In the given figure, ∆AMB ∼ ∆CMD; determine MD in terms of x, y and z.
Question: In the given figure, ∆AMB ∆CMD; determine MD in terms ofx,yandz. Solution: We are given ∆AMB ∆CMD We have to determine the value ofMDin terms ofx, yandz. $\Rightarrow \frac{B M}{M D}=\frac{A M}{C M}$ $\frac{x}{M D}=\frac{y}{z}$ By cross multiplication we get $M D=\frac{x z}{y}$ Hence, the value of $M D$ is $\frac{x z}{y}$....
Read More →If cos x
Question: If $\cos x=-\frac{3}{5}$ and $\pix\frac{3 \pi}{2}$ find the values of other five trigonometric functions and hence evaluate $\frac{\operatorname{cosec} x+\cot x}{\sec x-\tan x}$. Solution: We have: $\cos x=-\frac{3}{5}$ and $\pix\frac{3 \pi}{2}$ Thus, $x$ is in the third quadrant. In the third quadrant, $\tan x$ and $\cot x$ are positive And, all the other four $\mathrm{T}$ - ratios are negative. $\therefore \sin x=-\sqrt{1-\cos ^{2} x}=-\sqrt{1-\left(\frac{-3}{5}\right)^{2}}=\frac{-4}...
Read More →In ∆ABC, P and Q are points on sides AB and AC respectively such that PQ || BC. If AP = 3 cm, PB = 5 cm and AC = 8 cm, find AQ.
Question: In ∆ABC, P and Q are points on sides AB and AC respectively such that PQ || BC. If AP = 3 cm, PB = 5 cm and AC = 8 cm, find AQ. Solution: In $\triangle A B C, P$ and $Q$ are points on sides $A B$ and $A C$ respectively such that $P Q \| B C$ Then we have $\frac{A P}{A B}=\frac{A Q}{A C}$ $A P=3 \mathrm{~cm}, P B=5 \mathrm{~cm}, A C=8 \mathrm{~cm}$ and $A B=8 \mathrm{~cm}$ $\frac{3}{8}=\frac{A Q}{8}$ $3=A Q$ Hence the value of $A Q$ is $3 \mathrm{~cm}$...
Read More →Find the area enclosed by the parabola
Question: Find the area enclosed by the parabola $4 y=3 x^{2}$ and the line $2 y=3 x+12$ Solution: The area enclosed between the parabola, $4 y=3 x^{2}$, and the line, $2 y=3 x+12$, is represented by the shaded area OBAO as The points of intersection of the given curves are A (2, 3) and (4, 12). We draw AC and BD perpendicular tox-axis. Area OBAO = Area CDBA (Area ODBO + Area OACO) $=\int_{-2}^{1} \frac{1}{2}(3 x+12) d x-\int_{-2}^{1} \frac{3 x^{2}}{4} d x$ $=\frac{1}{2}\left[\frac{3 x^{2}}{2}+1...
Read More →If sin x
Question: If sinx+ cosx= 0 andxlies in the fourth quadrant, find sinxand cosx. Solution: We have : $\sin x+\cos x=0$ $\Rightarrow \sin x=-\cos x$ $\Rightarrow \frac{\sin x}{\cos x}=-1$ $\Rightarrow \tan x=-1$ Now, $x$ is in the fourth quadrant. In the fourth quadrant, $\cos x$ and $\sec x$ are positive and all the other four $\mathrm{T}$-ratios are negative. $\therefore \sec x=\sqrt{1+\tan ^{2} x}=\sqrt{1+(-1)^{2}}=\sqrt{2}$ $\cos x=\frac{1}{\sec x}=\frac{1}{\sqrt{2}}$ And, $\sin x=-\sqrt{1-\cos...
Read More →In ∆PQR, M and N are points on sides PQ and PR respectively
Question: In ∆PQR, M and N are points on sides PQ and PR respectively such that PM = 15 cm and NR = 8 cm. If PQ = 25 cm and PR = 20 cm state whether MN || QR. Solution: Given $P M=15 \mathrm{~cm}, M Q=10 \mathrm{~cm}, N R=8 \mathrm{~cm}$ and $P N=12 \mathrm{~cm}$. $\frac{P M}{P Q}=\frac{15 \mathrm{~cm}}{25 \mathrm{~cm}}=\frac{3}{5}$ $\frac{P N}{P R}=\frac{12 \mathrm{~cm}}{20 \mathrm{~cm}}=\frac{3}{5}$ $(P N=P R-N R=20-8=12 \mathrm{~cm})$ $\therefore \frac{P M}{P Q}=\frac{P N}{P R}$ So, by the co...
Read More →Construct a triangle XYZ in which ∠Y = 30°,
Question: Construct a triangle XYZ in which Y = 30, Z = 90 and XY + YZ + ZX = 11 cm. Solution: Steps of construction: 1. Draw a line segment AB of 11 cm. 2. DrawDAB = 30andFBA = 90. 3. Draw the angle bisectors ofDABandEBAwhich intersect each other at X. 4. Draw the perpendicular bisector of XA and XB which intersect AB at Y and Z respectively 5. Join XY and XZ. Hence △XYZ is the required triangle....
Read More →In each of the following figures, you find who triangles
Question: In each of the following figures, you find who triangles. Indicate whether the triangles are similar. Give reasons in support of your answer. Solution: (i) In two triangles, we observe that In similar triangle corresponding sides are proportional to each other. Therefore, by SSS-criterion of similarity, yes two triangles are similar PQ || BC (Corresponding angles formed are equal)In ΔAPQand ΔABC, $\angle A P Q=\angle B$(Corresponding angles) $\angle P A Q=\angle B A C$ (Common) So, $\t...
Read More →If sin x
Question: If $\sin x=\frac{3}{5}, \tan y=\frac{1}{2}$ and $\frac{\pi}{2}x\piy\frac{3 \pi}{2}$, find the value of $8 \tan x-\sqrt{5}$ sec $y$. Solution: We have: $\sin x=\frac{3}{5}, \tan y=\frac{1}{2}$ and $\frac{\pi}{2}x\piy\frac{3 \pi}{2}$ Thus, $x$ is in the second quadrant and $y$ is in the third quadrant. In the second quadrant, $\cos x$ and $\tan x$ are negative. In the third quadrant, $\sec y$ is negative. $\therefore \cos x=-\sqrt{1-\sin ^{2} x}=-\sqrt{1-\left(\frac{3}{5}\right)^{2}}=\fr...
Read More →Find the area enclosed between the parabola
Question: Find the area enclosed between the parabola $y^{2}=4 a x$ and the line $y=m x$ Solution: The area enclosed between the parabola, $y^{2}=4 a x$, and the line, $y=m x$, is represented by the shaded area $\mathrm{OABO}$ as The points of intersection of both the curves are $(0,0)$ and $\left(\frac{4 a}{m^{2}}, \frac{4 a}{m}\right)$. We draw AC perpendicular tox-axis. Area OABO = Area OCABO Area (ΔOCA) $=2 \sqrt{a}\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{\frac{4 a}{m^{2}}}-m\le...
Read More →Construct a right-angled triangle ABC whose base BC is 6 cm and
Question: Construct a right-angled triangle ABC whose base BC is 6 cm and the sum of the hypotenuse AC and other side AB is 10 cm. Solution: Steps of Construction: 1. Construct a line segment BC of 6 cm. 2. At the point B, drawXBC = 90. 3. Keeping B as center and radius 10cm draw an arc which intersects XB at D. 4. Join DC. 5. Draw the perpendicular bisector of DC which intersects DB at A. 6. Join AC. Hence △ABC is the required triangle....
Read More →If sin x
Question: If $\sin x=\frac{12}{13}$ and $x$ lies in the second quadrant, find the value of $\sec x+\tan x$. Solution: We have: $\sin x=\frac{12}{13}$ and $x$ lie in the second quadrant. In the second quadrant, $\sin x$ and $\operatorname{cosec} x$ are positive and all the other four $\mathrm{T}-$ ratios are negative. $\therefore \cos x=-\sqrt{1-\sin ^{2} x}$ $=-\sqrt{1-\left(\frac{12}{13}\right)^{2}}$ $=\frac{-5}{13}$ $\tan x=\frac{\sin x}{\cos x}$ $=\frac{\frac{12}{13}}{\frac{-5}{13}}$ $=\frac{...
Read More →Construct a triangle ABC such that BC = 6 cm,
Question: Construct a triangle ABC such that BC = 6 cm, AB = 6 cm and median AD = 4 cm. Solution: Steps of construction: 1. Draw a line segment BC of 6 cm. 2. Take mid-point O of side BC. 3. With center B and D and radii 6cm and 4cm, draw two arcs which intersect each other at A. 4. Join AB, AD and AC. Hence △ABC is the required triangle....
Read More →Construct a right-angled triangle whose perimeter is equal to 10 cm and one acute angle equal to 60°.
Question: Construct a right-angled triangle whose perimeter is equal to 10 cm and one acute angle equal to 60. Solution: Steps of Construction: 1. Draw a line segment XY of 10 cm. 2. DrawDXY = 90andEYX = 60. 3. Draw the angle bisectors ofDXYandEYXwhich intersect each other at A. 4. Draw the perpendicular bisector of AX and AY which intersect XY at B and C respectively. 5. Join AB and AC. Hence △ABC is the required triangle....
Read More →Find the area bounded by the curve
Question: Find the area bounded by the curve $y=\sin x$ between $x=0$ and $x=2 \pi$ Solution: The graph ofy= sinxcan be drawn as $\therefore$ Required area = Area OABO + Area BCDB $=\int_{0}^{\pi} \sin x d x+\left|\int_{\pi}^{2 \pi} \sin x d x\right|$ $=[-\cos x]_{0}^{\pi}+\left|[-\cos x]_{\pi}^{2 \pi}\right|$ $=[-\cos \pi+\cos 0]+|-\cos 2 \pi+\cos \pi|$ $=1+1+|(-1-1)|$ $=2+|-2|$ $=2+2=4$ units...
Read More →Find the values of the other five trigonometric functions in each of the following
Question: Find the values of the other five trigonometric functions in each of the following (i) $\cot x=\frac{12}{5}, x$ in quadrant III (ii) $\cos x=-\frac{1}{2}, x$ in quadrant II (iii) $\tan x=\frac{3}{4}, x$ in quadrant III (iv) $\sin x=\frac{3}{5}, x$ in quadrant । Solution: (i) We have: $\cot x=\frac{12}{5}$ and $x$ are in the third quadrant. In the third quadrant, $\tan x$ and $\cot x$ are positive A nd, $\sin x, \cos x, \sec x$ and $\operatorname{cosec} x$ are negative. $\therefore \tan...
Read More →In ∆ABC, P and Q are points on sides AB and AC respectively
Question: In ∆ABC, P and Q are points on sides AB and AC respectively such that PQ || BC. If AP = 4 cm, PB = 6 cm and PQ = 3 cm, determine BC. Solution: In triangle $A B C, P$ and $Q$ are points on sides $A B$ and $A C$ respectively such that $P Q \| B C$. In $\triangle A P Q$ and $\triangle A B C$, $\angle A P Q=\angle B$(Corresponding angles) $\angle P A Q=\angle B A C \quad$ (Common) So, $\triangle \mathrm{APQ}-\Delta \mathrm{ABC}$ (AASimilarity) $\frac{A P}{A B}=\frac{P Q}{B C}$ Substituting...
Read More →Using rulers and compasses only, construct a △ABC from the following data:
Question: Using rulers and compasses only, construct a △ABC from the following data: AB + BC + CA = 12 cm, B = 45 and C = 60 Solution: Steps of Construction: 1. Draw a line segment XY of 12 cm. 2. DrawDXY = 45andEYX = 60. 3. Draw the angle bisectors ofDXYandEYXwhich intersect each other at A. 4. Draw the perpendicular bisector of AX and AY which intersect XY at B and C respectively. 5. Join AB and AC. Hence △ABC is the required triangle....
Read More →Find the values of the other five trigonometric functions in each of the following
Question: Find the values of the other five trigonometric functions in each of the following (i) $\cot x=\frac{12}{5}, x$ in quadrant III (ii) $\cos x=-\frac{1}{2}, x$ in quadrant II (iii) $\tan x=\frac{3}{4}, x$ in quadrant III (iv) $\sin x=\frac{3}{5}, x$ in quadrant । Solution: (i) We have: $\cot x=\frac{12}{5}$ and $x$ are in the third quadrant. In the third quadrant, $\tan x$ and $\cot x$ are positive A nd, $\sin x, \cos x, \sec x$ and $\operatorname{cosec} x$ are negative. $\therefore \tan...
Read More →Construct a triangle whose perimeter is 6.4 cm,
Question: Construct a triangle whose perimeter is 6.4 cm, and angles at the base are 60 45. Solution: Steps of Construction: 1. Draw a line segment XY of 6.4 cm. 2. DrawDXY = 60andEYX = 45. 3. Draw the angle bisectors ofDXYandEYXwhich intersect each other at A. 4. Draw the perpendicular bisector of AX and AY which intersect XY at B and C respectively. 5. Join AB and AC. Hence △ABC is the required triangle....
Read More →Sketch the graph of
Question: Sketch the graph of $y=|x+3|$ and evaluate $\int_{-6}^{0}|x+3| d x$ Solution: The given equation is $y=|x+3|$ The corresponding values ofxandyare given in the following table. On plotting these points, we obtain the graph of $y=|x+3|$ as follows. It is known that, $(x+3) \leq 0$ for $-6 \leq x \leq-3$ and $(x+3) \geq 0$ for $-3 \leq x \leq 0$ $\therefore \int_{-6}^{0}|(x+3)| d x=-\int_{-6}^{-3}(x+3) d x+\int_{-3}^{0}(x+3) d x$ $=-\left[\frac{x^{2}}{2}+3 x\right]_{-6}^{-3}+\left[\frac{x...
Read More →Using rulers and compasses only, construct a △ABC,
Question: Using rulers and compasses only, construct a △ABC, given base BC = 7 cm, ABC = 60, AB + AC = 12 cm. Solution: Steps of Construction: 1. Construct a line segment BC of 7cm 2. At the point B, drawXBC = 60. 3. Keeping B as center and radius 12 cm draw an arc which intersects XB at D 4. Join DC 5. Draw the perpendicular bisector of DC which intersects DB at A 6. Join AC Hence △ABC is the required triangle....
Read More →In the given figure, given that ∆ABC ∼ ∆PQR and quad ABCD ∼ quad PQRS. Determine the value of x, y, z in each case.
Question: In the given figure, given that ∆ABC ∆PQR and quad ABCD quad PQRS. Determine the value ofx,y,zin each case. Solution: (i) We have, $\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$ So the ratio of the sides of the triangles will be proportional to each other. $\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{AC}}{\mathrm{PR}}$ Therefore put the values of the known terms in the above equation to get, $\frac{12}{9}=\frac{7}{x}=\frac{10}{y}$ On solving thes...
Read More →Construct a △ABC in which BC = 3.4 cm,
Question: Construct a △ABC in which BC = 3.4 cm, AB - AC = 1.5 cm and B = 45. Solution: Steps of Construction: 1. Construct a line segment BC of 3.4 cm. 2. At the point B, drawXBC = 45. 3. Keeping B as centre and radius 1.5cm draw an arc which intersects XB at D. 4. Join DC. 5. Draw the perpendicular bisector of DC which intersects DB at A. 6. Join AC. Hence △ABC is the required triangle....
Read More →Find the area of the region lying in the first quadrant and bounded by
Question: Find the area of the region lying in the first quadrant and bounded by $y=4 x^{2}, x=0, y=1$ and $y=4$ Solution: The area in the first quadrant bounded by $y=4 x^{2}, x=0, y=1$, and $y=4$ is represented by the shaded area $\mathrm{ABCDA}$ as Area of $\mathrm{ABCDA}=\int_{1}^{4} x d y$ $=\int_{1}^{4} \frac{\sqrt{y}}{2} d y \quad\left[\right.$ as, $\left.y=4 x^{2}\right]$ $=\frac{1}{2} \int_{1}^{4} \sqrt{y} d y$ $=\frac{1}{2} \times \frac{2}{3}\left[y^{3 / 2}\right]_{1}^{4}$ $=\frac{1}{3...
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