Is it possible to draw a triangle that is both obtuse and equilateral?

Is it possible to draw a triangle that is both obtuse and equilateral?
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1 Answer

Answer :

No because an equilateral triangle has 3 equal acute angles of 60 degrees
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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.

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Last Answer : Let each side of ㎝ equilateral triangle ABC be ′a′㎝ Now, ar△OAB=21 AB OP=21 a 14=7a㎠→1 ar△OBC= BC OQ =21 a 10=5a㎠→2 ar△OAC=21 AC OR=21 a 6=3a㎠→3 ∴ar△ABC=1+2+3=7a+5a+3a=15a㎠ Also area of equilateral ... ABC=43 a2 Now, 43 a2=15a⇒a=3 15 4 3 3 =3603 =203 ㎝ Now, ar△ABC=43 (203 )2=3003 ㎠

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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.

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

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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.

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.

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

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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.

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

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.

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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).

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

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

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Find the ratio of the diameter of the circles inscribed in and circumscribing an equilateral triangle to its height? -Maths 9th

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Last Answer : (b) 2 : 4 : 3.For an equilateral triangle of side a units,In-radius = \(rac{a}{2\sqrt3}\) units⇒ Diameter of inscribed circle = \(rac{a}{\sqrt3}\) unitsCircumradius = \(rac{a}{\sqrt3}\)⇒ Diameter of circumscrible circle = \( ... \(rac{2a}{\sqrt3}\): \(rac{\sqrt3}{2}a\) = 2a : 4a : 3a = 2 : 4 : 3.