solve the following If cos 2θ = 0,
Question: If $\cos 2 \theta=0$, then $\left|\begin{array}{ccc}0 \cos \theta \sin \theta \\ \cos \theta \sin \theta 0 \\ \sin \theta 0 \cos \theta\end{array}\right|^{2}=$___________ Solution: Given: $\cos 2 \theta=0$ $\cos 2 \theta=0$ $\Rightarrow \cos ^{2} \theta-\sin ^{2} \theta=0$ $\Rightarrow \cos ^{2} \theta=\sin ^{2} \theta$ $\Rightarrow \theta=\pm \frac{\pi}{4} \quad \ldots(1)$ Now, $\left|\begin{array}{ccc}0 \cos \theta \sin \theta \\ \cos \theta \sin \theta 0 \\ \sin \theta 0 \cos \theta...
Read More →Find the greatest common factor (GCF/HCF) of the following polynomial:
Question: Find the greatest common factor (GCF/HCF) of the following polynomial:2x2and 12x2 Solution: The numerical coefficients of the given monomials are 2 and 12. So, the greatest common factor of 2 and 12 is 2. The common literal appearing in the given monomials is x. The smallest power of x in the two monomials is 2. The monomial of the common literals with the smallest powers is x2. Hence, the greatest common factor is 2x2....
Read More →Diagonals of a parallelogram
Question: Diagonals of a parallelogram are perpendicular to each other. Is this statement true? Give reason for your answer. Solution: No, diagonals of a parallelogram are not perpendicular to each other, because they only bisect each other....
Read More →Resolve each of the following quadratic trinomial into factor:
Question: Resolve each of the following quadratic trinomial into factor:(2a b)2+ 2(2a b) 8 Solution: Assuming $x=2 a-b$, we have : $(2 a-b)^{2}+2(2 a-b)-8=x^{2}+2 x-8$ The given expression becomes $x^{2}+2 x-8 . \quad$ (Coefficient of $x^{2}=1$ and that of $x=2 ;$ constant term $=-8$ ) Now, we will split the coefficient of $x$ into two parts such that their sum is 2 and their product equals the product of the coefficient of $x^{2}$ and the cons $\tan t$ term, i.e., $1 \times(-8)=-8$. Clearly, $(...
Read More →Diagonals AC and BD of a parallelogram
Question: Diagonals AC and BD of a parallelogram ABCD intersect each other at O. If OA = 3 cm and OD = 2 cm, determine the lengths of AC and BD. Solution: Given, $A B C D$ is a parallelogram $O A=3 \mathrm{~cm}$ and $O D=2 \mathrm{~cm}$ We know that, diagonals of a parallelogram bisect each other. $\therefore \quad$ Diagonal $A C=2 O A=6 \mathrm{~cm} \quad[\because A O=O C]$ and Diagonal $B D=2 O D=4 \mathrm{~cm}$ $[\because B O=O D]$ Hence, the length of the diagonals $A C$ and $B D$ are $6 \ma...
Read More →A chord 10 cm long is drawn in a circle whose radius is
Question: A chord $10 \mathrm{~cm}$ long is drawn in a circle whose radius is $5 \sqrt{2} \mathrm{~cm}$. Find the areas of both the segments. Solution: Let O be the centre of the circle and AB be the chord. Consider $\Delta \mathrm{OAB}$. $\mathrm{OA}=\mathrm{OB}=5 \sqrt{2} \mathrm{~cm}$ $\mathrm{OA}^{2}+\mathrm{OB}^{2}=50+50=100$ Now, $\sqrt{100}=10 \mathrm{~cm}=A B$ Thus, $\triangle O A B$ is a right isosceles triangle. Thus, we have: Area of $\Delta O A B=\frac{1}{2} \times 5 \sqrt{2} \times ...
Read More →D and E are the mid-points of the sides AB and AC,
Question: D and E are the mid-points of the sides AB and AC, respectively, of ΔABC. DE is produced to F. To prove that CF is equal and parallel to DA, we need an additional information which is (a)DAE = EFC (b)AE = EF (c)DE = EF (d)ADE = ECF Solution: (c) in $\triangle A D E$ and $\triangle C F E$, suppose $D E=E F$ Now. $A E=C E$ [since, $E$ is the mid-point of $A C$ ] Suppose $D E=E F$ and $\quad \angle A E D=\angle F E C \quad$ [vertically opposite angles] $\begin{array}{lll}\therefore \Delta...
Read More →Solve this
Question: If $f(x)=\left|\begin{array}{lll}(1+x)^{17} (1+x)^{19} (1+x)^{23} \\ (1+x)^{23} (1+x)^{29} (1+x)^{34} \\ (1+x)^{41} (1+x)^{43} (1+x)^{47}\end{array}\right|$ $=A+B x+C x^{2}+$_________then $A=$____________ Solution: Let $f(x)=\left|\begin{array}{lll}(1+x)^{17} (1+x)^{19} (1+x)^{23} \\ (1+x)^{23} (1+x)^{29} (1+x)^{34} \\ (1+x)^{41} (1+x)^{43} (1+x)^{47}\end{array}\right|$ $f(x)=\left|\begin{array}{lll}(1+x)^{17} (1+x)^{19} (1+x)^{23} \\ (1+x)^{23} (1+x)^{29} (1+x)^{34} \\ (1+x)^{41} (1+x...
Read More →Resolve each of the following quadratic trinomial into factor:
Question: Resolve each of the following quadratic trinomial into factor:(x 2y)2 5(x 2y) + 6 Solution: The given expression is $\mathrm{a}^{2}-5 \mathrm{a}+6$. Assuming $\mathrm{a}=\mathrm{x}-2 \mathrm{y}$, we have : $(\mathrm{x}-2 \mathrm{y})^{2}-5(\mathrm{x}-2 \mathrm{y})+6=\mathrm{a}^{2}-5 \mathrm{a}+6 \quad$ (Coefficient of $\mathrm{a}^{2}=1$, coefficient of $\mathrm{a}=-5$ and constant term $\left.=6\right)$ Now, we will split the coefficient of a into two parts such that their sum is $-5$ a...
Read More →Find the lengths of the arcs cut off from a circle of radius 12 cm by a chord 12 cm long.
Question: Find the lengths of the arcs cut off from a circle of radius 12 cm by a chord 12 cm long. Also, find the area of the minor segment. Solution: Let AB be the chord. Joining A and B to O, we get an equilateral triangle OAB.Thus, we have: $\angle O=\angle A=\angle B=60^{\circ}$ Length of the arc ACB: $2 \pi \times 12 \times \frac{60}{360}$ $=4 \pi$ $=12.56 \mathrm{~cm}$ Length of the arc ADB: Circumference of the circle - Length of the arc ACB $=2 \pi \times 12-4 \pi$ $=20 \pi \mathrm{cm}$...
Read More →Find the lengths of the arcs cut off from a circle of radius 12 cm by a chord 12 cm long.
Question: Find the lengths of the arcs cut off from a circle of radius 12 cm by a chord 12 cm long. Also, find the area of the minor segment. Solution: Let AB be the chord. Joining A and B to O, we get an equilateral triangle OAB.Thus, we have: $\angle O=\angle A=\angle B=60^{\circ}$ Length of the arc ACB: $2 \pi \times 12 \times \frac{60}{360}$ $=4 \pi$ $=12.56 \mathrm{~cm}$ Length of the arc ADB: Circumference of the circle - Length of the arc ACB $=2 \pi \times 12-4 \pi$ $=20 \pi \mathrm{cm}$...
Read More →Which of the following is not true for a parallelogram?
Question: Which of the following is not true for a parallelogram? (a)Opposite sides are equal (b)Opposite angles are equal (c)Opposite angles are bisected by the diagonals (d)Diagonals bisect each other Solution: (c)We know that, in a parallelogram, opposite sides are equal, opposite angles are equal, opposite angles are not bisected by the diagonals and diagonals bisect each other....
Read More →Resolve each of the following quadratic trinomial into factor:
Question: Resolve each of the following quadratic trinomial into factor:15x2 16xyz 15y2z2 Solution: The given expression is $15 x^{2}-16 x y z-15 y^{2} z^{2}$. (Coefficient of $x^{2}=15$, coefficient of $x=-16 y z$ and constant term $=-15 y^{2} z^{2}$ ) Now, we will split the coefficient of $x$ into two parts such that their sum is $-16 y z$ and their product equals the product of the coefficient of $x^{2}$ and the cons $\tan t$ term, i.e., $15 \times\left(-15 y^{2} z^{2}\right)=-225 y^{2} z^{2}...
Read More →The radius of a circle with centre O is 7 cm.
Question: The radius of a circle with centre O is 7 cm. Two radii OA and OB are drawn at right angles to each other. Find the areas of minor and majorsegments. Solution: Area of minor segment = Area of sector AOBC Area of right triangle AOB $=\frac{90^{\circ}}{360^{\circ}} \pi(\mathrm{OA})^{2}-\frac{1}{2} \times \mathrm{OA} \times \mathrm{OB}$ $=\frac{1}{4} \times \frac{22}{7} \times(7)^{2}-\frac{1}{2} \times 7 \times 7$ $=\frac{1}{4} \times \frac{22}{7} \times(7)^{2}-\frac{1}{2} \times 7 \times...
Read More →The diagonals AC and BD of a parallelogram
Question: The diagonals AC and BD of a parallelogram ABCD intersect each other at the point 0. If DAC = 32 and AOB = 70, then DBC is equal to (a)24 (b)86 (c)38 (d)32 Solution: (c) Given, $\angle A O B=70^{\circ}$ and $\angle D A C=32^{\circ}$ $\therefore \quad \angle A C B=32^{\circ} \quad[A D \| B C$ and $A C$ is transversal $]$ Now, $\quad \angle A O B+\angle B O C=180^{\circ} \quad$ [linear pair axiom] $\Rightarrow \quad \angle B O C=180^{\circ}-\angle A O B=180^{\circ}-70^{\circ}=110^{\circ}...
Read More →If x, y, z ∈ R, the value of the determinant ∣∣∣∣
Question: If $x, y, z \in R$, the value of the determinant$\left|\begin{array}{lll}\left(2^{x}+2^{-x}\right)^{2} \left(2^{x}-2^{-x}\right)^{2} 1 \\ \left(3^{x}+3^{-x}\right)^{2} \left(3^{x}-3^{-x}\right)^{2} 1 \\ \left(4^{x}+4^{-x}\right)^{2} \left(4^{x}-4^{-x}\right)^{2} 1\end{array}\right|$is equal to_____________ Solution: Let $\Delta=\left|\begin{array}{lll}\left(2^{x}+2^{-x}\right)^{2} \left(2^{x}-2^{-x}\right)^{2} 1 \\ \left(3^{x}+3^{-x}\right)^{2} \left(3^{x}-3^{-x}\right)^{2} 1 \\ \left(...
Read More →The radius of a circle with centre O is 7 cm.
Question: The radius of a circle with centre O is 7 cm. Two radii OA and OB are drawn at right angles to each other. Find the areas of minor and majorsegments. Solution: Area of minor segment = Area of sector AOBC Area of right triangle AOB $=\frac{90^{\circ}}{360^{\circ}} \pi(\mathrm{OA})^{2}-\frac{1}{2} \times \mathrm{OA} \times \mathrm{OB}$ $=\frac{1}{4} \times \frac{22}{7} \times(7)^{2}-\frac{1}{2} \times 7 \times 7$ $=\frac{1}{4} \times \frac{22}{7} \times(7)^{2}-\frac{1}{2} \times 7 \times...
Read More →Resolve each of the following quadratic trinomial into factor:
Question: Resolve each of the following quadratic trinomial into factor:36a2+ 12abc 15b2c2 Solution: The given expression is $36 \mathrm{a}^{2}+12 \mathrm{abc}-15 \mathrm{~b}^{2} \mathrm{c}^{2} . \quad$ (Coefficient of $\mathrm{a}^{2}=36$, coefficient of $\mathrm{a}=12 \mathrm{bc}$ and constant term $=-15 \mathrm{~b}^{2} \mathrm{c}^{2}$ ) Now, we will split the coefficient of a into two parts such that their sum is 12 bc and their product equals the product of the coefficient of $\mathrm{a}^{2}$...
Read More →The length of an arc of a circle, subtending an angle of 54° at the centre, is 16.5 cm.
Question: The length of an arc of a circle, subtending an angle of 54 at the centre, is 16.5 cm. Calculate the radius, circumference and area of the circle. Solution: Length of the arc = 16.5 cm $\theta=54^{\circ}$ Radius = ?Circumference=?We know: Length of the $\operatorname{arc}=\frac{2 \pi \mathrm{r} \theta}{360}$ $\Rightarrow 16.5=\frac{2 \times \frac{22}{7} \times r \times 54}{360}$ $\Rightarrow r=\frac{16.5 \times 360 \times 7}{44 \times 54}$ $\Rightarrow r=17.5 \mathrm{~cm}$ $c=2 \pi \ma...
Read More →Resolve each of the following quadratic trinomial into factor:
Question: Resolve each of the following quadratic trinomial into factor:6a2+ 17ab3b2 Solution: The given expression is $6 \mathrm{a}^{2}+17 \mathrm{ab}-3 \mathrm{~b}^{2} . \quad$ (Coefficient of $\mathrm{a}^{2}=6$, coefficient of $\mathrm{a}=17 \mathrm{~b}$ and constant term $\left.=-3 b^{2}\right)$ Now, we will split the coefficient of a into two parts such that their sum is $17 \mathrm{~b}$ and their product equals the product of the coefficient of $\mathrm{a}^{2}$ and the constant term, i.e.,...
Read More →A chord PQ of a circle of radius 10 cm subtends an angle of 60° at the centre of the circle.
Question: A chord PQ of a circle of radius 10 cm subtends an angle of 60 at the centre of the circle. Find the area of major and minor segments of the circle. Solution: Radius of the circle,r = 10cm Area of sector $O P R Q$ $=\frac{60^{\circ}}{360^{\circ}} \times \pi r^{2}$ $=\frac{1}{6} \times 3.14 \times(10)^{2}$ $=52.33 \mathrm{~cm}^{2}$ In ΔOPQ,OPQ = OQP (As OP = OQ)OPQ + OQP + POQ = 1802OPQ = 120OPQ = 60ΔOPQ is an equilateral triangle.So, area of ΔOPQ $=\frac{\sqrt{3}}{4} \times(\text { Sid...
Read More →A chord PQ of a circle of radius 10 cm subtends an angle of 60° at the centre of the circle.
Question: A chord PQ of a circle of radius 10 cm subtends an angle of 60 at the centre of the circle. Find the area of major and minor segments of the circle. Solution: Radius of the circle,r = 10cm Area of sector $O P R Q$ $=\frac{60^{\circ}}{360^{\circ}} \times \pi r^{2}$ $=\frac{1}{6} \times 3.14 \times(10)^{2}$ $=52.33 \mathrm{~cm}^{2}$ In ΔOPQ,OPQ = OQP (As OP = OQ)OPQ + OQP + POQ = 1802OPQ = 120OPQ = 60ΔOPQ is an equilateral triangle.So, area of ΔOPQ $=\frac{\sqrt{3}}{4} \times(\text { Sid...
Read More →Resolve each of the following quadratic trinomial into factor:
Question: Resolve each of the following quadratic trinomial into factor:14x2+ 11xy 15y2 Solution: The given expression is $14 x^{2}+11 x y-15 y^{2} . \quad$ (Coefficient of $x^{2}=14$, coefficient of $x=11 y$ and constant term $=-15 \mathrm{y}^{2}$ ) Now, we will split the coefficient of $x$ into two parts such that their sum is $11 y$ and their product equals the product of the coefficient of $\mathrm{x}^{2}$ and the constant term, i.e., $14 \times\left(-15 \mathrm{y}^{2}\right)=-210 \mathrm{y}...
Read More →The value of the determinant
Question: The value of the determinant $\left|\begin{array}{ccc}\cos (x+y) -\sin (x+y) \cos 2 y \\ \sin x \cos x \sin y \\ -\cos x \sin x \sin y\end{array}\right|$ depends on_____________ only. Solution: Let $\Delta=\left|\begin{array}{ccc}\cos (x+y) -\sin (x+y) \cos 2 y \\ \sin x \cos x \sin y \\ -\cos x \sin x \sin y\end{array}\right|$ $\Delta=\left|\begin{array}{ccc}\cos (x+y) -\sin (x+y) \cos 2 y \\ \sin x \cos x \sin y \\ -\cos x \sin x \sin y\end{array}\right|$ Applying $R_{2} \rightarrow ...
Read More →The figure formed by joining the
Question: The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only, if (a)ABCD is a rhombus (b)diagonals of ABCD are equal (c)diagonals of ABCD are equal and perpendicular (d)diagonals of ABCD are perpendicular Solution: (c) Given, $A B C D$ is a quadrilateral and $P, Q, R$ and $S$ are the mid-points of sides of $A B, B C$, $C D$ and $D A$, respectively. Then, $P Q R S$ is a square. $\therefore \quad P Q=Q R=R S=P S \quad$ .......(i) and ...
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