QUADRILATERALS : Ganita Prakash | Grade 8 | Chapter-4 | Page 94 | Solutions

 "Figure it Out" 

1. Find all the other angles inside the following rectangles.

Key properties of a rectangle used for this problem:

• All four vertex angles are 90°.
• Diagonals are equal in length and bisect each other (meaning they cut each other into four equal segments).
• The four triangles formed by the intersection of the diagonals are isosceles.

---

(i) Rectangle ABCD

Given: In rectangle ABCD, the diagonals intersect at point O, and ∠OAB=30∘.

• In Triangle OAB:

  • Since diagonals of a rectangle are equal and bisect each other, OA=OB. Therefore, △OAB is an isosceles triangle.
  • This means ∠OBA=∠OAB=30∘.
  • The sum of angles in a triangle is 180°. So, ∠AOB=180∘−(30∘+30∘)=120∘.

• Angles at the intersection O:

  • ∠COD=∠AOB=120∘ (vertically opposite angles).
  • ∠BOC and ∠AOD are on a straight line with ∠AOB. So, ∠BOC=∠AOD=180∘−120∘=60∘.

• Other Triangles:

  • In △BOC: Since OB=OC and ∠BOC=60∘, it is an equilateral triangle. Thus, ∠OBC=∠OCB=60∘.
  • In △AOD: Since OA=OD and ∠AOD=60∘, it is also an equilateral triangle. Thus, ∠OAD=∠ODA=60∘.
  • In △COD: Since OC=OD, it is an isosceles triangle. ∠OCD=∠ODC. The vertex angle is ∠COD=120∘. So, ∠OCD=∠ODC=(180∘−120∘)/2=30∘.

Summary of all angles for (i):

• Angles at the intersection (O): ∠AOB=120∘, ∠BOC=60∘, ∠COD=120∘, ∠AOD=60∘.
• Angles along the sides:
  • ∠OBA=30∘
  • ∠OBC=60∘
  • ∠OCB=60∘
  • ∠OCD=30∘
  • ∠ODC=30∘
  • ∠ODA=60∘
  • ∠OAD=60∘
  • ∠OAB=30∘

---

(ii) Rectangle PQRS

Given: In rectangle PQRS, the diagonals intersect at point O, and ∠QOR=110∘.

• Angles at the intersection O:

  • ∠POS=∠QOR=110∘ (vertically opposite angles).
  • ∠POQ=∠ROS=180∘−110∘=70∘ (angles on a straight line).

• Analyzing the Triangles:

  • In △QOR: Since OQ=OR, it's an isosceles triangle. ∠OQR=∠ORQ=(180∘−110∘)/2=35∘.
  • In △POQ: Since OP=OQ, it's an isosceles triangle. ∠OPQ=∠OQP=(180∘−70∘)/2=55∘.
  • In △ROS: By symmetry with △POQ, ∠ORS=∠OSR=55∘.
  • In △POS: By symmetry with △QOR, ∠OPS=∠OSP=35∘.

Summary of all angles for (ii):

• Angles at the intersection (O): ∠QOR=110∘, ∠ROS=70∘, ∠SOP=110∘, ∠POQ=70∘.
• Angles along the sides:
  • ∠OPQ=55∘
  • ∠OQP=55∘
  • ∠OQR=35∘
  • ∠ORQ=35∘
  • ∠ORS=55∘
  • ∠OSR=55∘
  • ∠OSP=35∘
  • ∠OPS=35∘

2. Draw a quadrilateral whose diagonals have equal lengths of 8 cm that bisect each other, and intersect at a given angle.

A quadrilateral with diagonals that are equal and bisect each other is a rectangle. If these diagonals also intersect at a right angle (90°), it is a square.

General Drawing Method:

1. Draw a line segment of 8 cm. This will be the first diagonal (e.g., AC).
2. Find its midpoint (at the 4 cm mark).
3. Draw a second 8 cm line segment (e.g., BD) that passes through this midpoint, ensuring it intersects the first diagonal at the required angle (30°, 40°, 90°, or 140°).
4. Connect the four endpoints of the diagonals (A, B, C, D) to form the quadrilateral.

Resulting Shapes:

• For intersection angles of (i) 30°, (ii) 40°, and (iv) 140°, the resulting figure is a rectangle.
• For the intersection angle of (iii) 90°, the diagonals are equal, bisect each other, and are perpendicular. This is the definition of a square.

3. Consider a circle with centre O. Line segments PL and AM are two perpendicular diameters of the circle. What is the figure APML?

The figure APML is a square.

Reasoning:

1. Diagonals are Equal: PL and AM are both diameters of the same circle, so they are equal in length.
2. Diagonals Bisect Each Other: Both diameters pass through the center O, so they bisect each other.
3. Diagonals are Perpendicular: The problem states that the diameters are perpendicular.

A quadrilateral whose diagonals are equal, bisect each other, and are perpendicular is a square.

4. How do we make an exact 90° using two sticks of equal length and a thread?

This can be done by constructing a rectangle, as all angles in a rectangle are 90°.

Steps:

1. Sticks as Diagonals: Use the two sticks of equal length as the diagonals of a quadrilateral.
2. Placement and Intersection: Place the sticks so that they cross each other at their exact midpoints (i.e., they bisect each other).
3. Forming the Quadrilateral: Use the thread to connect the four endpoints of the sticks, forming the sides of the quadrilateral.

The resulting shape is a rectangle because its diagonals are equal in length and bisect each other. Therefore, all four angles of the figure formed by the thread will be exactly 90°.

5. Is every quadrilateral that has opposite sides parallel and equal, a rectangle?

No, that statement is not correct.

Explanation:

• A quadrilateral where opposite sides are parallel and equal is the definition of a parallelogram.
• A rectangle is a special type of parallelogram in which all four interior angles are right angles (90°).
• Therefore, while every rectangle is a parallelogram, not every parallelogram is a rectangle. The property of having opposite sides parallel and equal is necessary for a rectangle, but it is not sufficient to define one.

Counterexample:
A rhombus (that is not a square) or any slanted parallelogram has opposite sides that are parallel and equal, but its angles are not 90°. These are parallelograms, but not rectangles.

To define a rectangle, you must add another condition, such as:

• A parallelogram with one right angle.
• A parallelogram with equal diagonals.

Continue Your Grade 8 Maths Journey!

Congratulations on working through these challenging problems! These exercises build a strong foundation for understanding exponents and roots. If you are ready to move on, check out the next section of your curriculum.

➡️ Click here for Chapter 4 Solutions Page 102