The C.G. of a plane lamina will not be at its geometrical centre in the case of a (A) Right angled triangle (B) Equilateral triangle (C) Square (D) Circle

The C.G. of a plane lamina will not be at its geometrical centre in the case of a 

(A) Right angled triangle 

(B) Equilateral triangle 

(C) Square 

(D) Circle

by

1 Answer

Answer :

(A) Right angled triangle 

by

Related questions

The centre of gravity of a plane lamina will not be at its geometrical centre if it is a a.Square b.Equilateral triangle c.Circle d.Rectangle e.Right angled triangle

Description : The centre of gravity of a plane lamina will not be at its geometrical centre if it is a a.Square b.Equilateral triangle c.Circle d.Rectangle e.Right angled triangle

Last Answer : e. Right angled triangle

The hypotenuse of an isosceles right-angled triangle is q. If we describe equilateral triangles (outwards) on all its three sides, -Maths 9th

Description : The hypotenuse of an isosceles right-angled triangle is q. If we describe equilateral triangles (outwards) on all its three sides, -Maths 9th

Last Answer : (b) \(rac{q^2}{4}\) (2√3 + 1).AC = q, ∠ABC = 90º ⇒ q = \(\sqrt{AB^2+BC^2}\)⇒ q = \(\sqrt{2x^2}\)⇒ q2 = 2x2 ⇒ \(x\) = \(rac{q}{\sqrt2}\)∴ Area of the re-entrant hexagon = Sum of areas of (ΔABC + ΔADC ... (rac{\sqrt3}{4}\)q2 + \(rac{\sqrt3}{8}\)q2 + \(rac{\sqrt3q^2}{8}\) = \(rac{q^2}{4}\) (2√3 + 1).

Name the type of triangle formed. (a) Right angled (b) Equilateral (c) Isosceles (d) Scalene

Description : Name the type of triangle formed. (a) Right angled (b) Equilateral (c) Isosceles (d) Scalene

Last Answer : (d) Scalene

A circle is inscribed in an equilateral triangle of side a. What is the area of any square inscribed in this circle? -Maths 9th

Description : A circle is inscribed in an equilateral triangle of side a. What is the area of any square inscribed in this circle? -Maths 9th

Last Answer : (c) \(rac{a^2}{6}.\)If a' is length of the side of ΔABC, thenArea of ΔABC = \(rac{\sqrt3}{4}\,a^2\)semi-perimeter of ΔABC = \(rac{3a}{2}\)∴ Radius of in-circle = \(rac{ ext{Area}}{ ext{semi-perimeter}}\) = \( ... {( ext{diagonal})^2}{2}\) = \(rac{\big(rac{a}{\sqrt3}\big)^2}{2}\) = \(rac{a^2}{6}.\)

What is the ratio of the area of a circum circle of equilateral triangle to the area of the square with same side length as equilateral triangle? -Maths 9th

Description : What is the ratio of the area of a circum circle of equilateral triangle to the area of the square with same side length as equilateral triangle? -Maths 9th

Last Answer : (d) π : 3.Let each side of the equilateral Δ be a units. Then, circumradius of the circle = \(rac{ ext{side}}{\sqrt3}\) = \(rac{a}{\sqrt3}\) units∴ Area of circumcircle = \(\pi\bigg(rac{a}{\sqrt3}\bigg)^2\) = \( ... units∴ Required ratio = \(rac{rac{\pi{a}^2}{3}}{a^2}\) = \(rac{\pi}{3}\) = π : 3.

Find the area of an equilateral triangle inscribed in a circle circumscribed by a square made by joining the mid-points -Maths 9th

Description : Find the area of an equilateral triangle inscribed in a circle circumscribed by a square made by joining the mid-points -Maths 9th

Last Answer : (d) \(rac{3\sqrt3a^2}{32}\)Let AB = a be the side of the outermost square.Then AG = AH = \(rac{a}{2}\)⇒ GH = \(\sqrt{rac{a^2}{4}+rac{a^2}{4}}\) = \(rac{a}{\sqrt2}\)∴ Diameter of circle = \(rac{a} ... rac{\sqrt3}{2}\) = \(rac{\sqrt3a^2}{32}\)∴ Area of ΔPQR = 3 (Area of ΔPOQ) = \(rac{\sqrt3a^2}{32}\)

The area of a square and circle is same and the perimeter of square and equilateral triangle is same, -Maths 9th

Description : The area of a square and circle is same and the perimeter of square and equilateral triangle is same, -Maths 9th

Last Answer : (b) 9 : 4√3.Let each side of the square = a cm. Then, Area of square = a2 cm2 Also, let r be the radius of the circle. Then, πr2 = a2 Let each side of the equilateral triangle = b cm. Then 3b = 4a ⇒ ... ratio between area of circle and area of equilateral Δ is a2 : \(rac{4\sqrt3a^2}{9}\) = 9 : 4√3.

If the area of a circle, inscribed in an equilateral triangle is 4π cm^2, then what is the area of the triangle? -Maths 9th

Description : If the area of a circle, inscribed in an equilateral triangle is 4π cm^2, then what is the area of the triangle? -Maths 9th

Last Answer : (a) 12√3 cm2Since area of circle = 4π ⇒ πr2 = 4π ⇒ r = 2 cmIn ΔOAD,tan 30° = \(rac{OD}{AD}\) ⇒ \(rac{1}{\sqrt3}\) = \(rac{2}{AD}\)⇒ AD = 2√3 cm ∴ AB = 2AD = 4√3 cm∴ Area of equilateral ΔABC = \(rac{\sqrt3}{4}\) (AB)2= \(rac{\sqrt3}{4}\) (4√3)2 = 12√3 cm2.

O is a point in the interior of a square ABCD such that OAB is an equilateral triangle.Show that △OCD is an isosceles triangle. -Maths 9th

Description : O is a point in the interior of a square ABCD such that OAB is an equilateral triangle.Show that △OCD is an isosceles triangle. -Maths 9th

Last Answer : Solution :-

In the adjoining figure, ABC is an equilateral triangle inscribing a square of maximum possible area. Again in this squares -Maths 9th

Description : In the adjoining figure, ABC is an equilateral triangle inscribing a square of maximum possible area. Again in this squares -Maths 9th

Last Answer : (a) (873 - 504√3) cm2.Since ∠CPO = ∠COP = 60º, therefore, PCO is also an equilateral triangle. Let each side of the square MNOP be x cm. Then PC = CO = PO = x cm Then in ΔPAM,\(rac{PM}{PA}\) = sin 60º⇒ \(rac{x ... most square = y2= \(\big(3(7-4\sqrt2)\big)^2\)= 9(49 + 48 - 56√3) = (873 - 504√3) cm2.

A square and an equilateral triangle each have a perimeter of 72 cm. How much longer is one side of the triangle than one side of the square?

Description : A square and an equilateral triangle each have a perimeter of 72 cm. How much longer is one side of the triangle than one side of the square?

Last Answer : You see you can do is 72cm divide by 3 because an equilateral triangle has 3 sides 72cm/3 = 24cm(I am not sure if this is right)

A copper wire is bent to form an equilateral triangle of area 16√3 sq.cm. What will be the area (in sq.cm) if the same copper wire was bent to form a square? a) 16 b) 24 c) 32 d) 40 e) None of the above

Description : A copper wire is bent to form an equilateral triangle of area 16√3 sq.cm. What will be the area (in sq.cm) if the same copper wire was bent to form a square? a) 16 b) 24 c) 32 d) 40 e) None of the above

Last Answer : Area of equilateral triangle = 16√3 = √3 (side)2 /4 Or side2 = 16×4 = 64 So each side = 8cm Perimeter = 8×3 = 24 cm Since perimeter of the square is 24 cm, each side = 6cm So area = 6×6 = 36 sq cm Answer: e)

The base of a right-angled triangle measures 4 cm and its hypotenuse measures 5 cm. Find the area of the triangle. -Maths 9th

Description : The base of a right-angled triangle measures 4 cm and its hypotenuse measures 5 cm. Find the area of the triangle. -Maths 9th

Last Answer : In right-angled triangle ABC AB2 + BC2 = AC2 (By Pythagoras Theorem) ⇒ AB2 + 42 = 52 ⇒ AB2 = 25 – 16 = 9 5 cm ⇒ AB = 3 cm ∴ Area of △ABC = 1/2 BC x AB = 1/2 x 4 x 3 = 6cm2

If the length of hypotenuse of a right angled triangle is 5 cm and its area is 6 sq cm, then what are the lengths of the remaining sides? -Maths 9th

Description : If the length of hypotenuse of a right angled triangle is 5 cm and its area is 6 sq cm, then what are the lengths of the remaining sides? -Maths 9th

Last Answer : Let one of the remaining sides be x cm.Then, other side = \(\sqrt{5^2-x^2}\) cm∴ Area = \(rac{1}{2} imes{x} imes\sqrt{25-x^2}\) = 6⇒ \(x\sqrt{25-x^2}\) = 12 ⇒ x2(25 - x2) = 144⇒ 25x2 - x4 = 144 ⇒ x4 - 25x2 ... (x2 - 16) (x2 - 9) = 0 ⇒ x2 = 16 or x2 = 9 ⇒ x = 4 or 3∴ The two sides are 4 cm and 3 cm.

Let ABC be a right angled triangle with AC as its hypotenuse. Then, -Maths 9th

Description : Let ABC be a right angled triangle with AC as its hypotenuse. Then, -Maths 9th

Last Answer : answer:

Is it possible for a right angled triangle with sides 3 and 4 units long to have a hypotenuse 6 units in length?

Description : Is it possible for a right angled triangle with sides 3 and 4 units long to have a hypotenuse 6 units in length?

Last Answer : answer:I'm not quite getting you. It isn't actually a triangle when the hypotenuse has these indentations, right? The hypotenuse isn't a straight line as you describe it. If the other sides are 3 ... and 5.00001, you don't have a straight line. Unless I'm misunderstanding what you're suggesting.

ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC (ii) MD ⊥ AC (iii) CM = MA = ½ AB -Maths 9th

Description : ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC (ii) MD ⊥ AC (iii) CM = MA = ½ AB -Maths 9th

Last Answer : Solution: (i) In ΔACB, M is the midpoint of AB and MD || BC , D is the midpoint of AC (Converse of mid point theorem) (ii) ∠ACB = ∠ADM (Corresponding angles) also, ∠ACB = 90° , ∠ADM = 90° and MD ⊥ AC (iii ... SAS congruency] AM = CM [CPCT] also, AM = ½ AB (M is midpoint of AB) Hence, CM = MA = ½ AB

ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Description : ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Last Answer : Given: A △ABC , right - angled at C. A line through the mid - point M of hypotenuse AB parallel to BC intersects AC at D. To Prove: (i) D is the mid - point of AC (ii) MD | AC (iii) CM = MA = 1 / 2 ... congruence axiom] ⇒ AM = CM Also, M is the mid - point of AB [given] ⇒ CM = MA = 1 / 2 = AB.

ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Description : ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Last Answer : Given: A △ABC , right - angled at C. A line through the mid - point M of hypotenuse AB parallel to BC intersects AC at D. To Prove: (i) D is the mid - point of AC (ii) MD | AC (iii) CM = MA = 1 / 2 ... congruence axiom] ⇒ AM = CM Also, M is the mid - point of AB [given] ⇒ CM = MA = 1 / 2 = AB.

triangle ABC is right angled at A. AL is drawn perpendicular to BC. Prove that /_ BAL = /_ ACB -Maths 9th

Description : triangle ABC is right angled at A. AL is drawn perpendicular to BC. Prove that /_ BAL = /_ ACB -Maths 9th

Last Answer : NEED ANSWER

triangle ABC is right angled at A. AL is drawn perpendicular to BC. Prove that /_ BAL = /_ ACB -Maths 9th

Description : triangle ABC is right angled at A. AL is drawn perpendicular to BC. Prove that /_ BAL = /_ ACB -Maths 9th

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

ABC is a triangle right-angled at C. A line through the mid-point of hypotenuse AB and parallel to BC intersects AC at D. Show that -Maths 9th

Description : ABC is a triangle right-angled at C. A line through the mid-point of hypotenuse AB and parallel to BC intersects AC at D. Show that -Maths 9th

Last Answer : Solution :-

In a right-angled triangle ABC, D is the foot of the perpendicular from B on the hypotenuse AC -Maths 9th

Description : In a right-angled triangle ABC, D is the foot of the perpendicular from B on the hypotenuse AC -Maths 9th

Last Answer : Area of ΔABC = \(rac{1}{2}\) x 3 x 4 cm2 = 6 cm2. Also, AC = \(\sqrt{3^2+4^2}\) = 5 cm.∴ Area of ΔABC = \(rac{1}{2}\) x BD x AC ⇒ 6 = \(rac{1}{2}\) BD x 5 ⇒ BD = \(rac{12}{5}\) cm.Now in ΔABD, AD = \(\ ... \(rac{1}{2}\)x AD x BD = \(rac{1}{2}\) x \(rac{9}{5}\) x \(rac{12}{5}\) = \(rac{54}{25}\) cm2.

A piece of paper is in the shape of a right-angled triangle and is cut along a line that is parallel to the hypotenuse, leaving a smaller triangle. -Maths 9th

Description : A piece of paper is in the shape of a right-angled triangle and is cut along a line that is parallel to the hypotenuse, leaving a smaller triangle. -Maths 9th

Last Answer : (d) 14.365Given, ST || RQ∴ \(rac{ ext{Area of ΔSPT}}{ ext{Area of ΔRPQ}}\) = \(rac{ST^2}{RQ^2}\)Also, given ST = \(\bigg(1-rac{35}{100}\bigg)RQ\) = (0.65) RQ∴ \(rac{ST}{RQ}\) = 0.65 ⇒ \(\bigg(rac ... ΔRPQ}}\) = 0.4225 ⇒ \(rac{ ext{Area of ΔSPT}}{{34}}\) = 0.4225⇒ Area of ΔSPT = 0.4225 x 34 = 14.365

Let a, b, c be the lengths of the sides of a right angled triangle, the hypotenuse having the length c, then a + b is -Maths 9th

Description : Let a, b, c be the lengths of the sides of a right angled triangle, the hypotenuse having the length c, then a + b is -Maths 9th

Last Answer : answer:

If A is the area of the right angled triangle and b is one of the sides containing the right angle, then what is the length of the -Maths 9th

Description : If A is the area of the right angled triangle and b is one of the sides containing the right angle, then what is the length of the -Maths 9th

Last Answer : answer:

A city has a park shaped as a right angled triangle. The length of the longest side of this park is 80 m. The Mayor of the city -Maths 9th

Description : A city has a park shaped as a right angled triangle. The length of the longest side of this park is 80 m. The Mayor of the city -Maths 9th

Last Answer : answer:

Prove that the points (2, –2), (–2, 1) and (5, 2) are the vertices of a right angled triangle. Also find the length of the hypotenuse -Maths 9th

Description : Prove that the points (2, –2), (–2, 1) and (5, 2) are the vertices of a right angled triangle. Also find the length of the hypotenuse -Maths 9th

Last Answer : Let the co-ordinates of any point on the x-axis be (x, 0). Then distance between (x, 0) and (– 4, 8) is 10 units.⇒ \(\sqrt{(x+4)^2+(0-8)^2}\) = 10 ⇒ x2 + 8x + 16 + 64 = 100 ⇒ x2 + 8x – 20 = 0 ⇒ (x + 10) (x – 2) = 0 ⇒ x = –10 or 2 ∴ The required points are (– 10, 0) and (2, 0).

What is the value of h in a right angled triangle with 31° and 15cm?

Description : What is the value of h in a right angled triangle with 31° and 15cm?

Last Answer : Following the symbols in the image:Assuming 15cm corresponds to a (the line adjacent to the angle), then you need to use the cosine formulacos(ø) = a/hcos(31º) = 15cm/hh*cos(31º) = (15cm/h) * hh*cos(31º) = 15cm * 1h*cos(31º)/cos(31º) = 15cm/cos(31º)h*1 = 15cm/cos(31º)h = 16.398

What is a right-angled triangle ?

Description : What is a right-angled triangle ?

Last Answer : : A triangle which has three right angles is called a right angled triangle. The sum of any two angles of a right-angled triangle is always greater than 900. The arms of a right-angled ... the center , and the perpendicular center of a right -angled triangle are all located inside the triangle.

What is a right-angled triangle ?

Last Answer : : A triangle whose three angles are smaller than one right angle ( 90 0) is called a right-angled triangle.

Is right angled triangle a right angle?

Description : Is right angled triangle a right angle?

Last Answer : Yes and a right angle is 90 degrees

If A(5,2), B(2,-2) and C(-2,t) are the vertices of a right angled triangle with ∠B = 900 , then the value of t is: (a) -1 (b) 1 (c) -2 (d) 2

Description : If A(5,2), B(2,-2) and C(-2,t) are the vertices of a right angled triangle with ∠B = 900 , then the value of t is: (a) -1 (b) 1 (c) -2 (d) 2

Last Answer : (b) 1

In right triangle ABC, right angled at C, if tan A = 1, then the value of 2 sin A cos A is (a) 0 (b) 1 (c) – 1 (d) 2

Description : In right triangle ABC, right angled at C, if tan A = 1, then the value of 2 sin A cos A is (a) 0 (b) 1 (c) – 1 (d) 2

Last Answer : (b) 1

Find the area of a right angled triangle with sides of 90 degree unit and the functions described by L = cos y and M = sin x. a) 0 b) 45 c) 90 d) 180

Description : Find the area of a right angled triangle with sides of 90 degree unit and the functions described by L = cos y and M = sin x. a) 0 b) 45 c) 90 d) 180

Last Answer : d) 180

The 'fix' of a plane table station with three known points, is bad if the plane table station lies (A) In the great triangle (B) Outside the great triangle (C) On the circumference of the circumscribing circle (D) None of these

Description : The 'fix' of a plane table station with three known points, is bad if the plane table station lies  (A) In the great triangle  (B) Outside the great triangle  (C) On the circumference of the circumscribing circle  (D) None of these 

Last Answer : (C) On the circumference of the circumscribing circle 

A T-beam behaves as a rectangular beam of a width equal to its flange if its neutral axis (A) Remains within the flange (B) Remains below the slab (C) Coincides the geometrical centre of the beam (D) None of these

Description : A T-beam behaves as a rectangular beam of a width equal to its flange if its neutral axis (A) Remains within the flange (B) Remains below the slab (C) Coincides the geometrical centre of the beam (D) None of these

Last Answer : Answer: Option A

The sum of the interior angles of a geometrical figure laid on the surface of the earth differs from that of the corresponding plane figure only to the extent of one second for every (A) 100 sq. km of area (B) 150 sq. km of area (C) 200 sq. km of area (D) None of these

Description : The sum of the interior angles of a geometrical figure laid on the surface of the earth differs from that of the corresponding plane figure only to the extent of one second for every (A) 100 sq. km of area (B) 150 sq. km of area (C) 200 sq. km of area (D) None of these

Last Answer : (C) 200 sq. km of area

If the length of a wall on either side of a lintel opening is at least half of its effective span L, the load W carried by the lintel is equivalent to the weight of brickwork contained in an equilateral triangle, producing a maximum bending moment (A) WL/2 (B) WL/4 (C) WL/6 (D) WL/8

Description : If the length of a wall on either side of a lintel opening is at least half of its effective span L, the load W carried by the lintel is equivalent to the weight of brickwork contained in an equilateral triangle, producing a maximum bending moment (A) WL/2 (B) WL/4 (C) WL/6 (D) WL/8

Last Answer : Answer: Option C

Pick up the incorrect statement from the following: (A) The C.G. of a circle is at its center (B) The C.G. of a triangle is at the intersection of its medians (C) The C.G. of a rectangle is at the intersection of its diagonals (D) The C.G. of a semicircle is at a distance of r/2 from the center

Description : Pick up the incorrect statement from the following:  (A) The C.G. of a circle is at its center  (B) The C.G. of a triangle is at the intersection of its medians  (C) The C.G. of a ... intersection of its diagonals  (D) The C.G. of a semicircle is at a distance of r/2 from the center 

Last Answer : (D) The C.G. of a semicircle is at a distance of r/2 from the center 

A pointed arch which forms isosceles or equilateral triangle, is generally known as (A) Three centred arch (B) Two centred arch (C) Lancet arch (D) Bull's eye arch

Description : A pointed arch which forms isosceles or equilateral triangle, is generally known as (A) Three centred arch (B) Two centred arch (C) Lancet arch (D) Bull's eye arch

Last Answer : Answer: Option C

The base of a right prism is an equilateral triangle with a side 6 cm and its height is 18 cm. Find its volume, -Maths 9th

Description : The base of a right prism is an equilateral triangle with a side 6 cm and its height is 18 cm. Find its volume, -Maths 9th

Last Answer : Volume of a right prism = Area of base height. Since the base is an equilateral triangle of side 6 cm, Area of base = 3√434 x (side)2 = (3√4 62)(34 62)cm2 = 3√434 x 36 cm2 = 93-√93 cm2 ∴ Volume = (93-√93 x18) ... ) = (324 + 2 9√3 ) cm2 = (324 + 18√3 ) cm2 = (324 + 31.176) cm2 = 355.176 cm2.

In the given figure, equal chords AB and CD of a circle with centre O cut at right angles at E. If M and N are the mid-points of AB and CD respectively, prove that OMEN is a square. -Maths 9th

Description : In the given figure, equal chords AB and CD of a circle with centre O cut at right angles at E. If M and N are the mid-points of AB and CD respectively, prove that OMEN is a square. -Maths 9th

Last Answer : Join OE. In ΔOME and ΔONE, OM =ON [equal chords are equidistant from the centre] ∠OME = ∠ONE = 90° OE =OE [common sides] ∠OME ≅ ∠ONE [by SAS congruency] ⇒ ME = NE [by CPCT] In quadrilateral OMEN, ... =ON , ME = NE and ∠OME = ∠ONE = ∠MEN = ∠MON = 90° Hence, OMEN is a square. Hence proved.

In the given figure, equal chords AB and CD of a circle with centre O cut at right angles at E. If M and N are the mid-points of AB and CD respectively, prove that OMEN is a square. -Maths 9th

Description : In the given figure, equal chords AB and CD of a circle with centre O cut at right angles at E. If M and N are the mid-points of AB and CD respectively, prove that OMEN is a square. -Maths 9th

Last Answer : Join OE. In ΔOME and ΔONE, OM =ON [equal chords are equidistant from the centre] ∠OME = ∠ONE = 90° OE =OE [common sides] ∠OME ≅ ∠ONE [by SAS congruency] ⇒ ME = NE [by CPCT] In quadrilateral OMEN, ... =ON , ME = NE and ∠OME = ∠ONE = ∠MEN = ∠MON = 90° Hence, OMEN is a square. Hence proved.

In a beam, the neutral plane (A) May be its centre (B) Passes through the C.G. of the area of cross-section (C) Does not change during deformation (D) None of these

Description : In a beam, the neutral plane  (A) May be its centre  (B) Passes through the C.G. of the area of cross-section  (C) Does not change during deformation  (D) None of these 

Last Answer : (C) Does not change during deformation

The base in a right prism is an equilateral triangle of side 8 cm and the height of the prism is 10 cm. The volume of the prism is -Maths 9th

Description : The base in a right prism is an equilateral triangle of side 8 cm and the height of the prism is 10 cm. The volume of the prism is -Maths 9th

Last Answer : ⇒ Area of equilateral triangle =43 ( s i d e)2 =43 ( 8)2 =43 64 ... =3 3 2 . 5 5 4 cm3. =3 3 2 . 5 5 4 cc

The base of a right triangular prism is an equilateral triangle. If the height is halved and each side of the base is doubled, find the ratio of the -Maths 9th

Description : The base of a right triangular prism is an equilateral triangle. If the height is halved and each side of the base is doubled, find the ratio of the -Maths 9th

Last Answer : 1 : 2 Let each side of the base of the original prism be a units and the height of the prism be h units. Then Required ratio = Vol. of original prismVol. of new prismVol. of original ... )2×h3√4×(2a)2×h234×(a)2×h34×(2a)2×h2 = 2a2h4a2h2a2h4a2h = 1 : 2.

The points (-4, 0), (4, 0), (0, 3) are the vertices of a: (а) Right triangle (b) Isosceles triangle (c) Equilateral triangle (d) Scalene triangle

Description : The points (-4, 0), (4, 0), (0, 3) are the vertices of a: (а) Right triangle (b) Isosceles triangle (c) Equilateral triangle (d) Scalene triangle

Last Answer : (b) Isosceles triangle

0 Pick up the correct statement from the following: (A) The point through which the resultant of the shear stresses passes is known as shear centre(B) In the standard rolled channels, the shear centre is on the horizontal line passing through and away from the C.G. beyond web (C) In equal angles, the shear centre is on the horizontal plane and away from the C.G., outside of the leg projection (D) All the above

Description : 0 Pick up the correct statement from the following: (A) The point through which the resultant of the shear stresses passes is known as shear centre (B) In the standard rolled channels, the shear centre ... horizontal plane and away from the C.G., outside of the leg projection (D) All the above

Last Answer : (D) All the above

For a floating body to be in equilibrium (A) Meta centre should be above e.g.(B) Centre of buoyancy and e.g. must lie on same vertical plane (C) A righting couple should be formed (D) All of the above

Description : For a floating body to be in equilibrium (A) Meta centre should be above e.g . (B) Centre of buoyancy and e.g. must lie on same vertical plane (C) A righting couple should be formed (D) All of the above

Last Answer : Answer: Option D