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 = 90^o

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) 90^o

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

Ans: (b) 72^o

Solution: Let the measure of the angle be x^o.

Hence the one angle is 108^o.

Its adjacent = (180 – 108)^o = 72^o

Therefore, the smallest angle is 72^o.

(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 24^o less than twice the smallest angle then the largest angle of the parallelogram is

Ans: (c) 112^o

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

∴ x + (2x – 24) = 180

⇒ x + 2x = 180 + 24

⇒ 3x = 204

⇒ x = 68

Hence, the smallest angle is 68^o.

Its adjacent is = (180 – 68)^o = 112^o.

Therefore,  the largest angle is 112^o.