Corresponding sides of two triangles are in the ratio 2 : 3.
Question: Corresponding sides of two triangles are in the ratio $2: 3 .$ If the area of the smaller triangle is $48 \mathrm{~cm}^{2}$, determine the area of the larger triangle. Solution: The ratio of the areas of two similar triangles is equal to the ratio of the square of any two corresponding sides. $\frac{\text { Area of triangle }}{\text { Area of larger triangle }}=\frac{(\text { Corresponding side of smaller triangle })^{2}}{(\text { Corresponding side of larger triangle })^{2}}$ $\frac{\...
Read More →A 4 cm cube is cut into 1 cm cubes.
Question: A 4 cm cube is cut into 1 cm cubes. Calculate the total surface area of all the small cubes. Solution: Edge of the cube (a) = 4 cm Volume of the cube $=a^{3}$ $a=4^{3}$ $=64 \mathrm{~cm}^{3}$ Edge of the cube $=1 \mathrm{~cm}^{3}$ Therefore, Total number of small cubes $=64 \mathrm{~cm}^{3} / 1 \mathrm{~cm}^{3}=64$ Therefore, Total Surface area of all the cubes $=64 * 6 * 1=384 \mathrm{~cm}^{2}$...
Read More →ABCD is a trapezium having AB || DC. Prove that O,
Question: $A B C D$ is a trapezium having $A B \| D C$. Prove that $O$, the point of intersection of diagonals, divides the two diagonals in the same ratio. Also prove that $\frac{\operatorname{ar}(\Delta O C D)}{\operatorname{ar}(\Delta O A B)}=\frac{1}{9}$, if $A B=3 C D$. Solution: We are givenABCDis a trapezium withAB||DC $\angle A O B=\angle C O D$ $\angle A B O=\angle O D C$(alternative angle) $\angle B A O=\angle D C A$(alternative angle) Therefore, $\triangle O D C \sim \triangle O B A$ ...
Read More →Prove that:
Question: Prove that: (i) $\tan 225^{\circ} \cot 405^{\circ}+\tan 765^{\circ} \cot 675^{\circ}=0$ (ii) $\sin \frac{8 \pi}{3} \cos \frac{23 \pi}{6}+\cos \frac{13 \pi}{3} \sin \frac{35 \pi}{6}=\frac{1}{2}$ (iii) $\cos 24^{\circ}+\cos 55^{\circ}+\cos 125^{\circ}+\cos 204^{\circ}+\cos 300^{\circ}=\frac{1}{2}$ (iv) $\tan \left(-225^{\circ}\right) \cot \left(-405^{\circ}\right)-\tan \left(-765^{\circ}\right) \cot \left(675^{\circ}\right)=0$ (v) $\cos 570^{\circ} \sin 510^{\circ}+\sin \left(-330^{\circ...
Read More →Prove that:
Question: Prove that: (i) $\tan 225^{\circ} \cot 405^{\circ}+\tan 765^{\circ} \cot 675^{\circ}=0$ (ii) $\sin \frac{8 \pi}{3} \cos \frac{23 \pi}{6}+\cos \frac{13 \pi}{3} \sin \frac{35 \pi}{6}=\frac{1}{2}$ (iii) $\cos 24^{\circ}+\cos 55^{\circ}+\cos 125^{\circ}+\cos 204^{\circ}+\cos 300^{\circ}=\frac{1}{2}$ (iv) $\tan \left(-225^{\circ}\right) \cot \left(-405^{\circ}\right)-\tan \left(-765^{\circ}\right) \cot \left(675^{\circ}\right)=0$ (v) $\cos 570^{\circ} \sin 510^{\circ}+\sin \left(-330^{\circ...
Read More →Three equal cubes are placed adjacently in a row.
Question: Three equal cubes are placed adjacently in a row. Find the ratio of a total surface area of the new cuboid to that of the sum of the surface areas of the three cubes. Solution: Length of the new cuboid = 3a Breadth of the cuboid = a Height of the new cuboid = a The Total surface area of the new cuboid (TSA) = 2(lb + bh + hl) (TSA)1 =2(3a a + a a + a 3a) $(\mathrm{TSA})_{1}=14 \mathrm{a}^{2}$ The Total Surface area of three cubes $(\mathrm{TSA})_{2}=3 * 6 \mathrm{a}^{2}$ $(\text { TSA }...
Read More →Using the method of integration find the area of the region bounded by lines:
Question: Using the method of integration find the area of the region bounded by lines: 2x+y= 4, 3x 2y= 6 andx 3y+ 5 = 0 Solution: The given equations of lines are 2x+y= 4 (1) 3x 2y= 6 (2) And,x 3y+ 5 = 0 (3) The area of the region bounded by the lines is the area of ΔABC. AL and CM are the perpendiculars onx-axis. Area (ΔABC) = Area (ALMCA) Area (ALB) Area (CMB) $=\int_{1}^{1}\left(\frac{x+5}{3}\right) d x-\int_{1}^{2}(4-2 x) d x-\int_{2}^{4}\left(\frac{3 x-6}{2}\right) d x$ $=\frac{1}{3}\left[...
Read More →In ∆ABC, D and E are points on sides AB and AC respectively
Question: In ∆ABC, D and E are points on sides AB and AC respectively such that AD ✕ EC = AE ✕ DB. Prove that DE || BC. Solution: Given: In $\triangle A B C, D$ and $E$ are points on sides $A B$ and $A C$ such that $A D \times E C=A E \times D B$ To Prove: $D E \| B C$ Proof: Since $A D \times E C=A E \times D B$ $\Rightarrow \quad \frac{D B}{A D}=\frac{E C}{A E}$ $\Rightarrow \quad \frac{D B}{A D}+1=\frac{E C}{A E}+1$ $\Rightarrow \frac{D B+A D}{A D}=\frac{E C+A E}{A E}$ $\Rightarrow \quad \fra...
Read More →The length, breadth, and height of a room are 5 m, 4 m and 3 m respectively.
Question: The length, breadth, and height of a room are $5 \mathrm{~m}, 4 \mathrm{~m}$ and $3 \mathrm{~m}$ respectively. Find the cost of whitewashing the walls of the room and the ceiling at the rate of Rs $7.50 \mathrm{~m}$. Solution: Total Area to be washed = lb + 2(l + b)h Where, length (l) = 5 m breadth (b) = 4 m height (h) = 3 m Therefore, the total area to be white washed is = (5*4) + 2*(5 + 4)*3 $=74 \mathrm{~m}^{2}$ Now, The cost of white washing $1 \mathrm{~m}^{2}$ is Rs. $7.50$ Theref...
Read More →ABCD is a trapezium in which AB || DC. P and Q are points on sides
Question: ABCD is a trapezium in which AB || DC. P and Q are points on sides AD and BC such that PQ || AB. If PD = 18, BQ = 35 and QC = 15, find AD. Solution: In trapeziumABCD, AB || DC. P and Q are points on sides AD and BC such that PQ || AB.Join AC. Suppose AC intersects PQ in O. In $\Delta A C D, O P \| C D$ $\therefore \frac{A P}{P D}=\frac{A O}{O C} \quad \ldots \ldots(1) \quad(\mathrm{BPT})$ In $\Delta A B C, O Q \| A B$ $\therefore \frac{B Q}{Q C}=\frac{A O}{O C} \quad \ldots . .(2) \qua...
Read More →Using the method of integration find the area of the triangle
Question: Using the method of integration find the area of the triangle $A B C$, coordinates of whose vertices are $A(2,0), B(4,5)$ and $C(6,3)$ Solution: The vertices of $\triangle \mathrm{ABC}$ are $\mathrm{A}(2,0), \mathrm{B}(4,5)$, and $\mathrm{C}(6,3)$. Equation of line segment AB is $y-0=\frac{5-0}{4-2}(x-2)$ $2 y=5 x-10$ $y=\frac{5}{2}(x-2)$ ...(1) Equation of line segment BC is $y-5=\frac{3-5}{6-4}(x-4)$ $2 y-10=-2 x+8$ $2 y=-2 x+18$ $y=-x+9$ ...(2) Equation of line segment CA is $y-3=\f...
Read More →Mary wants to decorate her Christmas tree. She wants to place the tree on a wooden block covered with colored paper with a picture of Santa Claus on it.
Question: Mary wants to decorate her Christmas tree. She wants to place the tree on a wooden block covered with colored paper with a picture of Santa Claus on it. She must know the exact quantity of paper to buy for this purpose. If the box has length, breadth, and height as 80 cm, 40 cm and 20 cm respectively. How many square sheets of paper of side 40 cm would she require? Solution: Given that: Mary wants to paste a paper on the outer surface of the wooden block. The quantity of the paper requ...
Read More →Find the ratio of the total surface area and lateral surface area of a cube.
Question: Find the ratio of the total surface area and lateral surface area of a cube. Solution: Total Surface Area of the Cube $(T S A)=6 a^{2}$ Where, a = edge of the cube And, Lateral surface area of the Cube (LSA) $=4 \mathrm{a}^{2}$ Where, a = edge of the cube Hence, Ratio of TSA and LSA $=6 a^{2} / 4 a^{2}=3 / 2$ is $3: 2$....
Read More →In the given figure,
Question: In the given figure, (i) $\triangle \mathrm{DMU} \sim \triangle \mathrm{BMV}$ (ii) $\mathrm{DM} \times \mathrm{BV}=\mathrm{BM} \times \mathrm{DU}$ Solution: (i) Given $A B \| D C$ $\angle M U D=\angle M V B$ Each angle is equal to $90^{\circ}$ $\angle U M D=\angle V M B$ Each are vertically opposite angles. Therefore, by AA-criterion of similarity $\triangle D M U \sim \triangle B M V$ (ii)Since $\triangle D M U \sim \triangle B M V$ $\frac{D M}{B M}=\frac{M U}{M V}=\frac{D U}{B V}$ $\...
Read More →Find the values of the following trigonometric ratios:
Question: Find the values of the following trigonometric ratios: (i) $\sin \frac{5 \pi}{3}$ (ii) $\sin 17 \pi$ (iv) $\cos \left(-\frac{25 \pi}{4}\right)$ (v) $\tan \frac{7 \pi}{4}$ (vi) $\sin \frac{17 \pi}{6}$ (vii) $\cos \frac{19 \pi}{6}$ (viii) $\sin \left(-\frac{11 \pi}{6}\right)$ (ix) $\operatorname{cosec}\left(-\frac{20 \pi}{3}\right)$ (x) $\tan \left(-\frac{13 \pi}{4}\right)$ (xi) $\cos \frac{19 \pi}{4}$ (xii) $\sin \frac{41 \pi}{4}$ (xiii) $\cos \frac{39 \pi}{4}$ (xiv) $\sin \frac{151 \pi...
Read More →Find the lateral surface area and total surface area of a cube of edge 10 cm.
Question: Find the lateral surface area and total surface area of a cube of edge 10 cm. Solution: Cube of edge (a) = 10 cm We know that, Cube Lateral Surface Area $=4 a^{2}$ = 4(10*10) $=400 \mathrm{~cm}^{2}$ Total Surface Area $=6 a^{2}$ $=6^{*} 10^{2}$ $=600 \mathrm{~cm}^{2}$...
Read More →Find the area bounded by curves
Question: Find the area bounded by curves $\left\{(x, y): y \geq x^{2}\right.$ and $\left.y=|x|\right\}$ Solution: The area bounded by the curves, $\left\{(x, y): y \geq x^{2}\right.$ and $\left.y=|x|\right\}$, is represented by the shaded region as It can be observed that the required area is symmetrical abouty-axis. Required area $=2[$ Area $($ OCAO $)-$ Area $($ OCADO $)]$ $=2\left[\int_{0}^{1} x d x-\int_{0}^{1} x^{2} d x\right]$ $=2\left[\left[\frac{x^{2}}{2}\right]_{0}^{1}-\left[\frac{x^{3...
Read More →Find the values of the following trigonometric ratios:
Question: Find the values of the following trigonometric ratios: (i) $\sin \frac{5 \pi}{3}$ (ii) $\sin 17 \pi$ (iv) $\cos \left(-\frac{25 \pi}{4}\right)$ (v) $\tan \frac{7 \pi}{4}$ (vi) $\sin \frac{17 \pi}{6}$ (vii) $\cos \frac{19 \pi}{6}$ (viii) $\sin \left(-\frac{11 \pi}{6}\right)$ (ix) $\operatorname{cosec}\left(-\frac{20 \pi}{3}\right)$ (x) $\tan \left(-\frac{13 \pi}{4}\right)$ (xi) $\cos \frac{19 \pi}{4}$ (xii) $\sin \frac{41 \pi}{4}$ (xiii) $\cos \frac{39 \pi}{4}$ (xiv) $\sin \frac{151 \pi...
Read More →In the given figure, l || m
Question: In the given figure, $/ \| m$ (i) Name three pairs of similar triangles with proper correspondence; write similarities. (ii) Prove that $\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{BC}}{\mathrm{RQ}}$ Solution: (i) Three pair of similar triangles are- $\Delta A B K-\Delta P Q K$(AAA Similarity) $\Delta C B K-\Delta R Q K$(AAA Similarity) $\Delta A C K-\Delta P R K$(AAA Similarity) (ii) Since the pair of similar triangles mentioned above can give us the ...
Read More →Using the method of integration find the area bounded by the curve
Question: Using the method of integration find the area bounded by the curve $|x|+|y|=1$ [Hint: the required region is bounded by lines $x+y=1, x-y=1,-x+y=1$ and $-x-y=11$ ] Solution: The area bounded by the curve, $|x|+|y|=1$, is represented by the shaded region ADCB as The curve intersects the axes at points A (0, 1), B (1, 0), C (0, 1), and D (1, 0). It can be observed that the given curve is symmetrical aboutx-axis andy-axis. Area ADCB = 4 Area OBAO $=4 \int_{0}^{1}(1-x) d x$ $=4\left(x-\fra...
Read More →Find the lateral surface area and total surface area of a cuboid of length 80 cm,
Question: Find the lateral surface area and total surface area of a cuboid of length 80 cm, breadth 40 cm and height 20 cm. Solution: Given that: Cuboid length (l) = 80 cm Breadth (b) = 40 cm Height (h) = 20 cm We know that, Total Surface Area = 2[lb + bh + hl] = 2[(80)(40) + (40)(20) + (20)(80)] = 2[3200 + 800 + 1600] = 2[5600] $=11200 \mathrm{~cm}^{2}$ Lateral Surface Area = 2[l + b]h = 2[80 + 40]20 = 40[120] $=4800 \mathrm{~cm}^{2}$...
Read More →Find the area of the region enclosed by the parabola
Question: Find the area of the region enclosed by the parabola $x^{2}=y$, the line $y=x+2$ and $x$-axis Solution: The area of the region enclosed by the parabola, $x^{2}=y$, the line, $y=x+2$, and $x$-axis is represented by the shaded region $\mathrm{OACO}$ as The point of intersection of the parabola,x2=y, and the line,y=x+ 2, is A (1, 1) and C(2, 4). Area of OACO $=\int_{-1}^{2}(x+2) d x-\int_{-1}^{2} x^{2} d x$ $\Rightarrow$ Area of $\mathrm{OACO}=\left[\frac{x^{2}}{2}+2 x\right]_{-1}^{2}-\fr...
Read More →Find the area of the smaller region bounded by the ellipse
Question: Find the area of the smaller region bounded by the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ and the line $\frac{x}{a}+\frac{y}{b}=1$ Solution: The area of the smaller region bounded by the ellipse, $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, and the line, $\frac{x}{a}+\frac{y}{b}=1$, is represented by the shaded region BCAB as Area BCAB = Area (OBCAO) Area (OBAO) $=\int_{0}^{a} b \sqrt{1-\frac{x^{2}}{a^{2}}} d x-\int_{0}^{a} b\left(1-\frac{x}{a}\right) d x$ $=\frac{b}{a} \i...
Read More →Find the area of the smaller region bounded by the ellipse
Question: Find the area of the smaller region bounded by the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ and the line $\frac{x}{3}+\frac{y}{2}=1$ Solution: The area of the smaller region bounded by the ellipse, $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$, and the line, $\frac{x}{3}+\frac{y}{2}=1$, is represented by the shaded region BCAB as Area BCAB = Area (OBCAO) Area (OBAO) $=\int_{0}^{3} 2 \sqrt{1-\frac{x^{2}}{9}} d x-\int_{0}^{3} 2\left(1-\frac{x}{3}\right) d x$ $=\frac{2}{3}\left[\int_{0}^{3} \sqr...
Read More →In ∆ABC, the bisector of ∠A intersects BC in D.
Question: In $\triangle A B C$, the bisector of $\angle A$ intersects $B C$ in $D$. If $A B=18 \mathrm{~cm}, A C=15 \mathrm{~cm}$ and $B C=22 \mathrm{~cm}$, find $B D$. Solution: Given $A B=18 \mathrm{~cm}, A C=15 \mathrm{~cm}$ and $B C=22 \mathrm{~cm}$. In $\triangle A B C, A D$ the bisector of $\angle A$ $\frac{A B}{A C}=\frac{B D}{D C}$ $\frac{A B}{A C}=\frac{B D}{B C-B D}$ $\frac{18}{15}=\frac{B D}{22-B D}$ On cross multiplication, we get $6(22-B D)=5 \times B D$ $132-6 B D=5 B D$ $132=5 B D...
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