Construct a parallelogram ABCD in which AB = 5.2 cm,
Question: Construct a parallelogramABCDin whichAB= 5.2 cm,BC= 4.7 cm andAC= 7.6 cm. Solution: Steps of construction: Step 1: Draw $A B=5.2 \mathrm{~cm}$ Step 2: With $B$ as the centre, draw an $\operatorname{arc}$ of $4.7 \mathrm{~cm}$. Step 3: With $A$ as the centre, draw another arc of $7.6 \mathrm{~cm}$, cutting the previous arc at $C$. Step 4: Join $A$ and $C$. Step 5: We know that the opposite sides of a parallelogram are equal. Thus, with $C$ as the centre, draw an arc of $5.2 \mathrm{~cm}...
Read More →Construct a quadrilateral ABCD in which AB = 4 cm,
Question: Construct a quadrilateralABCDin whichAB= 4 cm,AC= 5 cm,AD= 5.5 cm and ABC= ACD= 90. Solution: Steps of construction: Step 1: Draw $A B=4 \mathrm{~cm}$ Step 2: Make $\angle B=90^{\circ}$ Step 3: $A C^{2}=A B^{2}+B C^{2}$ $5^{2}=4^{2}+B C^{2}$ $25-16=B C^{2}$ $B C=3 \mathrm{~cm}$ With $B$ as the centre, draw an arc equal to $3 \mathrm{~cm}$. Step 4: Make $\angle C=90^{\circ}$ Step 5: With $A$ as the centre and radius equal to $5.5 \mathrm{~cm}$, draw an arc and name that point as $D$. Th...
Read More →Construct a quadrilateral PQRS in which PQ = 5 cm,
Question: Construct a quadrilateralPQRSin whichPQ= 5 cm,QR= 6.5 cm, P= R= 100 and S= 75. Solution: Steps of construction: Step 1: DrawPQ=5cm5cm Step 2:$\angle P+\angle Q+\angle R+\angle S=360^{\circ}$ $100^{\circ}+\angle Q+100^{\circ}+75^{\circ}=360^{\circ}$ $275^{\circ}+\angle Q=360^{\circ}$ $\angle Q=360^{\circ}-275^{\circ}$ $\angle Q=85^{\circ}$ Step 3: Make $\angle P=100^{\circ}$ and $\angle Q=85^{\circ}$ Step 3: With $Q$ as the centre, draw an $\operatorname{arc}$ of $6.5 \mathrm{~cm}$. Ste...
Read More →Find the smallest positive integer n for which
Question: Find the smallest positive integer $n$ for which $(1+i)^{2 n}=(1-i)^{2 n} .$ Solution: Given: $(1+i)^{2 n}=(1-i)^{2 n}$ Consider the given equation $(1+i)^{2 n}=(1-i)^{2 n}$ $\Rightarrow \frac{(1+i)^{2 n}}{(1-i)^{2 n}}=1$ $\Rightarrow\left(\frac{1+i}{1-i}\right)^{2 n}=1$ Now, rationalizing by multiply and divide by the conjugate of (1 i) $\left(\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right)^{2 n}=1$ $\Rightarrow\left(\frac{(1+i)^{2}}{(1-i)(1+i)}\right)^{2 n}=1$ $\Rightarrow\left[\frac{1...
Read More →Construct a quadrilateral ABCD in which AB = 5.6 cm,
Question: Construct a quadrilateralABCDin whichAB= 5.6 cm,BC= 4 cm, A= 50, B= 105 and D= 80. Solution: Steps of construction: Step 1: Draw $A B=5.6 \mathrm{~cm}$ Step 2: Make $\angle A=50^{\circ}$ and $\angle B=105^{\circ}$ Step 3: With $B$ as the centre, draw an arc of $4 \mathrm{~cm}$. Step 3: Sum of all the angles of the quadrilateral is $360^{\circ}$. $\angle A+\angle B+\angle C+\angle D=360^{\circ}$ $50^{\circ}+105^{\circ}+\angle C+80^{\circ}=360^{\circ}$ $235^{\circ}+\angle C=360^{\circ}$ ...
Read More →Construct a quadrilateral PQRS in which PQ = 6 cm,
Question: Construct a quadrilateralPQRSin whichPQ= 6 cm,QR= 5.6 cm,RS= 2.7 cm, Q= 45 and R= 90. Solution: Steps of construction: Step 1: Draw $Q R=5.6 \mathrm{~cm}$ Step 2: Make $\angle Q=45^{\circ}$ and $\angle R=90^{\circ}$ Step 3: With $Q$ as the centre, draw an arc of $6 \mathrm{~cm}$. Name that point as $P$. Step 4: With $R$ as the centre, draw an arc of $2.7 \mathrm{~cm}$. Name that point as $S$. Step 6: Join $P$ and $S$. Then, $P Q R S$ is the required quadrilateral....
Read More →Construct a quadrilateral ABCD in which AB = 3.5 cm,
Question: Construct a quadrilateralABCDin whichAB= 3.5 cm,BC= 5 cm,CD= 4.6 cm, B= 125 and C= 60. Solution: Steps of construction: Step 1: Draw $B C=5 \mathrm{~cm}$ Step 2: Make $\angle B=125^{\circ}$ and $\angle C=60^{\circ}$ Step 3: With $B$ as the centre, draw an arc of $3.5 \mathrm{~cm}$. Name that point as $A$. Step 4: With $C$ as the centre, draw an arc of $4.6 \mathrm{~cm}$. Name that point as $D$. Step 5: Join $A$ and $D$. Then, $A B C D$ is the required quadrilateral....
Read More →Show that
Question: Show that $(1-i)^{n}\left(1-\frac{1}{i}\right)^{n}=2^{n}$ for all n N. Solution: To show: $(1-i)^{n}\left(1-\frac{1}{i}\right)^{n}=2^{n}$ Taking LHS, $(1-i)^{n}\left(1-\frac{1}{i}\right)^{n}$ $=(1-i)^{n}\left(1-\frac{1}{i} \times \frac{i}{i}\right)^{n} \quad$ [rationalize] $=(1-i)^{n}\left(1-\frac{i}{i^{2}}\right)^{n}$ $=(1-i)^{n}\left(1-\frac{i}{-1}\right)^{n}\left[\because i^{2}=-1\right]$ $=(1-i)^{n}(1+i)^{n}$ $=[(1-i)(1+i)]^{n}$ $=\left[(1)^{2}-(i)^{2}\right]^{n}\left[(a+b)(a-b)=a^...
Read More →Construct a quadrilateral ABCD in which AB = 2.9 cm,
Question: Construct a quadrilateralABCDin whichAB= 2.9 cm,BC= 3.2 cm,CD= 2.7 cm,DA= 3.4 cm and A= 70. Solution: Steps of construction: Step 1: Draw $A B=2.9 \mathrm{~cm}$ Step 2: Make $\angle A=70^{\circ}$ Step 3: With $A$ as the centre, draw an arc of $3.4 \mathrm{~cm}$. Name that point as $D$. Step 4: With $D$ as the centre, draw an arc of $2.7 \mathrm{~cm}$. Step 5: With $B$ as the centre, draw an arc of $3.2 \mathrm{~cm}$, cutting the previous arc at $C$. Step 6: Join $C D$ and $B C$. Then, ...
Read More →Construct a quadrilateral ABCD in which AB = BC = 3.5 cm,
Question: Construct a quadrilateralABCDin whichAB = BC= 3.5 cm,AD = CD= 5.2 cm and ABC= 120. Solution: Steps of construction: Step 1: Draw $A B=3.5 \mathrm{~cm}$. Step 2: Make $\angle A B C=120^{\circ}$. Step 3: With B as the centre, draw an $\operatorname{arc} 3.5 \mathrm{~cm}$ and name that point $C$. Step 4: With $C$ as the centre, draw an arc $5.2 \mathrm{~cm}$. Step 5: With $A$ as the centre, draw another $\operatorname{arc} 5.2 \mathrm{~cm}$, cutting the previous arc at $D$. Step 6: Join $...
Read More →construct a quadrilateral ABCD in which AB =3.4 cm,
Question: construct a quadrilateralABCDin whichAB=3.4 cm,CD= 3 cm,DA= 5.7 cm,AC= 8 cm andBD= 4 cm. Solution: Steps of construction: Step 1: Draw $A B=3.4 \mathrm{~cm} .$ Step 2: With $B$ as the centre and radius equal to $4 \mathrm{~cm}$, draw an arc. Step 3: With $A$ as the centre and radius equal to $5.7 \mathrm{~cm}$, draw another arc, cutting the previous arc at $D$. Step 4: Join $B D$ and $A D$. Step 5: With $A$ as the centre and radius equal to $8 \mathrm{~cm}$, draw an arc. Step 6: With $...
Read More →Construct a quadrilateral PQRS in which QR = 7.5 cm,
Question: Construct a quadrilateralPQRSin whichQR= 7.5 cm,PR=PS= 6 cm,RS= 5 cm andQS= 10 cm. Measure the fourth side. Solution: Steps of construction: Step 1: Draw $Q R=7.5 \mathrm{~cm}$. Step 2: With $Q$ as the centre and radius equal to $10 \mathrm{~cm}$, draw an arc. Step 3: With $R$ as the centre and radius equal to $5 \mathrm{~cm}$, draw another arc, cutting the previous arc at $S$. Step 4: Join $Q S$ and $R S$. Step 5: With $S$ as the centre and radius equal to $6 \mathrm{~cm}$, draw an ar...
Read More →Construct a quadrilateral ABCD in which AB = 3.6 cm,
Question: Construct a quadrilateralABCDin whichAB= 3.6 cm,BC= 3.3 cm,AD= 2.7 cm, diagonalAC= 4.6 cm and diagonalBD= 4 c Solution: Steps of construction: Step 1: Draw $\mathrm{AB}=3.6 \mathrm{~cm}$. Step 2: With B as the centre and radius equal to $4 \mathrm{~cm}$, draw an arc. Step 3: With A as the centre and radius equal to $2.7 \mathrm{~cm}$, draw another arc, cutting the previous arc at D. Step 4: Join BD and AD. Step 5: With A as the centre and radius equal to $4.6 \mathrm{~cm}$, draw an arc...
Read More →Construct a quadrilateral ABCD in which AB = 3.5 cm,
Question: Construct a quadrilateralABCDin whichAB= 3.5 cm,BC= 3.8 cm,CD=DA= 4.5 cm and diagonalBD= 5.6 cm. Solution: Steps of construction: Step 1: Draw $\mathrm{AB}=3.5 \mathrm{~cm}$. Step 2: With B as the centre and radius equal to $5.6 \mathrm{~cm}$, draw an arc. Step 3: With $A$ as the centre and radius equal to $4.5 \mathrm{~cm}$, draw another arc, cutting the previous arc at $D$. Step 4: Join BD and AD. Step 5: With D as the centre and radius equal to $4.5 \mathrm{~cm}$, draw an arc. Step ...
Read More →Solve this
Question: If $(a+i b)=\frac{c+i}{c-i}$, where $c$ is real, prove that $a^{2}+b^{2}=1$ and $\frac{b}{a}=\frac{2 c}{c^{2}-1}$ Solution: Consider the given equation, $a+i b=\frac{c+i}{c-i}$ Now, rationalizing $a+i b=\frac{c+i}{c-i} \times \frac{c+i}{c+i}$ $=\frac{(c+i)(c+i)}{(c-i)(c+i)}$ $=\frac{(c+i)^{2}}{(c)^{2}-(i)^{2}}$ $\left[(a-b)(a+b)=a^{2}-b^{2}\right]$ $=\frac{c^{2}+2 i c+i^{2}}{c^{2}-i^{2}}$ $a+i b=\frac{c^{2}+2 i c+(-1)}{c^{2}-(-1)} \quad\left[i^{2}=-1\right]$ $a+i b=\frac{c^{2}+2 i c-1}...
Read More →Construct a quadrilateral PQRS in which PQ = 5.4 cm,
Question: Construct a quadrilateralPQRSin whichPQ= 5.4 cm,QR= 4.6 cm,RS= 4.3 cm,SP= 3.5 cm and diagonalPR= 4 cm. Solution: Steps of construction: Step 1: Draw $\mathrm{PQ}=5.4 \mathrm{~cm}$. Step 2: With $P$ as the centre and radius equal to $4 \mathrm{~cm}$, draw an arc. Step 3: With $Q$ as the centre and radius equal to $4.6 \mathrm{~cm}$, draw another arc, cutting the previous arc at $R$. Step 4: Join QR. Step 5: With P as the centre and radius equal to $3.5 \mathrm{~cm}$, draw an arc. Step 6...
Read More →Construct a quadrilateral ABCD in which AB = 4.2 cm,
Question: Construct a quadrilateralABCDin whichAB= 4.2 cm,BC= 6 cm,CD= 5.2 cm,DA= 5 cm andAC= 8 cm. Solution: Steps of construction: Step 1: Draw $\mathrm{AB}=4.2 \mathrm{~cm}$. Step 2: With A as the centre and radius equal to $8 \mathrm{~cm}$, draw an arc. Step 3: With B as the centre and radius equal to $6 \mathrm{~cm}$, draw another arc, cutting the previous arc at C. Step 4: Join BC. Step 5: With A as the centre and radius equal to $5 \mathrm{~cm}$, draw an arc. Step 6: With C as the centre ...
Read More →Tick (✓) the correct answer
Question: Tick (✓) the correct answer: If one angle of a parallelogram is 24 less than twice the smallest angle then the largest angle of the parallelogram is (a) 68 (b) 102 (c) 112 (d) 176 Solution: (c) $112^{\circ}$ Let $x^{\circ}$ be the smallest angle of the parallelogram. $T$ he sum of adjacent angles of a parallelogram is $180^{\circ}$. $\therefore x+2 x-24=180$ $\Rightarrow 3 x-24=180$ $\Rightarrow 3 x=180+24$ $\Rightarrow 3 x=204$ $\Rightarrow x=\frac{204}{3}$ $\Rightarrow x=68$ $\theref...
Read More →Solve this
Question: If $x+i y=\frac{a+i b}{a-i b}$, prove that $\mathbf{x}^{2}+\mathbf{y}^{2}=1$ Solution: Consider the given equation, $x+i y=\frac{a+i b}{a-i b}$ Now, rationalizing $x+i y=\frac{a+i b}{a-i b} \times \frac{a+i b}{a+i b}$ $=\frac{(a+i b)(a+i b)}{(a-i b)(a+i b)}$ $=\frac{a(a+i b)+i b(a+i b)}{(a)^{2}-(i b)^{2}}$ $\left[(a-b)(a+b)=a^{2}-b^{2}\right]$ $=\frac{a^{2}+i a b+i a b+i^{2} b^{2}}{a^{2}-i^{2} b^{2}}$ $=\frac{a^{2}+i a b+i a b+(-1) b^{2}}{a^{2}-(-1) b^{2}}\left[i^{2}=-1\right]$ $x+i y=...
Read More →Tick (✓) the correct answer
Question: Tick (✓) the correct answer: In a squareABCD,AB= (2x+ 3) cm andBC= (3x 5) cm. Then, the value ofxis (a) 4 (b) 5 (c) 6 (d) 8 Solution: (d) 8 $A l l$ the sides of a square are equal. $\therefore A B=B C$ $\Rightarrow 2 x+3=3 x-5$ $\Rightarrow 3+5=3 x-2 x$ $\Rightarrow 8=x$ Therefore, the value of $x$ is 8 ....
Read More →Tick (✓) the correct answer
Question: Tick (✓) the correct answer: In a squareABCD,AB= (2x+ 3) cm andBC= (3x 5) cm. Then, the value ofxis (a) 4 (b) 5 (c) 6 (d) 8 Solution: (d) 8 $A l l$ the sides of a square are equal. $\therefore A B=B C$ $\Rightarrow 2 x+3=3 x-5$ $\Rightarrow 3+5=3 x-2 x$ $\Rightarrow 8=x$ Therefore, the value of $x$ is 8 ....
Read More →Tick (✓) the correct answer
Question: Tick (✓) the correct answer: The diagonals do not necessarily bisect the interior angles at the vertices in a (a) rectangle (b) square (c) rhombus (d) all of these Solution: (a) $r$ ectangle In a rectangle, the diagonals do not necessarily bisect the interior angles at the vertices....
Read More →Prove that
Question: If $\left(\frac{1+i}{1-i}\right)^{93}-\left(\frac{1-i}{1+i}\right)^{3}=x+i y$, find $x$ and $y$ Solution: Consider $x+i y=\left(\frac{1+i}{1-i}\right)^{93}-\left(\frac{1-i}{1+i}\right)^{3}$ Now, rationalizing $x+i y=\left(\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right)^{93}-\left(\frac{1-i}{1+i} \times \frac{1-i}{1-i}\right)^{3}$ $=\left(\frac{(1+i)^{2}}{(1-i)(1+i)}\right)^{93}-\left(\frac{(1-i)^{2}}{(1+i)(1-i)}\right)^{3}$ In denominator, we use the identity $(a-b)(a+b)=a^{2}-b^{2}$ $=\...
Read More →Tick (✓) the correct answer
Question: Tick (✓) the correct answer: If an angle of a parallelogram is two-thirds of its adjacent angle, the smallest angle of the parallelogram is (a) 54 (b) 72 (c) 81 (d) 108 Solution: (b) $72^{\circ}$ Let $x^{\circ}$ be the angle of the parallelogram. Sum of the adjacent angles of a parallelogram is $180^{\circ}$. $\therefore x+\left(\frac{2}{3} \times x\right)=180$ $\Rightarrow x+\frac{2 x}{3}=180$ $\Rightarrow\left(x+\frac{2 x}{3}\right)=180$ $\Rightarrow \frac{5 x}{3}=180$ $\Rightarrow x...
Read More →Tick (✓) the correct answer
Question: Tick (✓) the correct answer: The bisectors of any two adjacent angles of a parallelogram intersect at (a) 30 (b) 45 (c) 60 (d) 90 Solution: (d) $90^{\circ}$ The bisectors of any two adjacent angles of a parallelogram intersect at $90^{\circ}$....
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