Draw an equiliateral triangle of height 3.6 cm. Draw another triangle similar to it such that its side is of the side of the first. -Maths 9th

Draw an equiliateral triangle of height 3.6 cm. Draw another triangle similar to it such that its side is of the side of the first. -Maths 9th
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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

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 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

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).

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

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).

The perimeter of a triangle is 50 cm. One side of a triangle is 4 cm longer than the smaller side and the third side is 6 cm less than twice the smaller side. -Maths 9th

Description : The perimeter of a triangle is 50 cm. One side of a triangle is 4 cm longer than the smaller side and the third side is 6 cm less than twice the smaller side. -Maths 9th

Last Answer : Let the smallest side of the triangle be x cm long So, second side =(x+4)cm and third side =(2x−6)cm Given, perimeter =50cm Therefore, x+(x+4)+(2x−6)=50 ⇒4x=52 ⇒x=13cm So, first side of the triangle is ... =25cm Therefore, area of Δ=s(s−a)(s−b)(s−c) =25(25−13)(25−17)(25−20) =25 12 8 5 = 2030 cm2

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

The perimeter of a triangle is 50 cm. One side of a triangle is 4 cm longer than the smaller side and the third side is 6 cm less than twice the smaller side. -Maths 9th

Description : The perimeter of a triangle is 50 cm. One side of a triangle is 4 cm longer than the smaller side and the third side is 6 cm less than twice the smaller side. -Maths 9th

Last Answer : According to question find the area of triangle.

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

Find the area of a triangle with base =20cm and height are 10 cm. -Maths 9th

Description : Find the area of a triangle with base =20cm and height are 10 cm. -Maths 9th

Last Answer : Area of a triangle = 1/2 × Base × Altitude ( height ) therefore., Area of a triangle = 1/2 × 20 cm× 10cm = 10cm ×10cm = 100 cm^2

The base of a right triangular prism is an equilateral triangle. If the height is halved and each side of the base is doubled, find the ratio of the -Maths 9th

Description : The base of a right triangular prism is an equilateral triangle. If the height is halved and each side of the base is doubled, find the ratio of the -Maths 9th

Last Answer : 1 : 2 Let each side of the base of the original prism be a units and the height of the prism be h units. Then Required ratio = Vol. of original prismVol. of new prismVol. of original ... )2×h3√4×(2a)2×h234×(a)2×h34×(2a)2×h2 = 2a2h4a2h2a2h4a2h = 1 : 2.

Inside a triangular park, there is a flower bed forming a similar triangle. Around the flower bed runs a uniform path of such a width that the -Maths 9th

Description : Inside a triangular park, there is a flower bed forming a similar triangle. Around the flower bed runs a uniform path of such a width that the -Maths 9th

Last Answer : (d) 3 : 1Let ABC be the triangular flower bed of side lengths a, b and c respectively. Then Area of ΔABC = \(\sqrt{s(s-a)(s-b)(s-c)}\)where s = \(rac{a+b+c}{2}\)Now according to the given condition,ΔPQR ... of ΔABC - Area of ΔABC= 3 (Area of ΔABC)∴ Reqd. Ratio = Area of Path : Area of ΔABC = 3 : 1.

The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. -Maths 9th

Description : The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. -Maths 9th

Last Answer : Area of the triangle =

The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. -Maths 9th

Description : The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. -Maths 9th

Last Answer : Area of the triangle =

Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm. -Maths 9th

Description : Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm. -Maths 9th

Last Answer : Steps of Construction (i) Draw BC = 12 cm. (ii) Construct ÐCBY = 90°. (iii) From ray BY, cut-off line segment BD = 18 cm. (iv) Join CD. (v) Draw the perpendicular bisector of CD intersecting BD at A. (vi ... = AC Now, BD = BA + AD ⇒ BD = AB + AC Hence, △ABC is the required triangle.

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.

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.

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.

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.

Find the area of an isosceles triangle having base x cm and equal side y cm. -Maths 9th

Description : Find the area of an isosceles triangle having base x cm and equal side y cm. -Maths 9th

Last Answer : If h is the height of the triangle, then h 2 =y 2 − 4 x​ 2 ⇒h= 4 4y 2 −x 2 ​ ​ cm ∴Area= 2 1​ ×base×h = 2 x​ 4 4y 2 −x 2 ​ ​ cm 2

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

What is the radius of a circle inscribed in a triangle having side lengths 35 cm, 44 cm and 75 cm? -Maths 9th

Description : What is the radius of a circle inscribed in a triangle having side lengths 35 cm, 44 cm and 75 cm? -Maths 9th

Last Answer : (d) 6 cmLet a = 35 cm, b = 44 cm, c = 75 cm. Thens = \(rac{a+b+c}{2}\) = \(rac{34+44+75}{2}\) cm = 77 cm∴ Area if triangle = \(\sqrt{77(77-35)(77-44)(77-75)}\) cm2= \(\sqrt{77 imes42 ... ) cm2 = 462 cm2∴ Radius of incircle = \(rac{ ext{Area}}{ ext{semi-perimeter}}\) = \(rac{462}{77}\) cm = 6 cm.

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:

There are two congruent triangles each with area 198 cm^2. Triangle DEF is placed over triangle ABC in such a way that the centroid of -Maths 9th

Description : There are two congruent triangles each with area 198 cm^2. Triangle DEF is placed over triangle ABC in such a way that the centroid of -Maths 9th

Last Answer : answer:

A joker’s cap is in the form of right circular cone of base radius 7 cm and height 24cm. Find the area of the sheet required to make 10 such caps. -Maths 9th

Description : A joker’s cap is in the form of right circular cone of base radius 7 cm and height 24cm. Find the area of the sheet required to make 10 such caps. -Maths 9th

Last Answer : Radius of conical cap, r = 7 cm Height of conical cap, h = 24cm Slant height, l2 = (r2+h2) = (72+242) = (49+576) = (625) Or l = 25 cm CSA of 1 conical cap = πrl = (22/7)×7×24 = 550 CSA of 10 caps = (10×550) cm2 = 5500 cm2 Therefore, the area of the sheet required to make 10 such caps is 5500 cm2.

X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th

Description : X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th

Last Answer : Here, △XZM and △XZL are on the same base (XZ) and lie between the same parallels (XZ || LM). ∴ ar(△XZL) = ar( △XZM) Adding ar(△XZY) on both sides , we have ar(△XZL) + ar(△XZY) = ar(△XZM) + ar(△XZY) ⇒ ar(△LZY) = ar(quad.MZYX)

X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. -Maths 9th

Description : X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. -Maths 9th

Last Answer : Given X and Y are points on the side LN such that LX = XY = YN and XZ || LM To prove ar (ΔLZY) = ar (MZYX) Proof Since, ΔXMZ and ΔXLZ are on the same base XZ and between the same parallel lines LM and XZ. ... get ar (ΔXMZ) + ar (ΔXXZ) = ar (ΔXLZ) + ar (ΔXYZ) => ar (MZYX) = ar (ΔLZY) Hence proved.

X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th

Description : X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th

Last Answer : Here, △XZM and △XZL are on the same base (XZ) and lie between the same parallels (XZ || LM). ∴ ar(△XZL) = ar( △XZM) Adding ar(△XZY) on both sides , we have ar(△XZL) + ar(△XZY) = ar(△XZM) + ar(△XZY) ⇒ ar(△LZY) = ar(quad.MZYX)

X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. -Maths 9th

Description : X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. -Maths 9th

Last Answer : Given X and Y are points on the side LN such that LX = XY = YN and XZ || LM To prove ar (ΔLZY) = ar (MZYX) Proof Since, ΔXMZ and ΔXLZ are on the same base XZ and between the same parallel lines LM and XZ. ... get ar (ΔXMZ) + ar (ΔXXZ) = ar (ΔXLZ) + ar (ΔXYZ) => ar (MZYX) = ar (ΔLZY) Hence proved.

A point within an equilateral triangle whose perimeter is 30 m is 2 m from one side and 3 m from another side. Find its distance from third side. -Maths 9th

Description : A point within an equilateral triangle whose perimeter is 30 m is 2 m from one side and 3 m from another side. Find its distance from third side. -Maths 9th

Last Answer : answer:

A right triangular prism of height 18 cm and of base sides 5 cm, 12 cm and 13 cm is transformed into another right triangular prism on a base -Maths 9th

Description : A right triangular prism of height 18 cm and of base sides 5 cm, 12 cm and 13 cm is transformed into another right triangular prism on a base -Maths 9th

Last Answer : Vol. of △ ular prism = Area of △ ular base × height. ∴ Area of triangular base = area of triangle PQR By heron's formula. S=S(s−a)(s−b)(s−c)​where S=2a+b+c​∴Areaof△PQR= S=23+4+5​=6 S=6(6−3)(6−4)(6−5)​=3×2×3×2×1​=6cm2 ∴ vol. of Prism =6×10 =60cm3Answer.

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.

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 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

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

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 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°

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°

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.

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.

Draw a line segment of length 8.6 cm. Bisect it and measure the length of each part. -Maths 9th

Description : Draw a line segment of length 8.6 cm. Bisect it and measure the length of each part. -Maths 9th

Last Answer : Draw a line segment AB of length 8.6 cm. With A as centre and radius more than half of AB, draw arcs on both sides of AB. With the same radius and B as centre, draw arcs on the both sides of AB, ... line segment from E to F intersecting AB at C. On measuring AC and BC, we get: AC=BC=4.3 cm.

Draw a circle of diameter 6.4 cm. Then draw two tangents to the circle from a point P at a distance 6.4 cm from the centre of the circle. -Maths 9th

Description : Draw a circle of diameter 6.4 cm. Then draw two tangents to the circle from a point P at a distance 6.4 cm from the centre of the circle. -Maths 9th

Last Answer : clear .

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

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

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.

There are two prisms, one has equilateral triangle as a base and the other has a regular hexagon as a base. If both the prisms have equal height -Maths 9th

Description : There are two prisms, one has equilateral triangle as a base and the other has a regular hexagon as a base. If both the prisms have equal height -Maths 9th

Last Answer : answer:

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

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

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.