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Question 1: What is Heron's formula used for?
Solution:
Step 1: Heron's formula is used to find the area of a triangle.
Step 2: It is particularly useful when the height of the triangle is not known, but the lengths of all three sides are known.
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Question 2: What is the formula for the semi-perimeter (s) of a triangle with sides a, b, and c?
Solution:
Step 1: The perimeter of the triangle is the sum of its sides: $P = a + b + c$.
Step 2: The semi-perimeter is half of the perimeter. So, $s = \frac{a+b+c}{2}$.
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Question 3: Write down Heron's formula for the area of a triangle with sides a, b, and c.
Solution:
Step 1: First, calculate the semi-perimeter, $s = \frac{a+b+c}{2}$.
Step 2: The area ($\mathcal{A}$) is given by the formula: $\mathcal{A} = \sqrt{s(s-a)(s-b)(s-c)}$.
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Question 4: Find the semi-perimeter of a triangle with sides 13 cm, 14 cm, and 15 cm.
Solution:
Step 1: The sides are $a=13, b=14, c=15$.
Step 2: The semi-perimeter is $s = \frac{13+14+15}{2} = \frac{42}{2} = 21$ cm.
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Question 5: Can Heron's formula be used for an equilateral triangle?
Solution:
Step 1: Yes, it can. An equilateral triangle has all sides equal (let's say side 'a').
Step 2: So, $s = \frac{3a}{2}$.
Step 3: Area = $\sqrt{\frac{3a}{2}(\frac{3a}{2}-a)(\frac{3a}{2}-a)(\frac{3a}{2}-a)} = \sqrt{\frac{3a}{2} \cdot \frac{a}{2} \cdot \frac{a}{2} \cdot \frac{a}{2}} = \frac{\sqrt{3}a^2}{4}$. This is the standard formula for the area of an equilateral triangle.
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Question 6: The perimeter of a triangle is 30 cm and two of its sides are 10 cm and 12 cm. Find the third side.
Solution:
Step 1: Perimeter = $a+b+c = 30$ cm.
Step 2: Given $a=10, b=12$.
Step 3: $10+12+c = 30 \implies 22+c=30 \implies c = 8$ cm.
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Question 7: What is the area of a right-angled triangle with base 3 cm and height 4 cm?
Solution:
Step 1: The standard formula for the area of a right-angled triangle is $\frac{1}{2} \times \text{base} \times \text{height}$.
Step 2: Area = $\frac{1}{2} \times 3 \times 4 = 6$ cm$^2$.
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Question 8: Find the area of a triangle with sides 3, 4, and 5.
Solution:
Step 1: The sides are $a=3, b=4, c=5$.
Step 2: Calculate the semi-perimeter: $s = \frac{3+4+5}{2} = \frac{12}{2} = 6$.
Step 3: Use Heron's formula: Area = $\sqrt{6(6-3)(6-4)(6-5)} = \sqrt{6 \cdot 3 \cdot 2 \cdot 1} = \sqrt{36} = 6$ square units. (Note: this is a right-angled triangle).
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Question 9: The sides of a triangular plot are in the ratio 3:5:7 and its perimeter is 300 m. Find its area.
Solution:
Step 1: Let the sides be $3x, 5x, 7x$. Perimeter = $3x+5x+7x = 15x = 300 \implies x=20$.
Step 2: The sides are $a=60, b=100, c=140$.
Step 3: Semi-perimeter $s = 300/2 = 150$.
Step 4: Area = $\sqrt{150(150-60)(150-100)(150-140)} = \sqrt{150 \cdot 90 \cdot 50 \cdot 10} = \sqrt{6750000} = 1500\sqrt{3}$ m$^2$.
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Question 10: Find the area of an equilateral triangle with a side of 10 cm.
Solution:
Step 1: Using the direct formula for equilateral triangle area: $\frac{\sqrt{3}}{4} \times side^2$.
Step 2: Area = $\frac{\sqrt{3}}{4} \times 10^2 = \frac{100\sqrt{3}}{4} = 25\sqrt{3}$ cm$^2$.
Alternatively, using Heron's Formula:
Step 1: Sides are $a=10, b=10, c=10$.
Step 2: Semi-perimeter $s = \frac{10+10+10}{2} = 15$.
Step 3: Area = $\sqrt{15(15-10)(15-10)(15-10)} = \sqrt{15 \cdot 5 \cdot 5 \cdot 5} = \sqrt{3 \cdot 5 \cdot 5 \cdot 5 \cdot 5} = 25\sqrt{3}$ cm$^2$.