Pls give all the answers of the above attachment....
Answers
1) Shortest altitude of the triangle = 30cm.
Given :
Sides of triangle :
a = 50cm.
b = 78cm.
c = 112cm
To Find :
The smallest altitude of the triangle.
Solution :
Area of the triangle by Heron's formula.
★ √s(s-a)(s-b)(s-c)
Where,
★ s = a+b+c/2
We have,
• a = 50cm.
• b = 78 cm.
• c = 112cm.
=> s = 50+78+112/2 cm
=> 240/2 cm
=> 120cm
Area,
=> √120cm(120cm-50cm)(120cm-78cm)(120cm-112cm)
=> √120cm(70cm)(42cm)(8cm)
=> √8400cm² × 336cm²
=> √2822400cm⁴
=> 1680cm²
Area of triangle = 1/2 × base × height
We have,
• Area = 1680cm²
• base = 112cm.
[We take the longest base to find the shortest altitude.]
=> 1680cm² = 1/2 × 112cm × h
=> 1680cm² = 56cm × h
=> 1680cm²/56cm = h
=> 30cm = h
=> h = 30cm.
Hence, the shortest altitude is 30cm.
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2) Perimeter of an equilateral triangle = 36cm.
Given :
- Area of an equilateral triangle = 36√3cm²
To Find :
- Perimeter of an equilateral triangle.
Solution :
Area of an equilateral triangle = √3/4 (side)²
=> √3/4 (side)² = 36√3cm²
=> (side)² = 36√3cm² × 4/√3
=> (side)² = 36cm² × 4
=> (side)² = 144cm²
=> side = √144cm²
=> side = 12cm
Now,
Perimeter = 3 × side
=> 3 × 12cm
=> 36cm
Hence, the perimeter of an equilateral triangle is 36cm.
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3) Area of an isosceles triangle = 49.608cm².
Given :
- Perimeter of an isosceles triangle = 35cm.
- Length of equal sides = 10cm.
To Find :
- The area of an isosceles triangle.
Solution :
Perimeter = a + b + c
We have,
• Perimeter = 35cm.
• a = 10cm.
• b = 10cm.
• c = x.
=> 35cm = 10cm + 10cm + x
=> 35cm = 20cm + x
=> 35cm - 20cm = x
=> 15cm = x
=> c = 15cm
By heron's formula,
★ √s(s-a)(s-b)(s-c)
Where,
=> s = a+b+c/2
=> s = 10+10+15/2
=> s = 35/2
=> s = 17.5
Now,
=> √17.5cm(17.5cm-10cm)(17.5cm-10cm)(17.5cm-15cm)
=> √17.5cm(7.5cm)(7.5cm)(2.5cm)
=> √131.25cm²×18.75cm²
=> √2460.9375cm⁴
=> 49.608cm²
Hence, the area of an isosceles triangle is 49.608cm².
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4) The length of the altitudes of an equilateral triangle is 5√3cm.
Diagram in the attachment.
Given :
- The length of the sides of an equilateral triangle = 10cm.
To Find :
- The length of the altitudes of an equilateral triangle.
Solution :
By the Pythagoras theorem,
★ (H)² = (B)² + (P)²
Where,
• H = hypotenuse = AB = 10cm.
• B = base = BD = 5cm.
• P = perpendicular (altitude) = AD.
=> (AD)² = (BD)² + (AD)²
=> (10cm)² = (5cm)² + (AD)²
=> 100cm² = 25cm² + (AD)²
=> 100cm² - 25cm² = AD²
=> 75cm² = AD²
=> √75cm = AD
=> √3×5×5cm = AD
=> √3×(5)²cm = AD
=> 5√3cm = AD
Hence, the length of the altitudes of an equilateral triangle is 5√3cm.
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5) The area of the triangle = 20√14cm.
Given :
The difference between the semi perimeter and the sides of a triangle:
- s - a = 5cm.
- s - b = 7cm.
- s - c = 8cm.
To Find :
- The area of the triangle.
Solution :
First, we need to find the value of s.
We have,
• s - a = 5cm –––––(1)
• s - b = 7cm –––––(2)
• s - c = 8cm –––––(3)
Now, add all the equations.
=> (s - a) + (s - b) + (s - c) = 8cm + 7cm + 5cm
=> s - a + s - b + s - c = 20cm
=> 3s - a - b - c = 20cm
=> 3s - (a + b + c) = 20cm
[s = a+b+c/2 = a+b+c = 2s]
=> 3s - 2s = 20cm
=> s = 20cm
Using heron's formula,
★ √s(s-a)(s-b)(s-c)
=> √20cm(5cm)(7cm)(8cm)
=> √100cm² × 56cm²
=> √5600cm⁴
=> √2×2×2×2×2×5×5×7cm⁴
=> √(2)²×(2)²×2×(5)²×7cm⁴
=> 2×2×5√2×7cm²
=> 20√14cm²
Hence, the area of the triangle is 20√14cm².
