Factorize the following:
Question: Factorize the following:20a12b2 15a8b4 Solution: The greatest common factor of the terms $20 a^{12} b^{2}$ and $-15 a^{8} b^{4}$ of the expression $20 a^{12} b^{2}-15 a^{8} b^{4}$ is $5 a^{8} b^{2}$. $20 a^{12} b^{2}=5 \times 4 \times a^{8} \times a^{4} \times b^{2}=5 a^{8} \times b^{2} \times 4 a^{4}$ and $-15 a^{8} b^{4}=5 \times-3 \times a^{8} \times b^{2} \times b^{2}=5 a^{8} b^{2} \times-3 b^{2}$ Hence, the expression 20a12b2-15a8b4can be factorised as5a8b2(4a4-3b2)...
Read More →The determinant
Question: The determinant $A=\left|\begin{array}{ccc}\sqrt{23}+\sqrt{3} \sqrt{5} \sqrt{5} \\ \sqrt{15}+\sqrt{46} 5 \sqrt{10} \\ 3+\sqrt{115} \sqrt{15} 5\end{array}\right|$ is equal to is equal to Solution: Let $A=\left|\begin{array}{ccc}\sqrt{23}+\sqrt{3} \sqrt{5} \sqrt{5} \\ \sqrt{15}+\sqrt{46} 5 \sqrt{10} \\ 3+\sqrt{115} \sqrt{15} 5\end{array}\right|$ $A=\left|\begin{array}{ccc}\sqrt{23}+\sqrt{3} \sqrt{5} \sqrt{5} \\ \sqrt{15}+\sqrt{46} 5 \sqrt{10} \\ 3+\sqrt{115} \sqrt{15} 5\end{array}\right|...
Read More →Factorize the following:
Question: Factorize the following:5x 15x2 Solution: The greatest common factor of the terms 5xand 15x2of the expression5x - 15x2is 5x. Now, $5 x=5 x \times 1$ and $-15 x^{2}=5 x \times-3 x$ Hence, the expression 5x - 15x2can be factorised as 5x(1 - 3x)....
Read More →Factorize the following:
Question: Factorize the following:3x 9 Solution: The greatest common factor of the terms 3x and -9 of the expression 3x - 9 is 3. Now. 3x= 3x and -9 = 3.-3 Hence, the expression 3x - 9 can be factorised as 3(x - 3)....
Read More →Points P and Q have been taken on
Question: Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other. Solution: Given $A B C D$ is a parallelogram and $A P=C Q$ To show $A C$ and $P Q$ bisect each other. Proof In $\triangle A M P$ and $\Delta C M Q$, $\angle M A P=\angle M C Q$ [alternate interior angles] $A P=C Q$ [given] and $\angle A P M=\angle C Q M$ [alternate interior angles] $\therefore$ $\triangle A M P \cong \triangle C M Q$...
Read More →In the given figure, OPQR is a rhombus, three of whose vertices lie on a circle with centre O.
Question: In the given figure, $O P Q R$ is a rhombus, three of whose vertices lie on a circle with centre $O$. If the area of the rhombus is $32 \sqrt{3}$, find the radius of the circle. Solution: In a rhombus, all sides are congruent to each other.Thus, we have: $O P=P Q=Q R=R O$ Now, consider $\Delta Q O P$. $O Q=O P($ Both are radii. $)$ Therefore, $\Delta Q O P$ is equilateral. Similarly, $\Delta Q O R$ is also equilateral and $\Delta Q O P \cong \Delta Q O R$. Ar. $(Q R O P)=\operatorname{...
Read More →Find the greatest common factor of the terms in each of the following expression:
Question: Find the greatest common factor of the terms in each of the following expression:3a2b2+ 4b2c2+ 12a2b2c2 Solution: The expression has three monomials:3a2b2, 4b2c2and 12a2b2c2. The numerical coefficients of the given monomials are 3, 4 and 12. The greatest common factor of 3, 4 and 12 is 1. The common literal appearing in the three monomials is b. The smallest power of b in the three monomials is 2. The monomial of common literals with the smallest powers is b2. Hence, the greatest comm...
Read More →Find the greatest common factor of the terms in each of the following expression:
Question: Find the greatest common factor of the terms in each of the following expression:2xyz+ 3x2y+ 4y2 Solution: The expression has three monomials:2xyz, 3x2y and 4y2. The numerical coefficients of the given monomials are 2, 3 and 4. The greatest common factor of 2, 3 and 4 is 1. The common literal appearing in the three monomials is y. The smallest power of y in the three monomials is 1. The monomial of common literals with the smallest powers is y. Hence, the greatest common factor is y....
Read More →Find the greatest common factor of the terms in each of the following expression:
Question: Find the greatest common factor of the terms in each of the following expression:5a4+ 10a3 15a2 Solution: Terms are expressions separated by plus or minus signs. Here, the terms are 5a4, 10a3and 15a2. The numerical coefficients of the given monomials are 5, 10 and 15. The greatest common factor of 5, 10 and 15 is 5. The common literal appearing in the three monomials is a. The smallest power of a in the three monomials is 2. The monomial of common literals with the smallest powers is a...
Read More →If A, B, C are the angles of a triangle,
Question: If $A, B, C$ are the angles of a triangle, then $\left|\begin{array}{ccc}\sin ^{2} A \cot A 1 \\ \sin ^{2} B \cot B 1 \\ \sin ^{2} C \cot C 1\end{array}\right|=$___________ Solution: Given: $A, B, C$ are the angles of a triangle Then, $A+B+C=\pi$ Let $\Delta=\left|\begin{array}{lll}\sin ^{2} A \cot A 1 \\ \sin ^{2} B \cot B 1 \\ \sin ^{2} C \cot C 1\end{array}\right|$ $\Delta=\left|\begin{array}{lll}\sin ^{2} A \cot A 1 \\ \sin ^{2} B \cot B 1 \\ \sin ^{2} C \cot C 1\end{array}\right|$...
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:14x3y5, 10x5y3, 2x2y2 Solution: The numerical coefficients of the given monomials are 14, 10 and 2. The greatest common factor of 14, 10 and 2 is 2. The common literals appearing in the three monomials are x andy. The smallest power of x in the three monomials is 2. The smallest power of y in the three monomials is 2. The monomial of common literals with the smallest powers is x2y2. Hence, the greatest common factor ...
Read More →D, E and F are the mid-points of the sides BC,
Question: D, E and F are the mid-points of the sides BC, CA and AB, respectively of an equilateral ΔABC. Show that ΔDEF is also an equilateral triangle. Solution: Given In equilateral ΔABC, D, E and F are the mid-points of sides BC, CA and AB, respectively. To show ΔDEF is an equilateral triangle. Proof Since in ΔABC, E and F are the mid-points of AC and AB respectively, then EF || BC and EF = BC ,,.(i) DF || AC, DE || AB DE = AB and FD = AC [by mid-point theorem](ii) since ΔABC is an equilatera...
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:2x3y2, 10x2y3, 14xy Solution: The numerical coefficients of the given monomials are 2, 10 and 14. The greatest common factor of 2, 10 and 14 is 2. The common literals appearing in the three monomials are x andy. The smallest power of x in the three monomials is 1. The smallest power of y in the three monomials is 1. The monomial of common literals with the smallest powers is xy. Hence, the greatest common factor is 2...
Read More →Four cows are tethered at the four corners of a square field of side 50 m such that each can graze the maximum unshared area.
Question: Four cows are tethered at the four corners of a square field of side 50 m such that each can graze the maximum unshared area. What area will be left ungrazed? Solution: Each cow can graze a region that cannot be accessed by other cows. $\therefore$ Radius of the region grazed by each cow $=\frac{50}{2}=25 \mathrm{~m}$ Area that each cow grazes $=\frac{1}{4} \times \pi \times r^{2}$ $=\frac{1}{4} \times 3.14 \times 25 \times 25$ $=490.625 \mathrm{~cm}^{2}$ Total area grazed $=4 \times 4...
Read More →Through A, B and C lines RQ,
Question: Through A, B and C lines RQ, PR and QP have been drawn, respectively parallel to sides BC, CA and AB of a ΔABC as shown in figure. Show that BC = QR Solution: Given In ΔABC, PQ || AB and PR || AC and RQ || BC. To show BC = QR Proof In quadrilateral BCAR, BR || CA and BC|| RA So, quadrilateral, BCAR is a parallelogram. BC = AR (i) Now, in quadrilateral BCQA, BC || AQ and AB||QC So, quadrilateral BCQA is a parallelogram, BC = AQ (ii) On adding Eqs. (i) and(ii), we get 2 BC = AR+ AQ = 2 B...
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:15a3, 45a2, 150a Solution: The numerical coefficients of the given monomials are 15, -45 and -150. The greatest common factor of 15, -45 and -150 is 15. The common literal appearing in the three monomials is a. The smallest power of a in the three monomials is 1. Hence, the greatest common factor is 15a....
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:x3,yx2 Solution: The common literal appearing in the two monomials is x. The smallest power of x in both the monomials is 2. Hence, the greatest common factor is x2....
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:36a2b2c4, 54a5c2, 90a4b2c2 Solution: The numerical coefficients of the given monomials are 36, 54 and 90. The greatest common factor of 36, 54 and 90 is 18. The common literals appearing in the three monomials are a andc. The smallest power of a in the three monomials is 2. The smallest power of c in the three monomials is 2. The monomial of common literals with the smallest powers is a2c2. Hence, the greatest common...
Read More →A horse is tethered to one corner of a field which is in the shape of an equilateral triangle of side 12 m.
Question: A horse is tethered to one corner of a field which is in the shape of an equilateral triangle of side 12 m. If the length of the rope is 7 m, find the area of the field which the horse cannot graze. Write the answer correct to 2 places of decimal. Solution: Side of the equilateral triangle = 12 m Area of the equilateral triangle $=\frac{\sqrt{3}}{4} \times(\text { Side })^{2}$ $=\frac{\sqrt{3}}{4} \times 12 \times 12$ $=62.28 \mathrm{~m}^{2}$ Length of the rope = 7 mArea of the field t...
Read More →E is the mid-point of the side AD of the trapezium
Question: E is the mid-point of the side AD of the trapezium ABCD with AB || DC. A line through E drawn parallel to AB intersects BC at F. Show thatf is the mid- point of BC. Thinking Process Use the mid-point theorem i.e., the line segment joining the mid-points of two sides of a triangle is parallel to the third side and half of it. Further shown the required result. Solution: Given ABCD is a trapezium in which AB || DC and EF||AB|| CD. Construction Join, the diagonal AC which intersects EF at...
Read More →A horse is tethered to one corner of a field which is in the shape of an equilateral triangle of side 12 m.
Question: A horse is tethered to one corner of a field which is in the shape of an equilateral triangle of side 12 m. If the length of the rope is 7 m, find the area of the field which the horse cannot graze. Write the answer correct to 2 places of decimal. Solution: Side of the equilateral triangle = 12 m Area of the equilateral triangle $=\frac{\sqrt{3}}{4} \times(\text { Side })^{2}$ $=\frac{\sqrt{3}}{4} \times 12 \times 12$ $=62.28 \mathrm{~m}^{2}$ Length of the rope = 7 mArea of the field t...
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:a2b3,a3b2 Solution: The common literals appearing in the three monomials are a and b. The smallest power of x in the two monomials is 2. The smallest power of y in the two monomials is 2. The monomial of common literals with the smallest powers is a2b2. Hence, the greatest common factor is a2b2....
Read More →The maximum value of
Question: The maximum value of $\left|\begin{array}{ccc}1 1 1 \\ 1 1+\sin \theta 1 \\ 1 1 1+\cos \theta\end{array}\right|$ is________ Solution: Let $\Delta=\left|\begin{array}{ccc}1 1 1 \\ 1 1+\sin \theta 1 \\ 1 1 1+\cos \theta\end{array}\right|$ $\Delta=\left|\begin{array}{ccc}1 1 1 \\ 1 1+\sin \theta 1 \\ 1 1 1+\cos \theta\end{array}\right|$ Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1}$ $=\left|\begin{array}{ccc}1 1 1 \\ 1-1 1+\sin \theta-1 1-1 \\ 1-1 1-1 1+\cos ...
Read More →Factorize each of the following quadratic polynomials by using the method of completing the square:
Question: Factorize each of the following quadratic polynomials by using the method of completing the square:z2 4z 12 Solution: $z^{2}-4 z-12$ $=z^{2}-4 z+\left(\frac{4}{2}\right)^{2}-\left(\frac{4}{2}\right)^{2}-12 \quad\left[\right.$ Adding and subtracting $\left(\frac{4}{2}\right)^{2}$, that is, $\left.2^{2}\right]$ $=z^{2}-4 z+2^{2}-2^{2}-12$ $=(z-2)^{2}-16 \quad[$ Completing the square $]$ $=(z-2)^{2}-4^{2}$ $=[(z-2)-4][(z-2)+4]$ $=(z-6)(z+2)$...
Read More →E and F are points on diagonal
Question: E and F are points on diagonal AC of a parallelogram ABCD such that AE = CF. Show that BFDE is a parallelogram. Solution: Given $A B C D$ is a parallelogram and $A E=C F$ To show $O E=O F$ Construction Join $B D$, meet $A C$ at point $O$. Proof Since, diagonals of a parallelogram bisect each other. Thus, BFDE is a quadrilateral whose diagonals bisect each other. Hence, $B F D E$ is a parallelogram. Hence proved....
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