If the no of faces and vertex in a solid are 7 and 10 respectively the no of edges are?

If the no of faces and vertex in a solid are 7 and 10 respectively the no of edges are?
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A prism with an n-sided base will have 2n vertices, n + 2 faces,and 3n edges.15 edges
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Consider a Hamiltonian Graph (G) with no loops and parallel edges. Which of the following is true with respect to this Graph (G) ? (a) deg(v) ≥ n/2 for each vertex of G (b) |E(G)| ≥ 1/2 (n-1)(n-2)+2 edges (c) deg(v) + deg(w) ≥ n for every v and w not connected by an edge (A) (a) and (b) (B) (b) and (c) (C) (a) and (c) (D) (a), (b) and (c)

Description : Consider a Hamiltonian Graph (G) with no loops and parallel edges. Which of the following is true with respect to this Graph (G) ? (a) deg(v) ≥ n/2 for each vertex of G (b) |E(G)| ≥ 1/2 (n-1)(n-2)+2 edges (c) deg(v) + deg( ... edge (A) (a) and (b) (B) (b) and (c) (C) (a) and (c) (D) (a), (b) and (c)

Last Answer : (D) (a), (b) and (c)

Consider an undirected graph G where self-loops are not allowed. The vertex set of G is {(i, j) | 1 ≤ i ≤ 12, 1 ≤ j ≤ 12}. There is an edge between (a, b) and (c, d) if |a – c| ≤ 1 or |b–d| ≤ 1. The number of edges in this graph is (A) 726 (B) 796 (C) 506 (D) 616

Description : Consider an undirected graph G where self-loops are not allowed. The vertex set of G is {(i, j) | 1 ≤ i ≤ 12, 1 ≤ j ≤ 12}. There is an edge between (a, b) and (c, d) if |a – c| ≤ 1 or |b–d| ≤ 1. The number of edges in this graph is (A) 726 (B) 796 (C) 506 (D) 616

Last Answer : (D) 616

In STL files Euler’s rule for solids can be written as a.No. of faces – No. of edges + No. of vertices = 3 x No. of bodies b.No. of faces – No. of edges + No. of vertices = No. of bodies c.No. of faces – No. of edges + No. of vertices = 2 x No. of bodies d.No. of faces – No. of edges + No. of vertices = 4 x No. of bodies

Description : In STL files Euler’s rule for solids can be written as a.No. of faces †No. of edges + No. of vertices = 3 x No. of bodies b.No. of faces †No. of edges + No. of vertices = No. of bodies c.No. of ... = 2 x No. of bodies d.No. of faces †No. of edges + No. of vertices = 4 x No. of bodies

Last Answer : c.No. of faces – No. of edges + No. of vertices = 2 x No. of bodies

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Description : Write the coordinates of the vertices of a rectangle whose lenght and breadth are 7 and 4 units respectively,one vertex atthe the origin,the longer side lies on the x-axis and one of the vertices lies in the third quadrant. -Maths 9th

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Description : Write the coordinates of the vertices of a rectangle whose length and breadth are 6 and 3 units respectively, one vertex at the origin, the longer side lies on the y-axis and one of the vertices lies in the second quadrant. -Maths 9th

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Last Answer : (c) RhombusCo-ordinates of P are \(\bigg(rac{-1-1}{2},rac{-1+4}{2}\bigg)\)i.e, \(\big(-1,rac{3}{2}\big)\)Co-ordinates of Q are \(\bigg(rac{-1+5}{2},rac{4+4}{2}\bigg)\)i.e, (2, 4)Co-ordinates of R ... \sqrt{(2-2)^2+(4+1)^2}\) = \(\sqrt{25}\) = 5⇒ PR ≠ SQ ⇒ Diagonals are not equal ⇒ PQRS is a rhombus.

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Description : A cuboid shaped wooden block has 6 cm length, 4 cm breadth and 1 cm height. 7. Two faces measuring 4 cm x 1 cm are coloured in black. 8. Two faces measuring 6 cm x 1 cm are coloured in red. 9. Two faces measuring ... 1 cm(from 4 cm side). How many small cubes will be formed? (a)6 (b)12 (c)16 (d)24

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Description : Two dice are rolling simultaneously .What is the probability that the sum of the number on the two faces is divided by 5 Or 7. A) 13/36 B) 14/36 C) 11/36 D) 9/36 

Last Answer :  Answer: C) Clearly, n(S) = 6 x 6 = 36 Let E be the event that the sum of the numbers on the two faces is divided by 5or 7. Then,E = {(1,4),(1,6),(2,3),(2,5),(3,2),(3,4),(4,1),(4,3),(4,6),(5,2),(5,5),(6,1),(6,4)} n(E) = 11. Hence, P(E) = n(E)/n(S) = 11/36

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The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 60°. -Maths 9th

Description : The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 60°. -Maths 9th

Last Answer : Let the parallelogram be ABCD, in which ∠ADC and ∠ABC are obtuse angles. Now, DE and DF are two altitudes of parallelogram and angle between them is 60°.

The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 60°. -Maths 9th

Description : The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 60°. -Maths 9th

Last Answer : Let the parallelogram be ABCD, in which ∠ADC and ∠ABC are obtuse angles. Now, DE and DF are two altitudes of parallelogram and angle between them is 60°.

Find the area of a parallelogram given in the figure. Also, find the length of the altitude from vertex A on the side DC. -Maths 9th

Description : Find the area of a parallelogram given in the figure. Also, find the length of the altitude from vertex A on the side DC. -Maths 9th

Last Answer : Weknowthatthediagonalofaparallelogram(∥gm)dividesitintotwocongruenttriangles.SoAreaof∥gmABCD=2 Areaof△BCD.AccordingtoHeron′sformulathearea(A)oftrianglewithsidesa,b&cisgivenasA=2[s(s−a)(s−b) ... 90=180Areaof∥gm=base heightHeightofaltitudefromvertexAonsideCDoftheof∥gm=baseCDareaof∥gmABCD =12180 =15cm

Find the area of a parallelogram given in the figure. Also, find the length of the altitude from vertex A on the side DC. -Maths 9th

Description : Find the area of a parallelogram given in the figure. Also, find the length of the altitude from vertex A on the side DC. -Maths 9th

Last Answer : =3 x 3 x 5 x 2 cm2 Area of parallelogram ABCD = 2 x 90 = 180 cm2 (ii) Let altitude of a parallelogram be h. Also, area of parallelogram ABCD =Base x Altitude ⇒ 180 = DC x h [from Eq. (ii)] ... h ∴ h = 180/12= 15 cm Hence, the area of parallelogram is 180 cm2 and the length of altitude is 15 cm.

Let the vertex of an angle ABC be located outside a circle -Maths 9th

Description : Let the vertex of an angle ABC be located outside a circle -Maths 9th

Last Answer : Given: AD = CE To prove: ∠ABC = 1/2(∠DOE - ∠AOC) In △AOD and △COE AD = CE (Given) AO = OC and DO = OE (Radii of same circle) ∴ △AOD ≅ △COE (By SSS congruence criterion) ⇒ ∠1 = ∠3, ∠2 = ∠4 (CPCT) ... = 1/2(4z + 4x - 360°) ....(vii) From (iv) and (vii), we have ∠BAC = 1/2(∠DOE - ∠AOC)

a square is inscribed in an isosceles triangle so that the square and the triangle have one angle common. show that the vertex of the square opposite the vertex of the common angle bisect the hypotenuse. -Maths 9th

Description : a square is inscribed in an isosceles triangle so that the square and the triangle have one angle common. show that the vertex of the square opposite the vertex of the common angle bisect the hypotenuse. -Maths 9th

Last Answer : This answer was deleted by our moderators...

If the distance from the vertex to the centroid of an equilateral triangle is 6 cm, then what is the area of the triangle? -Maths 9th

Description : If the distance from the vertex to the centroid of an equilateral triangle is 6 cm, then what is the area of the triangle? -Maths 9th

Last Answer : (b) 27√3 cm2.Let G be the centroid of ΔPQR. Then, PG = 6 cm.Now, \(rac{PG}{GS}\) = \(rac{2}{1}\) ⇒ GS = 3 cm∴PS = PG + GS = 9 cm (i)∴ If a is the length of a side of ΔPQR, then ... = 6√3 cm∴ Area of equilateral ΔPQR = \(rac{\sqrt3}{4}\) (a)2= \(rac{\sqrt3}{4}\) x (6√3)2 cm2 = 27√3 cm2.

In a triangle, the ratio of the distance between a vertex and the orthocentre and the distance of the circumcentre from the side -Maths 9th

Description : In a triangle, the ratio of the distance between a vertex and the orthocentre and the distance of the circumcentre from the side -Maths 9th

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Prove that any straight line drawn from the vertex of a triangle to the base is bisected by the straight line which -Maths 9th

Description : Prove that any straight line drawn from the vertex of a triangle to the base is bisected by the straight line which -Maths 9th

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Perpendiculars are drawn from the vertex of the obtuse angles of a rhombus to its sides. The length of each perpendicular is equal to a units. -Maths 9th

Description : Perpendiculars are drawn from the vertex of the obtuse angles of a rhombus to its sides. The length of each perpendicular is equal to a units. -Maths 9th

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The area of a triangle is 5. Two of its vertices are (2, 1) and (3, –2). The third vertex is (x, y) -Maths 9th

Description : The area of a triangle is 5. Two of its vertices are (2, 1) and (3, –2). The third vertex is (x, y) -Maths 9th

Last Answer : Let A(x1, y1) = (3, 4), B(x2, y2) ≡ (0, 5), C(x3, y3) ≡ (2, -1)and D(x4, y4) ≡ (3, -2) be the vertices of quadrilateral ABCD.Area of quad. ABCD = \(rac{1}{2}\) |{(x1 y2 - x2 y1) + (x2y3 - x3y2) + (x3y4 - x4y3) ... ) + (12 + 6)}|= \(rac{1}{2}\) |{15 - 11 + 0 + 18}| = \(rac{1}{2}\)x 22 = 11 sq. units.

The two vertices of a triangle are (2, –1), (3, 2) and the third vertex lies on the line x + y = 5. The area of the triangle is 4 units. -Maths 9th

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Last Answer : (c) (5, 0) or (1, 4) Let the third vertex of the triangle be P(a, b). Since it lies on the line x + y = 5, a + b = 5 ...(i) Also, given area of triangle formed by the points (2, -1), (3, 2) and (a, b) = 4 ... b) - (-3a + b) = 5 + 15⇒ 4a = 20 ⇒ a = 5 ⇒ b = 0. ∴ The points are (1, 4) and (5, 0).

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Last Answer : D=slid ht of BC D≅(20+4​,20+2​) =(2,1) ∴ Length of median = Light of AD =root(−2−2)2+(4−1)2​=root42+32​=5 hope it helps thank u

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