In Fig. 4.6, if ABC and ABD are equilateral triangles then find the coordinates of C and D. -Maths 9th

In Fig. 4.6, if ABC and ABD are equilateral triangles then find the coordinates of C and D. -Maths 9th
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In the given figure, ABC is a triangle in which CDEFG is a pentagon. Triangles ADE and BFG are equilateral -Maths 9th

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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.

In the fig, D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. -Maths 9th

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Last Answer : Since line segment joining the mid-points of two sides of a triangle is half of the third side. Therefore, D and E are mid-points of BC and AC respectively. ⇒ DE = 1 / 2 AB --- (i) E and F are the mid - ... CA ⇒ DE = EF = FD [using (i) , (ii) , (iii) ] Hence, DEF is an equilateral triangle .

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Last Answer : Solution :- △ ABC ≅ △PQR

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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).

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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 : Given, AC is the common hypotenuse. ∠B = ∠D = 90°. To prove, ∠CAD = ∠CBD Proof: Since, ∠ABC and ∠ADC are 90°. These angles are in the semi circle. Thus, both the triangles are lying in the semi ... D are concyclic. Thus, CD is the chord. ⇒ ∠CAD = ∠CBD (Angles in the same segment of the circle)

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Last Answer : Given, AC is the common hypotenuse. ∠B = ∠D = 90°. To prove, ∠CAD = ∠CBD Proof: Since, ∠ABC and ∠ADC are 90°. These angles are in the semi circle. Thus, both the triangles are lying in the semi ... D are concyclic. Thus, CD is the chord. ⇒ ∠CAD = ∠CBD (Angles in the same segment of the circle)

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Description : Let ABCD be a cyclic quadrilateral. Show that the incentres of the triangles ABC, BCD, CDA and DAB form a rectangle. -Maths 9th

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Description : ABC is an isosceles triangle in which altitude BE and CF are drawn to equal sides AC and AB respectively (Fig. 7.15). Show that these altitudes are equal. -Maths 9th

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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.

A relation R={A,B,C,D,E,F,G} is given with following set of functional dependencies: F={AD→E, BE→F, B→C, AF→G} Which of the following is a candidate key? (A) A (B) AB (C) ABC (D) ABD

Description : A relation R={A,B,C,D,E,F,G} is given with following set of functional dependencies: F={AD→E, BE→F, B→C, AF→G} Which of the following is a candidate key? (A) A (B) AB (C) ABC (D) ABD

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Last Answer : Solution :- As, any point on x-axis has coordinates (,)x0 where x is the distance from origin, so required coordinates are (6, 0).

In Fig.5.6, if AC = BD, then prove that AB = CD. -Maths 9th

Description : In Fig.5.6, if AC = BD, then prove that AB = CD. -Maths 9th

Last Answer : if equals are subtracted to equals ,the remainders are equal subtract bc on both sides ab-bc=bd-bc ab=cd hence proved

In Fig. 6.7, if l||m, then find the value of x. -Maths 9th

Description : In Fig. 6.7, if l||m, then find the value of x. -Maths 9th

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In Fig. 6.10, if m|n, then find the value of x. -Maths 9th

Description : In Fig. 6.10, if m|n, then find the value of x. -Maths 9th

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In Fig. 6.14, if x+y = w+z,then then prove that AOB is a line. -Maths 9th

Description : In Fig. 6.14, if x+y = w+z,then then prove that AOB is a line. -Maths 9th

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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

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 a, b, c are the lengths of the sides of a non-equilateral triangle, then -Maths 9th

Description : If a, b, c are the lengths of the sides of a non-equilateral triangle, then -Maths 9th

Last Answer : https://discuss.aiforkids.in/36748/if-are-the-lengths-of-the-sides-non-equilateral-triangle-then

In an equilateral triangle if a, b and c denote the lengths of the perpendicular from A, B and C respectively on the opposite sides, then -Maths 9th

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Last Answer : b=2c=3​⇒⇒b=3​c=23​​cosA=2bcb2+c2−a2​⇒23​​=33+43​−a2​⇒233​​=415​−a2 ⇒a2=415​−43​​⇒a=1.673278 We know sinaa​=2R1​⇒R=2asina​=221​​=41​

In the given figure, ΔABC is an equilateral triangle and ABDC is a cyclic quadrilateral, then find the measure of ∠BDC. -Maths 9th

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Last Answer : △ABC is an equilateral triangle. ∠BAE=60 ABEC is a cyclic quadrilateral. We know, ∠BAC+∠BEC=180 =>∠BEC=180−60 =120 We know that the angles subtended by an arc on the circumference on the same side are equal. Therefore, ∠ABC=∠BDC =>∠BDC=60

What shape has 8 congruent equilateral triangles as faces and 6 vertices?

Description : What shape has 8 congruent equilateral triangles as faces and 6 vertices?

Last Answer : The Platonic solid known as an octahedron would fit the givendescription

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 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.

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

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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.

‘If two sides and an angle of one triangle are equal to two sides and an angle of another triangle , then the two triangles must be congruent’. -Maths 9th

Description : ‘If two sides and an angle of one triangle are equal to two sides and an angle of another triangle , then the two triangles must be congruent’. -Maths 9th

Last Answer : No, because in the congruent rule, the two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle i.e., SAS rule.

‘If two sides and an angle of one triangle are equal to two sides and an angle of another triangle , then the two triangles must be congruent’. -Maths 9th

Description : ‘If two sides and an angle of one triangle are equal to two sides and an angle of another triangle , then the two triangles must be congruent’. -Maths 9th

Last Answer : No, because in the congruent rule, the two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle i.e., SAS rule.

If the corresponding angles of two triangles are equal, then they are always. State true or false and justify your answer. -Maths 9th

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Last Answer : Solution :- False, because two equilateral triangles with sides 3 cm and 6 cm respectively have all angles equal, but the triangles are not congruent.

Prove that if in two triangles,two angles and the included side of one triangle are equal to two angles and the included side of the other triangle,then two triangles are congruent. -Maths 9th

Description : Prove that if in two triangles,two angles and the included side of one triangle are equal to two angles and the included side of the other triangle,then two triangles are congruent. -Maths 9th

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Let R be a relation defined on the set A of all triangles such that R = {(T1, T2) : T1 is similar to T2}. Then R is -Maths 9th

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Last Answer : (d) An equivalence relation.Every triangle is similar to itself, so (T1, T1) ∈ R ⇒ R is reflexive. (T1, T2) ∈ R ⇒ T1 ~ T2 ⇒T2 ~ T1, ⇒ (T2, T1) ∈ R ⇒ R is symmetrictransitive. ∴ R is an equivalence relation.

If vertices of a triangles are (1, k), (4, -3) and (-9, 7) and its area is 15 sq. units then find then the value of k. -Maths 9th

Description : If vertices of a triangles are (1, k), (4, -3) and (-9, 7) and its area is 15 sq. units then find then the value of k. -Maths 9th

Last Answer : hope it helps if the vertices of a triangle are (1,k),(4,−3)(−9,7) area = 15 sq.units. find the value of k. Area of △ 21 [x1 (y2 −y3 )+x2 (y3 −y1 )+x3 (y1 −y2 )]=15 21 [1(−3−7)+ ... k+3)]=15 21 [(−10+28−4k−9k−27)]=15 −10+28−4k−9k−27=30 −10+28−13k−27=30 −13k=30+10+27−28 −13k=39 k=1339 k=−3 thank u

A (5,-1) , B (-3,-2) and C (-1,8) are the vertices of ∆ ABC . The median from A meets BC at D ,then the coordinates of the point D are: (a) (-4,6) (b) (2,7/2) (c) (1,-3/2) (d) (-2,3)

Description : A (5,-1) , B (-3,-2) and C (-1,8) are the vertices of ∆ ABC . The median from A meets BC at D ,then the coordinates of the point D are: (a) (-4,6) (b) (2,7/2) (c) (1,-3/2) (d) (-2,3)

Last Answer : (d) (-2,3)

PQRS is a square. A is a point on PS ,B is a point on PQ,C is a point on QR. ABC is a triangle inside square PQRS. Angle abc = 90° . If AP=BQ=CR then prove that angle BAC =45° -Maths 9th

Description : PQRS is a square. A is a point on PS ,B is a point on PQ,C is a point on QR. ABC is a triangle inside square PQRS. Angle abc = 90° . If AP=BQ=CR then prove that angle BAC =45° -Maths 9th

Last Answer : This is the sketch of the question but its hard to answer.

PQRS is a square. A is a point on PS ,B is a point on PQ,C is a point on QR. ABC is a triangle inside square PQRS. Angle abc = 90° . If AP=BQ=CR then prove that angle BAC =45° -Maths 9th

Description : PQRS is a square. A is a point on PS ,B is a point on PQ,C is a point on QR. ABC is a triangle inside square PQRS. Angle abc = 90° . If AP=BQ=CR then prove that angle BAC =45° -Maths 9th

Last Answer : This is the sketch of the question but its hard to answer.