If the area of an equilateral tirangle is `A=12sqrt(3)cm^(2)`, then the length of its altitude h is?

If the area of an equilateral tirangle is `A=12sqrt(3)cm^(2)`, then the length of its altitude h is?
by

1 Answer

Answer :

If the area of an equilateral tirangle is `A=12sqrt(3)cm^(2)`, then the length of its altitude h is?
by
selected by

Related questions

Find the area of an equilateral triangle having altitude h cm -Maths 9th

Description : Find the area of an equilateral triangle having altitude h cm -Maths 9th

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

An equilateral triangle, if its altitude is 3.2 cm. -Maths 9th

Description : An equilateral triangle, if its altitude is 3.2 cm. -Maths 9th

Last Answer : First observe that the altitudes from any vertex to the opposite sides of an equilateral triangle are all of equal length. Hence we can define the height of an equilateral triangle as this common value of three ... ∠MAB=30∘ and ∠MAC=30∘, with B and C on XY. Then ABC is the required triangle.

An equilateral triangle, if its altitude is 3.2 cm. -Maths 9th

Description : An equilateral triangle, if its altitude is 3.2 cm. -Maths 9th

Last Answer : We know that, in an equilateral triangle all sides are equal and all angles are equal i.e., each angle is of 60°. Given, altitude of an equilateral triangle say ABC is 3.2 cm. To construct the ΔABC ... ∠DBA = 60° Similary, ∠DCA = 60° Thus, ∠A = ∠B=∠C = 60° Hence, ΔABC is an equilateral triangle.

Construct an equilateral triangle if its altitude is 6 cm. -Maths 9th

Description : Construct an equilateral triangle if its altitude is 6 cm. -Maths 9th

Last Answer : Steps of Construction (i) Draw a line XY. (ii) Construct perpendicular PD at any point D on the line XY. (iii) From point D, cut-off line segment AD = 6 cm. (iv) Construct ∠BAD = ∠CAD ... 30 °+ 30° = 60° and AD perpendicular BC therefore, △ABC is an equilateral triangle with altitude AD = 6 cm.

Construct an equilateral triangle of altitude 6 cm. -Maths 9th

Description : Construct an equilateral triangle of altitude 6 cm. -Maths 9th

Last Answer : 1.Draw any line l. 2.Take any point M on it and draw a line p perpendicular to l. 3.With M as centre, cut off MC = 6 cm 4.At C, with initial line CM construct angles of measures 30° on both sides and let these lines intersect line l in A and B. Thus, ΔABC is the required triangle.

Construct an equilateral triangle of altitude 6 cm. -Maths 9th

Description : Construct an equilateral triangle of altitude 6 cm. -Maths 9th

Last Answer : 1.Draw any line l. 2.Take any point M on it and draw a line p perpendicular to l. 3.With M as centre, cut off MC = 6 cm 4.At C, with initial line CM construct angles of measures 30° on both sides and let these lines intersect line l in A and B. Thus, ΔABC is the required triangle.

Construct an equilateral triangle whose altitude is 7 cm -Maths 9th

Description : Construct an equilateral triangle whose altitude is 7 cm -Maths 9th

Last Answer : NEED ANSWER

Construct an equilateral triangle whose altitude is 7 cm -Maths 9th

Description : Construct an equilateral triangle whose altitude is 7 cm -Maths 9th

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

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.

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 the given figure, ABC is an equilateral triangle of side length 30 cm. XY is parallel to BC, XP is parallel to AC and YQ is parallel to AB. -Maths 9th

Description : In the given figure, ABC is an equilateral triangle of side length 30 cm. XY is parallel to BC, XP is parallel to AC and YQ is parallel to AB. -Maths 9th

Last Answer : answer:

One side of an equilateral triangle is 4 cm.Find its area. -Maths 9th

Description : One side of an equilateral triangle is 4 cm.Find its area. -Maths 9th

Last Answer : Area of equilateral triangle = √3/4a2 = √3/4 x 42 = 4√3 cm2.

Each side of an equilateral triand 12 is \( 24 cm \) the mid - point of its sides are joined to form another tranpla this brress is doins centinuously infinite46. The perimeter of 7 th triangle is \( ( \) in \( cm ) \)a) \( \frac{3}{4} \)b) \( \frac{5}{8} \)c) \( \frac{9}{8} \)d) \( \frac{7}{8} \)47. The sum of perimeter of all triangle is (in \( cm \) )a) 144b) 625c) 400d) 16948. The area of all the triangle is (in sq cm)a) 576b) \( 144 \sqrt{3} \)c) \( 169 \sqrt{3} \)d) \( 192 \sqrt{3} \)49. The sum of perimeter of first 6 triangle is (in \( cm \) )a) 120b) \( \frac{567}{4} \)c) \( \frac{569}{4} \)d) 14450. The side of the 5th triangle is (in \( cm \) )a) 6b) \( 1.5 \)c) \( 0.75 \)d) 3

Description : Each side of an equilateral triand 12 is \( 24 cm \) the mid - point of its sides are joined to form another tranpla this brress is doins centinuously infinite46. The perimeter of 7 th triangle is \( ( \) in \( cm ) \)a) \( \ ... of the 5th triangle is (in \( cm \) )a) 6b) \( 1.5 \)c) \( 0.75 \)d) 3

Last Answer : Each side of an equilateral triand 12 is \( 24 cm \) the mid - point of its sides are joined to form another tranpla ... ( 1.5 \) c) \( 0.75 \) d) 3

In the figure, arcs and drawn by taking vertices A, B and C of an equilateral triangle of side 10 cm to intersect the sides BC, CA and AB at their respective mid-points D, E and F. Find the area of teh shaded region. [use π = 3.14] -Maths 10th

Description : In the figure, arcs and drawn by taking vertices A, B and C of an equilateral triangle of side 10 cm to intersect the sides BC, CA and AB at their respective mid-points D, E and F. Find the area of teh shaded region. [use π = 3.14] -Maths 10th

Last Answer : Step-by-step explanation: We have been provided that, Triangle ABC is an Equilateral triangle. Side of triangle is = 10 cm The arcs are drawn from each vertices of the triangle. We get three sectors ... portion is, Remaining area = Area of triangle ABC - Area of all the sectors 39.25cm square

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 sides of a triangle are 35 cm, 54 cm and 61 cm, respectively. The length of its longest altitude -Maths 9th

Description : The sides of a triangle are 35 cm, 54 cm and 61 cm, respectively. The length of its longest altitude -Maths 9th

Last Answer : s= 2 a+b+c​ = 2 35+54+61​ =75 Area, A= s(s−a)(s−b)(s−c)​ = 75(75−35)(75−54)(75−61)​ =420 5​ cm 2 Now, Area of the triangle is also given as A= 2 1​ ×a×h Where, h is the longest altitude. Therefore, 2 1​ ×a×h=420 5​ Hence, h=24 5​ cm

The sides of a triangle are 35 cm, 54 cm and 61 cm, respectively. The length of its longest altitude -Maths 9th

Description : The sides of a triangle are 35 cm, 54 cm and 61 cm, respectively. The length of its longest altitude -Maths 9th

Last Answer : The length of its longest altitude

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.

The base and the corresponding altitude of a parallelogram are 10 cm and 7 cm, respectively. Find its area. -Maths 9th

Description : The base and the corresponding altitude of a parallelogram are 10 cm and 7 cm, respectively. Find its area. -Maths 9th

Last Answer : Area of parallelogram = Base × Corresponding altitude = 10 × 7 = 70 cm2.

Following are the characteristics of an area in India: [IAS 2010] 1. Hot and humid climate 2. Annual rainfall 200 cm 3. Hill slopes up to an altitude of 1100 metres 4. Annual range of temperature 15 degree C to 30 degree C Which one among the following crops are you most likely to find in the area described above? (a) Mustard (b) Cotton (c) Pepper (d) Virginia tobacco

Description : Following are the characteristics of an area in India: [IAS 2010] 1. Hot and humid climate 2. Annual rainfall 200 cm 3. Hill slopes up to an altitude of 1100 metres 4. Annual range of temperature 15 ... to find in the area described above? (a) Mustard (b) Cotton (c) Pepper (d) Virginia tobacco

Last Answer : Ans: (c)

One side of an equilateral triangle is 24 cm. The mid-points of its sides are joined to form another triangle whose mid-points -Maths 9th

Description : One side of an equilateral triangle is 24 cm. The mid-points of its sides are joined to form another triangle whose mid-points -Maths 9th

Last Answer : Perimeter of the largest (outermost) equilateral triangle = 3 24 = 72 cm. Now, the perimeter of the triangle formed by joining the midpoints of a given triangle will be half the perimeter of the original triangle. ∴ Required sum = 72 + ... -rac{1}{2}}\) = \(rac{72}{rac{1}{2}}\) = 72 x 2 = 144 cm.

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.

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

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)

If each side of an equilateral triangle is 9 cm and you must find the height how would you solve in pythagorean theorem?

Description : If each side of an equilateral triangle is 9 cm and you must find the height how would you solve in pythagorean theorem?

Last Answer : Divide the base in half and draw the median from the apex. Thismedian is also the altitude and so its length is the requiredheight. Also, since it is the altitude, it forms a right angledtriangle. Using Pythagoras on this triangle, the height is9*sqrt(3)/2.

If the perimeter of an equilateral triangle is 24cm, then find the area of the triangle -Maths 9th

Description : If the perimeter of an equilateral triangle is 24cm, then find the area of the triangle -Maths 9th

Last Answer : Perimeter of equilateral triangle = 24cm ( there are three equal sides in an e quilateral triangle ) Each side = 24 / 3 = 8cm Area of an equilateral triangle = root 3 / 4 s square = root 3 / 4 × 8 square = root 3 / 4 x 64 = root 3 × 16 = 16 root 3

If the perimeter of an equilateral triangle is 24cm, then find the area of the triangle -Maths 9th

Description : If the perimeter of an equilateral triangle is 24cm, then find the area of the triangle -Maths 9th

Last Answer : Perimeter of equilateral triangle = 24cm ( there are three equal sides in an e quilateral triangle ) Each side = 24 / 3 = 8cm Area of an equilateral triangle = root 3 / 4 s square = root 3 / 4 × 8 square = root 3 / 4 x 64 = root 3 × 16 = 16 root 3

If the side of an equilateral triangle is x unit, then find the area of the triangle. -Maths 9th

Description : If the side of an equilateral triangle is x unit, then find the area of the triangle. -Maths 9th

Last Answer : Solution :- √3/4.x2 sq. unit

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.

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

A rectangle inscribed in a triangle has its base coinciding with the base b of the triangle. If the altitude of the triangle is h, and the -Maths 9th

Description : A rectangle inscribed in a triangle has its base coinciding with the base b of the triangle. If the altitude of the triangle is h, and the -Maths 9th

Last Answer : answer:

An equilateral triangle has a side of length 4.68m. What it is perimeter ?

Description : An equilateral triangle has a side of length 4.68m. What it is perimeter ?

Last Answer : Since each side is equal, just multiply by 3.

State the formula for acceleration due to gravity at depth ‘d’ and altitude ‘h’ Hence show that their ratio is equal to ((R - d)/(R - 2h))

Description : State the formula for acceleration due to gravity at depth ‘d’ and altitude ‘h’ Hence show that their ratio is equal to ((R - d)/(R - 2h))

Last Answer : State the formula for acceleration due to gravity at depth d' and altitude h' Hence show that ... small as compared to the radius of the Earth.

Length of cut for silage ranges from E. 2-4 cm F. 5-10 cm G. 15-20 cm H. 20-25 cm

Description : Length of cut for silage ranges from E. 2-4 cm F. 5-10 cm G. 15-20 cm H. 20-25 cm

Last Answer : E. 2-4 cm

If length of weld is l and leg h, then area of throat can be given by a) 0.707 hl b) 1.414hl c) hl d) None of the listed

Description : If length of weld is l and leg h, then area of throat can be given by a) 0.707 hl b) 1.414hl c) hl d) None of the listed

Last Answer : a) 0.707 hl

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.

Two concentric circles have radii of 23 cm and 9 cm. A rectangle has an area equal to the annular space between the above two circles. If the breadth of the rectangle is 22 cm then find its length. 1 : 32 cm 2 : 48 cm 3 : 76 cm 4 : 64 cm 5 : None of these

Description : Two concentric circles have radii of 23 cm and 9 cm. A rectangle has an area equal to the annular space between the above two circles. If the breadth of the rectangle is 22 cm then find its length. 1 : 32 cm 2 : 48 cm 3 : 76 cm 4 : 64 cm 5 : None of these

Last Answer : 4 : 64 cm

An equilateral triangle is cut from its three vertices to form a regular hexagon. What is the percentage of area wasted? -Maths 9th

Description : An equilateral triangle is cut from its three vertices to form a regular hexagon. What is the percentage of area wasted? -Maths 9th

Last Answer : (c) 33.33%When an equilateral triangle is cut from its three vertices to form a regular hexagon then out of the 9 equilateral triangles that form ΔABC, three triangle, ΔADE, ΔFCG,ΔIHB are cut off and 6 remain in the ... to get the hexagon.∴ Area wasted = \(\bigg(rac{1}{3} imes100\bigg)\)% = 33.33%

The area of parallelogram PQRS is 88 cm sq. A perpendicular from S is drawn to intersect PQ at M. If SM = 8 cm, then find the length of PQ. -Maths 9th

Description : The area of parallelogram PQRS is 88 cm sq. A perpendicular from S is drawn to intersect PQ at M. If SM = 8 cm, then find the length of PQ. -Maths 9th

Last Answer : Given area of parallelogram = 88 cm² And SM = 8cm Area of a parallelogram = height × base (Height is the measurement of a perpendicular drawn from one side to other) Here, Area of PQRS = SM × PQ 88cm² = 8cm × PQ 11cm = PQ

The external length, breadth and height of a closed box are 10cm, 9cm and 7cm respectively. The total inner surface area of the box is 262 sq.cm. If the walls of the box are of uniform thickness “t” cm, then “t” equals a) 1cm b) 23/3 cm c) 28/9 cm d) 42.5 cm

Description : The external length, breadth and height of a closed box are 10cm, 9cm and 7cm respectively. The total inner surface area of the box is 262 sq.cm. If the walls of the box are of uniform thickness “t” cm, then “t” equals a) 1cm b) 23/3 cm c) 28/9 cm d) 42.5 cm

Last Answer : a) 1cm

A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side a. Find the area of the signal board, using Heron’s formula. -Maths 9th

Description : A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side a. Find the area of the signal board, using Heron’s formula. -Maths 9th

Last Answer : Let each side of the equilateral triangle be a. Semi-perimeter of the triangle,

The perimeter of an equilateral triangle is 60 m. The area is -Maths 9th

Description : The perimeter of an equilateral triangle is 60 m. The area is -Maths 9th

Last Answer : Let each side of an equilateral be x. Then, perimeter of an equilateral triangle = 60 m x + x + x = 60 ⇒ 3x = 60 ⇒ x = 60/3 = 20 m Area of an equilateral triangle = √3/4 (Side)2 = (√3/4) x 20 x 20 = 100 √3 m2 Thus, the area of triangle is 100√3 m2.

The perimeter of an equilateral triangle is 60 m. The area is -Maths 9th

Description : The perimeter of an equilateral triangle is 60 m. The area is -Maths 9th

Last Answer : (d) Let each side of an equilateral be x. Then, perimeter of an equilateral triangle = 60 m x + x + x = 60 ⇒ 3x = 60 ⇒ x = 60/3 = 20 m Area of an equilateral triangle = √3/4 (Side)2 = (√3/4) x 20 x 20 = 100 √3 m2 Thus, the area of triangle is 100√3 m2.

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 = \(rac{ ext{Area}}{ ext{semi-perimeter}}\) = \( ... {( ext{diagonal})^2}{2}\) = \(rac{\big(rac{a}{\sqrt3}\big)^2}{2}\) = \(rac{a^2}{6}.\)

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) = \(rac{\sqrt3a^2}{32}\)

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.

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.

What is the formula for finding the area of an equilateral triangle ?

Last Answer : Area of equilateral triangle = ( root3 * a ^ 2) / 4

What is the perimeter and area of an equilateral triangle with a side of 8?

Description : What is the perimeter and area of an equilateral triangle with a side of 8?

Last Answer : Perimeter of equilateral triangle: 24 units Area of equilateral triangle: 27.713 square units rounded to three decimal places