathrm{BC}$. Now we will use Pythagora...
Read More →Area lying in the first quadrant and bounded by the circle
Question: Area lying in the first quadrant and bounded by the circle $x^{2}+y^{2}=4$ and the lines $x=0$ and $x=2$ is A. $\pi$ B. $\frac{\pi}{2}$ C. $\frac{\pi}{3}$ D. $\frac{\pi}{4}$ Solution: The area bounded by the circle and the lines,x= 0 andx= 2, in the first quadrant is represented as $\therefore$ Area $\mathrm{OAB}=\int_{0}^{2} y d x$ $=\int_{0}^{2} \sqrt{4-x^{2}} d x$ $=\left[\frac{x}{2} \sqrt{4-x^{2}}+\frac{4}{2} \sin ^{-1} \frac{x}{2}\right]_{0}^{2}$ $=2\left(\frac{\pi}{2}\right)$ $=\...
Read More →Draw a line segment AB and by ruler and compasses, obtain a line segment of length (3/4)(AB).
Question: Draw a line segment AB and by ruler and compasses, obtain a line segment of length (3/4)(AB). Solution: Steps of construction: 1. Draw a line segment AB. 2. With A as center and radius more than half of AB draw arcs, one on each side of AB. 3. With B as the center and same radius draw arcs cutting the previous arcs at points P and Q respectively. 4. Join P and Q which intersects AB at C. 5. With A as center and radius more than half of AC draw arcs, one on each side of AC. 6. With C as...
Read More →Draw a line segment AB bisect it.
Question: Draw a line segment AB bisect it. Bisect one of the equal parts to obtain a line segment of length (1/4) (AB). Solution: Steps of construction: 1. Draw a line segment AB. 2. With A as center and radius more than half of AB draw arcs, one on each side of AB. 3. With B as the center and same radius draw arcs cutting the previous arcs at points P and Q respectively. 4. Join P and Q which intersects AB at C. 5. With A as center and radius more than half of AC draw arcs, one on each side of...
Read More →∆ABD is a right triangle right-angled at A and AC ⊥ BD. Show that
Question: $\triangle \mathrm{ABD}$ is a right triangle right-angled at $\mathrm{A}$ and $\mathrm{AC} \perp \mathrm{BD}$. Show that (i) $\mathrm{AB}^{2}=\mathrm{BC} \cdot \mathrm{BD}$ (ii) $\mathrm{AC}^{2}=\mathrm{BC} . \mathrm{DC}$ (iii) $\mathrm{AD}^{2}=\mathrm{BD} \cdot \mathrm{CD}$ (iv) $\frac{\mathrm{AB}^{2}}{\mathrm{AC}^{2}}=\frac{\mathrm{BD}}{\mathrm{DC}}$ Solution: In $\triangle A B D$ and $\triangle A B C$, $\angle A C B=\angle A=90^{\circ}$ $\angle B=\angle B$ (Common angle) So, by AA c...
Read More →Find the area of the region bounded by the curve
Question: Find the area of the region bounded by the curve $y^{2}=4 x$ and the line $x=3$ Solution: The region bounded by the parabola, $y^{2}=4 x$, and the line, $x=3$, is the area $\mathrm{OACO}$. The area OACO is symmetrical aboutx-axis. Area of OACO = 2 (Area of OAB) Area $\mathrm{OACO}=2\left[\int_{0}^{3} y d x\right]$ $=2 \int_{0}^{3} 2 \sqrt{x} d x$ $=4\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{3}$ $=\frac{8}{3}\left[(3)^{\frac{3}{2}}\right]$ $=8 \sqrt{3}$ Therefore, the requir...
Read More →Draw a line segment of length 10 cm and bisect it. Further, bisect one of the equal parts and measure its length.
Question: Draw a line segment of length 10 cm and bisect it. Further, bisect one of the equal parts and measure its length. Solution: Steps of construction: 1. Draw a line segment AB of length 10 cm. 2. Keeping A as center and radius more than half of AB draw arcs one on each side of AB. 3. Keeping B as center and same radius draw arcs cutting the previous arc at point P and Q respectively. 4. Join P and Q which intersects AB at C. 5. Keeping A as center and radius more than half of AC draw arcs...
Read More →Draw a circle with center at point O. Draw its two chords AB and CD such that AB is not parallel to CD.
Question: Draw a circle with center at point O. Draw its two chords AB and CD such that AB is not parallel to CD. Draw the perpendicular bisectors of AB and CD. At what point do they intersect? Solution: Steps of construction: 1. Keeping O as the centre and any radius draw a circle. 2. Draw two chords AB and CD. 3. Keeping A as center and radius more than half of AB draw arcs, one on each side of AB. 4. Keeping B as center and the same radius draw arcs cutting the previous arcs at point P and Q ...
Read More →If x
Question: If $x=\frac{2 \sin x}{1+\cos x+\sin x}$, then prove that $\frac{1-\cos x+\sin x}{1+\sin x}$ is also equal to $a$. Solution: Disclaimer: There is some error in the given question. The question should have been Question: If $a=\frac{2 \sin x}{1+\cos x+\sin x}$, then prove that $\frac{1-\cos x+\sin x}{1+\sin x}$ is also equal to $a .$ So, the solution is done accordingly. Solution: $a=\frac{2 \sin x}{1+\sin x+\cos x}$ Rationalising the denominator : $\frac{2 \sin x}{1+\sin x+\cos x} \time...
Read More →Draw a circle with center at point O and radius 5 cm.
Question: Draw a circle with center at point O and radius 5 cm. Draw its chord AB, the perpendicular bisector of line segment AB. Does it pass through the center of the circle? Solution: Steps of construction: 1. Keeping O as center and radius as 5 cm draw a circle. 2. Draw a chord AB. 3. Keeping A as center and radius more than half of AB draw arcs one on each side of chord AB. 4. Keeping B as center and the same radius draw arcs cutting the previous arcs at points P and Q respectively. 5. Join...
Read More →Find the area bounded by the curve
Question: Find the area bounded by the curve $x^{2}=4 y$ and the line $x=4 y-2$ Solution: The area bounded by the curve, $x^{2}=4 y$, and line, $x=4 y-2$, is represented by the shaded area $\mathrm{OBAO}$. Let A and B be the points of intersection of the line and parabola. Coordinates of point $A$ are $\left(-1, \frac{1}{4}\right)$. Coordinates of point $\mathrm{B}$ are $(2,1)$. We draw AL and BM perpendicular to $x$-axis. It can be observed that, Area $\mathrm{OBAO}=$ Area $\mathrm{OBCO}+$ Area...
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