Find the area of a regular hexagon ABCDEF in which each side measures 13 cm and whose height is 23 cm,
Question: Find the area of a regular hexagonABCDEFin which each side measures 13 cm and whose height is 23 cm, as shown in the given figure. Solution: Let $\mathrm{AL}=\mathrm{DM}=\mathrm{x} \mathrm{cm}$ $\mathrm{LM}=\mathrm{BC}=13 \mathrm{~cm}$ $\therefore x+13+x=23$ $\Rightarrow 2 x+13=23$ $\Rightarrow 2 x=(23-13)$ $\Rightarrow 2 x=10$ $\Rightarrow x=5$ $\therefore \mathrm{AL}=5 \mathrm{~cm}$ From the right $\Delta \mathrm{AFL}$, we have $:$ $F L^{2}=A F^{2}-A L^{2}$ $\Rightarrow F L^{2}=\left...
Read More →The range of the data 30,
Question: The range of the data 30, 61, 55, 56, 60, 20, 26, 46, 28, 56 is (a) 26 (b) 30 (c) 41 (d) 61 Solution: (c) 41 The difference between the lowest and the highest observation in a given data is called its Range. Then, Range = Highest observation Lowest observation = 61 20 = 41...
Read More →Prove that
Question: Express $(1-2 i)^{-3}$ in the form $(a+i b)$ Solution: We have, $(1-2 i)^{-3}$ $\Rightarrow \frac{1}{(1-2 i)^{3}}=\frac{1}{1-8 i^{3}-6 i+12 i^{2}}=\frac{1}{1+8 i-6 i-12}=\frac{1}{2 i-11}$ $\Rightarrow \frac{1}{-11+2 i}$ $=\frac{1}{-11+2 i} \times \frac{-11-2 i}{-11-2 i}$ $=\frac{-11-2 i}{(-11)^{2}-(2 i)^{2}}=\frac{-11-2 i}{121+4}$ $=\frac{-11-2 i}{125}$ $=\frac{-11}{125}-\frac{2 i}{125}$...
Read More →In a pie chart, the total angle at
Question: In a pie chart, the total angle at the centre of the circle is (a) 180o (b) 360o (c) 270o (d) 90o Solution: (b) 360o...
Read More →A geometric representation showing
Question: A geometric representation showing the relationship between a whole and its parts is a (a) Pie chart (b) Histogram (c) Bar graph (d) Pictograph Solution: (a) Pie chart A geometric representation showing the relationship between a whole and its parts is pie chart....
Read More →Find the area of given figure ABCDEFGH as per dimensions given in it.
Question: Find the area of given figure ABCDEFGH as per dimensions given in it. Solution: We will find the length of $\mathrm{AC}$. From the right triangles $\mathrm{ABC}$ and HGF, we have: $\mathrm{AC}^{2}=\mathrm{HF}^{2}=\left\{(5)^{2}-(4)^{2}\right\} \mathrm{cm}$ $=(25-16) \mathrm{cm}$ $=9 \mathrm{~cm}$ $\mathrm{AC}=\mathrm{HF}=\sqrt{9} \mathrm{~cm}$ $=3 \mathrm{~cm}$ Area of the given figure ABCDEFGH = (Area of rectangle ADEH) + (Area of $\Delta \mathrm{ABC})+($ Area of $\Delta \mathrm{HGF})...
Read More →The height of a rectangle in a histogram shows the
Question: The height of a rectangle in a histogram shows the (a) width of the class (b) upper limit of the class (c) lower limit of the class (d) frequency of the class Solution: (d) Frequency of the class...
Read More →Prove that
Question: If $(x+i y)^{1 / 3}=(a+i b)$ then prove that $\left(\frac{x}{a}+\frac{y}{b}\right)=4\left(a^{2}-b^{2}\right)$ Solution: Given that, $(x+i y)^{1 / 3}=(a+i b)$ $\Rightarrow(x+i y)=(a+i b)^{3}$ $\Rightarrow(a+i b)^{3}=x+i y$ $\Rightarrow a^{3}+(i b)^{3}+3 a^{2} i b+3 a i^{2} b^{2}=x+i y$ $\Rightarrow a^{3}-i b^{3}+3 a^{2} i b-3 a b^{2}=x+i y$ $\Rightarrow a^{3}-3 a b^{2}+i\left(3 a^{2} b-b^{3}\right)=x+i y$ On equating real and imaginary parts, we get $x=a^{3}-3 a b^{2}$ and $y=3 a^{2} b-...
Read More →Find the area enclosed by the given figure ABCDEF as per dimensions given herewith.
Question: Find the area enclosed by the given figureABCDEFas per dimensions given herewith. Solution: Area enclosed by the given figure $=$ (Area of trapezium FEDC) $+$ (Area of square $\mathrm{ABCF}$ ) $=\left[\left\{\frac{1}{2} \times(6+20) \times 8\right\}+(20 \times 20)\right] \mathrm{cm}^{2}$ $=(104+400) \mathrm{cm}^{2}$ $=504 \mathrm{~cm}^{2}$ Hence, the area enclosed by the figure is $504 \mathrm{~cm}^{2}$....
Read More →Find the area of pentagon ABCDE in which BL ⊥ AC,
Question: Find the area of pentagonABCDEin whichBLAC,CMADandENADsuch thatAC= 10 cm,AD= 12 cm,BL= 3 cm,CM= 7 cm andEN= 5 cm. Solution: Area of pentagon $\mathrm{ABCDE}=($ Area of $\Delta \mathrm{ABC})+($ Area of $\Delta \mathrm{ACD})+($ Area of $\Delta \mathrm{ADE})$ $=\left(\frac{1}{2} \times \mathrm{AC} \times \mathrm{BL}\right)+\left(\frac{1}{2} \times \mathrm{AD} \times \mathrm{CM}\right)+\left(\frac{1}{2} \times \mathrm{AD} \times \mathrm{EM}\right)$ $=\left[\left(\frac{1}{2} \times 10 \time...
Read More →Solve this
Question: If $(x+i y)^{3}=(u+i v)$ then prove that $\left(\frac{u}{x}+\frac{v}{y}\right)=4\left(x^{2}-y^{2}\right)$ Solution: Given that, $(x+i y)^{3}=(u+i v)$ $\Rightarrow x^{3}+(i y)^{3}+3 x^{2} y+3 x i^{2} y^{2}=u+i v$ $\Rightarrow x^{3}-i y^{3}+3 x^{2} y-3 x y^{2}=u+i v$ $\Rightarrow x^{3}-3 x y^{2}+i\left(3 x^{2} y-y^{3}\right)=u+i v$ On equating real and imaginary parts, we get $U=x^{3}-3 x y^{2}$ and $v=3 x^{2} y-y^{3}$ Now, $\frac{u}{x}+\frac{v}{y}=\frac{\mathrm{x}^{3}-3 \mathrm{xy}^{2}}...
Read More →Find the area of hexagon ABCDEF in which BL ⊥ AD,
Question: Find the area of hexagonABCDEFin whichBLAD,CMAD,ENADandFPADsuch thatAP= 6 cm,PL= 2 cm,LN= 8 cm,NM= 2 cm,MD= 3 cm,FP= 8 cm,EN= 12 cm,BL= 8 cm andCM= 6 cm. Solution: Area of hexagon ABCDEF $\begin{aligned}=(\text { Area of } \Delta \text { AFP })+(\text { Area of trapezium FENP })+(\text { Area of } \Delta \text { END }) \\ +\text { (Area of } \Delta \text { CMD) }+(\text { Area of trapezium BCML })+(\text { Area of } \Delta \text { ALB }) \end{aligned}$ $\begin{aligned}=\left(\frac{1}{2...
Read More →Express each of the following in the form (a + ib) and find its multiplicative
Question: Express each of the following in the form (a + ib) and find its multiplicative inverse: (i) $\frac{1+2 \mathrm{i}}{1-3 \mathrm{i}}$ (ii) $\frac{(1+7 \mathrm{i})}{(2-\mathrm{i})^{2}}$ (iii) $\frac{-4}{(1+i \sqrt{3})}$ Solution: (i) Let $Z=\frac{1+2 i}{1-3 i}$ $=\frac{1+2 i}{1-3 i} \times \frac{1+3 i}{1+3 i}=\frac{1+3 i+2 i+6 i^{2}}{1-9 i^{2}}$ $=\frac{1+5 i+6 i^{2}}{1+9}=\frac{-5+5 i}{10}$ $z=\frac{-1}{2}+\frac{1}{2} i$ $\Rightarrow \bar{z}=\frac{-1}{2}-\frac{1}{2} i$ $\Rightarrow|z|^{2...
Read More →d the area of pentagon ABCDE in which BL ⊥ AC,
Question: Find the area of pentagonABCDEin whichBLAC,DMACandENACsuch thatAC= 18 cm,AM= 14 cm,AN= 6 cm,BL= 4 cm,DM= 12 cm andEN= 9 cm. Solution: Area of pentagon $\mathrm{ABCDE}=($ Area of $\Delta \mathrm{AEN})+($ Area of trapezium EDMN $)+($ Area of $\Delta \mathrm{DMC})$ $+($ Area of $\Delta \mathrm{ACB})$ $=\left(\frac{1}{2} \times \mathrm{AN} \times \mathrm{EN}\right)+\left(\frac{1}{2} \times(\mathrm{EN}+\mathrm{DM}) \times \mathrm{NM}\right)+\left(\frac{1}{2} \times \mathrm{MC} \times \mathr...
Read More →In the given figure, ABCD is a quadrilateral-shaped field in which diagonal BD is 36 m,
Question: In the given figure,ABCDis a quadrilateral-shaped field in which diagonalBDis 36 m,ALBDandCMBDsuch thatAL= 19 m andCM= 11 m. Find the area of the field. Solution: Area of quadrilateral $\mathrm{ABCD}=($ Area of $\Delta \mathrm{ABD})+($ Area of $\Delta \mathrm{BCD})$ $=\left(\frac{1}{2} \times \mathrm{BD} \times \mathrm{AL}\right)+\left(\frac{1}{2} \times \mathrm{BD} \times \mathrm{CM}\right)$ $=\left[\left(\frac{1}{2} \times 36 \times 19\right)+\left(\frac{1}{2} \times 36 \times 11\rig...
Read More →In the given figure, ABCD is a quadrilateral in which AC = 24 cm,
Question: In the given figure,ABCDis a quadrilateral in whichAC= 24 cm,BLACandDMACsuch thatBL= 8 cm andDM= 7 cm. Find the area of quad.ABCD. Solution: Area of quadrilateral $\mathrm{ABCD}=\left(\begin{array}{ll}\text { Area of } \Delta \mathrm{ADC} \\ \text { tarea of } \Delta \mathrm{ACB}\end{array}\right)$ $=\left(\frac{1}{2} \times \mathrm{AC} \times \mathrm{DM}\right)+\left(\frac{1}{2} \times \mathrm{AC} \times \mathrm{BL}\right)$ $=\left[\left(\frac{1}{2} \times 24 \times 7\right)+\left(\fr...
Read More →Express each of the following in the form (a + ib) and find its conjugate.
Question: Express each of the following in the form (a + ib) and find its conjugate. (i) $\frac{1}{(4+3 \mathrm{i})}$ (ii) $(2+3 i)^{2}$ (iii) $\frac{(2-i)}{(1-2 i)^{2}}$ (iv) $\frac{(1+\mathrm{i})(1+2 \mathrm{i})}{(1+3 \mathrm{i})}$ (v) $\left(\frac{1+2 i}{2+i}\right)^{2}$ (vi) $\frac{(2+i)}{(3-i)(1+2 i)}$ Solution: (i) Let $Z=\frac{1}{4+3 i}=\frac{1}{4+3 i} \times \frac{4-3 i}{4-3 i}$ $=\frac{4-3 i}{16+9}=\frac{4}{25}-\frac{3}{25} i$ $\Rightarrow \quad \bar{z}=\frac{4}{25}+\frac{3}{25} i$ (ii)...
Read More →Find the complex number z for which
Question: Find the complex number $z$ for which $|z|=z+1+2 i$ Solution: Given: $|z|=z+1+2 i$ Consider, $|z|=(z+1)+2 i$ Squaring both the sides, we get $|z|^{2}=[(z+1)+(2 i)]^{2}$ $\Rightarrow|z|^{2}=|z+1|^{2}+4 i^{2}+2(2 i)(z+1)$ $\Rightarrow|z|^{2}=|z|^{2}+1+2 z+4(-1)+4 i(z+1)$ $\Rightarrow 0=1+2 z-4+4 i(z+1)$ $\Rightarrow 2 z-3+4 i(z+1)=0$ Let $z=x+i y$ $\Rightarrow 2(x+i y)-3+4 i(x+i y+1)=0$ $\Rightarrow 2 x+2 i y-3+4 i x+4 i^{2} y+4 i=0$ $\Rightarrow 2 x+2 i y-3+4 i x+4(-1) y+4 i=0$ $\Righta...
Read More →The parallel sides of a trapezium are 25 cm and 11 cm,
Question: The parallel sides of a trapezium are 25 cm and 11 cm, while its nonparallel sides are 15 cm and 13 cm. Find the area of the trapezium. Solution: Let $\mathrm{ABCD}$ be the trapezium in which $\mathrm{AB} \| \mathrm{DC}, \mathrm{AB}=25 \mathrm{~cm}, \mathrm{CD}=11 \mathrm{~cm}, \mathrm{AD}=13 \mathrm{~cm}$ and $\mathrm{BC}=15 \mathrm{~cm}$. Draw $\mathrm{CL} \perp \mathrm{AB}$ and $\mathrm{CM} \| \mathrm{DA}$ meeting $\mathrm{AB}$ at $\mathrm{L}$ and $\mathrm{M}$, respectively. Clearly...
Read More →The parallel sides of a trapezium are 25 cm and 11 cm,
Question: The parallel sides of a trapezium are 25 cm and 11 cm, while its nonparallel sides are 15 cm and 13 cm. Find the area of the trapezium. Solution: Let $\mathrm{ABCD}$ be the trapezium in which $\mathrm{AB} \| \mathrm{DC}, \mathrm{AB}=25 \mathrm{~cm}, \mathrm{CD}=11 \mathrm{~cm}, \mathrm{AD}=13 \mathrm{~cm}$ and $\mathrm{BC}=15 \mathrm{~cm}$. Draw $\mathrm{CL} \perp \mathrm{AB}$ and $\mathrm{CM} \| \mathrm{DA}$ meeting $\mathrm{AB}$ at $\mathrm{L}$ and $\mathrm{M}$, respectively. Clearly...
Read More →Manavi and Kuber each receives an equal allowance.
Question: Manavi and Kuber each receives an equal allowance. The table shows the fraction of their allowance each deposits into his/her saving account and the fraction each spends at the mall. If allowance of each is ₹ 1260 find the amount left with each. Solution: From the question, Manavi and Kuber each receives and equal allowance = ₹ 1260 Let us assume total cost be ₹ 1 For Manavi, left over = Total cost Total spends = 1 ( + ) = 1 (2 + 1)/4 = 1 (3/4) = (4 3)/4 = So, Amount = 1260 = ₹ 315 For...
Read More →A skirt that is
Question: A skirt that iscm long has a hem ofcm. How long will the skirt be if the hem is let down? Solution: From the question it is given that, Length of the skirt =cm = 287/8 cm Dimension of hem =cm = 25/8 cm Length of skirt, if hem is let down = ((287/8) + (25/8)) cm = 312/8 cm = 39 cm...
Read More →The average life expectancies of males
Question: The average life expectancies of males for several states are shown in the table. Express each decimal in the form p/q and arrange the states from the least to the greatest male life expectancy. State-wise data are included below; more indicators can be found in the FACTFILE section on the homepage for each state. Source: Registrar General of India (2003) SRS Based Abridged Lefe Tables. SRS Analytical Studies, Report No. 3 of 2003, New Delhi: Registrar General of India. The data are fo...
Read More →Solve the system of equations
Question: Solve the system of equations, $\operatorname{Re}\left(z^{2}\right)=0,|z|=2$ Solution: Given: $\operatorname{Re}\left(z^{2}\right)=0$ and $|z|=2$ Let $z=x+i y$ $\therefore|z|=\sqrt{x^{2}+y^{2}}$ $\Rightarrow 2=\sqrt{x^{2}+y^{2}}$ [Given] Squaring both the sides, we get $x^{2}+y^{2}=4 \ldots$ (i) Since, $z=x+i y$ $\Rightarrow z^{2}=(x+i y)^{2}$ $\Rightarrow z^{2}=x^{2}+i^{2} y^{2}+2 i x y$ $\Rightarrow z^{2}=x^{2}+(-1) y^{2}+2 i x y$ $\Rightarrow z^{2}=x^{2}-y^{2}+2 i x y$ It is given t...
Read More →The parallel sides of a trapezium are 20 cm and 10 cm.
Question: The parallel sides of a trapezium are 20 cm and 10 cm. Its nonparallel sides are both equal, each being 13 cm. Find the area of the trapezium. Solution: Let $\mathrm{ABCD}$ be the given trapezium in which $\mathrm{AB} \| \mathrm{DC}, \mathrm{AB}=20 \mathrm{~cm}, \mathrm{DC}=10 \mathrm{~cm}$ and $\mathrm{AD}=\mathrm{BC}=13 \mathrm{~cm}$. Draw $\mathrm{CL} \perp \mathrm{AB}$ and $\mathrm{CM} \| \mathrm{DA}$ meeting $\mathrm{AB}$ at $\mathrm{L}$ and $\mathrm{M}$, respectively. Clearly, AM...
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