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

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

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

Last Answer : The area of the triangle is

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.

A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, -Maths 9th

Description : A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, -Maths 9th

Last Answer : For the given triangle, we have a = 28 cm, b = 30 cm, c = 26 cm Area of the given parallelogram = Area of the given triangle ∴ Area of the parallelogram = 336 cm2 ⇒ base x height = 336 ⇒ ... be the height of the parallelogram. ⇒ h = 33628 = 12 Thus, the required height of the parallelogram = 12 cm

Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area. -Maths 9th

Description : Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area. -Maths 9th

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

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

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

Last Answer : Step-by-step explanation: ◾As we have given the two sides of triangle, let the three sides of triangle are (a) , (b), (c) . ◾And perimeter of given triangle is 10.5 cm ◾were, let us assume the sides are, ... . ◾So, the area of a triangle whose sides are 4.5 cm and 10 cm and perimeter 10.5 [Area ]=

Find the area of a triangle whose sides are 12 cm, 6 cm and 15 cm. -Maths 9th

Description : Find the area of a triangle whose sides are 12 cm, 6 cm and 15 cm. -Maths 9th

Last Answer : Using the formulas A=s(s﹣a)(s﹣b)(s﹣c) s=a+b+c 2Solving forA A=1 4﹣a4+2(ab)2+2(ac)2﹣b4+2(bc)2﹣c4=1 4·﹣124+2·(12·6)2+2·(12·15)2﹣64+2·(6·15)2﹣154≈34.19704cm²

The area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm, is -Maths 9th

Description : The area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm, is -Maths 9th

Last Answer : s= 2 4+4+2​ =5 Area of the triangle Δ= s(s−a)(s−b)(s−c)​ = 5(5−4)(5−4)(5−2)​ = 15​ cm 2

The area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm, is -Maths 9th

Description : The area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm, is -Maths 9th

Last Answer : s= 2 4+4+2​ =5 Area of the triangle Δ= s(s−a)(s−b)(s−c)​ = 5(5−4)(5−4)(5−2)​ = 15​ cm 2

Find the area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm. -Maths 9th

Description : Find the area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm. -Maths 9th

Last Answer : s= 2 4+4+2​ =5 Area of the triangle Δ= s(s−a)(s−b)(s−c)​ = 5(5−4)(5−4)(5−2)​ = 15​ cm 2

The sides of a triangle are 8 cm, 15 cm and 17 cm. Find its area. -Maths 9th

Description : The sides of a triangle are 8 cm, 15 cm and 17 cm. Find its area. -Maths 9th

Last Answer : Let a = 8cm, b = 15cm, c = 17cm s = (a + b + c)/2 = (8 + 15/ + 17)/2 = 40/2 = 20cm ∴ Area = root under √s(s - a)(s - b)(s - c) = root under √20(20 - 8)(20 - 15)(20 - 17) = root under √20 x 12 x 5 x 3 = 60 cm2

Find the area of an isosceles triangle, whose equal sides are of length 15 cm each and third side is 12 cm. -Maths 9th

Description : Find the area of an isosceles triangle, whose equal sides are of length 15 cm each and third side is 12 cm. -Maths 9th

Last Answer : We have, Three sides13cm,13cm and 20cm. By using Heron's formula We need to get the semi-perimeter s= 2 a+b+c​ = 2 13+13+20​ = 2 46​ =23 Now, put the heron's formula, s= s(s−a)(s−b)(s−c)​ = 23(23−13)(23−13)(23−20)​ = 23×10×10×3​ =10 23×3​ =83.07cm 2

In an isosceles triangle, the measure of each of equal sides is 10 cm and the angle between them is 45º. The area of the triangle is: -Maths 9th

Description : In an isosceles triangle, the measure of each of equal sides is 10 cm and the angle between them is 45º. The area of the triangle is: -Maths 9th

Last Answer : (c) 25√2 cm2.ΔABC is an isosceles triangle with AB = AC = 10 cm. ∠A = 45° ∴ Area of ΔABC= \(rac{1}{2}\) x 10 x 10 x sin 45°[Using Δ = \(rac{1}{2}\) bc sin A]= \(rac{50}{\sqrt2}\) = \(rac{50}{\sqrt2}\) x \(rac{\sqrt2}{\sqrt2}\) = 25√2 cm2.

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

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

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

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.

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

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.

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

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

A spherical ball is divided into two equal halves. If the curved surface area of each half is 56.57 cm?, find the volume of the spherical ball.11531/cylinder-radius-halved-and-height-doubled-then-find-volume-with-respect-original-volume -Maths 9th

Description : A spherical ball is divided into two equal halves. If the curved surface area of each half is 56.57 cm?, find the volume of the spherical ball.11531/cylinder-radius-halved-and-height-doubled-then-find-volume-with-respect-original-volume -Maths 9th

Last Answer : since curved surface of half of the spherical ball = 56.57 cm2 ∴ 2πr2 = 56.57 ⇒ r2 = 56.57 / 2 × 3.14 = 9 ⇒ r = 3 cm Now, volume of spherical ball = 4 / 3 πr3 = 4 / 3 × 3.14 × 3 × 3 × 3 = 113.04 cm3

A spherical ball is divided into two equal halves. If the curved surface area of each half is 56.57 cm?, find the volume of the spherical ball.11531/cylinder-radius-halved-and-height-doubled-then-find-volume-with-respect-original-volume -Maths 9th

Description : A spherical ball is divided into two equal halves. If the curved surface area of each half is 56.57 cm?, find the volume of the spherical ball.11531/cylinder-radius-halved-and-height-doubled-then-find-volume-with-respect-original-volume -Maths 9th

Last Answer : since curved surface of half of the spherical ball = 56.57 cm2 ∴ 2πr2 = 56.57 ⇒ r2 = 56.57 / 2 × 3.14 = 9 ⇒ r = 3 cm Now, volume of spherical ball = 4 / 3 πr3 = 4 / 3 × 3.14 × 3 × 3 × 3 = 113.04 cm3

A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see figure). -Maths 9th

Description : A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see figure). -Maths 9th

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

A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in figure. -Maths 9th

Description : A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in figure. -Maths 9th

Last Answer : Each shade of paper is divided into 3 triangles i.e., I, II, III 8 cm For triangle I: ABCD is a square [Given] ∵ Diagonals of a square are equal and bisect each other. ∴ AC = BD = 32 cm Height of AABD ... are: Area of shade I = 256 cm2 Area of shade II = 256 cm2 and area of shade III = 17.92 cm2

Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm ? -Maths 9th

Description : Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm ? -Maths 9th

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.

Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm ? -Maths 9th

Description : Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm ? -Maths 9th

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.

Construct a triangle whose sides are 3.6 cm , 3.0 cm and 4. 8 cm. Bisect the smallest angle and .measure each part. -Maths 9th

Description : Construct a triangle whose sides are 3.6 cm , 3.0 cm and 4. 8 cm. Bisect the smallest angle and .measure each part. -Maths 9th

Last Answer : To construct a triangle ABC in which AB = 3.6 cm, AC = 3.0 cm and BC = 4. 8 cm, use the following steps. Draw a line segment BC of length 4.8 cm. From B, point A is at a distance of 3.6 cm. ... at P. Joining BP, we obtain angle bisector of ∠B. Flere, ∠ABC=39° Thus, ∠ABD = ∠DBC = ½ x 139° = 19.5°

A right triangle when one side is 3.5 cm and sum of other sides and the hypotenuse is 5.5 cm. -Maths 9th

Description : A right triangle when one side is 3.5 cm and sum of other sides and the hypotenuse is 5.5 cm. -Maths 9th

Last Answer : Let given right triangle be ABC. Then, given BC = 3.5 cm, ∠B = 90° and sum of other side and hypotenuse i.e., AB + AC = 5.5 cm To construct ΔABC use the following steps 1.Draw the base BC = 3.5 cm 2.Make ... AB = BD - AD = BD - AC [from Eq. (i)] => BD = AB + AC Thus, our construction is justified.

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

A right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm. -Maths 9th

Description : A right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm. -Maths 9th

Last Answer : Since, the given right angled triangle is revolved about the side 8 cm, it will form a Cone of radius 6cm and height 8cm. Volume of a cone = 1/3∏r2h = 1/3 3.14 6 6 8 = 301.44 cm3 Curved Surface area of a cone ... value of l in (i), we get Curved Surface area of a cone = 3.14 6 10 = 188.4 cm2

Construct a triangle whose sides are 3.6 cm , 3.0 cm and 4. 8 cm. Bisect the smallest angle and .measure each part. -Maths 9th

Description : Construct a triangle whose sides are 3.6 cm , 3.0 cm and 4. 8 cm. Bisect the smallest angle and .measure each part. -Maths 9th

Last Answer : To construct a triangle ABC in which AB = 3.6 cm, AC = 3.0 cm and BC = 4. 8 cm, use the following steps. 1.Draw a line segment BC of length 4.8 cm. 2.From B, point A is at a distance of 3.6 ... 3.Joining BP, we obtain angle bisector of ∠B. 4.Flere, ∠ABC=39° Thus, ∠ABD = ∠DBC = 1/2 x 139° = 19.5°

A right triangle when one side is 3.5 cm and sum of other sides and the hypotenuse is 5.5 cm. -Maths 9th

Description : A right triangle when one side is 3.5 cm and sum of other sides and the hypotenuse is 5.5 cm. -Maths 9th

Last Answer : Let given right triangle be ABC. Then, given BC = 3.5 cm, ∠B = 90° and sum of other side and hypotenuse i.e., AB + AC = 5.5 cm To construct ΔABC use the following steps 1.Draw the base BC = 3.5 cm 2.Make ... AB = BD - AD = BD - AC [from Eq. (i)] => BD = AB + AC Thus, our construction is justified.

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

A right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm. -Maths 9th

Description : A right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm. -Maths 9th

Last Answer : Since, the given right angled triangle is revolved about the side 8 cm, it will form a Cone of radius 6cm and height 8cm. Volume of a cone = 1/3∏r2h = 1/3 3.14 6 6 8 = 301.44 cm3 Curved Surface area of a cone ... value of l in (i), we get Curved Surface area of a cone = 3.14 6 10 = 188.4 cm2

Is it possible to construct a triangle with lengths of its sides as 7 cm, 8 cm and 5 cm? Give reason for your answer. -Maths 9th

Description : Is it possible to construct a triangle with lengths of its sides as 7 cm, 8 cm and 5 cm? Give reason for your answer. -Maths 9th

Last Answer : Solution :- Yes, because in each case sum of two sides is greater than the third side.

The lengths of the sides of a triangle are 7 cm, 13 cm and 12 cm. -Maths 9th

Description : The lengths of the sides of a triangle are 7 cm, 13 cm and 12 cm. -Maths 9th

Last Answer : Let, a = 7 cm, b = 13 cm, c = 12 cm ∴ s = (a + b + c)/2 = (7 +13 +12)/2 = 32/2 = 16 cm Area of △ ABC = under root( √s(s -a) (s - b)(s -c)) = under root( √16(16 - 7)(16 - 13)(16 - 12) = ... 24 √3 cm2 Also, Area of △ ABC = 1/2AC.BD 24 √3 = 1/2 x 12 x BD ⇒ BD = (24 √3 x 2)/12 = 4 √3 cm

A right triangle ABC with sides 5 cm, -Maths 9th

Description : A right triangle ABC with sides 5 cm, -Maths 9th

Last Answer : Let ABC be a right triangle with AB = 12 cm, BC = 5 cm and AC = 13 cm. When △ABC is revolved about AB, it forms a right circular cone of radius BC = 5 cm and height AB = 12 cm. Thus, volume of cone formed = 1/3 πr2h = 1/3 x π x 52 x 12 = 100π cm3

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 lengths of a sides of a triangle are in the ratio 4 : 5 : 6 and the in-radius of the triangle is 3 cm, -Maths 9th

Description : If the lengths of a sides of a triangle are in the ratio 4 : 5 : 6 and the in-radius of the triangle is 3 cm, -Maths 9th

Last Answer : (b) 7.5 cm.Area of a triangle = \(rac{1}{2}\)x base x height= In-radius x semi-perimeter of the Δ \(\big[ ext{Using r =}rac{\Delta}{s}\big]\)Let the sides of triangle be 4x, 5x and 6x respectively. Given: In-radius = 3 ... x \(rac{15x}{2}\) = \(rac{6x}{2}\) x h ⇒ h = \(rac{45}{6}\) = 7.5 cm.

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:

The sides of a triangle are in the ratio of 1/2:1/3:1/4. If its perimeter is 52 cm, the length of the smallest side is : (A) 9 cm (B) 10 cm (C) 11 cm (D) 12 cm

Description : The sides of a triangle are in the ratio of 1/2:1/3:1/4. If its perimeter is 52 cm, the length of the smallest side is : (A) 9 cm (B) 10 cm (C) 11 cm (D) 12 cm

Last Answer : (D) 12 cm Answer: D Explanation: Sides of a triangle are in the ratio of a:b:c = 1/2:1/3:1/4 = 12 /2 : 12 /3 : 12 /4 = 6:4:3 Let the lengths of three sides of the triangle be 6x, 4x, 3x Perimeter of the ... ⇒ 52 cm = 6x + 4x + 3x x = 52/13 = 4 cm length of the smallest side = 3x = 3 x 4 = 12 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.

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.

Let ABC be a triangle of area 16 cm^2 . XY is drawn parallel to BC dividing AB in the ratio 3 : 5. If BY is joined, then the area of triangle BXY is -Maths 9th

Description : Let ABC be a triangle of area 16 cm^2 . XY is drawn parallel to BC dividing AB in the ratio 3 : 5. If BY is joined, then the area of triangle BXY is -Maths 9th

Last Answer : answer:

The area of triangle ABC is 15 cm sq. If ΔABC and a parallelogram ABPD are on the same base and between the same parallel lines then what is the area of parallelogram ABPD. -Maths 9th

Description : The area of triangle ABC is 15 cm sq. If ΔABC and a parallelogram ABPD are on the same base and between the same parallel lines then what is the area of parallelogram ABPD. -Maths 9th

Last Answer : area of parallelogram=2× area of triangle ABC =2×15=30sq cm theorem on area

The median of the data 26,56,32,33,60,17,34,29,45 is 33. If 26 is replaced by 62, then find the new median. -Maths 9th

Description : The median of the data 26,56,32,33,60,17,34,29,45 is 33. If 26 is replaced by 62, then find the new median. -Maths 9th

Last Answer : Here, the given data in ascending order is 17, 29, 32, 33, 34, 45, 56, 60, 62 Now median is (9 + 1 / 2)th term i.e. , 5th term Hence, new median is 34.

The median of the data 26,56,32,33,60,17,34,29,45 is 33. If 26 is replaced by 62, then find the new median. -Maths 9th

Description : The median of the data 26,56,32,33,60,17,34,29,45 is 33. If 26 is replaced by 62, then find the new median. -Maths 9th

Last Answer : Here, the given data in ascending order is 17, 29, 32, 33, 34, 45, 56, 60, 62 Now median is (9 + 1 / 2)th term i.e. , 5th term Hence, new median is 34.

Find the area of a trapezium whose parallel sides are 25 cm and 13 cm long and the distance between them is 8 cm. -Maths 9th

Description : Find the area of a trapezium whose parallel sides are 25 cm and 13 cm long and the distance between them is 8 cm. -Maths 9th

Last Answer : Area of a trapezium = 1/2(sum of parallel sides) x (perpendicular diatance between them) = 1/2(25 + 13) x 8 = 152cm2

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

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

Last Answer : Solution of this question

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

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

Last Answer : Solution of this question

How many times area is changed, when sides of a triangle are doubled. -Maths 9th

Description : How many times area is changed, when sides of a triangle are doubled. -Maths 9th

Last Answer : Four times area is changed, when sides of a triangle are doubled.

The sides of a triangle are 15cm, 8cm and 17cm. Find its area -Maths 9th

Description : The sides of a triangle are 15cm, 8cm and 17cm. Find its area -Maths 9th

Last Answer : Answer:- As we know herons formula = √[s(s-a)(s-b)(s-c)] Now here a=15cm, b=8cm, c=17cm And, s=20cm Then, area is=√[20*5*12*3] =√3600 =60cm² Therefore the Area is 60cm². Regards, Rishikesh

ABCD is a trapezium with AB and CD as parallel sides. The diagonals intersect at O. The area of the triangle ABO is p and that of triangle CDO is q. -Maths 9th

Description : ABCD is a trapezium with AB and CD as parallel sides. The diagonals intersect at O. The area of the triangle ABO is p and that of triangle CDO is q. -Maths 9th

Last Answer : answer:

If two sides of one triangle are equal to two sides of another triangle and the contained angles are supplementary, show that the two sides are equal in area -Maths 9th

Description : If two sides of one triangle are equal to two sides of another triangle and the contained angles are supplementary, show that the two sides are equal in area -Maths 9th

Last Answer : 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.