AP State Syllabus AP Board 7th Class Maths Solutions Chapter 12 Quadrilaterals Ex 2 Textbook Questions and Answers.

## AP State Syllabus 7th Class Maths Solutions 12th Lesson Quadrilaterals Exercise 2

Question 1.

State whether true or false

(i) All rectangles are squares ( )

(ii) All rhombuses are parallelogram ( )

(iii) All squares arc rhombuses and also rectangles ( )

(iv) All squares are not parallelograms ( )

(v) All kites arc rhombuses ( )

(vi) All rhombuses are kites ( )

(vii) All parallelograms are trapeziums ( )

(viii) All squares are trapeziums ( )

Solution:

(i) All rectangles are squares ( False)

(ii) All rhombuses are parallelogram ( True)

(iii) All squares arc rhombuses and also rectangles ( True)

(iv) All squares are not parallelograms (False )

(v) All kites arc rhombuses ( False)

(vi) All rhombuses are kites (True )

(vii) All parallelograms are trapeziums ( True)

(viii) All squares are trapeziums ( True)

Question 2.

Explain how a square is a

(i) quadrilateral

(ii) parallelogram

(iii) rhombus

(iv) rectangle.

Solution:

i) quadrilateral : A square is a closed figure bounded by four line segments and hence it is a

quadrilateral.

ii) Parallelogram : In a square both pairs of opposite sides are parallel and hence it is a parallelogram.

iii) Rhombus : All four sides of a square are equal, thus it is a Rhombus.

iv) Rectangle : In a square each angle ¡s a right angle and hence it is a rectangle.

Question 3.

In a rhombus ABCD, ∠CBA = 40°.

Find the other angles.

Given that ∠CBA , ∠B = 40°

We know that rhombus is a parallelogram and

adjacent angles are supplementary.

Thus ∠A + ∠B = 180°

∠A + 40° = 180°

∠A = 180° – 40° = 140°

Also opposite angles are equal.

∴ ∠A =∠C = 140°

∠D = ∠B = 40°

Question 4.

The adjacent angels of a parallelogram arex° and (2x + 30)°.

Find all the angles of the parallelogram.

Solution:

Given that the adjacent angles of a parallelogram are x° ; (2x + 30)°

Adjacent angles of a parallelogram are supplementary.

∴ x + (2x + 30) = 180°

3x + 30 = 180°

3x = 180° – 30° = 150°

x = \(\frac{150^{\circ}}{3}\) = 50°

∴ The adjacent angles are x = 50°

2x + 30°= 2 x 50 + 30 = 130°

∴ The other two angles are 50°, 130°

∴ The four angles are 50°, 130°, 50°, 130°

Question 5.

Explain how DEAR is a trapezium. Which of its two sides are parallel?

Solution:

In quadrilateral DEAR

∠D = 80° and ∠E = 100°

Also ∠D + ∠E = 80°+ 100° = 180°

(∵ D, E are interior angles on the same side of \(\overline{\mathrm{DE}}\))

∴ DR //EA

Also ∠A = ∠R = 90° and ∠A + ∠R = 180°

∴ DEAR is a trapezium.

Question 6.

BASE is a rectangle. Its diagonals intersect atO. Findx, if OB = 5x+1 and OE = 2x + 4.

Solution:

Given BASE is a rectangle.

OB = 5x + 1 and OA = 2x + 4

In a rectangle diagonals are equal.

BS = EA

(i.e.) 2 x BO = 2 x OA

2(5x + 1) = 2(2x + 4 )

10x + 2 = 4x + 8

10x – 4x = 8 – 2

6x = 6

x = 1

Question 7.

Is quadrilateral ABCD a parallelogram, if ∠A = 70° and ∠C = 65° ? Give reason.

Solution:

Given that in quadrilateral ABCD, ∠A = 70° and ∠C = 65°

∠A and ∠C are opposite angles and are not equal.

Hence quadrilateral ABCD is not a parallelogram.

Question 8.

Two adjacent sides of a parallelogram are in the ratio 5:3 the perimeter of the parallelogram is 48cm. Find the length of each of its sides.

Solution:

Given that the ratio of two adjacent sides of a parallelogram = 5 : 3

Sum of the terms of the ratio = 5 + 3 = 8

Sum of two sides = half of the perimeter = \(\frac { 1 }{ 2 }\) x 48 = 24 cm

∴ one side = \(\frac { 5 }{ 8 }\) x 24 = 15 cm

other side = \(\frac { 3 }{ 8 }\) x 24 = 9 cm

∴ The four sides are 15 cm, 9 cm, 15 cm, 9 cm ( opposite sides are equal)

Question 9.

The diagonals of the quadrilateral are perpendicular to each other. Is such a quadrilateral

always a rhombus? Draw a rough figure to justify your answer.

Solution:

Given : Diagonals are perpendicular.

If the diagonals are perpendicular then the

quadrilateral need not necessarily be a rhombus.

Look at the figure,

the diagonals are perpendicular to each other.

But all the sides are not equal.

PR ⊥ QS but PQ ≠ QR ≠ RS ≠ SP

Question 10.

ABCD is a trapezium in which \(\overline{\mathrm{AB}} \| \overline{\mathrm{DC}}\) If ∠A = ∠B =30°, what are the measures of the other two angles?

Solution:

Given : ABCD is a trapezium.

∠A = ∠B = 30°

Since AB//CD,

∠A + ∠D = ∠B + ∠C = 180° (interior angles on the same side of a

transversal are supplementary)

30° + ∠D = 180°

∠D = 180° – 300°

= 150°

and 30° + ∠C = 180°

∠C = 180° – 30’

= 150°

Question 11.

Fill in the blanks.

(i) A parallelogram in which two adjacent sides are equal is a ………………….

(ii) A parallelogram in which one angle is 90° and two adjacent sides are equal is a ………………….

(iii) IntrapeziurnABCD,\(\overline{\mathrm{AB}} \| \overline{\mathrm{DC}}\) . If ∠D = x° then ∠A = ………………

(iv) Every diagonal in a parallelogram divides it into …………………. triangles.

(v) In parallelogram ABCD, its diagonals \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{BD}}\) intersect atO. IfAO = 5cm then AC = …………….. cm.

(vi) In a rhombus ABCD, its diagonals intersect at ‘O’. Then ∠AOB = …………. degrees.

(vii) ABCD is a parallelogram then ∠A — ∠C = ………………… degrees.

(viii) In a rectangle ABCD, the diagonal AC =10cm then the diagonal BD = …………… cm.

(ix) In a square ABCD, the diagonal \(\overline{\mathrm{AC}}\) is drawn. Then ∠BAC = …………

Solution:

(i) A parallelogram in which two adjacent sides are equal is a rhombus

(ii) A parallelogram in which one angle is 90° and two adjacent sides are equal is a square

(iii) IntrapeziurnABCD,\(\overline{\mathrm{AB}} \| \overline{\mathrm{DC}}\) . If ∠D = x° then ∠A = 180° – x

(iv) Every diagonal in a parallelogram divides it into two congruent triangles.

(v) In parallelogram ABCD, its diagonals \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{BD}}\) intersect atO. If AO = 5cm then AC = 10 cm.

(vi) In a rhombus ABCD, its diagonals intersect at ‘O’. Then ∠AOB = 90°

(vii) ABCD is a parallelogram then ∠A — ∠C = zero degrees.

(viii) In a rectangle ABCD, the diagonal AC =10cm then the diagonal BD = 10 cm.

(ix) In a square ABCD, the diagonal \(\overline{\mathrm{AC}}\) is drawn. Then ∠BAC = 45°