The figure shows the front view of a convex lens, which originally had only one edge. Five holes of different shapes, namely triangle, square, pentagon, hexagon and circle, were drilled through it at points P, Q, R, S and T in such a way that the holes were parallel to each other, perpendicular to the edge and all the holes were still within the lens edge. What is the total number of edges in the lens after the holes were drilled?

The figure shows the front view of a convex lens, which originally had only one edge. Five holes of different shapes, namely triangle, square, pentagon, hexagon and circle, were drilled through it at points P, Q, R, S and T in such a way that the holes were parallel to each other, perpendicular to the edge and all the holes were still within the lens edge. What is the total number of edges in the lens after the holes were drilled? 

Image 1022 - .jpg

by

1 Answer

Answer :

57
by

Related questions

Triangle P and pentagon Q have markings on them as shown in the figure. If they are placed over each other, which of the following arrangements is/are possible?

Description : Triangle P and pentagon Q have markings on them as shown in the figure. If they are placed over each other, which of the following arrangements is/are possible? 

Last Answer : A, B, D

Figure P shows an isometric view of an object made with a single sheet of cardboard. Figure Q shows the front view of the same object after colouring. Which option shows the object after it is cut open?

Description : Figure P shows an isometric view of an object made with a single sheet of cardboard. Figure Q shows the front view of the same object after colouring. Which option shows the object after it is cut open?

Last Answer : C

In figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. -Maths 9th

Description : In figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. -Maths 9th

Last Answer : Given ABCDE is a pentagon. BP || AC and EQ|| AD. To prove ar (ABCDE) = ar (APQ) Proof We know that, triangles on the same base and between the same parallels are equal in area. Here, ΔADQ and ΔADE lie on the ... ar (ΔACD) = ar (ΔADE) + ar (ΔACB) + ar (ΔACD) ⇒ ar (ΔAPQ) = ar (ABCDE) Hence proved.

In figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. -Maths 9th

Description : In figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. -Maths 9th

Last Answer : Given ABCDE is a pentagon. BP || AC and EQ|| AD. To prove ar (ABCDE) = ar (APQ) Proof We know that, triangles on the same base and between the same parallels are equal in area. Here, ΔADQ and ΔADE lie on the ... ar (ΔACD) = ar (ΔADE) + ar (ΔACB) + ar (ΔACD) ⇒ ar (ΔAPQ) = ar (ABCDE) Hence proved.

If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Description : If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Last Answer : According to question prove that P, Q, R and D are concyclic.

If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Description : If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Last Answer : According to question prove that P, Q, R and D are concyclic.

Write the coordinates of each of the points P, Q, R, S, T and 0 from the figure . -Maths 9th

Description : Write the coordinates of each of the points P, Q, R, S, T and 0 from the figure . -Maths 9th

Last Answer : Here, points P and S lie in I quadrant so their both coordinates will be positive. Now, perpendicular distance of P from both axes is 1, so coordinates of P are (1, 1). Also, perpendicular distance of S ... 0 is the intersection of both axes, so it is the origin and its coordinates are O (0,0).

Write the coordinates of each of the points P, Q, R, S, T and 0 from the figure . -Maths 9th

Description : Write the coordinates of each of the points P, Q, R, S, T and 0 from the figure . -Maths 9th

Last Answer : Here, points P and S lie in I quadrant so their both coordinates will be positive. Now, perpendicular distance of P from both axes is 1, so coordinates of P are (1, 1). Also, perpendicular distance of S ... 0 is the intersection of both axes, so it is the origin and its coordinates are O (0,0).

What mathematical plane figure is ideal in cellular system design and engineering it graphically and functionally depicts overlapping radio coverage between and among adjacent cell base stations? A. Hexagon B. Octagon C. Pentagon D. Nonagon

Description : What mathematical plane figure is ideal in cellular system design and engineering it graphically and functionally depicts overlapping radio coverage between and among adjacent cell base stations? A. Hexagon B. Octagon C. Pentagon D. Nonagon

Last Answer : A. Hexagon

Pick up the correct specification of Ramsden eyepiece from the following: (A) It consists of two equal piano convex lenses (B) The curved surfaces of Plano-convex lenses face each other (C) The distance between the diaphragm and the front lens of the eyepiece is kept equal to 1/4 th of the focal length of a lens so that rays from a point on the diaphragm enter the eye as a parallel beam (D) All the above

Description : Pick up the correct specification of Ramsden eyepiece from the following:  (A) It consists of two equal piano convex lenses  (B) The curved surfaces of Plano-convex lenses face each other  (C) ... from a point on the diaphragm enter the eye as a  parallel beam  (D) All the above 

Last Answer : (D) All the above 

In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

Description : In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

Last Answer : In △PAD, ∠A = 90° and DA = PA = PB ⇒ ∠ADP = ∠APD = 90° / 2 = 45° Similarly, in △QBC, ∠B = 90° and BQ = BC = AB ⇒∠BCQ = ∠BQC = 90° / 2 = 45° In △PAD and △QBC , we have PA = QB [given] ∠A = ... [each = 90° + 45° = 135°] ⇒ △PDC = △QCD [by SAS congruence rule] ⇒ PC = QD or DQ = CP

In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

Description : In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

Last Answer : In △PAD, ∠A = 90° and DA = PA = PB ⇒ ∠ADP = ∠APD = 90° / 2 = 45° Similarly, in △QBC, ∠B = 90° and BQ = BC = AB ⇒∠BCQ = ∠BQC = 90° / 2 = 45° In △PAD and △QBC , we have PA = QB [given] ∠A = ... [each = 90° + 45° = 135°] ⇒ △PDC = △QCD [by SAS congruence rule] ⇒ PC = QD or DQ = CP

The table gives the populations (in thousands) of ten species (A-J) in four areas (p-s) consisting of the number of habitats given within brackets against each. Study the table and answer the question which follows. Which area out of p - s shows maximum species diversity? (a) s (b) p (c) q (d) r

Description : The table gives the populations (in thousands) of ten species (A-J) in four areas (p-s) consisting of the number of habitats given within brackets against each. Study the table and answer the question which follows. ... out of p - s shows maximum species diversity? (a) s (b) p (c) q (d) r

Last Answer : (a) s

In the above figure, AB is parallel to CD. P and R are the points on AB and CD respectively. Q is in between AB and CD. Find the value of x in degrees

Description : In the above figure, AB is parallel to CD. P and R are the points on AB and CD respectively. Q is in between AB and CD. Find the value of x in degrees

Last Answer : In the above figure, AB is parallel to CD. P and R are the points on AB and CD respectively. Q is in between AB and CD. Find the value of x in degrees

Draw the ray diagram in each case to show the position and nature of the image formed when the object is placed: (i) at the centre of curvature of a concave mirror (ii) between the pole P and focus F of a concave mirror (iii) in front of a convex mirror (iv) at 2F of a convex lens (v) in front of a concave lens -Physics

Description : Draw the ray diagram in each case to show the position and nature of the image formed when the object is placed: (i) at the centre of curvature of a concave mirror (ii) between the pole P and focus F of ... front of a convex mirror (iv) at 2F of a convex lens (v) in front of a concave lens -Physics

Last Answer : Answer. Nature of image: Real, inverted and same size image is formed at the centre of curvature. Nature of image: Virtual, enlarged and erect image is formed behind the mirror. Nature of image: Virtual ... : Virtual, erect and diminished image is formed between O and F on the same side of object.

When light enters a closed room through a small hole in the door, the image of an outside building appears as inverted on the opposite wall. This is because (a) The hole acts as a convex lens (b) Light takes curvature at the edges of the hole (c) Of rectilinear propagation of light (d) The hole acts as a concave lens

Description : When light enters a closed room through a small hole in the door, the image of an outside building appears as inverted on the opposite wall. This is because (a) The hole acts as a convex lens (b) ... the edges of the hole (c) Of rectilinear propagation of light (d) The hole acts as a concave lens

Last Answer : Ans:(b)

In Fig. 9.30, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. Prove that ar (ABCDE) = ar(APQ). -Maths 9th

Description : In Fig. 9.30, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. Prove that ar (ABCDE) = ar(APQ). -Maths 9th

Last Answer : hope its clearhope its clear

Plot the points P(1, 0), Q(4, 0) and 5(1, 3). Find the coordinates of the point R such that PQRS is a square. -Maths 9th

Description : Plot the points P(1, 0), Q(4, 0) and 5(1, 3). Find the coordinates of the point R such that PQRS is a square. -Maths 9th

Last Answer : see the below answer

Plot the points P(1, 0), Q(4, 0) and 5(1, 3). Find the coordinates of the point R such that PQRS is a square. -Maths 9th

Description : Plot the points P(1, 0), Q(4, 0) and 5(1, 3). Find the coordinates of the point R such that PQRS is a square. -Maths 9th

Last Answer : see the below answer

PQ and RS are two equal and parallel line segments.Any points M not lying on PQ or RS is joined to Q and S and lines through P parallel to SM meet at N.Prove that line segments MN and PQ are equal and parallel to each other. -Maths 9th

Description : PQ and RS are two equal and parallel line segments.Any points M not lying on PQ or RS is joined to Q and S and lines through P parallel to SM meet at N.Prove that line segments MN and PQ are equal and parallel to each other. -Maths 9th

Last Answer : hope its clear

Figure P shows how a point on a circle traces a path when it is rolled on the ground. The point in which of the polygons shown in the options creates the path in Figure Q?

Description : Figure P shows how a point on a circle traces a path when it is rolled on the ground. The point in which of the polygons shown in the options creates the path in Figure Q?  

Last Answer : D 

P is the mid - point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R (see figure). -Maths 9th

Description : P is the mid - point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R (see figure). -Maths 9th

Last Answer : (i) In △ARB,P is the mid point of AB and PD || BR. ∴ D is a mid - point of AR [converse of mid - point theorem] ∴ AR = 2AD But BC = AD [opp sides of ||gm ABCD] Thus, AR = 2BC (ii) ∴ ABCD is a ... a mid - point of AR and DQ || AB ∴ Q is a mid point of BR [converse of mid - point theorem] ⇒ BR = 2BQ

P is the mid - point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R (see figure). -Maths 9th

Description : P is the mid - point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R (see figure). -Maths 9th

Last Answer : (i) In △ARB,P is the mid point of AB and PD || BR. ∴ D is a mid - point of AR [converse of mid - point theorem] ∴ AR = 2AD But BC = AD [opp sides of ||gm ABCD] Thus, AR = 2BC (ii) ∴ ABCD is a ... a mid - point of AR and DQ || AB ∴ Q is a mid point of BR [converse of mid - point theorem] ⇒ BR = 2BQ

P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square. -Maths 9th

Description : P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square. -Maths 9th

Last Answer : Given In quadrilateral ABCD, P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Also, AC = BD and AC ⊥ BD. To prove PQRS is a square. Proof Now, in ΔADC, S and R are the mid-points of the sides AD and DC respectively, then by mid-point theorem,

P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square. -Maths 9th

Description : P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square. -Maths 9th

Last Answer : Given In quadrilateral ABCD, P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Also, AC = BD and AC ⊥ BD. To prove PQRS is a square. Proof Now, in ΔADC, S and R are the mid-points of the sides AD and DC respectively, then by mid-point theorem,

ABCD is a square. P, Q, R, S are the mid-points of AB, BC, CD and DA respectively. By joining AR, BS, CP, DQ, we get a quadrilateral which is a -Maths 9th

Description : ABCD is a square. P, Q, R, S are the mid-points of AB, BC, CD and DA respectively. By joining AR, BS, CP, DQ, we get a quadrilateral which is a -Maths 9th

Last Answer : According to the given statement, the figure will be a shown alongside; using mid-point theorem: In △ABC,PQ∥AC and PQ=21 AC .......(1) In △ADC,SR∥AC and SR=21 AC .... ... are perpendicular to each other) ∴PQ⊥QR(angle between two lines = angle between their parallels) Hence PQRS is a rectangle.

Let P(–3, 2), Q(–5, –5), R(2, –3) and S(4, 4) be four points in a plane. Then show that PQRS is a rhombus. Is it a square ? -Maths 9th

Description : Let P(–3, 2), Q(–5, –5), R(2, –3) and S(4, 4) be four points in a plane. Then show that PQRS is a rhombus. Is it a square ? -Maths 9th

Last Answer : Let P(1, -1), Q \(\big(rac{-1}{2},rac{1}{2}\big)\) and R(1,2) be the vertices of the ΔPQR.Then, PQ = \(\sqrt{\big(rac{-1}{2}-1\big)^2+\big(rac{1}{2}+1\big)^2}\) = \(\sqrt{rac{9}{4}+rac{9}{4}} ... {3\sqrt2}{2}\)PR = \(\sqrt{(1-1)^2+(2+1)^2}\) = \(\sqrt9\) = 3∵ PQ = QR, the triangle PQR is isosceles.

Points P, Q, R and S divide a line segment joining A (2, 6) and B (7, -4) in five equal parts. Find the coordinates of P and R. -Maths 9th

Description : Points P, Q, R and S divide a line segment joining A (2, 6) and B (7, -4) in five equal parts. Find the coordinates of P and R. -Maths 9th

Last Answer : this is the ans hope its clear

Compute the charge enclosed by a cube of 2m each edge centered at the origin and with the edges parallel to the axes. Given D = 10y 3 /3 j. a) 20 b) 70/3 c) 80/3 d) 30

Description : Compute the charge enclosed by a cube of 2m each edge centered at the origin and with the edges parallel to the axes. Given D = 10y 3 /3 j. a) 20 b) 70/3 c) 80/3 d) 30

Last Answer : c) 80/3

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)

Which of the points P(0, 3), Q(l, 0), R(0, – 1), S(-5, 0) and T(1, 2) do not lie on the X-axis ? -Maths 9th

Description : Which of the points P(0, 3), Q(l, 0), R(0, – 1), S(-5, 0) and T(1, 2) do not lie on the X-axis ? -Maths 9th

Last Answer : (c) We know that, if a point is of the form (x, 0)i.e., its y-coordinate is zero, then it will lie on X-axis otherwise not. Here, y-coordinates of points P(0, 3), R (0, -1) and T (1,2) are not zero, so these points do not lie on the X-axis.

Which of the points P(0, 3), Q(l, 0), R(0, – 1), S(-5, 0) and T(1, 2) do not lie on the X-axis ? -Maths 9th

Description : Which of the points P(0, 3), Q(l, 0), R(0, – 1), S(-5, 0) and T(1, 2) do not lie on the X-axis ? -Maths 9th

Last Answer : (c) We know that, if a point is of the form (x, 0)i.e., its y-coordinate is zero, then it will lie on X-axis otherwise not. Here, y-coordinates of points P(0, 3), R (0, -1) and T (1,2) are not zero, so these points do not lie on the X-axis.

In the given figure, ABC is a triangle in which CDEFG is a pentagon. Triangles ADE and BFG are equilateral -Maths 9th

Description : In the given figure, ABC is a triangle in which CDEFG is a pentagon. Triangles ADE and BFG are equilateral -Maths 9th

Last Answer : (b) 7√3 cm2.AB = 6 cm, ∠C = 60º (∴ ∠A = ∠B = 60º) ∴ ΔABC is an equilateral triangle Area of ΔABC = \(rac{\sqrt3}{4}\) × (6)2 = 9√3 Area of (ΔADE + ΔBFG) = 2 x \(\bigg(rac{\sqrt3}{4} imes(2)^2\bigg)\) = 2√3 ∴ Area of pentagon = 9√3 - 2√3 = 7√3 cm2.

__________ shape is used to make codebook in structure vector quantization. a. Square b. Rectangle c. Circle d. Hexagon

Description : __________ shape is used to make codebook in structure vector quantization. a. Square b. Rectangle c. Circle d. Hexagon

Last Answer : d. Hexagon

What shape has a front view of a rectangle side view of a triangle and a top view of a circle?

Description : What shape has a front view of a rectangle side view of a triangle and a top view of a circle?

Last Answer : It is a shape that is like a toothpaste tube which has acircular lid which is the same size as the tube below it.

What shape has a front view of a rectangle side view of a triangle and a top view of a circle?

Description : What shape has a front view of a rectangle side view of a triangle and a top view of a circle?

Last Answer : It is a shape that is like a toothpaste tube which has acircular lid which is the same size as the tube below it.

In the figure, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR= RS and PA || QB || RC. -Maths 9th

Description : In the figure, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR= RS and PA || QB || RC. -Maths 9th

Last Answer : Given In a parallelogram PSDA, points 0 and R are on PS such that PQ = QR = RS and PA || QB || RC. To prove ar (PQE) = ar (CFD) Proof In parallelogram PABQ, and PA||QB [given] So, ... = ΔDCF [by ASA congruence rule] ∴ ar (ΔPQE) = ar (ΔCFD) [since,congruent figures have equal area] Hence proved.

In the figure, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR= RS and PA || QB || RC. -Maths 9th

Description : In the figure, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR= RS and PA || QB || RC. -Maths 9th

Last Answer : Given In a parallelogram PSDA, points 0 and R are on PS such that PQ = QR = RS and PA || QB || RC. To prove ar (PQE) = ar (CFD) Proof In parallelogram PABQ, and PA||QB [given] So, ... = ΔDCF [by ASA congruence rule] ∴ ar (ΔPQE) = ar (ΔCFD) [since,congruent figures have equal area] Hence proved.

l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Description : l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Last Answer : Though E, draw a line parallel to p intersecting L at G and n at H respectively. Since l | | m ⇒ AG | | BE and AB | | GE [by construction] ∴ Opposite sides of quadrilateral AGEB are ... ∠DGE = ∠FHE [alternate interior angles] By ASA congruence axiom, we have △DEG ≅ △FEH Hence, DE = EF

l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Description : l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Last Answer : Though E, draw a line parallel to p intersecting L at G and n at H respectively. Since l | | m ⇒ AG | | BE and AB | | GE [by construction] ∴ Opposite sides of quadrilateral AGEB are ... ∠DGE = ∠FHE [alternate interior angles] By ASA congruence axiom, we have △DEG ≅ △FEH Hence, DE = EF

If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, prove that arc PXA = arc PYB. -Maths 9th

Description : If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, prove that arc PXA = arc PYB. -Maths 9th

Last Answer : Let AB be a chord of a circle having centre at OPQ be the perpendicular bisector of the chord AB, which intersects at M and it always passes through O. To prove arc PXA ≅ arc PYB Construction Join AP and BP. Proof In ... ΔBPM, AM = MB ∠PMA = ∠PMB PM = PM ∴ ΔAPM s ΔBPM ∴PA = PB ⇒arc PXA ≅ arc PYB

If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, prove that arc PXA = arc PYB. -Maths 9th

Description : If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, prove that arc PXA = arc PYB. -Maths 9th

Last Answer : Let AB be a chord of a circle having centre at OPQ be the perpendicular bisector of the chord AB, which intersects at M and it always passes through O. To prove arc PXA ≅ arc PYB Construction Join AP and BP. Proof In ... ΔBPM, AM = MB ∠PMA = ∠PMB PM = PM ∴ ΔAPM s ΔBPM ∴PA = PB ⇒arc PXA ≅ arc PYB

If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, then prove that arc PXA ≅ arc PYB. -Maths 9th

Description : If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, then prove that arc PXA ≅ arc PYB. -Maths 9th

Last Answer : Solution :- Let AB be a chord of a circle having centre at O. Let PQ be the perpendicular bisector of the chord AB intersect it say at M. Perpendicular bisector of the chord passes through the centre of the circle,i. ... = PM (Common) ∴ △APM ≅ △BPM (SAS) PA = PB (CPCT) Hence, arc PXA ≅ arc PYB

AB and AC are two chords of a circle of radius r such that AB = 2AC. If p and q are the distances of AB and AC from the centre. Prove that -Maths 9th

Description : AB and AC are two chords of a circle of radius r such that AB = 2AC. If p and q are the distances of AB and AC from the centre. Prove that -Maths 9th

Last Answer : Draw OM perpendicular AB and ON perpendicular AC Join OA. In right △OAM, OA2 = OM2 + AM2 ⇒ r2 = p2 + (1/2AB)2 (Since,OM perpendicular AB, ∴ OM bisects AB ) ⇒ 1/4AB2 = r2 - p2 or AB2 = 4r2 - 4p2 ... ) and (ii), we have 4r2 - 4p2 = 16r2 - 16q2 or r2 - p2 = 4r2 - 4q2 or 4q2 = 3r2 + p2

Which of the following statement(s) is/are correct with reference to curve generation? I. Hermite curves are generated using the concepts of interpolation. II. Bezier curves are generated using the concepts of approximation. III. The Bezier curves lies entirely within the convex hull of its control points. IV. The degree of Bezier curve does not depend on the number of control points. (A) I, II and IV only (B) II and III only (C) I and II only (D) I, II and III only

Description : Which of the following statement(s) is/are correct with reference to curve generation? I. Hermite curves are generated using the concepts of interpolation. II. Bezier curves are generated using the concepts of approximation. III. The ... (B) II and III only (C) I and II only (D) I, II and III only

Last Answer : (D) I, II and III only

When load is applied on concrete pavement (A) Away from edges, the maximum bending moment is negative (B) Away from edges, the maximum bending moment cause compression (C) On the edges, the maximum stress is parallel to the edge of the slab (D) None of these

Description : When load is applied on concrete pavement (A) Away from edges, the maximum bending moment is negative (B) Away from edges, the maximum bending moment cause compression (C) On the edges, the maximum stress is parallel to the edge of the slab (D) None of these

Last Answer : Answer: Option C

The main reinforcement in the toe of a T-shaped R C. retaining wall is provided on (i) Top face parallel to the wall (ii) Top face perpendicular to the wall (iii) Bottom face parallel to the wall (iv) Bottom face perpendicular to the wall The correct answer is (A) Only (ii) is correct (B) (i) and (ii) are correct (C) (iii) and (iv) are correct (D) Only (iv) is correct

Description : The main reinforcement in the toe of a T-shaped R C. retaining wall is provided on (i) Top face parallel to the wall (ii) Top face perpendicular to the wall (iii) Bottom face parallel to the wall (iv) ... (i) and (ii) are correct (C) (iii) and (iv) are correct (D) Only (iv) is correct

Last Answer : Answer: Option D

The distance between a convex lens and a plane mirror is 15 cm. The parallel rays incident on the convex lens, after reflection from the mirror, form

Description : The distance between a convex lens and a plane mirror is 15 cm. The parallel rays incident on the convex lens, after reflection from the mirror, form

Last Answer : The distance between a convex lens and a plane mirror is 15 cm. The parallel rays incident on the ... and find out the focal length of the lens.

3. Where should an object be placed in front of a convex lens to get a real image of the size of the object? -Physics-10

Description : 3. Where should an object be placed in front of a convex lens to get a real image of the size of the object? -Physics-10

Last Answer : (b) The object should be placed at twice the focal length