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 line segment AB of length 5.8 cm.
Question: Draw a line segment AB of length 5.8 cm. Draw the perpendicular bisector of this line segment. Solution: Steps of construction: 1. Draw a line segment AB of 5.8 cm. 2. Keeping A as center and radius more than half of AB draw arcs on each side of AB. 3. Keeping B as center and the same radius draw arcs on each side of AB cutting the previous arcs at P and Q respectively. 4. Join the points P and Q. Hence PQ is the perpendicular bisector of AB...
Read More →Draw a line segment of length 8.6 cm. Bisect it and measure the length of each part.
Question: Draw a line segment of length 8.6 cm. Bisect it and measure the length of each part. Solution: Steps of construction: 1. Draw a line segment AB of 8.6 cm. 2. Keeping A as center and radius more than half of AB draw arcs on each side of AB. 3. Keeping B as center and the same radius draw arcs on each side of AB cutting the previous arcs at P and Q respectively. 4. Join the points P and Q which intersects AB at C. Therefore AC = BC = 4.3 cm...
Read More →In an equilateral ∆ABC, AD ⊥ BC prove that AD2 = 3BD2.
Question: In an equilateral $\triangle \mathrm{ABC}, \mathrm{AD} \perp \mathrm{BC}$ prove that $\mathrm{AD}^{2}=3 \mathrm{BD}^{2}$. Solution: We have to prove that $A D^{2}=3 B D^{2}$. In right angled $\triangle A B D$, using Pythagoras theorem we get, $A B^{2}=A D^{2}+B D^{2}$...(1) We know that in an equilateral triangle every altitude is also median. Therefore, AD bisects BC. Therefore, we have $B D=D C$ Since $\triangle A B C$ is an equilateral triangle, $A B=B C=A C$ Therefore, we can write...
Read More →Prove the following identities (1-16)
Question: Prove the following identities (1-16) $\cos x(\tan x+2)(2 \tan x+1)=2 \sec x+5 \sin x$ Solution: $\mathrm{LHS}=\cos x(\tan x+2)(2 \tan x+1)$ $=\cos x\left(2 \tan ^{2} x+5 \tan x+2\right)$ $=\cos x\left(\frac{2 \sin ^{2} x}{\cos ^{2} x}+\frac{5 \sin x}{\cos x}+2\right)$ $=\frac{2 \sin ^{2} x+5 \sin x \cos x+2 \cos ^{2} x}{\cos x}$ $=\frac{2+5 \sin x \cos x}{\cos x}$ $=2 \sec x+5 \sin x$ $=$ RHS Hence proved....
Read More →A chord of a circle is equal to the radius of the circle.
Question: A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc. Solution: We have, Radius OA = Chord AB ⟹ OA = OB = AB Then triangle OAB is an equilateral triangle. AOB = 60 [one angle of equilateral triangle] By degree measure theorem AOB = 2APB ⇒ 60 = 2APB ⇒ APB = 60/2 = 30 Now, APB + AQB = 180 [opposite angles of cyclic quadrilateral] ⇒ 300 + AQB = 180 ⇒ AQB = 180 30 = 150. Therefore, A...
Read More →Find the area of the region bounded by the parabola
Question: Find the area of the region bounded by the parabola $y=x^{2}$ and $y=|x|$ Solution: The area bounded by the parabola, $x^{2}=y$, and the line, $y=|x|$, can be represented as The given area is symmetrical abouty-axis. Area OACO = Area ODBO The point of intersection of parabola, $x^{2}=y$, and line, $y=x$, is $\mathrm{A}(1,1)$. Area of $\mathrm{OACO}=$ Area $\triangle \mathrm{OAM}-$ Area OMACO Area of $\Delta \mathrm{OAM}=\frac{1}{2} \times \mathrm{OM} \times \mathrm{AM}=\frac{1}{2} \tim...
Read More →In a quadrilateral ABCD, ∠B = 90°, AD2 = AB2 + BC2 + CD2,
Question: In a quadrilateral $A B C D, \angle B=90^{\circ}, A D^{2}=A B^{2}+B C^{2}+C D^{2}$, prove that $\angle A C D=90^{\circ}$. Solution: In order to prove angle $\angle A C D=90^{\circ}$ it is enough to prove that $A D^{2}=A C^{2}+C D^{2}$. Given, $A D^{2}=A B^{2}+B C^{2}+C D^{2}$ $A D^{2}-C D^{2}=A B^{2}+B C^{2}$.....(1) Since $\angle B=90^{\circ}$, so applying Pythagoras theorem in the right angled triangle $\mathrm{ABC}$, we get $A C^{2}=A B^{2}+B C^{2}$......(2) From (1) and (2), we get...
Read More →Prove the following identities (1-16)
Question: Prove the following identities (1-16) $\frac{2 \sin x \cos x-\cos x}{1-\sin x+\sin ^{2} x-\cos ^{2} x}=\cot x$ Solution: $\mathrm{LHS}=\frac{2 \sin x \cos x-\cos x}{1-\sin x+\sin ^{2} x-\cos ^{2} x}$ $=\frac{\cos x(2 \sin x-1)}{2 \sin ^{2} x-\sin x} \quad\left(\because 1-\cos ^{2} x=\sin ^{2} x\right)$ $=\frac{\cos x(2 \sin x-1)}{\sin x(2 \sin x-1)}$ $=\cot x$ = RHS Hence proved....
Read More →In figure, if ∠ACB = 40°, ∠DPB = 120°, find ∠CBD.
Question: In figure, if ACB = 40, DPB = 120, find CBD. Solution: We have, ACB = 40; DPB = 120 APB = DCB = 40 [Angle in same segment] In ΔPOB, by angle sum property PDB + PBD + BPD = 180 ⇒ 40 + PBD + 120 = 180 ⇒ PBD = 180 40 120 ⇒ PBD = 20 CBD = 20...
Read More →Prove the following identities (1-16)
Question: Prove the following identities (1-16) $\frac{(1+\cot x+\tan x)(\sin x-\cos x)}{\sec ^{3} x-\operatorname{cosec}^{3} x}=\sin ^{2} x \cos ^{2} x$ Solution: $\frac{(1+\cot x+\tan x)(\sin x-\cos x)}{\sec ^{3} x-\operatorname{cosec}^{3} x}=\sin ^{2} x \cos ^{2} x$ $\mathrm{LHS}=\frac{(1+\cot x+\tan x)(\sin x-\cos x)}{\sec ^{3} x-\operatorname{cosec}^{3} x}$ $=\frac{\left(1+\frac{\cos x}{\sin x}+\frac{\sin x}{\cos x}\right)(\sin x-\cos x)}{\frac{1}{\cos ^{3} x}-\frac{1}{\sin ^{3} x}}$ $=\fra...
Read More →In a right ∆ABC right-angled at C, if D is the mid-point of BC,
Question: In a right $\triangle A B C$ right-angled at $C$, if $D$ is the mid-point of $B C$, prove that $B C^{2}=4\left(A D^{2}-A C^{2}\right)$. Solution: It is given that ∆ABC is a right-angled at C and D is the mid-point of BC. In the right angled triangle ADC, we will use Pythagoras theorem, $A D^{2}=D C^{2}+A C^{2}$ Since $D$ is the midpoint of $B C$, we have $D C=\frac{B C}{2}$ Substituting $D C=\frac{B C}{2}$ in equation (1), we get $A D^{2}=\left(\frac{B C}{2}\right)^{2}+A C^{2}$ $A D^{2...
Read More →In figure, O is the centre of a circle and PQ is a diameter.
Question: In figure, O is the centre of a circle and PQ is a diameter. If ROS = 40, find RTS. Solution: Since PQ is diameter Then, PRQ = 90 [Angle in semicircle] PRQ + TRQ = 180 [Linear pair of angle] 900 + TRQ = 180 TRQ = 180 90 = 90. By degree measure theorem ROS = 2RQS ⇒ 40 = 2RQS ⇒ RQS = 40/2 = 20 InΔRQT, by Angle sum property RQT + QRT + RTS = 180 ⇒ 20 + 90 + RTS = 180 ⇒ RTS = 180 20 90 = 70...
Read More →The area between
Question: The area between $x=y^{2}$ and $x=4$ is divided into two equal parts by the line $x=a$, find the value of $a$. Solution: The line, $x=a$, divides the area bounded by the parabola and $x=4$ into two equal parts. $\therefore$ Area $\mathrm{OAD}=$ Area $\mathrm{ABCD}$ It can be observed that the given area is symmetrical aboutx-axis. ⇒ Area OED = Area EFCD Area of EFCD $=\int_{a}^{4} \sqrt{x} d x$ $=\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{a}^{4}$ $=\frac{2}{3}\left[8-a^{\frac{3}...
Read More →In figure, O and O' are centers of two circles intersecting at B and C.
Question: In figure, O and O' are centers of two circles intersecting at B and C. ACD is a straight line, find x. Solution: By degree measure theorem AOB = 2ACB ⇒ 130 = 2ACB ⇒ ACB = 130/2 = 65 ACB + BCD = 180 [Liner a pair of angles] ⇒ 65 + BCD = 180 ⇒ BCD = 180 65 = 115 By degree measure theorem reflex BOD = 2BCD ⇒ reflex BOD = 2 115 = 230 Now, reflex BOD + BOD = 360 [Complex angle] ⇒ 230 + x = 360 ⇒ x = 360 230 x = 130...
Read More →In ∆ABC, ∠A is obtuse, PB ⊥ AC and QC ⊥ AB. Prove that:
Question: In $\triangle A B C, \angle A$ is obtuse, $P B \perp A C$ and $Q C \perp A B$. Prove that: (i) $\mathrm{AB} \times \mathrm{AQ}=\mathrm{AC} \times \mathrm{AP}$ (ii) $B C^{2}=(A C \times C P+A B \times B Q)$ Solution: Given: ΔABC whereBAC is obtuse. PBAC and QCAB. To prove: (i) $\mathrm{AB} \times \mathrm{AQ}=\mathrm{AC} \times \mathrm{AP}$ and (ii) $\mathrm{BC}^{2}=\mathrm{AC} \times \mathrm{CP}+\mathrm{AB} \times \mathrm{BQ}$ Proof: In $\triangle \mathrm{ACQ}$ and $\triangle \mathrm{AB...
Read More →In figure, O is the centre of the circle, then prove that ∠x = ∠y + ∠z.
Question: In figure, O is the centre of the circle, then prove that x = y + z. Solution: We have, 3 = 4 [Angles in same segment] x = 23 [By degree measure theorem] ⇒ x = 3 + 3 ⇒ x = 3 + 4 ... (i) [3 = angle 4] But y = 3 + 1 [By exterior angle property] ⇒ 3 = y 1 .... (ii) from (i) and (ii) x = y 1 + 4 ⇒ x = y + 4 1 ⇒ x = y + z + 1 1 [By exterior angle property] ⇒ x = y + z...
Read More →Prove the following identities (1-17)
Question: Prove the following identities (1-17) $(1+\tan \alpha \tan \beta)^{2}+(\tan \alpha-\tan \beta)^{2}=\sec ^{2} \alpha \sec ^{2} \beta$ Solution: $(1+\tan \alpha \tan \beta)^{2}+(\tan \alpha-\tan \beta)^{2}=\sec ^{2} \alpha \sec ^{2} \beta$ $\mathrm{LHS}=(1+\tan \alpha \tan \beta)^{2}+(\tan \alpha-\tan \beta)^{2}$ $=1+\tan ^{2} \alpha \tan ^{2} \beta+2 \tan \alpha \tan \beta+\tan ^{2} \alpha+\tan ^{2} \beta-2 \tan \alpha \tan \beta$ $=1+\tan ^{2} \alpha \tan ^{2} \beta+\tan ^{2} \alpha+\t...
Read More →In figure, O is the centre of the circle, BO is the bisector of ∠ABC. Show that AB = AC.
Question: In figure, O is the centre of the circle, BO is the bisector of ABC. Show that AB = AC. Solution: Given, BO is the bisector ofABC To prove AB = BC Proof: Since, BO is the bisector ofABC. Then,ABO = CBO ... (i) Since, OB = OA [Radius of circle] Then,ABO = DAB... (ii) [opposite angles to equal sides] Since OB = OC [Radius of circle] Then,OAB = OCB... (iii)[opposite angles to equal sides] Compare equations (i), (ii) and (iii) OAB = OCB ... (iv) In ΔOAB and ΔOCB OAB = OCB From (iv)] OBA = ...
Read More →Prove the following identities (1-16)
Question: Prove the following identities (1-16) $\left(\frac{1}{\sec ^{2} x-\cos ^{2} x}+\frac{1}{\operatorname{cosec}^{2} x-\sin ^{2} x}\right) \sin ^{2} x \cos ^{2} x=\frac{1-\sin ^{2} x \cos ^{2} x}{2+\sin ^{2} x \cos ^{2} x}$ Solution: $\left(\frac{1}{\sec ^{2} x-\cos ^{2} x}+\frac{1}{\operatorname{cosec}^{2} x-\sin ^{2} x}\right) \sin ^{2} x \cos ^{2} x=\frac{1-\sin ^{2} x \cos ^{2} x}{2+\sin ^{2} x \cos ^{2} x}$ LHS $=\left(\frac{1}{\sec ^{2} x-\cos ^{2} x}+\frac{1}{\operatorname{cosec}^{2...
Read More →Find the area of the smaller part of the circle
Question: Find the area of the smaller part of the circle $x^{2}+y^{2}=a^{2}$ cut off by the line $x=\frac{a}{\sqrt{2}}$ Solution: The area of the smaller part of the circle, $x^{2}+y^{2}=a^{2}$, cut off by the line, $x=\frac{a}{\sqrt{2}}$, is the area ABCDA. It can be observed that the area ABCD is symmetrical aboutx-axis. Area ABCD = 2 Area ABC Area of $A B C=\int_{\frac{a}{\sqrt{2}}}^{a} y d x$ $=\int_{\frac{a}{\sqrt{2}}}^{\pi} \sqrt{a^{2}-x^{2}} d x$ $=\left[\frac{x}{2} \sqrt{a^{2}-x^{2}}+\f...
Read More →O is the circumference of the triangle ABC and OD is perpendicular on BC. Prove that ∠BOD = ∠A.
Question: O is the circumference of the triangle ABC and OD is perpendicular on BC. Prove that BOD = A. Solution: Given O is the circum centre of triangle ABC andODBC To prove BOD = 2A Proof: InΔOBD and ΔOCD ODB = ODC [Each 90] OB = OC [Radius of circle] OD = OD [Common] Then ΔOBD ΔOCD [By RHS Condition]. BOD = COD .... (i) [PCT]. By degree measure theorem BOC = 2BAC ⇒ 2BOD = 2BAC [By using (i)] ⇒ BOD = BAC....
Read More →In the given figure, ∠B < 90° and segment AD ⊥ BC, show that
Question: In the given figure, $\angle \mathrm{B}90^{\circ}$ and segment $\mathrm{AD} \perp \mathrm{BC}$, show that Solution: (i) Since AD perpendicular to BC we obtained two right angled triangles, triangle ADB and triangle ADC. We will use Pythagoras theorem in the right angled triangle ADC $A C^{2}=A D^{2}+D C^{2}$.....(1) Let us substitute AD =h, AC =band DC = (a x) in equation (1) we get, $b^{2}=h^{2}+(a-x)^{2}$ $b^{2}=h^{2}+a^{2}-2 a x+x^{2}$ $b^{2}=h^{2}+a^{2}+x^{2}-2 a x$.....(2) (ii) Le...
Read More →If O is the centre of the circle, find the value of x in each of the following figures.
Question: If O is the centre of the circle, find the value of x in each of the following figures. Solution: (i) AOC = 135 AOC + BOC = 180 [Linear pair of angles] ⇒ 135 + BOC = 180 ⇒ BOC = 180 135 ⇒ BOC = 45 By degree measure theorem BOC = 2CPB ⇒ 45 = 2x ⇒ x = 45/2 = 22 (ii) We have ABC = 40 ACB = 90 [Angle in semi circle] In ΔABC, by angle sum property CAB + ACB + ABC = 180 ⇒ CAB + 90 + 40 = 180 ⇒ CAB = 180 90 40 ⇒ CAB = 50 Now, CDB = CAB [Angle is same in segment] ⇒ x = 50 (iii) We have AOC = 1...
Read More →Find the area of the region in the first quadrant enclosed by
Question: Find the area of the region in the first quadrant enclosed by $x$-axis, line $x=\sqrt{3} y$ and the circle $x^{2}+y^{2}=4$ Solution: The area of the region bounded by the circle, $x^{2}+y^{2}=4, x=\sqrt{3} y$, and the $x$-axis is the area $\mathrm{OAB}$. The point of intersection of the line and the circle in the first quadrant is $(\sqrt{3}, 1)$. Area $\mathrm{OAB}=$ Area $\triangle \mathrm{OCA}+$ Area $\mathrm{ACB}$ Area of $O A C=\frac{1}{2} \times O C \times A C=\frac{1}{2} \times ...
Read More →