**Calculating the Areas and Perimeters of Triangles**

**Area of a triangle: A = (1/2)bh**

**b is the base of the triangle **

**h is the height of the triangle**

**Area of an equilateral triangle: A = (sqrt(3)/4)s²**

**Perimeter of a triangle: P = a + b + c**

**perimeter of an equilateral triangle: P = 3 x side length**

**Question 1:**

**Find the area of a triangle with base 6 cm and height 8 cm.**

**Solution: **

**A = (1/2)bh **

**A= (1/2)(6 cm)(8 cm) = 24 cm².**

**Requested by: Zaib**

**Question 2:**

**Find the perimeter of a triangle with sides of length 3 cm, 4 cm, and 5 cm.**

**Solution: **

**P = a + b + c **

**P = 3 cm + 4 cm + 5 cm = 12 cm.**

**Requested by: Zaib**

**Question 3:**

**Find the perimeter of a triangle with sides of length 12 cm, 15 cm, and 18 cm.**

**Solution: **

**P = a + b + c = **

**P = 12 cm + 15 cm + 18 cm = 45 cm.**

**Requested by: Ameer**

**Question 4:**

**Find the area of an equilateral triangle with side length 10 cm.**

**Solution: **

**A = (sqrt(3)/4)s² **

**A = (sqrt(3)/4)(10 cm)² = 25(sqrt(3)) cm².**

**Requested by: Saroop**

**Question 5:**

**Find the perimeter of an isosceles triangle with sides of length 6 cm, 6 cm, and 8 cm.**

**Solution: **

**P = a + b + c **

**P= 6 cm + 6 cm + 8 cm = 20 cm.**

**Requested by: Naeem**

**Question 6:**

**Find the area of an isosceles triangle with base 8 cm and height 6 cm.**

**Solution:**

**Area = (1/2) x base x height**

**Area = (1/2) x 8 cm x 6 cm**

**Area = 24 cm²**

**Requested by: Naeem**

**Question 7:**

**Find the perimeter of a right triangle with hypotenuse 10 cm and one leg of length 6 cm.**

**Solution:**

**Let the other leg of the triangle be x cm.**

**Using Pythagorean theorem, we have:**

**x² + 6² = 10²**

**x² + 36 = 100**

**x² = 64**

**x = 8 cm**

**Therefore, the perimeter of the triangle is:**

**Perimeter = 6 cm + 8 cm + 10 cm**

**Perimeter = 24 cm**

**Requested by: Kantesh**

**Question 8: **

**Find the area of a triangle with a base of 10 cm and a height of 6 cm.**

**Solution:**

**Area = (1/2)bh = (1/2) * 10 cm * 6 cm = 30 cm^2**

**Requested by: Ali Raza**

**Question 9: **

**Find the area of a triangle with side lengths of 6 cm, 8 cm, and 10 cm.**

**Solution:**

**This is a special triangle known as a “right triangle”, since one of its angles measures 90 degrees. **

**We can use the formula for the area of a triangle and the Pythagorean theorem to find its height:**

**Area = (1/2)bh**

**b = 6 cm (the length of the base)**

**c = 10 cm (the length of the hypotenuse)**

**a = sqrt(c^2 – b^2) = sqrt(10^2 – 6^2) = 8 cm (the length of the height)**

**Therefore, the area of the triangle is:**

**Area = (1/2)bh = (1/2) * 6 cm * 8 cm = 24 cm^2**

**Requested by: Saramd**

**Question 10: **

**Find the perimeter of a triangle with side lengths of 7 cm, 8 cm, and 9 cm.**

**Solution:**

**Perimeter = sum of all sides = 7 cm + 8 cm + 9 cm = 24 cm**

**Requested by: Sarmad**

**Question 11:**

**Find the perimeter of an isosceles triangle with base length 6 cm and height 8 cm.**

**Solution:**

**Since this is an isosceles triangle, we know that the two other sides are equal. **

**Let’s call their length x.**

**Using the Pythagorean theorem, we can find x:**

**x^2 = 8^2 + (1/2 * 6)^2 = 64 + 9 = 73**

**x = sqrt(73) ≈ 8.5 cm**

**The perimeter of the triangle is:**

**Perimeter = base + 2 equal sides = 6 cm + 2(8.5 cm) = 23 cm.**

**Requested by: Sarmad**

**Question 12:**

**Find the perimeter of a triangle with sides of length 6 cm, 8 cm, and 10 cm.**

**Solution:**

**Perimeter = 6 cm + 8 cm + 10 cm**

**Perimeter = 24 cm**

**Requested by: Hallar**

**Question 13:**

**Find the area of a right triangle with legs of length 3 cm and 4 cm.**

**Solution:**

**Area = (1/2) x base x height**

**Area = (1/2) x 3 cm x 4 cm**

**Area = 6 cm²**

**Requested by: Sarmad**

**Question 14:**

**Find the perimeter of an equilateral triangle with sides of length 5 cm.**

**Solution:**

**Perimeter = 3 x side length**

**Perimeter = 3 x 5 cm**

**Perimeter = 15 cm**

**Requested by: Sarmad**

**Question 15:**

**Find the perimeter of a right triangle with legs of length 5 cm and 12 cm.**

**Solution:**

**Using Pythagorean theorem, we have:**

**hypotenuse = sqrt(5² + 12²) = 13 cm**

**Perimeter = 5 cm + 12 cm + 13 cm**

**Perimeter = 30 cm**

**Requested by: Kabeer**

**Question 16:**

**Find the area of a triangle with sides of length 3 cm, 4 cm, and 5 cm.**

**Solution:**

**This is a right triangle, so we can use the formula for the area of a right triangle:**

**Area = (1/2) x base x height**

**Area = (1/2) x 3 cm x 4 cm**

**Area = 6 cm²**

**Requested by: Kabeer**

**Question 17:**

**Find the perimeter of a triangle with sides of length 12 cm, 16 cm, and 20 cm.**

**Solution:**

**Perimeter = 12 cm + 16 cm + 20 cm**

**Perimeter = 48 cm**

**Requested by: Faizan**

**Question 18:**

**Find the perimeter of a triangle with sides of length 9 cm, 12 cm, and 15 cm.**

**Solution: P = a + b + c = 9 cm + 12 cm + 15 cm = 36 cm.**

**Requested by: Faizan**

**Question 19: **

**Find the area of an equilateral triangle with side length 8 cm.**

**Solution: An equilateral triangle has all sides equal, so the height can be found using the Pythagorean theorem: **

**h = sqrt(8 cm² – (4 cm²)) = sqrt(48) cm = 4sqrt(3) cm. **

**Therefore, A = (1/2)bh = (1/2)(8 cm)(4sqrt(3) cm) **

**A = 16sqrt(3) cm².**

**Requested by: Faizan**

**Question 20:**

**Find the area of a right triangle with legs of length 3 cm and 4 cm.**

**Solution: A = (1/2)bh = (1/2)(3 cm)(4 cm) = 6 cm².**

**Requested by: Hamid**

**Question 21:**

**Find the perimeter of a right triangle with legs of length 5 cm and 12 cm.**

**Solution: **

**The hypotenuse can be found using the Pythagorean theorem: **

**c = sqrt(5 cm² + 12 cm²) = 13 cm. **

**Therefore, P = a + b + c = 5 cm + 12 cm + 13 cm **

**P = 30 cm.**

**Requested by: Hamid**

**Question 22:**

**Find the area of a triangle with base 12 cm and height 9 cm.**

**Solution: A = (1/2)bh = (1/2)(12 cm)(9 cm) = 54 cm².**

**Requested by: Ali Nawaz**

**Question 23:**

**Find the perimeter of a triangle with sides of length 8 cm, 10 cm, and 12 cm.**

**Solution: P = a + b + c = 8 cm + 10 cm + 12 cm = 30 cm.**

**Requested by: Ali Nawaz**

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