Mensuration | Chapter Assessment | English | Grade 8

What is the area of the given quadrilateral?

• 1668 m²

• 1664 m²

• 988 m²

• 3328 m²

Answer explanation

Area of given quadrilateral = (1/2) × 52 × 38 + (1/2) × 52 × 26 = 988 + 676 = 1664 m² So, option 2 is correct.

Choose correct option showing the area of given figure, if each side of square is of 6 cm and all triangles are isosceles with equal sides 5 cm long.

• 36 cm²

• 48 cm²

• 84 sq. cm

• 89 sq. cm

Answer explanation

Area of the square = l² = 6² = 36 cm² Consider one isosceles triangle. Draw an altitude from the common vertex of equal sides to the opposite side which will also be a median of the opposite side. Now, consider a right triangle with hypotenuse = 5 cm, base = 6/2= 3 cm. We can find the height (or altitude) by using Pythagoras Theorem as below, Height of a right angled triangle = √ (5² - 3²) = √ (25 - 9) = √16 = 4 cm Now, base of an isosceles triangle = 6 cm and its height = 4 cm Area of 1 isosceles triangle = (1/2) × b × h = (1/2) × 6 × 4 = 12 cm² Area of 4 isoseles triangles = 4 × 12 cm² = 48 cm² Area of given figure = Area of square + Area of 4 isosceles triangles = 36 cm² + 48 cm² = 84 cm² So, option 3 is correct.

A cuboidal container of dimensions 54 m × 41 m × 32 m is filled with cubes of side 4 m each. How many such cubes can be accomodated in the container?

• 4428

• 1107

• 17712

• 17721

Answer explanation

Volume of cuboidal container = l × b × h = 54 × 41 × 32 = 70848 m³ Volume of cube = l³ = 4³ = 64 m³ Number of small boxes accomodated in container = Volume of cuboidal container/Volume of cube = 70848 / 64 = 1107

A road roller of radius 14 m and height of 21 m takes 21 revolutions to cover a road. What is the area of the road?

• 38818 m²

• 38805 m²

• 38808 m²

• 38812 m²

Answer explanation

Lateral surface area of cylinder (road roller) = 2πrh = 2 × (22/7) × 14 × 21 = 1848 m² Area covered by road roller in one revolution = 1848 m² Total area of road = Area covered in one revolution x total number of revolutions required to cover the road = 21 × 1848 = 38808 m² So, option 3 is correct.

In the given figure, PB = 46 cm, TQ = 72 cm, AT = 10 cm, AS = 25 cm, CQ = 30 cm CR = 34 cm. Area of the polygon PQRST is ________.

• 3237 cm²

• 3239 cm²

• 3242 cm²

• 3235 cm²

Answer explanation

AC = TQ - AT - CQ = 72 - 10 - 30 = 32 cm Area of given polygon = ar(△ PQT) + ar(□ ACRS) + ar(△ TAS) + ar(△ QCR) ar(△ PQT) = (1/2)× TQ × PB = (1/2) × 72 × 46 = 1656 cm² ar(□ ACRS) = (1/2) × AC ( AS + CR) = (1/2) × 32 × (25 + 34) = 944 cm² ...(ACRS is a trapezium) ar(△ TAS) = (1/2) × AT × AS = (1/2) × 10 × 25 = 125 cm² ar(△ QCR) = (1/2)× CQ × CR = (1/2) × 30 × 34 = 510 cm² Hence Area of given polygon = ar(△ PQT) + ar(□ ACRS) + ar(△ TAS) + ar(△ QCR) = 1656 + 125 + 510 + 944 = 3235 cm² So, option 4 is correct.

Manoj gifted a mobile to her sister. It is packed in a cuboidal box of dimensions 21 cm × 15 cm × 9 cm. If he wants to wrap the box using gift wrapping paper, how much paper will he require?

• 1278 cm²

• 1275 cm²

• 648 cm²

• 2835 cm²

Answer explanation

Total paper required = Total surface area of cuboid = 2(lb +bh +lh) = 2(21 × 15 + 15 × 9 + 21 × 9) = 1278 cm² So, option 1 is correct.

State whether the statement given below if is true or false. The total surface area of a cuboid formed by joining face to face 3 cubes of side 5 cm each is 3 times the total surface area of a cube of side 5 cm.

• TRUE

• FALSE

Answer explanation

For the cuboid formed by joining 3 cubes, l = 15 cm, b = 5 cm, h = 5 cm Total surface area of cuboid = 2(lb +bh +lh) = 2(15 × 5 + 5 × 5 + 15 × 5) = 2(75 + 25 + 75) = 350 cm² Total surface area of one cube with side 5 cm = 6 (l)² = 6 (5)² = 6 (25) = 150 cm² So 3 times the total surface area of the cube = 3 × 150 = 450 cm² Which is not equal to the total surface area of the cuboid i. e 350 cm² Therefore, the given statement is false.