The circumference of a circle exceeds its diameter by 45 cm.
Question: The circumference of a circle exceeds its diameter by 45 cm. Find the circumference of the circle. Solution: Let the radius of the circle ber.Now,Circumference = Diameter + 45 $\Rightarrow 2 \pi r=2 r+45$ $\Rightarrow 2 \pi r-2 r=45$ $\Rightarrow 2 r\left(\frac{22}{7}-1\right)=45$ $\Rightarrow 2 r \times \frac{15}{7}=45$ $\Rightarrow r=10.5 \mathrm{~cm}$ Circumference = Diameter + 45 = 2(10.5) + 45 = 66 cmHence, the circumference of the circle is 66 cm....
Read More →The circumference of a circle exceeds its diameter by 45 cm.
Question: The circumference of a circle exceeds its diameter by 45 cm. Find the circumference of the circle. Solution: Let the radius of the circle ber.Now,Circumference = Diameter + 45 $\Rightarrow 2 \pi r=2 r+45$ $\Rightarrow 2 \pi r-2 r=45$ $\Rightarrow 2 r\left(\frac{22}{7}-1\right)=45$ $\Rightarrow 2 r \times \frac{15}{7}=45$ $\Rightarrow r=10.5 \mathrm{~cm}$ Circumference = Diameter + 45 = 2(10.5) + 45 = 66 cmHence, the circumference of the circle is 66 cm....
Read More →Solve this
Question: If $A=\left[\begin{array}{cc}\ln x -1 \\ -\ln x 2\end{array}\right]$ and if $\operatorname{det}(A)=2$, then $x=$ _______ Solution: Given: $A=\left[\begin{array}{cc}\ln x -1 \\ -\ln x 2\end{array}\right]$ $\operatorname{det}(A)=2$ Now, $|A|=2$ $\Rightarrow\left|\begin{array}{cc}\ln x -1 \\ -\ln x 2\end{array}\right|=2$ $\Rightarrow 2 \ln x-\ln x=2$ $\Rightarrow \ln x=2$ $\Rightarrow x=\mathrm{e}^{2}$ Hence, $x=\underline{e}^{2}$....
Read More →If ABCD is quadrilateral such that AB = AD
Question: If ABCD is quadrilateral such that AB = AD and CB = CD, then prove that AC is the perpendicular bisector of BD. Solution: Given In quadrilateral ABCD, AB= AD and CB = CD. Construction Join AC and BD. To prove AC is the perpendicular bisector of BD....
Read More →Factorize each of the following algebraic expression:
Question: Factorize each of the following algebraic expression:x2 22x+ 120 Solution: To factorise $\mathrm{x}^{2}-22 \mathrm{x}+120$, we will find two numbers $\mathrm{p}$ and $\mathrm{q}$ such that $\mathrm{p}+\mathrm{q}=-22$ and $\mathrm{pq}=120$. Now, $(-12)+(-10)=-22$ and $(-12) \times(-10)=120$ Splitting the middle term $-22 \mathrm{x}$ in the given quadratic as $-12 \mathrm{x}-10 \mathrm{x}$, we get: $\mathrm{x}^{2}-22 \mathrm{x}+12=\mathrm{x}^{2}-12 \mathrm{x}-10 \mathrm{x}+120$ $=\left(\...
Read More →The area of a circle is 98.56 cm2.
Question: The area of a circle is 98.56 cm2. Find its circumference. Solution: Let the radius of the circle ber.Now, Area $=98.56$ $\Rightarrow \pi r^{2}=98.56$ $\Rightarrow \frac{22}{7} \times r^{2}=98.56$ $\Rightarrow r=5.6$ Now, Circumference $=2 \pi r=2 \times \frac{22}{7} \times 5.6=35.2 \mathrm{~cm}$ Hence, the circumference of the circle is 35.2 cm....
Read More →Prove that in a triangle,
Question: Prove that in a triangle, other than an equilateral triangle, angle opposite the longest side is greater than 2/3 of a right angle. Solution: Consider $\triangle A B C$ in which $B C$ is the longest side. To prove $\angle A=\frac{2}{3}$ right angle Proof $\ln \triangle A B C, \quad B CA B$. [consider $B C$ is the largest side] $\Rightarrow \quad \angle A\angle C \quad \ldots(i)$ [angle opposite the longest side is greatest] and $\quad B CA C$ $\Rightarrow \quad \angle A\angle B .$........
Read More →Solve the following equations
Question: If $A=\left[\begin{array}{cc}\alpha 2 \\ 2 \alpha\end{array}\right]$ and $\left|A^{3}\right|=125$, then $\alpha=$ Solution: Given: $A=\left[\begin{array}{ll}\alpha 2 \\ 2 \alpha\end{array}\right]$ $\left|A^{3}\right|=125$ Now, $\left|A^{3}\right|=125$ $\Rightarrow|A|^{3}=5^{3}$ $\left(\because\left|A^{n}\right|=|A|^{n}\right)$ $\Rightarrow|A|=5$ $\Rightarrow\left|\begin{array}{ll}\alpha 2 \\ 2 \alpha\end{array}\right|=5$ $\Rightarrow \alpha^{2}-4=5$ $\Rightarrow \alpha^{2}=5+4$ $\Right...
Read More →Factorize each of the following algebraic expression:
Question: Factorize each of the following algebraic expression:x2+ 14x+ 45 Solution: To factorise $\mathrm{x}^{2}+14 \mathrm{x}+45$, we will find two numbers $\mathrm{p}$ and $\mathrm{q}$ such that $\mathrm{p}+\mathrm{q}=14$ and $\mathrm{pq}=45$. Now, $9+5=14$ and $9 \times 5=45$ Splitting the middle term $14 \mathrm{x}$ in the given quadratic as $9 \mathrm{x}+5 \mathrm{x}$, we get: $\mathrm{x}^{2}+14 \mathrm{x}+45=\mathrm{x}^{2}+9 \mathrm{x}+5 \mathrm{x}+45$ $=\left(\mathrm{x}^{2}+9 \mathrm{x}\...
Read More →The circumference of a circle is 39.6 cm. Find its area.
Question: The circumference of a circle is 39.6 cm. Find its area. Solution: Circumference = 39.6 cmWe know: Circumference of a circle $=2 \pi r$ $\Rightarrow 39.6=2 \times \frac{22}{7} \times r$ $\Rightarrow \frac{39.6 \times 7}{2 \times 22}=r$ $\Rightarrow r=6.3 \mathrm{~cm}$ Also, Area of the circle $=\pi r^{2}$ $=\frac{22}{7} \times 6.3 \times 6.3$ $=124.74 \mathrm{~cm}^{2}$...
Read More →Factorize each of the following algebraic expression:
Question: Factorize each of the following algebraic expression:a2 14a 51 Solution: To factorise $\mathrm{a}^{2}-14 \mathrm{a}-51$, we will find two numbers $\mathrm{p}$ and $\mathrm{q}$ such that $\mathrm{p}+\mathrm{q}=-14$ and $\mathrm{pq}=-51$. Now, $3+(-17)=-14$ and $3 \times(-17)=-51$ Splitting the middle term $-14 \mathrm{a}$ in the given quadratic as $3 \mathrm{a}-17 \mathrm{a}$, we get: $a^{2}-14 a-51=a^{2}+3 a-17 a-51$ $=\left(a^{2}+3 a\right)-(17 a+51)$ $=a(a+3)-17(a+3)$ $=(a-17)(a+3)$...
Read More →If A and B are square matrices
Question: If $A$ and $B$ are square matrices of order 3 such that $|A|=-1,|B|=3$, then $|3 A B|=$________ Given: $A$ and $B$ are square matrices of order 3 $|A|=-1$ $|B|=3$ Solution: Now, $|3 A B|=|3 A||B| \quad(\because|A B|=|A||B|$, if they are square matrices of same order $) $=|3 A| \times 3 \quad(\because|B|=3)$ $=3^{3}|A| \times 3 \quad(\because$ Order of $A$ is $3 \times 3)$ $=81|A|$ $=81 \times(-1) \quad(\because|A|=-1)$ $=-81$ Hence, $|3 A B|=\underline{-81}$....
Read More →If AB and CD are the smallest and
Question: If AB and CD are the smallest and largest sides of a quadrilateral ABCD, out of B and D decide which is greater. Solution: Given In quadrilateral $A B C D, A B$ is the smallest and $C D$ is the largest side To find $\angle B\angle D$ or $\angle D\angle B$....
Read More →If ABC is a right angled triangle
Question: If ABC is a right angled triangle such that AB = AC and bisector of angle C intersects the side AB at D, then prove that AC + AD = BC. Solution: Given in right angled $\triangle A B C, A B=A C$ and $C D$ is the bisector of $\angle C$. Construction Draw $D E \perp B C$. To prove $A C+A D=B C$ Proof In right angled $\triangle A B C, A B=A C$ and $B C$ is a hypotenuse [given] $\therefore \quad \angle A=90^{\circ}$ In $\triangle D A C$ and $\triangle D E C$, $\angle A=\angle 3=90^{\circ}$...
Read More →Factorize each of the following algebraic expression:
Question: Factorize each of the following algebraic expression:a2+ 3a 88 Solution: To factorise $a^{2}+3 a-88$, we will find two numbers $p$ and $q$ such that $p+q=3$ and $p q=-88$. Now, $11+(-8)=3$ And $11 \times(-8)=-88$ Splitting the middle term $3 \mathrm{a}$ in the given quadratic as $11 \mathrm{a}-8 \mathrm{a}$, we get: $a^{2}+3 a-88=a^{2}+11 a-8 a-88$ $=\left(a^{2}+11 a\right)-(8 a+88)$ $=a(a+11)-8(a+11)$ $=\left(a^{2}+11 a\right)-(8 a+88)$ $=a(a+11)-8(a+11)$ $=(a-8)(a+11)$...
Read More →In a four-sides field, the length of the longer diagonal is 128 m.
Question: In a four-sides field, the length of the longer diagonal is 128 m. The lengths of perpendiculars from the opposite vertices upon this diagonal are 22.7 m and 17.3 m. Find the area of the field. Solution: The field, which is represented as ABCD, is given below. The area of the field is the sum of the areas of triangles ABC and ADC. Area of the triangle $\mathrm{ABC}=\frac{1}{2}(\mathrm{AC} \times \mathrm{BF})=\frac{1}{2}(128 \times 22.7)=1452.8 \mathrm{~m}^{2}$ Area of the triangle $\ma...
Read More →If A is a 2 × 2 matrix such that
Question: If $A$ is a $2 \times 2$ matrix such that $|A|=5$, then $|5 A|=$______ Given: $A$ is a $2 \times 2$ matrix $|A|=5$ Solution: Now, $|5 A|=5^{2}|A| \quad(\because$ Order of $A$ is $2 \times 2)$ $=25|A|$ $=25 \times 5 \quad(\because|A|=5)$ $=125$ Hence, $|5 A|=\underline{125}$....
Read More →If ABCD is a quadrilateral such that diagonal
Question: If ABCD is a quadrilateral such that diagonal AC bisects the angles A and C,then prove that AB = AD and CB = CD. Solution: Given in a quadrilateral $A B C D$, diagonal $A C$ bisects the angles $A$ and $C$. To prove $A B=A D$ and $C B=C D$ Proof in $\triangle A D C$ and $\triangle A B C$,...
Read More →The adjacent sides of a parallelogram are 36 cm and 27 cm in length.
Question: The adjacent sides of a parallelogram are 36 cm and 27 cm in length. If the distance between the shorter sides is 12 cm, find the distance between the longer sides. Solution: Area of a parallelogram $=$ Base $\times$ Height $\therefore A B \times D E=B C \times D F$ $\Rightarrow D E=\frac{B C \times D F}{A B}$ $=\frac{27 \times 12}{36}$ $=9 \mathrm{~cm}$ Distance between the longer sides = 9 cm...
Read More →Factorize each of the following algebraic expression:
Question: Factorize each of the following algebraic expression:40 + 3xx2 Solution: We have: $40+3 x-x^{2}$ $\Rightarrow-\left(x^{2}-3 x-40\right)$ To factorise $\left(\mathrm{x}^{2}-3 \mathrm{x}-40\right)$, we will find two numbers $\mathrm{p}$ and $\mathrm{q}$ such that $\mathrm{p}+\mathrm{q}=-3$ and $\mathrm{pq}=-40$. Now, $5+(-8)=-3$ And $5 \times(-8)=-40$ Splitting the middle term $-3 \mathrm{x}$ in the given quadratic as $5 \mathrm{x}-8 \mathrm{x}$, we get: $40+3 x-x^{2}=-\left(x^{2}-3 x-40...
Read More →The set of real values of a for which the matrix
Question: The set of real values of a for which the matrix $A=\left[\begin{array}{ll}a 2 \\ 2 4\end{array}\right]$ is non-singular is_________ Solution: Given: The matrix $A=\left[\begin{array}{ll}a 2 \\ 2 4\end{array}\right]$ is non-singular $A$ is non-singular $\Rightarrow|A| \neq 0$ Thus, $\left|\begin{array}{ll}a 2 \\ 2 4\end{array}\right| \neq 0$ $\Rightarrow 4 a-4 \neq 0$ $\Rightarrow 4 a \neq 4$ $\Rightarrow a \neq 1$ $\Rightarrow a \in R-\{1\}$ Hence, the set of real values of a for whic...
Read More →Line segment joining the mid-points
Question: Line segment joining the mid-points M and N of parallel sides AB and DC, respectively of a trapezium ABCD is perpendicular to both the sides AB and DC. Prove that AD = BC. Solution: Given In trapezium $A B C D$, points $M$ and $N$ are the mid-points of parallel sides $A B$ and $D C$ respectively and join $M N$, which is perpendicular to $A B$ and $D C$. To prove $A D=B C$ Proof Since, $M$ is the mid-point of $A B$. On multiplying both sides of above equation by $-1$ and than adding $90...
Read More →The diagonals of a rhombus are 48 cm and 20 cm long.
Question: The diagonals of a rhombus are 48 cm and 20 cm long. Find the perimeter of the rhombus. Solution: Diagonals of a rhombus perpendicularly bisect each other. The statement can help us find a side of the rhombus. Consider the following figure. ABCDis the rhombus andACandBDare the diagonals. The diagonals intersect at pointO.We know: $\angle D O C=90^{\circ}$ $D O=O B=\frac{1}{2} D B=\frac{1}{2} \times 48=24 \mathrm{~cm}$ Similarly, $A O=O C=\frac{1}{2} A C=\frac{1}{2} \times 20=10 \mathrm...
Read More →The diagonals of a rhombus are 48 cm and 20 cm long.
Question: The diagonals of a rhombus are 48 cm and 20 cm long. Find the perimeter of the rhombus. Solution: Diagonals of a rhombus perpendicularly bisect each other. The statement can help us find a side of the rhombus. Consider the following figure. ABCDis the rhombus andACandBDare the diagonals. The diagonals intersect at pointO.We know: $\angle D O C=90^{\circ}$ $D O=O B=\frac{1}{2} D B=\frac{1}{2} \times 48=24 \mathrm{~cm}$ Similarly, $A O=O C=\frac{1}{2} A C=\frac{1}{2} \times 20=10 \mathrm...
Read More →If A and B are non-singular square matrices
Question: If $A$ and $B$ are non-singular square matrices of order $n$ such that $A=k B$, then $\frac{|A|}{|B|}=$_______ Solution: Given: $A$ and $B$ are non-singular square matrices of order $n$. $A=k B$ $A=k B$ Taking determinant on both sides, we get $\Rightarrow|A|=|k B|$ $\Rightarrow|A|=k^{n}|B| \quad(\because$ Order of $B$ is $n \times n)$ $\Rightarrow \frac{|A|}{|B|}=k^{n}$ Hence, $\frac{|A|}{|B|}=\underline{k}^{n}$....
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