Two adjacent angles of a parallelogram are in the ratio 2 : 3.
Question: Two adjacent angles of a parallelogram are in the ratio 2 : 3. Find the measure of each of its angles. Solution: Let the two adjacent angles of the parallelogram be $(2 x)^{\circ}$ and $(3 x)^{\circ}$. Sum of any two adjacent angles of a parallelogram is $180^{\circ}$. $\therefore 2 x+3 x=180$ $\Rightarrow 5 x=180$ $\Rightarrow x=36$ $(2 x)^{\circ}=(2 \times 36)^{\circ}=72^{\circ}$ $(3 x)^{\circ}=(3 \times 36)^{\circ}=108^{\circ}$ Measures of the angles are $72^{\circ}$ and $108^{\circ...
Read More →The angles of a quadrilateral are in the ratio 1 : 2 : 3 : 4.
Question: The angles of a quadrilateral are in the ratio 1 : 2 : 3 : 4. Find the measure of each angle. Solution: Let the angles be $(x)^{\circ},(2 x)^{\circ},(3 x)^{\circ}$ and $(4 x)^{\circ}$. Sum of the angles of a quadrilateral is $360^{\circ}$. $x+2 x+3 x+4 x=360$ $10 x=360$ $x=\frac{360}{10}$ $x=36$ $(2 x)^{\circ}=(2 \times 36)^{\circ}=72^{\circ}$ $(3 x)^{\circ}=(3 \times 36)^{\circ}=108^{\circ}$ $(4 x)^{\circ}=(4 \times 36)^{\circ}=144^{\circ}$ The angles of the quadrilateral are $36^{\ci...
Read More →Define the terms:
Question: Define the terms: (i) Open curve (ii) Closed curve (iii) Simple closed curve Solution: (i) Open curve: An open curve is a curve where the beginning and end pointsare different. Example: Parabola(ii) Closed Curve: A curve that joins up so there are no end points.Example: Ellipse(iii) Simple closed curve: A closed curve that does not intersect itself....
Read More →Use the distributivity of multiplication of rational numbers
Question: Use the distributivity of multiplication of rational numbers over addition to simplify. Solution: (a) (3/5) [(35/24) + (10/1)] We know that the distributivity of multiplication of rational numbers over addition, a (b + c) = a b + a c Where, a =3/5, b =35/24, c = 10/1 Then, (3/5) [(35/24) + (10/1)] = ((3/5) (35/24)) + ((3/5) (10/1)) = ((1/1) (7/8)) + ((3/1) (2/1)) = (7/8) + (6/1) = (7 + 48)/8 = 55/8 = (b) (-5/4) [(8/5) + (16/15)] Solution:- We know that the distributivity of multiplicat...
Read More →Draw a trapezium ABCD in which AB||DC, AB = 7 cm,
Question: Draw a trapeziumABCDin whichAB||DC,AB= 7 cm,BC= 5 cm,AD= 6.5 cm and B= 60. Solution: Steps of construction: Step1: Draw $A B$ equal to $7 \mathrm{~cm}$. Step2: Make an angle, $\angle A B X$, equal to $60^{\circ}$. Step3: With $B$ as the centre, draw an arc of $5 \mathrm{~cm}$. Name that point as $C$. Join $B$ and $C$. Step4: $A B \| D C$ $\therefore \angle A B X+\angle B C Y=180^{\circ}$ $\Rightarrow \angle B C Y=180^{\circ}-60^{\circ}=120^{\circ}$ Draw an angle, $\angle B C Y$, equal ...
Read More →Verify the property x × (y × z) = (x × y) × z
Question: Verify the property x (y z) = (x y) z of rational numbers by using Solution: (a) x = 1, y = - and z = In the question is given to verify the property x (y z) = (x y) z The arrangement of the given rational number is as per the rule of associative property for multiplication. Then, 1 (- ) = (1 -) LHS = 1 (- ) = 1 (-1/8) = -1/8 RHS = (1 -) = (-) = -1/8 By comparing LHS and RHS LHS = RHS -1/8 = -1/8 Hence x (y z) = (x y) z (b) x = 2/3, y = -3/7 and z = Solution:- In the question is given ...
Read More →Construct a trapezium ABCD in which AB = 6 cm,
Question: Construct a trapeziumABCDin whichAB= 6 cm,BC= 4 cm,CD= 3.2 cm, B= 75 andDC||AB. Solution: Steps of construction: Step 1: Draw $A B=6 \mathrm{~cm}$ Step 2: Make $\angle A B X=75^{\circ}$ Step 3: With $B$ as the centre, draw an arc at $4 \mathrm{~cm}$. Name that point as $C$. Step 4: $A B \| C D$ $\therefore \angle A B X+\angle B C Y=180^{\circ}$ $\Rightarrow \angle B C Y=180^{\circ}-75^{\circ}=105^{\circ}$ Make $\angle B C Y=105^{\circ}$ At $C$, draw an arc of length $3.2 \mathrm{~cm}$....
Read More →Draw a rhombus whose side is 7.2 cm
Question: Draw a rhombus whose side is 7.2 cm and one angle is 60. Solution: Steps of construction: Step1: Draw $A B=7.2 \mathrm{~cm}$ Step2: Draw $\angle A B Y=60^{\circ}$ $\angle B A X=120^{\circ}$ Sum of the adjacent angles is 180 $\angle B A X+\angle A B Y=180^{\circ}$ $=\angle B A X=180^{\circ}-60^{\circ}=120^{\circ}$ Step 3: Set off $A D(7.2 \mathrm{~cm})$ along $A X$ and $B C(7.2 \mathrm{~cm})$ along $B Y$. Step 4: Join $C$ and $D$. Then,ABCDis the required rhombus....
Read More →Construct a rhombus ABCD in which AB = 4 cm
Question: Construct a rhombusABCDin whichAB= 4 cm and diagonalACis 6.5 cm. Solution: Steps of construction: Step 1: DrawAB= 4 cm Step 2: WithBas the centre, draw an arc of4 cm. Step 3: WithAas the centre, draw another arc of 6.5 cm, cutting the previous arc atC. Step 4: JoinACandBC. Step 5: WithCas the centre, draw an arc of 4 cm. Step 6:WithAas the centre, draw another arc of 4 cm,cutting the previous arc atD. Step 7: JoinADandCD. ABCDis the required rhombus....
Read More →Verify the property x × y = y × z
Question: Verify the property x y = y z of rational numbers by using Solution: (a) x = 7 and y = In the question is given to verify the property = x y = y x Where, x = 7, y = Then, 7 = 7 LHS = 7 = 7/2 RHS = 7 = 7/2 By comparing LHS and RHS LHS = RHS 7/2 = 7/2 Hence x y = y x (b) x = 2/3 and y = 9/4 Solution:- In the question is given to verify the property = x y = y x Where, x = 2/3, y = 9/4 Then, (2/3) (9/4) = (9/4) (2/3) LHS = (2/3) (9/4) = (1/1) (3/2) = 3/2 RHS = (9/4) (2/3) = (3/2) (1/1) = 3...
Read More →Construct a rhombus the lengths of whose diagonals are 6 cm and 8 cm.
Question: Construct a rhombus the lengths of whose diagonals are 6 cm and 8 cm. Solution: We know that the diagonals of a rhombus bisect each other. Steps of construction: Step 1: Draw AC= 6cm Step 2:Draw a perpendicular bisector(XY) of AC, which bisects AC at O. Step 3:$O B=\frac{1}{2}(8) \mathrm{cm}$ $O B=4 \mathrm{~cm}$ and $O D=\frac{1}{2}(8) \mathrm{cm}$ $O D=4 \mathrm{~cm}$ Draw an arc of length 4 cm onOXand name that point asB. Draw an arc of length 4 cm onOYand name that point asD. Step ...
Read More →Construct a rectangle PQRS in which QR = 3.6 cm and diagonal PR = 6 cm.
Question: Construct a rectanglePQRSin whichQR= 3.6 cm and diagonalPR= 6 cm. Measure the other side of the rectangle. Solution: Steps of construction: Step 1: Draw $Q R=3.6 \mathrm{~cm}$ Step 2: Make $\angle Q=90^{\circ}$ $\angle R=90^{\circ}$ Step 3: $P R^{2}=P Q^{2}+Q R^{2}$ $6^{2}=P Q^{2}+3.6^{2}$ $P Q^{2}=36-12.96$ $P Q^{2}=23.04$ $P Q=4.8 \mathrm{~cm}$ Step 3: Draw an arc of length 4.8 cm from pointQand name that point asP. Step 4: Draw an arc of length 6 cm from pointR, cutting the previou...
Read More →Construct a square, each of whose diagonals measures 5.8 cm.
Question: Construct a square, each of whose diagonals measures 5.8 cm. Solution: We know that the diagonals of a square bisect each other at right angles. Steps of construction: Step 1: Draw $A C=5.8 \mathrm{~cm}$ Step 2: Draw the perpendicular bisector $X Y$ of $A C$, meeting it at $O$. Step 3:From $O$ : $O B=\frac{1}{2}(5.8) \mathrm{cm}=2.9 \mathrm{~cm}$ $O D=\frac{1}{2}(5.8) \mathrm{cm}=2.9 \mathrm{~cm}$ Step 4: JoinAB, BC, CDandDA. ABCDis the required square....
Read More →Tell which property allows you to compute
Question: Tell which property allows you to compute (1/5) [(5/6) (7/9)] as [(1/5) (5/6)] (7/9) Solution: The arrangement of the given rational number is as per the rule of Associative property for Multiplication....
Read More →Construct a square, each of whose sides measures 6.4 cm.
Question: Construct a square, each of whose sides measures 6.4 cm. Solution: All the sides of a square are equal. Steps of construction: Step 1: Draw $A B=6.4 \mathrm{~cm}$ Step 2: Make $\angle A=90^{\circ}$ $\angle B=90^{\circ}$ Step 3: Draw an arc of length 6.4 cm from pointAand name that point asD. Step 4: Draw an arc of length 6.4 cm from pointBand name that point asC. Step 5: JoinCandD. Thus,ABCDis a required square....
Read More →Simplify each of the following by using suitable property.
Question: Simplify each of the following by using suitable property. Also name the property. (a) [() ()] + [() 6] Solution: (a) [() ()] + [() 6] The arrangement of the given rational number is as per the rule of distributive law over addition. Now take out as common. Then, = [ + 6] = [(1 + 24)/4] = [25/24] = (25/24) = 25/8 (b)[(1/5) (2/15)] [(1/5) (2/5)] Solution:- The arrangement of the given rational number is as per the rule of distributive law over subtraction. Now take out 1/5 as common. Th...
Read More →If a = (cosθ + i sinθ), prove that
Question: If $\mathrm{a}=(\cos \theta+\mathrm{i} \sin \theta)$, prove that $\frac{1+\mathrm{a}}{1-\mathrm{a}}=\left(\cot \frac{\theta}{2}\right) \mathrm{i}$ Solution: Given: a = cos + isin To prove: $\frac{1+a}{1-a}=\left(\cot \frac{\theta}{2}\right) i$ Taking LHS, $\frac{1+a}{1-a}$ Putting the value of a, we get $=\frac{1+\cos \theta+i \sin \theta}{1-(\cos \theta+i \sin \theta)}$ $=\frac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta}$ We know that, $1+\cos 2 \theta=2 \cos ^{2} \theta...
Read More →Construct a rectangle ABCD whose adjacent sides are 11 cm and 8.5 cm.
Question: Construct a rectangleABCDwhose adjacent sides are 11 cm and 8.5 cm. Solution: Steps of construction: Step 1: Draw $A B=11 \mathrm{~cm}$ Step 2: Make $\angle A=90^{\circ}$ $\angle B=90^{\circ}$ Step 3: Draw an arc of 8.5 cm from pointAand name that point asD.Step 4: Draw an arc of 8.5 cm from pointBand name that point asC.Step 5: JoinCandD.Thus,ABCDis the required rectangle....
Read More →Construct a parallelogram ABCD, in which diagonal AC = 3.8 cm,
Question: Construct a parallelogramABCD,in which diagonalAC= 3.8 cm, diagonalBD= 4.6 cm and the angle betweenACandBDis 60. Solution: We know that the diagonals of a parallelogram bisect each other. Steps of construction: Step 1: Draw $A C=3.8 \mathrm{~cm}$ Step 2: Bisect $A C$ at $O$. Step 3: Make $\angle C O X=60^{\circ}$ Produce $X O$ to $Y$. Step 4: $O B=\frac{1}{2}(4.6) \mathrm{cm}$ $O B=2.3 \mathrm{~cm}$ and $O D=\frac{1}{2}(4.6) \mathrm{cm}$ $O D=2.3 \mathrm{~cm}$ Sten 5. Join $A B, B C, C...
Read More →Construct a parallelogram ABCD in which AB = 6.5 cm,
Question: Construct a parallelogramABCDin whichAB= 6.5 cm,AC= 3.4 cm and the altitudeALfromAis 2.5 cm. Draw the altitude fromCand measure it. Solution: Steps of construction: Step 1: DrawAB= 6.5cm Step 2: Draw a perpendicular at pointA. Name that ray asAX. From pointA,draw an arc of length 2.5 cm on the rayAXand name that point as L. Step 3: On pointL,make a perpendicular. Draw a straight line YZ passing through L, which is perpendicular to the ray AX. Step 4: Cut an arc of length 3.4 cm on the ...
Read More →Construct a parallelogram, one of whose sides is 4.4 cm and whose diagonals are 5.6 cm and 7 cm.
Question: Construct a parallelogram, one of whose sides is 4.4 cm and whose diagonals are 5.6 cm and 7 cm. Measure the other side. Solution: We know that the diagonals of a parallelogram bisect each other. Steps of construction: Step 1: Draw $A B=4.4 \mathrm{~cm}$ Step 2: With $A$ as the centre and radius $2.8 \mathrm{~cm}$, draw an arc. Step 3: With $B$ as the centre and radius $3.5 \mathrm{~cm}$, draw another arc, cutting the previous arc at point $O$. Step 4: Join $O A$ and $O B$. Step 5: Pro...
Read More →Construct a parallelogram ABCD in which BC = 5 cm,
Question: Construct a parallelogramABCDin whichBC= 5 cm,BCD= 120 andCD= 4.8 cm. Solution: Steps of construction: Step 1: Draw BC= $5 \mathrm{~cm}$ Step 2: Make an $\angle B C D=120^{\circ}$ Step 2: With $C$ as centre draw an arc of $4.8 \mathrm{~cm}$, name that point as $D$ Step 3: With $D$ as centre draw an $\operatorname{arc} 5 \mathrm{~cm}$, name that point as $A$ Step 4: With $B$ as centre draw another $\operatorname{arc} 4.8 \mathrm{~cm}$ cutting the previous arc at $A$. Step 5: Join $A D$ ...
Read More →Construct a parallelogram PQRS in which QR = 6 cm,
Question: Construct a parallelogramPQRSin whichQR= 6 cm,PQ= 4 cm andPQR= 60 cm. Solution: Steps of construction: Step 1: Draw $P Q=4 \mathrm{~cm}$ Step 2: Make $\angle P Q R=60^{\circ}$ Step 3: With Qas the centre, draw an arc of 6 cm and name that point asR. Step 4: With Ras the centre, draw an arc of 4 cm and name that point asS. Step 5: Join SRandPS. Then,PQRSis the required parallelogram....
Read More →Prove that
Question: Prove that $(x+1+i)(x+1-i)(x-1-i)(x-1-i)=\left(x^{4}+4\right)$ Solution: To Prove: $(x+1+i)(x+1-i)(x-1+i)(x-1-i)=\left(x^{4}+4\right)$ Taking LHS $(x+1+i)(x+1-i)(x-1+i)(x-1-i)$ $=[(x+1)+i][(x+1)-i][(x-1)+i][(x-1)-i]$ Using $(a-b)(a+b)=a^{2}-b^{2}$ $=\left[(x+1)^{2}-(i)^{2}\right]\left[(x-1)^{2}-(i)^{2}\right]$ $=\left[x^{2}+1+2 x-i^{2}\right]\left(x^{2}+1-2 x-i^{2}\right]$ $=\left[x^{2}+1+2 x-(-1)\right]\left(x^{2}+1-2 x-(-1)\right]\left[\because i^{2}=-1\right]$ $=\left[x^{2}+2+2 x\ri...
Read More →Construct a parallelogram ABCD in which AB = 4.3 cm,
Question: Construct a parallelogramABCDin whichAB= 4.3 cm,AD= 4 cm andBD= 6.8 cm. Solution: Steps of construction: Step 1: Draw $A B=4.3 \mathrm{~cm}$ Step 2: With $B$ as the centre, draw an arc of $6.8 \mathrm{~cm}$. Step 3: With $A$ as the centre, draw another $\operatorname{arc}$ of $4 \mathrm{~cm}$, cutting the previous arc at $D$. Step 4: Join $B D$ and $A D$. Step 5: We know that the opposite sides of a parallelogram are equal. Thus, with $D$ as the centre, draw an arc of $4.3 \mathrm{~cm}...
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