Which of the following cannot be
Question: Which of the following cannot be true for a polyhedron? (a) V = 4, F = 4, E= 6 (b) V=6,F=8,E=12 (c) V = 20,F = 12, E = 30 (d) V = 4, F = 6, E = 6 Solution: (d) We know that, Eulers formula for any polyhedron isF+V-E = 2 where, F = faces, V = vertices and E =edges (a) $V=4, F=4$ and $E=6$ $\mathrm{LHS}=F+V-E$ $=4+4-6$ $=8-6=2$ $=\mathrm{RHS}$ $\therefore$ Option (a) is true for a polyhedron. (b) $V=6, F=8$ and $E=12$ $\mathrm{LHS}=F+V-E$ $=8+6-12$ $=14-12=2$ $=\mathrm{RHS}$ $\therefore$...
Read More →find the problem
Question: Find $A$ and $B$ so that $y=A \sin 3 x+B \cos 3 x$ satisfies the equation $\frac{d^{2} y}{d x^{2}}+4 \frac{d y}{d x}+3 y=10 \cos 3 x$. Solution: Formula: - (i) $\frac{\mathrm{dy}}{\mathrm{dx}}=y_{1}$ and $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=y_{2}$ (ii) $\frac{\mathrm{d}}{\mathrm{dx}} \cos \mathrm{x}=\sin \mathrm{x}$ (iii) $\frac{\mathrm{d}}{\mathrm{dx}} \sin \mathrm{x}=-\cos \mathrm{x}$ (iv) chain rule $\frac{\mathrm{df}}{\mathrm{dx}}=\frac{\mathrm{d} \text { (wou) }}{\ma...
Read More →Side of a square garden is 30 m.
Question: Side of a square garden is 30 m. If the scale used to draw its picture is 1cm: 5m, the perimeter of the square in the picture is (a) 20 cm (b) 24 cm (c) 28 cm (d) 30 cm Solution: (b) 24 cm Given, side of a square garden = 30m Scale to draw garden picture is 1cm: 5m So, perimeter of the square garden is = 4 30 = 120m Then, Perimeter to draw garden in picture = 120/5 = 24 cm...
Read More →We have 4 congruent equilateral triangles.
Question: We have 4 congruent equilateral triangles. What do we need more to make a pyramid? (a) An equilateral triangle. (b) A square with same side length as of triangle. (c) 2 equilateral triangles with side length same as triangle. (d) 2 squares with side length same as triangle. Solution: (b) We know that, a pyramid is a polyhedron, whose base is a polygon and lateral faces are triangles. But in the question, we have only 4 congruent equilateral triangles. Thus, we have to add a polygon in ...
Read More →If a, b, c are in AP, show that
Question: If a, b, c are in AP, show that $(a+2 b-c)(2 b+c-a)(c+a-b)=4 a b c$ Solution: To prove: $(a+2 b-c)(2 b+c-a)(c+a-b)=4 a b c$. Given: a, b, c are in A.P. Proof: Since a, b, c are in A.P. $\Rightarrow 2 b=a+c \ldots$ (i) Taking $L H S=(a+2 b-c)(2 b+c-a)(c+a-b)$ Substituting the value of 2b from eqn. (i) $=(a+a+c-c)(a+c+c-a)(c+a-b)$ $=(2 a)(2 c)(c+a-b)$ Substituting the value of (a + c) from eqn. (i) $=(2 a)(2 c)(2 b-b)$ $=(2 a)(2 c)(b)$ $=4 a b c$ = RHS Hence Proved...
Read More →In a solid if F = V = 5,
Question: In a solid if F = V = 5, then the number of edges in this shape is (a) 6 (b) 4 (c) 8 (d) 2 Solution: (c) 8 We have, Eulers formula for any polyhedron is, F + V E = 2 Given, F = V = 5 Where, face (F) = 5, Vertex (V) = 5, Edge (E) =? Then, 5 + 5 E = 2 10 E = 2 10 2 = E Edges (E) = 8...
Read More →Solid having only line segments
Question: Solid having only line segments as its edges is a (a) Polyhedron (b) Cone (c) Cylinder (d) Polygon Solution: (a) In polyhedron, the faces meet at edges which are line segments and edges meet at vertex....
Read More →Which of the following 3-D
Question: Which of the following 3-D shapes does not have a vertex? (a) Pyramid (b) Prism (c) Cone (d) Sphere Solution: (d) As we kndw that, a vertex is a meeting point of two or more edges. Since, a sphere has only one curved face, so it has no vertex and no edges....
Read More →If a, b, c are in AP, prove that
Question: If a, b, c are in AP, prove that (i) $(a-c)^{2}=4(a-b)(b-c)$ (ii) $a^{2}+c^{2}+4 a c=2(a b+b c+c a)$ (iii) $a^{3}+c^{3}+6 a b c=8 b^{3}$ Solution: (i) $(a-c)^{2}=4(a-b)(b-c)$ To prove: $(a-c)^{2}=4(a-b)(b-c)$ Given: $a, b, c$ are in A.P. Proof: Since $a, b, c$ are in A.P. $\Rightarrow \mathrm{c}-\mathrm{b}=\mathrm{b}-\mathrm{a}=$ common difference $\Rightarrow \mathrm{b}-\mathrm{c}=\mathrm{a}-\mathrm{b} \ldots$ (i) And, $2 b=a+c(a, b, c$ are in A.P. $)$ $\Rightarrow 2 b-c=a \ldots$ (ii...
Read More →Which of the following can be
Question: Which of the following can be the base of a pyramid? (a) Line segment (b) Circle (c) Octagon (d) Oval Solution: (c) Since, a pyramid is a polyhedron whose base is a polygon and lateral faces are triangles. Hence, octagon can be the base of a pyramid....
Read More →Which of the following is a two
Question: Which of the following is a two dimensional figure? (a) Rectangle (b) Rectangle prism (c) Square pyramid (d) Square prism Solution: (a) A two dimensional figure have two dimensions (measurements) like length and breadth. In the given options, only rectangle has two dimensions, i.e. length and breadth....
Read More →If the solve the problem
Question: If $x=a \sin t-b \cos t, y=a \cos t+b \sin t$, prove that $\frac{d^{2} y}{d x^{2}}=-\frac{x^{2}+y^{2}}{y^{3}}$ Solution: Formula: - (i) $\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}_{1}$ and $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\mathrm{y}_{2}$ (ii) $\frac{\mathrm{d}}{\mathrm{dx}} \cos \mathrm{x}=\sin \mathrm{x}$ (iii) $\frac{d}{d x} \sin x=-\cos x$ (iv) $\frac{d}{d x} x^{n}=n x^{n-1}$ (v) chain rule $\frac{\mathrm{df}}{\mathrm{dx}}=\frac{\mathrm{d}(\mathrm{wou})}{\mathrm{dt...
Read More →Which of the following is a regular
Question: Which of the following is a regular polyhedron? (a) Cuboid (b) Triangular prism (c) Cube (d) Square prism Solution: (c) A polyhedron is regular, if its faces are congruent regular polygons and the same number of faces meet at each vertex. Hence, a cube satisfies- all the properties for a regular polyhedron....
Read More →Which of the following will not form
Question: Which of the following will not form a polyhedron? (a) 3 triangles (b) 2 triangles and 3 parallelograms (c) 8 triangles (d) 1 pentagon and 5 triangles Solution: (a) A polyhedron is bounded by more than four polygonal faces. But in case of 3 triangles, it is not possible. So, option (a) does not form a polyhedron....
Read More →Prove that the ratio of sum of m arithmetic means between
Question: Prove that the ratio of sum of m arithmetic means between the two numbers to the sum of n arithmetic means between them is m:n. Solution: To prove: ratio of sum of m arithmetic means between the two numbers to the sum of n arithmetic means between them is m:n Formula used: (i) $d=\frac{b-a}{n+1}$, where, $d$ is the common difference n is the number of arithmetic means (ii) $S_{n}=\frac{n}{2}[a+1]$ , Where n = Number of terms $a=$ First term $I=$ Last term Let the first series of arithm...
Read More →If the solve the problem
Question: If $x=3 \cot t-2 \cos ^{3} t, y=3 \sin t-2 \sin ^{3} t$, find $\frac{d^{2} y}{d x^{2}}$ Solution: Formula: - (i) $\frac{\mathrm{dy}}{\mathrm{dx}}=y_{1}$ and $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=y_{2}$ (ii) $\frac{\mathrm{d}}{\mathrm{dx}} \cos \mathrm{x}=\sin \mathrm{x}$ (iii) $\frac{\mathrm{d}}{\mathrm{dx}} \sin \mathrm{x}=-\cos \mathrm{X}$ (iv) $\frac{\mathrm{d}(\cot \mathrm{x})}{\mathrm{dx}}=-\operatorname{cosec}^{2} \mathrm{x}$ (v) $\frac{\mathrm{d}}{\mathrm{dx}} \math...
Read More →Is it possible to construct a quadrilateral ABCD
Question: Is it possible to construct a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 5.4 cm, DA = 5.9 cm and diagonal AC = 8 cm? If not, why? Solution: No, Given measures are AS = 3 cm, SC = 4 cm,CD = 5.4 cm, DA = 59cmand AC = 8cm Here, we observe that AS + SC = 3 + 4 = 7 cm and AC = 8 cm i.e. the sum of two sides of a triangle is less than the third side, which is absurd. Hence, we cannot construct such a quadrilateral....
Read More →If the solve the problem
Question: If $x=a(\cos 2 t+2 t \sin 2 t)$ and $y=a(\sin 2 t-2 t \cos 2 t)$, then find $\frac{d^{2} y}{d x^{2}}$ Solution: Formula: - (i) $\frac{d y}{d x}=y_{1}$ and $\frac{d^{2} y}{d x^{2}}=y_{2}$ (ii) $\frac{\mathrm{d}}{\mathrm{dx}} \cos \mathrm{x}=\sin \mathrm{x}$ (iii) $\frac{\mathrm{d}}{\mathrm{dx}} \sin \mathrm{x}=-\cos \mathrm{x}$ (iv) $\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{x}^{\mathrm{n}}=\mathrm{n} \mathrm{x}^{\mathrm{n}-1}$ (v) chain rule $\frac{\mathrm{df}}{\mathrm{dx}}=\frac{\mathrm{...
Read More →Find maximum number of acute angles
Question: Find maximum number of acute angles which a convex quadrilateral, a pentagon and a hexagon can have. Observe the pattern and generalise the result for any polygon. Solution: If an angle is acute, then the corresponding exterior angle is greater than 90. Now, suppose a convex polygon has four or more acute angles. Since, the polygon is convex, all the exterior angles are positive, so the sum of the exterior angle is at least the sum of the interior angles. Now, supplementary of the four...
Read More →A playground in the town is in the form of a kite.
Question: A playground in the town is in the form of a kite. The perimeter is 106 m. If one of its sides is 23 m, what are the lengths of other three sides? Solution: Let the length of other non-consecutive side be x cm. Then, we have, perimeter of playground = 23 + 23+ x + x = 106 = 2 (23+ x) =46 + 2x = 106 2x = 106 46 =2x = 60 =x = 30 m Hence, the lengths of other three sides are 23m, 30m and 30m. As a kite has two pairs of equal consecutive sides....
Read More →Two sticks each of length 7 cm are crossing
Question: Two sticks each of length 7 cm are crossing each other such that they bisect each other at right angles. What shape is formed by joining their end points? Give reason. Solution: Sticks can be treated as the diagonals of a quadrilateral. Now, since the diagonals (sticks) are bisecting each other at right angles, therefore the shape formed by joining their end points will be a rhombus....
Read More →Two sticks each of length 5 cm are crossing
Question: Two sticks each of length 5 cm are crossing each other such that they bisect each other. What shape is formed by joining their end points? Give reason. Solution: Sticks can be taken as the diagonals of a quadrilateral. Now, since they are bisecting each other, therefore the shape formed by joining their end points will be a parallelogram. Hence, it may be a rectangle or a square depending on the angle between the sticks....
Read More →The point of intersection of diagonals
Question: The point of intersection of diagonals of a quadrilateral divides one diagonal in the ratio 1: 2. Can it be a parallelogram? Why or why not? Solution: No, it can never be a parallelogram, as the diagonals of a parallelogram intersect each other in the ratio 1 : 1....
Read More →A photo frame is in the shape of a quadrilateral,
Question: A photo frame is in the shape of a quadrilateral, with one diagonal longer than the other. Is it a rectangle? Why or why not? Solution: No, it cannot be a rectangle, as in rectangle, both the diagonals are of equal lengths....
Read More →Of the four quadrilaterals – square, rectangle,
Question: Of the four quadrilaterals square, rectangle, rhombus and trapezium-one is somewhat different from the others because of its design. Find it and give justification. Solution: In square, rectangle and rhombus, opposite sides are parallel and equal. Also, opposite angles are equal, i.e. they all are parallelograms. But in trapezium, there is only one pair of parallel sides, i.e. it is not a parallelogram. Therefore, trapezium has different design....
Read More →