What is the area of a triangle and square?

What is the area of a triangle and square?
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Square: b*hTriangle:1/2 b*h*** means multiply**
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In trapezium ABCD, AB|| DC and diagonals AC and BD intersect at O. If area of triangle AOD is 30cm square , find the area of triangle BOC -Maths 9th

Description : In trapezium ABCD, AB|| DC and diagonals AC and BD intersect at O. If area of triangle AOD is 30cm square , find the area of triangle BOC -Maths 9th

Last Answer : In the given figure: Area of triangle ADC = Area of triangle BCD (Triangles on the same and parallel) Now subtract the area of triangle DOC from both of them so... (Area of triangle ADC - Area of ... => Area of triangle AOD = Area of triangle BOC Hence the area of triangle BOC is 30 cm square.

In trapezium ABCD, AB|| DC and diagonals AC and BD intersect at O. If area of triangle AOD is 30cm square , find the area of triangle BOC -Maths 9th

Description : In trapezium ABCD, AB|| DC and diagonals AC and BD intersect at O. If area of triangle AOD is 30cm square , find the area of triangle BOC -Maths 9th

Last Answer : In the given figure: Area of triangle ADC = Area of triangle BCD (Triangles on the same and parallel) Now subtract the area of triangle DOC from both of them so... (Area of triangle ADC - Area of ... => Area of triangle AOD = Area of triangle BOC Hence the area of triangle BOC is 30 cm square.

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

What is the ratio of the area of a circum circle of equilateral triangle to the area of the square with same side length as equilateral triangle? -Maths 9th

Description : What is the ratio of the area of a circum circle of equilateral triangle to the area of the square with same side length as equilateral triangle? -Maths 9th

Last Answer : (d) π : 3.Let each side of the equilateral Δ be a units. Then, circumradius of the circle = \(rac{ ext{side}}{\sqrt3}\) = \(rac{a}{\sqrt3}\) units∴ Area of circumcircle = \(\pi\bigg(rac{a}{\sqrt3}\bigg)^2\) = \( ... units∴ Required ratio = \(rac{rac{\pi{a}^2}{3}}{a^2}\) = \(rac{\pi}{3}\) = π : 3.

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 adjoining figure, ABC is an equilateral triangle inscribing a square of maximum possible area. Again in this squares -Maths 9th

Description : In the adjoining figure, ABC is an equilateral triangle inscribing a square of maximum possible area. Again in this squares -Maths 9th

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.

The area of a square and circle is same and the perimeter of square and equilateral triangle is same, -Maths 9th

Description : The area of a square and circle is same and the perimeter of square and equilateral triangle is same, -Maths 9th

Last Answer : (b) 9 : 4√3.Let each side of the square = a cm. Then, Area of square = a2 cm2 Also, let r be the radius of the circle. Then, πr2 = a2 Let each side of the equilateral triangle = b cm. Then 3b = 4a ⇒ ... ratio between area of circle and area of equilateral Δ is a2 : \(rac{4\sqrt3a^2}{9}\) = 9 : 4√3.

A copper wire is bent to form an equilateral triangle of area 16√3 sq.cm. What will be the area (in sq.cm) if the same copper wire was bent to form a square? a) 16 b) 24 c) 32 d) 40 e) None of the above

Description : A copper wire is bent to form an equilateral triangle of area 16√3 sq.cm. What will be the area (in sq.cm) if the same copper wire was bent to form a square? a) 16 b) 24 c) 32 d) 40 e) None of the above

Last Answer : Area of equilateral triangle = 16√3 = √3 (side)2 /4 Or side2 = 16×4 = 64 So each side = 8cm Perimeter = 8×3 = 24 cm Since perimeter of the square is 24 cm, each side = 6cm So area = 6×6 = 36 sq cm Answer: e)

What is the square of a triangle?

Description : What is the square of a triangle?

Last Answer : The only way to answer the verbal form is by carefully being sure of the meanings of each word. I used to be very strong in math, but I'm rusty. When I hear square I think of a 2D ... approach, as opposed to the usual one based on concise verbal definitions, which seems to be the weak point.

The centre of gravity of a plane lamina will not be at its geometrical centre if it is a a.Square b.Equilateral triangle c.Circle d.Rectangle e.Right angled triangle

Description : The centre of gravity of a plane lamina will not be at its geometrical centre if it is a a.Square b.Equilateral triangle c.Circle d.Rectangle e.Right angled triangle

Last Answer : e. Right angled triangle

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

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.

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)

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.

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)

O is a point in the interior of a square ABCD such that OAB is an equilateral triangle.Show that △OCD is an isosceles triangle. -Maths 9th

Description : O is a point in the interior of a square ABCD such that OAB is an equilateral triangle.Show that △OCD is an isosceles triangle. -Maths 9th

Last Answer : Solution :-

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...

A square and an equilateral triangle each have a perimeter of 72 cm. How much longer is one side of the triangle than one side of the square?

Description : A square and an equilateral triangle each have a perimeter of 72 cm. How much longer is one side of the triangle than one side of the square?

Last Answer : You see you can do is 72cm divide by 3 because an equilateral triangle has 3 sides 72cm/3 = 24cm(I am not sure if this is right)

If a right triangle has legs of length 6 and 8 then we know the length of the hypotenuse is the square root of 6 plus 8 which equals 10?

Description : If a right triangle has legs of length 6 and 8 then we know the length of the hypotenuse is the square root of 6 plus 8 which equals 10?

Last Answer : Using Pythagoras theorem: 6^2 + 8^2 = 100 and its square root is10 which is the length of the hypotenuse

If the square of the length of the longest side of a triangle is less than the sum of the squares of the lengths of the other two sides then the triangle is an acute triangle?

Description : If the square of the length of the longest side of a triangle is less than the sum of the squares of the lengths of the other two sides then the triangle is an acute triangle?

Last Answer : Need answer

The schematic symbol for an operational amplifier in an analog circuit is a _________. A. circle B. square C. trapezoid D. triangle

Description : The schematic symbol for an operational amplifier in an analog circuit is a _________. A. circle B. square C. trapezoid D. triangle

Last Answer : Answer: D

By method of images, the problem can be easily calculated by replacing the boundary with which polygon? a) Rectangle b) Trapezoid c) Square d) Triangle

Description : By method of images, the problem can be easily calculated by replacing the boundary with which polygon? a) Rectangle b) Trapezoid c) Square d) Triangle

Last Answer : d) Triangle

The C.G. of a plane lamina will not be at its geometrical centre in the case of a (A) Right angled triangle (B) Equilateral triangle (C) Square (D) Circle

Description : The C.G. of a plane lamina will not be at its geometrical centre in the case of a  (A) Right angled triangle  (B) Equilateral triangle  (C) Square  (D) Circle

Last Answer : (A) Right angled triangle 

When sine wave is given as input to Schmitt trigger then its generates a) Sine wave b) Saw tooth wave c) Triangle wave d) Square wave

Description : When sine wave is given as input to Schmitt trigger then its generates a) Sine wave b) Saw tooth wave c) Triangle wave d) Square wave

Last Answer : Ans: Square wave

In a mid point loaded fixed beam,the free bending moment diagram is a a.square b.rectangle c.triangle d.trapezium

Description : In a mid point loaded fixed beam,the free bending moment diagram is a a.square b.rectangle c.triangle d.trapezium

Last Answer : c.triangle

In a mid point loaded fixed beam,the fixed bending moment diagram is a a.square b.rectangle c.triangle d.trapezium

Description : In a mid point loaded fixed beam,the fixed bending moment diagram is a a.square b.rectangle c.triangle d.trapezium

Last Answer : b.rectangle

In an off centrepoint loaded fixed beam the fixed bending moment diagram is a a.square b.rectangle c.triangle d.trapezium

Description : In an off centrepoint loaded fixed beam the fixed bending moment diagram is a a.square b.rectangle c.triangle d.trapezium

Last Answer : d.trapezium

In an off centre point loaded fixed beam the free bending moment diagram is a a.square b.rectangle c.triangle d.trapezium

Description : In an off centre point loaded fixed beam the free bending moment diagram is a a.square b.rectangle c.triangle d.trapezium

Last Answer : c.triangle

If a rectangle were called a circle, a circle a point, a point a triangle and a triangle a square, the shape of a wheel is (A) Rectangle (B) Circle (C) Point (D) Triangle

Description : If a rectangle were called a circle, a circle a point, a point a triangle and a triangle a square, the shape of a wheel is (A) Rectangle (B) Circle (C) Point (D) Triangle

Last Answer : (C) Point

The figure shows the front view of a convex lens, which originally had only one edge. Five holes of different shapes, namely triangle, square, pentagon, hexagon and circle, were drilled through it at points P, Q, R, S and T in such a way that the holes were parallel to each other, perpendicular to the edge and all the holes were still within the lens edge. What is the total number of edges in the lens after the holes were drilled?

Description : The figure shows the front view of a convex lens, which originally had only one edge. Five holes of different shapes, namely triangle, square, pentagon, hexagon and circle, were drilled through it at points P ... . What is the total number of edges in the lens after the holes were drilled? 

Last Answer : 57