Solve math problems for Grade 5 page 43, 44 VBT Exercise 2, General practice, exercise 119

Solving math problems for Grade 5 page 43, 44 VBT Exercise 2, General practice, exercise 119, guide to solving and providing detailed answers for exercises 1, 2, 3. Students can refer to learn how to apply the knowledge to exercises, do them correctly and quickly.

Solve exercise 1 - Math exercise book for Grade 5 exercise 2 page 43

Problem:

Given trapezoid ABCD (see diagram) with AB = 20cm, AD = 30cm, DC = 40cm.

Joining A with C forms two triangles ABC and ADC. Calculate:

a. The area of each triangle.

b. The percentage ratio of the area of triangle ABC to triangle ADC.

Solution Method

- Area of trapezoid ABCD = (large base + small base) ⨯ height : 2 = (AB + DC) ⨯ AD : 2 .

- Area of triangle ADC = AD ⨯ DC : 2.

- The area of triangle ABC = the area of trapezoid ABCD - the area of triangle ADC.

- To find the percentage ratio of the area of triangle ABC to triangle ADC, divide the area of triangle ABC by the area of triangle ADC, then multiply the quotient by 100 and add the % sign to the right.

a. The area of trapezoid ABCD is :

(20+40) x 30/2 = 900 (cm²)

The area of triangle ADC is :

(40x30)/2 = 600 (cm²)

The area of triangle ABC is :

900 - 600 = 300 (cm²)

b. The percentage ratio of the area of triangle ABC to triangle ADC is :

300/600 x 100 = 50%

Given square ABCD with side length 4cm. On the sides of the square, take the midpoints M, N, P, Q respectively. Connect these four points to form quadrilateral MNPQ (see diagram). Calculate the ratio of the area of quadrilateral MNPQ to square ABCD.

- Area of square ABCD = side x side.

- Triangles AMQ, BMN, CPN, DPQ are right triangles with equal areas.

The area of each triangle is the product of the lengths of the two sides of the right angle divided by 2.

- The area of quadrilateral MNPQ is equal to the area of square ABCD minus the sum of the areas of triangles AMQ, BMN, CPN, DPQ.

- To find the ratio of the area of quadrilateral MNPQ to square ABCD, divide the area of quadrilateral MNPQ by the area of square ABCD.

Because M, N, P, Q are the midpoints of AB, BC, CD, and AD respectively, we have:

AM = MB = BN = NC = CP = PD = DQ = QA = 4 : 2 = 2cm

The area of square ABCD is:

The area of square ABCD is :

4 ⨯ 4 = 16 (cm²)

The area of triangle AMQ is :

(2x2)/2 = 2 (cm²)

The area of quadrilateral MNPQ is :

16 - (4 ⨯ 2) = 8 (cm²)

The ratio of the area of quadrilateral MNPQ to square ABCD is :

8/16 = 1/2

Given the figure consisting of rectangle ABCD with AD = 2dm and a semicircle center O with radius 2dm. Calculate the area of the shaded part of rectangle ABCD.

- Find the length of the rectangle = OD ⨯ 2.

- Area of rectangle ABCD = length ⨯ width.

- Area of semicircle center O = (radius ⨯ radius ⨯ 3.14) : 2.

- Area of shaded part = area of rectangle ABCD - area of semicircle center O.

Area of rectangle ABCD is :

2 ⨯ 4 = 8 (dm²)

Area of semicircle center O is :

(2x2)/2 x 3.14 = 6.28 (dm²)

Area of shaded part is :

8 - 6.28 = 1.72 (dm²)

Answer : 1.72dm²

You are viewing the guide for solving Grade 5 math problems on pages 43 and 44 of VBT Volume 2, Concentrated Practice, Exercise 119. You can review the instructions for solving Grade 5 math problems on pages 41 and 42 of VBT Volume 2 Introduction to the cylinder. Introduction to the sphere or preview the instructions for solving Grade 5 math problems on page 45 of VBT Volume 2 Concentrated Practice to understand more about the lesson.

Wishing you all the best in your math studies.