How many face edges and vertices does a rectangular prism have?

How many face edges and vertices does a rectangular prism have?
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A rectangular prism has 6 faces, 12 edges and 8 vertices
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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

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Description : Coordinate of â- ABCD is WCS are: lowermost corner A(2,2) & diagonal corner are C(8,6). W.r.t MCS. The coordinates of origin of WCS system are (5,4). If the axes of WCS are at 600 in CCW w.r.t. the axes of MCS. Find new ... in MCS. a.(4.268, 6.732) b.(5.268, 6.732) c.(4.268, 4.732) d.(6.268, 4.732)

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In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22). Show that (i) quadrilateral ABED is a parallelogram (ii) quadrilateral BEFC is a parallelogram (iii) AD || CF and AD = CF (iv) quadrilateral ACFD is a parallelogram (v) AC = DF (vi) ΔABC ≅ ΔDEF. -Maths 9th

Description : In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22). Show that (i) quadrilateral ABED is a parallelogram ( ... CF and AD = CF (iv) quadrilateral ACFD is a parallelogram (v) AC = DF (vi) ΔABC ≅ ΔDEF. -Maths 9th

Last Answer : . Solution: (i) AB = DE and AB || DE (Given) Two opposite sides of a quadrilateral are equal and parallel to each other. Thus, quadrilateral ABED is a parallelogram (ii) Again BC = EF and BC || EF ... (Given) BC = EF (Given) AC = DF (Opposite sides of a parallelogram) , ΔABC ≅ ΔDEF [SSS congruency]

ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.21). Show that (i) ΔAPB ≅ ΔCQD (ii) AP = CQ -Maths 9th

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Last Answer : Q Solution: (i) In ΔAPB and ΔCQD, ∠ABP = ∠CDQ (Alternate interior angles) ∠APB = ∠CQD (= 90o as AP and CQ are perpendiculars) AB = CD (ABCD is a parallelogram) , ΔAPB ≅ ΔCQD [AAS congruency] (ii) As ΔAPB ≅ ΔCQD. , AP = CQ [CPCT]

ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD . -Maths 9th

Description : ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD . -Maths 9th

Last Answer : In gm ABCD , AP and CQ are perpendicular from the vertices A and C on diagonal BD. Show that : (i) AAPB ≅ ACQD (ii) AP = CQ .

Points A(5, 3), B(-2, 3) and 0(5, – 4) are three vertices of a square ABCD. -Maths 9th

Description : Points A(5, 3), B(-2, 3) and 0(5, – 4) are three vertices of a square ABCD. -Maths 9th

Last Answer : The graph obtained by plotting the points A, B and C and D is given below. Take a point C on the graph such that ABCD is a square i.e., all sides AB, BC, CD, and AD are equal. So, abscissa of C should be ... of C should be equal to ordinate of D i.e., -4. Hence, the coordinates of C are (-2, - 4).

Write the coordinates of the vertices of a rectangle whose length and breadth are 5 and 3 units respectively, -Maths 9th

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Last Answer : Given, length of a rectangle = 5 units and breadth of a rectangle = 3 units One vertex is at origin i.e., (0, 0) and one of the other vertices lies in III quadrant. So, the length of the rectangle is 5 ... negative,direction of y-axis and then vertex is C(0, -3). The fourth vertex B is (-5, - 3).

The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to -Maths 9th

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Last Answer : Solution of this question

ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD . -Maths 9th

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Last Answer : In gm ABCD , AP and CQ are perpendicular from the vertices A and C on diagonal BD. Show that : (i) AAPB ≅ ACQD (ii) AP = CQ .

Points A(5, 3), B(-2, 3) and 0(5, – 4) are three vertices of a square ABCD. -Maths 9th

Description : Points A(5, 3), B(-2, 3) and 0(5, – 4) are three vertices of a square ABCD. -Maths 9th

Last Answer : The graph obtained by plotting the points A, B and C and D is given below. Take a point C on the graph such that ABCD is a square i.e., all sides AB, BC, CD, and AD are equal. So, abscissa of C should be ... of C should be equal to ordinate of D i.e., -4. Hence, the coordinates of C are (-2, - 4).