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# Karl benedic, the president of mathematic 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 the member of the club, what will you suggest to karl benedic if you want a maximum lot area? you must convince him with a mathematical solution. consider the following guidelines: 1. make an illustration of the lot with the needed labels 2. solve the problem. 3. make a second illustration that satisfies the findings in the solution made in number 2 4. submit your paper on a sheet of paper with recommendations

• Réponse publiée par: hellcrack777

ADD MO MUNA AKO SA PEYS BOOK ❤✨

Name:Jhay Hanson Saycon

CHÀT TAYO SA MËŚŠÈNGÉR TANONG KA LAHAŤ NG GUSŤO MØNG ÌTÀŃØŃG SAKÉN SÁSÂGUTIN KO ❤✨

• Réponse publiée par: snow01

• Réponse publiée par: batopusong81

Selina solutions

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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°

• Réponse publiée par: jbaningzzz 