How many more sides does a nonagon have then a triangle?

1 Answer

Answer :

6 more sides

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Description : The perimeter of a right triangle is 30 cm. If its hypotenuse is 13 cm, then what are two sides? -Maths 9th

Last Answer : The other two sides of the triangle are 12 cm and 5 cm Explanation: Let the other two sides of triangle be x and y It's hypotenuse is 13 cm Perimeter of triangle = Sum of all sides ... When y = 12 x=17-y = 17-12 =5 So, the other two sides of the triangle are 12 cm and 5 cm

Description : If the lengths of the sides of a triangle are in the ratio 6:11:15 and it's perimeter is 96cm , then the height corresponding to the longest side is -Maths 9th

Last Answer : LET EACH SIDE BE X 6X+11X+15X=96 32X=96 X=3 SIDES=6 3=18 11 3=33 15 3=45 AREA OF TRIANGLE BY HERONS FORMULA=S=96/2=48 WHOLE UNDERROOT 48 48-18 48-33 48-45 UNDERROOT=12 4 30 15 3 4 3 15ROOT2 180 ... bh/2 180root2=18 h/2 360root2=18h h=20 root2 But root 2=1.4(approx) h=20 1.4(approx) h=28cm(approx).

Description : If two sides of a triangle are of lengths 5 cm and 1.5 cm, then the length of third side of the triangle cannot be -Maths 9th

Last Answer : (d) Given, the length of two sides of a triangle are 5 cm and 1.5 cm, respectively. Let sides AB = 5 cm and CA = 1.5 cm We know that, a closed figure formed by three intersecting lines ( ... options (a), (b) and (c) satisfy the above inequality but option (d) does not satisfy the above inequality.

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.

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

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Description : If the lengths of the sides of a triangle are in the ratio 6:11:15 and it's perimeter is 96cm , then the height corresponding to the longest side is -Maths 9th

Last Answer : LET EACH SIDE BE X 6X+11X+15X=96 32X=96 X=3 SIDES=6 3=18 11 3=33 15 3=45 AREA OF TRIANGLE BY HERONS FORMULA=S=96/2=48 WHOLE UNDERROOT 48 48-18 48-33 48-45 UNDERROOT=12 4 30 15 3 4 3 15ROOT2 180 ... bh/2 180root2=18 h/2 360root2=18h h=20 root2 But root 2=1.4(approx) h=20 1.4(approx) h=28cm(approx).

Description : If two sides of a triangle are of lengths 5 cm and 1.5 cm, then the length of third side of the triangle cannot be -Maths 9th

Last Answer : (d) Given, the length of two sides of a triangle are 5 cm and 1.5 cm, respectively. Let sides AB = 5 cm and CA = 1.5 cm We know that, a closed figure formed by three intersecting lines ( ... options (a), (b) and (c) satisfy the above inequality but option (d) does not satisfy the above inequality.

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.

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

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Description : The sides of a triangle are 56 cm, 60 cm and 52 cm long. Then, the area of the triangle is -Maths 9th

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Description : The sides of a triangle are 56 cm, 60 cm and 52 cm long. Then, the area of the triangle is -Maths 9th

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Description : If the sides of a triangle are 3 cm, 4 cm and 5 cm, then what is the radius of the circum-circle? -Maths 9th

Last Answer : Semi-perimeter of triangle (s) = \(rac{3+4+5}{2}\)cm = 6 cm∴ Area of triangle A = \(\sqrt{s(s-a)(s-b)(s-c)}\) = \(\sqrt{6 imes3 imes2 imes1}\) cm2 = 6 cm2∴ Radius of circum-circle = \(rac{abc}{4( ext{Area of}\,\Delta)}\) = \(rac{3+4+5}{4 imes60}\) cm = 2.5 cm

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.

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

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Description : If a, b, c are the sides of a triangle, then a/(b+c-a) + b/(c+a-b)+ c/(a+b-c) is -Maths 9th

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

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Description : If a, b, c are the sides of a non-equilateral triangle, then the expression (b + c – a) (c + a – b) (a + b – c) – abc is -Maths 9th

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Description : If the angles of a triangle are in the ratio 1 : 2 : 3, then find the ratio of the corresponding opposite sides. -Maths 9th

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Description : 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

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​

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

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Description : If a, b, c are the sides of a triangle and a^2 + b^2 + c^2 = bc + ca + ab, then the triangle is: -Maths 9th

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Description : Is this statement true or falseIf two sides and one angle of one triangle are congruent to two sides and one angle of another triangle, then the triangles are congruent by the Side-Angle-Side Postulate?

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Description : In what way does knowing the measure of the three sides of a triangle tell you more than just knowing the measures of the three angles?

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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.

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Last Answer : NCERT Solutions for Class 9 Maths Chapter 12 Heron's Formula NCERT Solutions for Class 9 Maths Chapter 12 Heron's Formula Ex 12.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths ... = 48 m Sides ∆ABC are a = AB = 30m, b = AD = 30m, c = BD = 48m S

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Last Answer : Let the sides of the triangle be a = 12x cm, b = 17x cm, c = 25x cm Perimeter of the triangle = 540 cm Now, 12x + 17x + 25x = 540 ⇒ 54x = 54 ⇒ x = 10 ∴ a = (12 x10)cm = 120cm, b = (17 x 10) cm = 170 cm and c = (25 x 10)cm = 250 cm Now, semi-perimeter, s = 5402cm = 270 cm

Description : The sides of a triangle are in the ratio of 3 : 4 : 5 and its perimeter is 510 m. What is the measure of its greatest side? -Maths 9th

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Description : Find the area of a triangle whose sides are 4.5 cm and 10 cm and perimeter 10.5 cm. -Maths 9th

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Description : State and prove-line joining the midpoint of any two sides of a triangle is parallel to throw side and is equal to 1/2 of it -Maths 9th

Last Answer : Here, In △△ ABC, D and E are the midpoints of sides AB and AC respectively. D and E are joined. Given: AD = DB and AE = EC. To Prove: DE ∥∥ BC and DE = 1212 BC. Construction: Extend line segment DE to ... we have DF ∥∥ BC and DF = BC DE ∥∥ BC and DE = 1212BC (DE = EF by construction) Hence proved.

Description : D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. -Maths 9th

Last Answer : Since the segment joining the mid points of any two sides of a triangle is half the third side and parallel to it. DE = 1 / 2 AC ⇒ DE = AF = CF EF = 1 / 2 AB ⇒ EF = AD = BD DF = 1 ... △DEF ≅ △AFD Thus, △DEF ≅ △CFE ≅ △BDE ≅ △AFD Hence, △ABC is divided into four congruent triangles.

Description : 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

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 .

Description : The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it. -Maths 9th

Last Answer : Given = A △ABC in which D and E are the mid-points of side AB and AC respectively. DE is joined . To Prove : DE || BC and DE = 1 / 2 BC. Const. : Produce the line segment DE to F , such that DE = ... of ||gm are equal and parallel] Also, DE = EF [by construction] Hence, DE || BC and DE = 1 / 2 BC

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Last Answer : No, it is not possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm because here we see that sum of the lengths of two sides is equal to third side i.e., 4+3 ... , the sum of any two sides of a triangle is greater than its third side, so given statement is not correct.

Description : The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to -Maths 9th

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Description : If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th

Last Answer : Given ΔABC is an isosceles triangle such that AB = AC and also DE || SC. To prove Quadrilateral BCDE is a cyclic quadrilateral. Construction Draw a circle passes through the points B, C, D and E.

Description : State and prove-line joining the midpoint of any two sides of a triangle is parallel to throw side and is equal to 1/2 of it -Maths 9th

Last Answer : Here, In △△ ABC, D and E are the midpoints of sides AB and AC respectively. D and E are joined. Given: AD = DB and AE = EC. To Prove: DE ∥∥ BC and DE = 1212 BC. Construction: Extend line segment DE to ... we have DF ∥∥ BC and DF = BC DE ∥∥ BC and DE = 1212BC (DE = EF by construction) Hence proved.