# 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:

AC^{2} = AB^{2} + BC^{2}

⇒ (25)^{2} = (4x)^{2} + (3x)^{2}

⇒ 16x^{2} + 9x^{2} = 625

⇒ 25x^{2} = 625

⇒ x^{2} = 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}.

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