Karl Benedic, the president of Mathematics Club, proposed a project to put up a rectangular Math Garden whose lot perimeter is 36 meters. He was soliciting suggestions from the members for feasible dimensions of the lot.
Suppose you are a member of the club, what will you suggest to Karl
Benedic if you want a maximum lot area? You must convince him through a mathematical solution.
Consider the following guidelines:
1. Make an illustration of the lot with the needed labels.
2. Solve the problem.
Hint: Consider the formulas P = 21 + 2w for perimeter and A= lw for the area of the rectangle. Use the formula for P and the given information in the problem to express A in terms of either l or w.
3. Make a second illustration that satisfies the findings in the solution made in number 2.
4. Submit your solution on a sheet of paper with recommendations.
Answers
Answer:
Selina solutions
Grade 7
Biology
Mathematics
Physics
Chapters in CONCISE Mathematics - Middle School - 7
Exercises in Triangles
Question 2
Q2) Apply the properties of isosceles and equilateral triangles to find the unknown angles in the given figures :
Solution:
A = 70° (Angles opposite to equal sides)
But a + 70° + x = 180° (Angles of a triangle)
⇒ 70° + 70° + x = 180°
⇒ 140° + x = 180°
⇒ x = 180° - 140° = 40°
y = b (Angles opposite to equal sides)
But a = y + b
(Exterior angle of a triangle is equal to sum of its interior opposite angles)
⇒ 70° = y + y ⇒ 2y = 70°
⇒ y = 70°/2 = 35°
Hence x = 40°, y = 35°
(ii) In Fig. (ii),
In an equilateral triangle,
Each angle = 60°
In isosceles triangle.,
Let each base angle = a
∴ a + a + 100° = 180°
⇒ 2a + 100° = 180°
⇒ 2a = 180° – 100° = 80°
∴ a = 80°/2 = 40° ∴ x = 60° + 40° = 100°
And y = 60° + 40° = 100°
(iii) In Fig. (iii), 130° = x + p
(Exterior angle of a triangle is equal to the sum of its interior opposite angles)
∵ Lines are parallel (Given)
∴ p = 60° (Alternate angle)
And y = a
But a + 130° = 180° (Linear pair)
⇒ a = 180° – 130° = 50v ∴ y = 50°
And x + p = 130°
⇒ x + 60° = 130° ⇒ x = 130° – 60° = 70°
Hence, x = 70°, y = 50° and p = 60°
(iv) In Fig. (iv),
x = a + b
but b = y (Angles opposite to equal sides)
Similarly a = c
But a + c + 30° = 180°
⇒ a + a + 30° = 180° ⇒ 2a + 30° = 180°
⇒ 2a = 180° – 30° = 150°
⇒ a = 150°/2 = 75° and b + y = 90°
⇒ y + y = 90° ⇒ 2y = 90°
⇒ y = 90°/2 = 45° ⇒ b = 45°
Hence, x = a + b = 75° + 45° = 120°
And y = 45°
(v) In Fig.(v),
a + b + 40° = 180° (Angles of a triangle)
⇒ a + b = 180° – 40° = 140°
But a = b (Angles opposite to equal sides)
∴ a = b = 140°/2 = 70°
∴ x = b + 40° = 70° + 40° = 110°
(Exterior angle of a triangle is equal to the sum of its interior opposite angles)
Similarly y = a + 40°
= 70° + 40° = 110°
Hence, x = 110°, y = 110°
(vi) In the Fig. (vi)
a = b (Angles opposite to equal sides)
∴ y = 120°
But a + 120° = 180° (Linear pair)
⇒ a = 180° – 120° = 60°
∴ b = 60°
But x + a + b = 180° (Angles of a triangle)
⇒ x + 60° + 60° = 180°
⇒ x + 120° = 180°
∴ x = 180° – 120° = 60°
B = z + 25
(Exterior angle of a triangle is equal to the sum of its interior opposite angles)
⇒ 60° = z + 25°
⇒ z = 60° – 25° = 35°
Hence x = 60°, y = 120° and z = 35°
Answer:
your answer ppl mark me as brainlist
