Question

Two identical circles with same inside design as shown in the figure are to be made at the entrance. The identical triangular leaves are to be painted red and the remaining are to be painted green. Find the total area to be painted red.​

Answer 1

$\huge\mathfrak{Bonjour!!}$

$\huge\bold\purple{Solution:-}$

⏩ For one identical triangular leaf, let

a= 28 cm, b= 15 cm and c= 41cm.

Also,

s= a + b + c/2

= 28 + 15 + 41/2

= 84/2

= 42 cm.

Now, using Heron's formula,

Area of one triangle leaf

= √s(s-a)(s-b)(s-c)

= √42(42-28) (42-15) (42-41)

=√42 × 14 × 27 × 1

=√3 × 14 × 14 × 3 × 9

= 9 × 14

= 126 cm²

Now, as we see, there are 6 leaves in the circle.

So, total number of leaves in 2 circles

= 2 × 6 = 12

Therefore,

Area of 12 leaves = (12 × 126) cm²

= 1512 cm²

→ Hence, the total area to be painted red

= 1512 cm². [The required solution]

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Hope it helps...:-)

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Answer 2

Step-by-step explanation:

Solution:−

For one identical triangular leaf, let

a= 28 cm, b= 15 cm and c= 41cm.

Also,

s= a + b + c/2

= 28 + 15 + 41/2

= 84/2

= 42 cm.

Now, using Heron's formula,

Area of one triangle leaf

= √s(s-a)(s-b)(s-c)

= √42(42-28) (42-15) (42-41)

=√42 × 14 × 27 × 1

=√3 × 14 × 14 × 3 × 9

= 9 × 14

= 126 cm²

Now, as we see, there are 6 leaves in the circle.

So, total number of leaves in 2 circles

= 2 × 6 = 12

Therefore,

Area of 12 leaves = (12 × 126) cm²

= 1512 cm²

→ Hence, the total area to be painted red

= 1512 cm². [The required solution]