A rectangle inscribed in a triangle has its base coinciding with the base b of the triangle. If the altitude of the triangle is h, and the -Maths 9th

A rectangle inscribed in a triangle has its base coinciding with the base b of the triangle. If the altitude of the triangle is h, and the -Maths 9th
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Find the area of an equilateral triangle having altitude h cm -Maths 9th

Description : Find the area of an equilateral triangle having altitude h cm -Maths 9th

Last Answer : This answer was deleted by our moderators...

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.

A square is inscribed in an isosceles right triangle, so that the square and the triangle have one angle common. -Maths 9th

Description : A square is inscribed in an isosceles right triangle, so that the square and the triangle have one angle common. -Maths 9th

Last Answer : Given In isosceles triangle ABC, a square ΔDEF is inscribed. To prove CE = BE Proof In an isosceles ΔABC, ∠A = 90° and AB=AC …(i) Since, ΔDEF is a square. AD = AF [all sides of square are equal] … (ii) On subtracting Eq. (ii) from Eq. (i), we get AB – AD = AC- AF BD = CF ….(iii)

A square is inscribed in an isosceles right triangle, so that the square and the triangle have one angle common. -Maths 9th

Description : A square is inscribed in an isosceles right triangle, so that the square and the triangle have one angle common. -Maths 9th

Last Answer : Given In isosceles triangle ABC, a square ΔDEF is inscribed. To prove CE = BE Proof In an isosceles ΔABC, ∠A = 90° and AB=AC …(i) Since, ΔDEF is a square. AD = AF [all sides of square are equal] … (ii) On subtracting Eq. (ii) from Eq. (i), we get AB – AD = AC- AF BD = CF ….(iii)

a square is inscribed in an isosceles triangle so that the square and the triangle have one angle common. show that the vertex of the square opposite the vertex of the common angle bisect the hypotenuse. -Maths 9th

Description : a square is inscribed in an isosceles triangle so that the square and the triangle have one angle common. show that the vertex of the square opposite the vertex of the common angle bisect the hypotenuse. -Maths 9th

Last Answer : This answer was deleted by our moderators...

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.

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

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 diagram AB and AC are the equal sides of an isosceles triangle ABC, in which is inscribed equilateral triangle DEF. -Maths 9th

Description : In the diagram AB and AC are the equal sides of an isosceles triangle ABC, in which is inscribed equilateral triangle DEF. -Maths 9th

Last Answer : answer:

An equilateral triangle, if its altitude is 3.2 cm. -Maths 9th

Description : An equilateral triangle, if its altitude is 3.2 cm. -Maths 9th

Last Answer : First observe that the altitudes from any vertex to the opposite sides of an equilateral triangle are all of equal length. Hence we can define the height of an equilateral triangle as this common value of three ... ∠MAB=30∘ and ∠MAC=30∘, with B and C on XY. Then ABC is the required triangle.

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

An equilateral triangle, if its altitude is 3.2 cm. -Maths 9th

Description : An equilateral triangle, if its altitude is 3.2 cm. -Maths 9th

Last Answer : We know that, in an equilateral triangle all sides are equal and all angles are equal i.e., each angle is of 60°. Given, altitude of an equilateral triangle say ABC is 3.2 cm. To construct the ΔABC ... ∠DBA = 60° Similary, ∠DCA = 60° Thus, ∠A = ∠B=∠C = 60° Hence, ΔABC is an equilateral triangle.

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

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.

abcd is a rectangle and bd is one of its diagonals if area of triangle abd is 8cm^2 find are of triangle bcd -Maths 9th

Description : abcd is a rectangle and bd is one of its diagonals if area of triangle abd is 8cm^2 find are of triangle bcd -Maths 9th

Last Answer : This answer was deleted by our moderators...

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.

Prove that altitude of a triangle are equal -Maths 9th

Description : Prove that altitude of a triangle are equal -Maths 9th

Last Answer : 1 Answer. Proof: To prove this, I must first prove that the three perpendicular bisectors of a triangle are concurrent. From Figure 1, consider triangle ABC, I know that the perpendicular bisector of AB, passing ... midpoint M of AB, is the set of all points that have equal distances to A and B.

Construct an equilateral triangle whose altitude is 7 cm -Maths 9th

Description : Construct an equilateral triangle whose altitude is 7 cm -Maths 9th

Last Answer : NEED ANSWER

Prove that altitude of a triangle are equal -Maths 9th

Description : Prove that altitude of a triangle are equal -Maths 9th

Last Answer : Proof: To prove this, I must first prove that the three perpendicular bisectors of a triangle are concurrent. From Figure 1, consider triangle ABC, I know that the perpendicular bisector of AB, passing ... on the perpendicular bisector. Because, PM is congruent to PM, MA is congruent to MB and

Construct an equilateral triangle whose altitude is 7 cm -Maths 9th

Description : Construct an equilateral triangle whose altitude is 7 cm -Maths 9th

Last Answer : This answer was deleted by our moderators...

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

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

Last Answer : In △ABE and △ACF, we have ∠BEA=∠CFA (Each 90 0 ) ∠A=∠A (Common angle) AB=AC (Given) ∴△ABE≅△ACF (By SAS congruence criteria) ∴BF=CF [C.P.C.T]

ABC is an acute angled triangle. CD is the altitude through C. If AB = 8 units, CD = 6 units, find the distance -Maths 9th

Description : ABC is an acute angled triangle. CD is the altitude through C. If AB = 8 units, CD = 6 units, find the distance -Maths 9th

Last Answer : answer:

The base and the corresponding altitude of a parallelogram are 10 cm and 7 cm, respectively. Find its area. -Maths 9th

Description : The base and the corresponding altitude of a parallelogram are 10 cm and 7 cm, respectively. Find its area. -Maths 9th

Last Answer : Area of parallelogram = Base × Corresponding altitude = 10 × 7 = 70 cm2.

If a rectangle and a parallelogram are equal in area and have the same base and are situated on the same side, and the ratio of the perimeter -Maths 9th

Description : If a rectangle and a parallelogram are equal in area and have the same base and are situated on the same side, and the ratio of the perimeter -Maths 9th

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If a sphere is inscribed in a cube, then find the ratio of the volume of the cube to the volume of the sphere. -Maths 9th

Description : If a sphere is inscribed in a cube, then find the ratio of the volume of the cube to the volume of the sphere. -Maths 9th

Last Answer : The ratio of the volume of the cube to the volume of the sphere are as

If a sphere is inscribed in a cube, then find the ratio of the volume of the cube to the volume of the sphere. -Maths 9th

Description : If a sphere is inscribed in a cube, then find the ratio of the volume of the cube to the volume of the sphere. -Maths 9th

Last Answer : The ratio of the volume of the cube to the volume of the sphere are as

In a sphere of radius 2 cm a cone of height 3 cm is inscribed. What is the ratio of volumes of the cone and sphere ? -Maths 9th

Description : In a sphere of radius 2 cm a cone of height 3 cm is inscribed. What is the ratio of volumes of the cone and sphere ? -Maths 9th

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A trapezium ABCD in which AB || CD is inscribed in a circle with centre O. Suppose the diagonals AC and BD of the trapezium intersect at M -Maths 9th

Description : A trapezium ABCD in which AB || CD is inscribed in a circle with centre O. Suppose the diagonals AC and BD of the trapezium intersect at M -Maths 9th

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The ellipse `x^2+""4y^2=""4` is inscribed in a rectangle aligned with the coordinate axes, which in turn is inscribed in another ellipse that passes t

Description : The ellipse `x^2+""4y^2=""4` is inscribed in a rectangle aligned with the coordinate axes, which in turn is inscribed in another ellipse that passes t

Last Answer : The ellipse `x^2+""4y^2=""4` is inscribed in a rectangle aligned with the coordinate axes, which in turn is ... 4x^2+48y^2=48` D. `4x^2+64y^2=48`

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

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.

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

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.

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.

The base of a right-angled triangle measures 4 cm and its hypotenuse measures 5 cm. Find the area of the triangle. -Maths 9th

Description : The base of a right-angled triangle measures 4 cm and its hypotenuse measures 5 cm. Find the area of the triangle. -Maths 9th

Last Answer : In right-angled triangle ABC AB2 + BC2 = AC2 (By Pythagoras Theorem) ⇒ AB2 + 42 = 52 ⇒ AB2 = 25 – 16 = 9 5 cm ⇒ AB = 3 cm ∴ Area of △ABC = 1/2 BC x AB = 1/2 x 4 x 3 = 6cm2

The base of an isosceles triangle is 24cm and its area is 192cm^2. Find its perimeter -Maths 9th

Description : The base of an isosceles triangle is 24cm and its area is 192cm^2. Find its perimeter -Maths 9th

Last Answer : Given, base of an isosceles triangle =24 cm Area of isosceles triangle =192 sq.cm Area = 21​×b×h ∴192=224.h​ ∴h=16 cm Side=h2+122​=256+144​=20 cm Perimeter of triangle =2a+b =2(20)+24=64 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

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.

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

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

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

If a triangle and a parallelogram are on same base and between same parallels, then find the ratio of the area of the triangle to the area of parallelogram. -Maths 9th

Description : If a triangle and a parallelogram are on same base and between same parallels, then find the ratio of the area of the triangle to the area of parallelogram. -Maths 9th

Last Answer : If a triangle and a parallelogram are on the same base and between the same parallels, the area of the triangle is equal to half of the parallelogram.

If a triangle and a parallelogram are on the same base and between same parallels, then the ratio of the area of the triangle to the area of parallelogram is -Maths 9th

Description : If a triangle and a parallelogram are on the same base and between same parallels, then the ratio of the area of the triangle to the area of parallelogram is -Maths 9th

Last Answer : (b) We know that, if a parallelogram and a triangle are on the same base and between the same parallels, then area of the triangle is half the area of the parallelogram.

If a triangle and a parallelogram are on same base and between same parallels, then find the ratio of the area of the triangle to the area of parallelogram. -Maths 9th

Description : If a triangle and a parallelogram are on same base and between same parallels, then find the ratio of the area of the triangle to the area of parallelogram. -Maths 9th

Last Answer : If a triangle and a parallelogram are on the same base and between the same parallels, the area of the triangle is equal to half of the parallelogram.

If a triangle and a parallelogram are on the same base and between same parallels, then the ratio of the area of the triangle to the area of parallelogram is -Maths 9th

Description : If a triangle and a parallelogram are on the same base and between same parallels, then the ratio of the area of the triangle to the area of parallelogram is -Maths 9th

Last Answer : (b) We know that, if a parallelogram and a triangle are on the same base and between the same parallels, then area of the triangle is half the area of the parallelogram.

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

If a triangle and a parallelogram are on same base and between the same parallels,then find the ratio of the area of the triangle to the area of parallelogram -Maths 9th

Description : If a triangle and a parallelogram are on same base and between the same parallels,then find the ratio of the area of the triangle to the area of parallelogram -Maths 9th

Last Answer : Solution :-

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