RS Aggarwal Class 8 Math Sixteenth Chapter Parallelograms Exercise 16B Solution

RS Aggarwal Class 8 Math Sixteenth Chapter Parallelograms Exercise 16B Solution

EXERCISE 16B

OBJECTIVE QUESTIONS

Tick (√) the correct answer in each of the following:

(1) The two diagonals are not necessarily equal in a

Ans: (c) Rhombus

(2) The lengths of the diagonals of a rhombus are 16 cm and 12 cm. The length of each side of the rhombus is

Ans: (c) 10 cm

(3) Two adjacent angles of a parallelogram are (2x + 25)o and (3x – 5)o. The value of x is

Ans: (b) 32

∴ (2x + 25) + (3x – 5) = 180

⇒ 2x + 25 + 3x – 5 = 180

⇒ 5x = 180 – 20

⇒ 5x = 160

⇒ x = 32

(4) The diagonals do not necessarily intersect at right angles in a

Ans: (a) Parallelogram.

(5) The length and breadth of a rectangle are in the ratio 4 : 3. If the diagonal measures 25 cm then the perimeter of the rectangle is

Ans: (c) 70 cm

Solution: Let the length AB be 4x and Breadth BC be 3x.

Each angle of a rectangle is a right angle. We have,

∴ ∠ABC = 90o

From the right ∆ABC:

AC2 = AB2 + BC2

⇒ (25)2 = (4x)2 + (3x)2

⇒ 16x2 + 9x2 = 625

⇒ 25x2 = 625

⇒ x2 = 25

⇒ x = 5

Therefore, length = 4 × 5 = 20 cm and breadth = 3 × 5 = 15 cm.

(6) The bisectors of any two adjacent angles of a parallelogram intersect at

Ans: (d) 90o

(7) If an angle of a parallelogram is two-thirds of its adjacent angle, the smallest angle of the parallelogram is

Ans: (b) 72o

Solution: Let the measure of the angle be xo.

Hence the one angle is 108o.

Its adjacent = (180 – 108)o = 72o

Therefore, the smallest angle is 72o.

(8) The diagonals do not necessarily bisect the interior angles at the vertices in a

Ans: (a) rectangle

(9) In a square ABCD, AB = (2x + 3) cm and BC = (3x – 5) cm. Then, the value of x is

Ans: (d) 8

Solution: We know, all sides are equal of a square. Then,

∴ AB = BC

⇒ 2x + 3 = 3x – 5

⇒ 3x – 2x = 3 + 5

⇒ x = 8

(10) If one angle of a parallelogram is 24o less than twice the smallest angle then the largest angle of the parallelogram is

Ans: (c) 112o

Solution: Let the measure of smallest angle be xo and other is (2x – 24)o.

∴ x + (2x – 24) = 180

⇒ x + 2x = 180 + 24

⇒ 3x = 204

⇒ x = 68

Hence, the smallest angle is 68o.

Its adjacent is = (180 – 68)o = 112o.

Therefore,  the largest angle is 112o.

 


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  1. My name is krish and I am really your fan because I read in 8class and from 6. 7.8.
    Class I am pass out from the big help of you and your chanel

  2. Thanks for the email request

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