Using the factor theorem it is found that
Question: Using the factor theorem it is found that $a+b, b+c$ and $c+a$ are three factors of the determinant $\left|\begin{array}{ccc}-2 a a+b a+c \\ b+a -2 b b+c \\ c+a c+b -2 c\end{array}\right|$ The other factor in the value of the determinant is (a) 4 (b) 2 (c) $a+b+c$ (d) none of these Solution: (a) 4 $\Delta=\mid-2 a \quad a+b \quad a+c$ $b+a-2 b \quad b+c$ $c+a \quad c+b \quad-2 c \mid$ Let $a+b=2 C, b+c=2 A$ and $c+a=2 B$. $\Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{b}+\mathrm{c}+\mathrm...
Read More →The parallel sides of a trapezium are 12 cm and 9 cm and the distance between them is 8 cm.
Question: The parallel sides of a trapezium are 12 cm and 9 cm and the distance between them is 8 cm. Find the area of the trapezium. Solution: Area of the trapezium $=\frac{1}{2} \times$ (sum of the parallel sides) $\times$ distance between the parallel sides $=\frac{1}{2} \times(12+9) \times 8$ $=21 \times 4$ $=84 \mathrm{~cm}^{2}$ So, the area of the trapezium is 84 cm2....
Read More →Simplify each of the following:
Question: Simplify each of the following: (i) $x^{2}-3 x+5-\frac{1}{2}\left(3 x^{2}-5 x+7\right)$ (ii) $[5-3 x+2 y-(2 x-y)]-(3 x-7 y+9)$ (iii) $\frac{11}{2} x^{2} y-\frac{9}{4} x y^{2}+\frac{1}{4} x y-\frac{1}{14} y^{2} x+\frac{1}{15} y x^{2}+\frac{1}{2} x y$ (iv) $\left(\frac{1}{3} y^{2}-\frac{4}{7} y+11\right)-\left(\frac{1}{7} y-3+2 y^{2}\right)-\left(\frac{2}{7} y-\frac{2}{3} y^{2}+2\right)$ (v) $-\frac{1}{2} a^{2} b^{2} c+\frac{1}{3} a b^{2} c-\frac{1}{4} a b c^{2}-\frac{1}{5} c b^{2} a^{2}...
Read More →Simplify each of the following:
Question: Simplify each of the following: (i) $x^{2}-3 x+5-\frac{1}{2}\left(3 x^{2}-5 x+7\right)$ (ii) $[5-3 x+2 y-(2 x-y)]-(3 x-7 y+9)$ (iii) $\frac{11}{2} x^{2} y-\frac{9}{4} x y^{2}+\frac{1}{4} x y-\frac{1}{14} y^{2} x+\frac{1}{15} y x^{2}+\frac{1}{2} x y$ (iv) $\left(\frac{1}{3} y^{2}-\frac{4}{7} y+11\right)-\left(\frac{1}{7} y-3+2 y^{2}\right)-\left(\frac{2}{7} y-\frac{2}{3} y^{2}+2\right)$ (v) $-\frac{1}{2} a^{2} b^{2} c+\frac{1}{3} a b^{2} c-\frac{1}{4} a b c^{2}-\frac{1}{5} c b^{2} a^{2}...
Read More →The area of a rhombus is 480 cm2, and one of its diagonals measures 48 cm.
Question: The area of a rhombus is 480 cm2, and one of its diagonals measures 48 cm. Find (i) the length of the other diagonal, (ii) the length of each of its sides, and (iii) its perimeter. Solution: i) Area of a rhombus $=\frac{1}{2} \times d_{1} \times d_{2}$, where $d_{1}$ and $d_{2}$ are the lengths of the diagonals. $\Rightarrow 480=\frac{1}{2} \times 48 \times d_{2}$ $\Rightarrow d_{2}=\frac{480 \times 2}{48}$ $\Rightarrow d_{2}=20 \mathrm{~cm}$ Length of the other diagonal = 20 cm (ii) S...
Read More →ABC is an isosceles triangle
Question: ABC is an isosceles triangle with AB = AC and BD, CE are its two medians. Show that BD = CE. Solution: Given ΔABC is an isosceles triangle in which AB = AC and BD, CE are its two medians. To show BD = CE. Proof in $\triangle A B D$ and $\triangle A C E$, $A B=A C$ [given] $\angle A=\angle A$ [common angle] and $A D=A E$ $\because$ $A B=A C$ $\Rightarrow$ $\frac{1}{2} A B=\frac{1}{2} A C$ $\Rightarrow$ $A E=A D$ As $D$ is the mid-point of $A C$ and $E$ is the mid-point of $A B$. $\there...
Read More →Solve this
Question: Let $\left|\begin{array}{ccc}x^{2}+3 x x-1 x+3 \\ x+1 -2 x x-4 \\ x-3 x+4 3 x\end{array}\right|=a x^{4}+b x^{3}+c x^{2}+d x+e$ be an identity in $x$, where $a, b, c, d$, e are independent of $x$. Then the value of $e$ is (a) 4 (b) 0 (c) 1 (d) none of these Solution: (b) 0 Let $\Delta=\mid x^{2}+3 x \quad x-1 \quad x+3$ $x+1 \quad-2 x \quad x-4$ $x-3 \quad x+4 \quad 3 x$ $=\left(x^{2}+3 x\right) \mid-2 x \quad x-4$ $x+4 \quad 3 x|-(x-1)| x+1 \quad x-4$ $x-3 \quad 3 x|+(x+3)| x+1 \quad-2...
Read More →Subtract the sum of
Question: Subtract the sum of 2xx2+ 5 and 4x 3 + 7x2from 5. Solution: We have to subtract the sum of $\left(2 x-x^{2}+5\right)$ and $(-4 x-3+7 x)$ from 5 . $5-\left\{\left(2 x-x^{2}+5\right)+\left(-4 x-3+7 x^{2}\right)\right\}$ $=5-\left(2 x-4 x-x^{2}+7 x^{2}+5-3\right)$ $=5-2 x+4 x+x^{2}-7 x^{2}-5+3$ $=5-5+3-2 x+4 x+x^{2}-7 x^{2}$ (Combining like terms) $=3+2 x-6 x^{2}$ (Combining like terms) Thus, the answer is $3+2 x-6 x^{2}$....
Read More →The perimeter of a rhombus is 60 cm.
Question: The perimeter of a rhombus is 60 cm. If one of its diagonals is 18 cm long, find (i) the length of the other diagonal, and (ii) the area of the rhombus. Solution: Perimeter of a rhombus = 4a(Here,ais the side of the rhombus.) $\Rightarrow 60=4 a$ $\Rightarrow a=15 \mathrm{~cm}$ (i) Given:One of the diagonals is 18 cm long. $d_{1}=18 \mathrm{~cm}$ Thus, we have: Side $=\frac{1}{2} \sqrt{d_{1}^{2}+d_{2}^{2}}$ $\Rightarrow 15=\frac{1}{2} \sqrt{18^{2}+d_{2}^{2}}$ $\Rightarrow 30=\sqrt{18^{...
Read More →Subtract the sum of
Question: Subtract the sum of 3l 4m 7n2and 2l+ 3m 4n2from the sum of 9l+ 2m 3n2and 3l+m+ 4n2..... Solution: We have to subtract the sum of $\left(3 /-4 m-7 n^{2}\right)$ and $\left(2 l+3 m-4 n^{2}\right)$ from the sum of $\left(9 /+2 m-3 n^{2}\right)$ and $\left(-3 /+m+4 n^{2}\right)$ $\left\{\left(9 l+2 m-3 n^{2}\right)+\left(-3 l+m+4 n^{2}\right)\right\}-\left\{\left(3 l-4 m-7 n^{2}\right)+\left(2 l+3 m-4 n^{2}\right)\right\}$ $=\left(9 l-3 l+2 m+m-3 n^{2}+4 n^{2}\right)-\left(3 l+2 l-4 m+3 m-...
Read More →Solve the following equations
Question: If $D_{k}=\left|\begin{array}{ccc}1 n n \\ 2 k n^{2}+n+2 n^{2}+n \\ 2 k-1 n^{2} n^{2}+n+2\end{array}\right|$ and $\sum_{k=1}^{n} D_{k}=48$, then $n$ equals (a) 4 (b) 6 (c) 8 (d) none of these Solution: (a) 4 $D_{k}=\mid \begin{array}{lll}1 n n\end{array}$ $2 k \quad n^{2}+n+2 \quad n^{2}+n$ $\begin{array}{lll}2 k-1 n^{2} n^{2}+n+2 \mid\end{array}$ $=\mid 1 \quad n \quad n$ $\begin{array}{lll}1 n+2 -2\end{array}$ $\begin{array}{lll}2 k-1 n^{2} n^{2}+n+2 \mid\end{array}$ $\left[\right.$ ...
Read More →Is it possible to construct a triangle
Question: Is it possible to construct a triangle with lengths of its sides as 8 cm, 7 cm and 4 cm? Give reason for your answer. Solution: Yes, because in each case the sum of two sides is greater than the third side. i.e., 7 + 48, 8+ 4 7, 7 + 8 4 Hence, it is possible to construct a triangle with given sides....
Read More →Is it possible to construct a triangle
Question: Is it possible to construct a triangle with lengths of its sides as 9 cm, 7 cm and 17 cm? Give reason for your answer. Solution: No. Here, we see that 9 + 7 =16 17 i.e., the sum of two sides of a triangle is less than the third side. Hence, it contradicts the property that the sum of two sides of a triangle is greater than the third side. Therefore, it is not possible to construct a triangle with given sides....
Read More →Subtract 3x − 4y − 7z from the sum of
Question: Subtract 3x 4y 7zfrom the sum ofx 3y+ 2zand 4x+ 9y 11z. Solution: Let first add the expressions $x-3 y+2 z$ and $-4 x+9 y-11 z$. We get: $(x-3 y+2 z)+(-4 x+9 y-11 z)$ $=x-3 y+2 z-4 x+9 y-11 z$ $=x-4 x-3 y+9 y+2 z-11 z$ (Combining like terms) $=-3 x+6 y-9 z$ (Combining like terms) Now, subtracting the expression $3 x-4 y-7 z$ from the above sum; we get: $(-3 x+6 y-9 z)-(3 x-4 y-7 z)$ $=-3 x+6 y-9 z-3 x+4 y+7 z$ $=-3 x-3 x+6 y+4 y-9 z+7 z$ (Combining like terms) $=-6 x+10 y-2 z$ (Combini...
Read More →M is a point on side BC of a triangle
Question: M is a point on side BC of a triangle ABC such that AM is the bisector of BAC. Is it true to say that perimeter of the triangle is greater than 2AM? Give reason for your answer? Solution: Yes, in $\triangle A B C, M$ is a point of side $B C$ such that $A M$ is the bisector of $\angle B A C$. $\ln \triangle A B M, \quad A B+B MA M \quad \ldots$ (i) [sum of two sides of a triangle is greater than the third side] $\ln \triangle A C M$ $A C+C MA M$ ... (ii) [sum of two sides of a triangle ...
Read More →Take away:
Question: Take away: (i) $\frac{6}{5} x^{2}-\frac{4}{5} x^{3}+\frac{5}{6}+\frac{3}{2} x$ from $\frac{x^{3}}{3}-\frac{5}{2} x^{2}+\frac{3}{5} x+\frac{1}{4}$ (ii) $\frac{5 a^{2}}{2}+\frac{3 a^{3}}{2}+\frac{a}{3}-\frac{6}{5}$ from $\frac{1}{3} a^{3}-\frac{3}{4} a^{2}-\frac{5}{2}$ (iii) $\frac{7}{4} x^{3}+\frac{3}{5} x^{2}+\frac{1}{2} x+\frac{9}{2}$ from $\frac{7}{2}-\frac{x}{3}-\frac{x^{2}}{5}$ (iv) $\frac{y^{3}}{3}+\frac{7}{3} y^{2}+\frac{1}{2} y+\frac{1}{2}$ from $\frac{1}{3}-\frac{5}{3} y^{2}$ (...
Read More →Take away:
Question: Take away: (i) $\frac{6}{5} x^{2}-\frac{4}{5} x^{3}+\frac{5}{6}+\frac{3}{2} x$ from $\frac{x^{3}}{3}-\frac{5}{2} x^{2}+\frac{3}{5} x+\frac{1}{4}$ (ii) $\frac{5 a^{2}}{2}+\frac{3 a^{3}}{2}+\frac{a}{3}-\frac{6}{5}$ from $\frac{1}{3} a^{3}-\frac{3}{4} a^{2}-\frac{5}{2}$ (iii) $\frac{7}{4} x^{3}+\frac{3}{5} x^{2}+\frac{1}{2} x+\frac{9}{2}$ from $\frac{7}{2}-\frac{x}{3}-\frac{x^{2}}{5}$ (iv) $\frac{y^{3}}{3}+\frac{7}{3} y^{2}+\frac{1}{2} y+\frac{1}{2}$ from $\frac{1}{3}-\frac{5}{3} y^{2}$ (...
Read More →AD is a median of the ΔABC.
Question: AD is a median of the ΔABC. Is it true that AB + BC +CA 2AD? Give reason for your answer. Solution: Yes, In $\triangle A B D$, we have $A B+B DA D$$\ldots($ i) [sum of any two sides of a triangle is greater than third side] In $\triangle A C D$, we have $A C+C DA D$ ....(ii) [sum of any two sides of a triangle is greater than third side] On adding Eqs. (i) and (ii), we get $(A B+B D+A C+C D)2 A D$ $\Rightarrow \quad(A B+B D+C D+A C)2 A D$ Hence, $A B+B C+A C2 A D$ $[\because B C=B D+C ...
Read More →Find the area of the rhombus, the lengths of whose diagonals are 30 cm
Question: Find the area of the rhombus, the lengths of whose diagonals are 30 cm and 16 cm. Also find the perimeter of the rhombus. Solution: Area of the rhombus $=\frac{1}{2} \times d_{1 \times} d_{2}$, where $d_{1}$ and $d_{2}$ are the lengths of the diagonals. $=\frac{1}{2} \times 30 \times 16$ $=240 \mathrm{~cm}^{2}$ Side of the rhombus $=\frac{1}{2} \sqrt{{d_{1}}^{2}+{d_{2}}^{2}}$ $=\frac{1}{2} \sqrt{30^{2}+16^{2}}$ $=\frac{1}{2} \sqrt{1156}$ $=\frac{1}{2} \times 34$ $=17 \mathrm{~cm}$ Peri...
Read More →In ΔPQR, ∠P = 70° and ∠R = 30°.
Question: In ΔPQR, P = 70 and R = 30. Which side of this triangle is the longest? Give reason for your answer. Solution: Given, in ΔPQR, P = 70 and R = 30. We know that, sum of all the angles of a triangle is 180. P + Q + R = 180 We know that here Q is longest, so side PR is longest. [ since in a triangle, the side opposite to the largest angle is the longest]...
Read More →Solve this
Question: If $\Delta_{1}=\left|\begin{array}{ccc}1 1 1 \\ a b c \\ a^{2} b^{2} c^{2}\end{array}\right|, \Delta_{2}=\left|\begin{array}{ccc}1 b c a \\ 1 c a b \\ 1 a b c\end{array}\right|$, then (a) $\Delta_{1}+\Delta_{2}=0$ (b) $\Delta_{1}+2 \Delta_{2}=0$ (c) $\Delta_{1}=\Delta_{2}$ (d) none of these Solution: (a) $\Delta_{1}+\Delta_{2}=0$ $\Delta_{2}=\mid \begin{array}{lll}1 b c a\end{array}$ $1 \quad c a \quad b$ $\begin{array}{lll}1 a b c\end{array}$ $=\frac{1}{a b c} \mid \begin{array}{lll}a...
Read More →If ΔPQR ≅ ΔEOF,
Question: If ΔPQR ΔEOF, then is it true to say that PR = EF? Give reason for your answer. Solution: Yes, if ΔPQR ΔEDF, then it means that corresponding angles and their sides are equal because we know that, two triangles are congruent, if the sides and angles of one triangle are equal to the corrosponding sides and angles of other triangle. Here, ΔPQR ΔEDF PQ = ED, QR = DF and PR = EF Hence, it is true to say that PR = EF....
Read More →The adjacent sides of a parallelogram ABCD measure 34 cm and 20 cm,
Question: The adjacent sides of a parallelogramABCDmeasure 34 cm and 20 cm, and the diagonalACmeasures 42 cm. Find the area of the parallelogram. Solution: Parallelogram $A B C D$ is made up of congruent $\triangle A B C$ and $\triangle A D C$. Area of triangle $A B C=\sqrt{s(s-a)(s-b)(s-c)}$ (Here, $s$ is the semiperimeter.) Thus, we have: $s=\frac{a+b+c}{2}$ $s=\frac{34+20+42}{2}$ $s=48 \mathrm{~cm}$ Area of $\Delta A B C=\sqrt{48(48-34)(48-20)(48-42)}$ $=\sqrt{48 \times 14 \times 28 \times 6}...
Read More →It is given that ΔABC ≅ ΔRPQ.
Question: It is given that ΔABC ΔRPQ. Is it true to say that BC = QR? Why? Solution: No, we know that two triangles are congruent, if the sides and angles of one triangle are equal to the corresponding side and angles of other triangle. Here ΔABC ΔRPQ AB = RP, BC = PQ and AC = RQ Hence, it is not true to say that BC = QR....
Read More →Is it possible to construct a triangle
Question: Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm? Give reason for your answer. Solution: No, it is not possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm because here we see that sum of the lengths of two sides is equal to third side i.e., 4+3 = 7. As we know that, the sum of any two sides of a triangle is greater than its third side, so given statement is not correct....
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