Mathematics
General
Easy

Question

What is the area of the trapezoid?

  1. 26 straight m squared
  2. 25 straight m squared
  3. 28 straight m squared
  4. 22 straight m squared

The correct answer is: 26 straight m squared

Related Questions to study

General
General

The radius of a circle is 7 meters. What is area of the circle?

The radius of a circle is 7 meters. What is area of the circle?

GeneralGeneral
General
General

Multiply: 9 × 4 = 

Multiply: 9 × 4 = 

GeneralGeneral
General
General

Solve the quadratic equation below using the Quadratic Formula x2+5x14=0

Solve the quadratic equation below using the Quadratic Formula x2+5x14=0

GeneralGeneral
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General
Maths-

In the figure, C A B equals 90 to the power of ring operator end exponent text  and  end text A D perpendicular B C. If AC = 75 cm AB = 100 cm and BD = 1.25 cm then AD = 

we can use the property that ratio of sides remains same in similar triangles.

In the figure, C A B equals 90 to the power of ring operator end exponent text  and  end text A D perpendicular B C. If AC = 75 cm AB = 100 cm and BD = 1.25 cm then AD = 

Maths-General

we can use the property that ratio of sides remains same in similar triangles.

General
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text In the figure  end text fraction numerator A O over denominator O C end fraction equals fraction numerator B O over denominator O D end fraction equals fraction numerator 1 over denominator 2 end fraction text  and  end text A B equals 5 c m. text  The value of  end text D C text  is end text

we can use the property that ratio of sides remains same in similar triangles.

text In the figure  end text fraction numerator A O over denominator O C end fraction equals fraction numerator B O over denominator O D end fraction equals fraction numerator 1 over denominator 2 end fraction text  and  end text A B equals 5 c m. text  The value of  end text D C text  is end text

Maths-General

we can use the property that ratio of sides remains same in similar triangles.

General
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In the figuretext ,  end text stack D E with bar on top divided by divided by stack B C with bar on top text end textand areatext end text left parenthesis triangle A D E right parenthesis equals text  area  end text left parenthesis triangle B C E D right parenthesis. text end textThe value oftext end text fraction numerator B D over denominator A B end fraction equals

In the figuretext ,  end text stack D E with bar on top divided by divided by stack B C with bar on top text end textand areatext end text left parenthesis triangle A D E right parenthesis equals text  area  end text left parenthesis triangle B C E D right parenthesis. text end textThe value oftext end text fraction numerator B D over denominator A B end fraction equals

maths-General
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General
Maths-

In the figure, angle A equals angle C E D, AB = 9 cm, AD = 7 cm, CD = 8 cm and CE = 10 cm Then DE = ?

if 2 angles are same in a triangle, then the triangles are similar. we can use the property that ratio of sides remains same in similar triangles.

In the figure, angle A equals angle C E D, AB = 9 cm, AD = 7 cm, CD = 8 cm and CE = 10 cm Then DE = ?

Maths-General

if 2 angles are same in a triangle, then the triangles are similar. we can use the property that ratio of sides remains same in similar triangles.

General
Maths-

In the figuretext ,  end text stack A B with bar on top divided by divided by stack Q R with bar on top text  . If  end text A B equals 3 c m comma P B equals 2 c m text  and end text PR = 6 cm then QR = ? cm

we can use the property that ratio of sides remains same in similar triangles

In the figuretext ,  end text stack A B with bar on top divided by divided by stack Q R with bar on top text  . If  end text A B equals 3 c m comma P B equals 2 c m text  and end text PR = 6 cm then QR = ? cm

Maths-General

we can use the property that ratio of sides remains same in similar triangles

General
Maths-

In the figuretext ,  end text stack A B with bar on top divided by divided by stack C D with bar on top text  and  end text stack A C with bar on top intersection stack B D with bar on top equals 0. text  If  end text O A equals 3 x minus 1 text  , end text OB = 2x + 1, OC = 5x – 3, OD = 6x – 5 then AC = ? units.

solving the quadratic equations by the factorization method is used. in this method, the linear term is broken down into 2 terms so that we can take out the common factors from the terms and convert the equation into product form.

In the figuretext ,  end text stack A B with bar on top divided by divided by stack C D with bar on top text  and  end text stack A C with bar on top intersection stack B D with bar on top equals 0. text  If  end text O A equals 3 x minus 1 text  , end text OB = 2x + 1, OC = 5x – 3, OD = 6x – 5 then AC = ? units.

Maths-General

solving the quadratic equations by the factorization method is used. in this method, the linear term is broken down into 2 terms so that we can take out the common factors from the terms and convert the equation into product form.

parallel
General
Maths-

Two line segments text end text stack A B with rightwards arrow on top text  and  end text stack C D with bar on top text end text intersect at E such that triangle A C E tilde operator triangle B D E. If AE = 4cm, BE = 3cm, CE=6cm and DE = x cm then x = ?

We can also use trigonometry to solve this question since all the angles are same in similar triangles. Angle AEC = angle BED.

Two line segments text end text stack A B with rightwards arrow on top text  and  end text stack C D with bar on top text end text intersect at E such that triangle A C E tilde operator triangle B D E. If AE = 4cm, BE = 3cm, CE=6cm and DE = x cm then x = ?

Maths-General

We can also use trigonometry to solve this question since all the angles are same in similar triangles. Angle AEC = angle BED.

General
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From the adjacent figure ,the values of x and y are

From the adjacent figure ,the values of x and y are

maths-General
General
Maths-

In the given figure, ABC is an Isosceles Triangle in which AB = AC, then

we can also use trigonometry to solve this problem. For the same perpendicular distance from the vertex, the length of side increases with increase in the vertex angle since the perpendicular is the cosine component of the side.
height = side x cos(theta). we know that cosine is a decreasing function. therefore, to keep the height same, the side has to increase.

In the given figure, ABC is an Isosceles Triangle in which AB = AC, then

Maths-General

we can also use trigonometry to solve this problem. For the same perpendicular distance from the vertex, the length of side increases with increase in the vertex angle since the perpendicular is the cosine component of the side.
height = side x cos(theta). we know that cosine is a decreasing function. therefore, to keep the height same, the side has to increase.

parallel
General
maths-

In the figure, AB = BC = CD = DE = EF and AF = AE. Then angle A = ?

In the figure, AB = BC = CD = DE = EF and AF = AE. Then angle A = ?

maths-General
General
Maths-

In the figure, AB = AC, angle A = 48° and angle A C D = 18° . Then

an isosceles triangle is one that has two of its sides equal and the base angles equal to each other.

In the figure, AB = AC, angle A = 48° and angle A C D = 18° . Then

Maths-General

an isosceles triangle is one that has two of its sides equal and the base angles equal to each other.

General
Maths-

In the figure, ‘x° ’ and ‘y° ’ are two exterior angle measures of triangle A B C . Then x° + y° is

if one of the angles were acute then we'd not have been able to conclude whether the sum of the angles were greater than or less than 180 degrees.

In the figure, ‘x° ’ and ‘y° ’ are two exterior angle measures of triangle A B C . Then x° + y° is

Maths-General

if one of the angles were acute then we'd not have been able to conclude whether the sum of the angles were greater than or less than 180 degrees.

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