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Area and Properties of Parallelograms

Area and Properties of Parallelograms

Assessment

Interactive Video

Mathematics

5th - 6th Grade

Practice Problem

Hard

Created by

Thomas White

FREE Resource

Presto Walker presents a challenging math problem involving a parallelogram and a red triangle. The problem, used to identify gifted students, requires basic arithmetic and understanding of geometric areas. Walker explains the solution, which involves calculating the area of the red triangle using given areas of yellow regions. He reviews the principles of calculating areas of parallelograms and triangles, and demonstrates two methods to solve the problem. The video concludes with an invitation to subscribe and explore more math content.

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8 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Who originally presented the problem discussed in the video?

A Japanese math teacher

A French math teacher

A Chinese math teacher

An American math teacher

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What shape is central to the problem discussed in the video?

Square

Rectangle

Parallelogram

Circle

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the area of the red triangle according to the solution?

72

9

10

8

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the formula for the area of a parallelogram?

Base times height

Length times width

Base times width

Height times width

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How is the area of a triangle related to the area of a parallelogram?

It is half the area of the parallelogram

It is equal to the area of the parallelogram

It is one-third the area of the parallelogram

It is double the area of the parallelogram

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What principle is used to solve the problem involving the red triangle?

Applying the quadratic formula

Using trigonometric identities

Finding triangles with half the area of the parallelogram

Using the Pythagorean theorem

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the alternative method to solve the problem?

Using different sets of triangles

Using a different parallelogram

Using a circle instead

Using a rectangle instead

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