find the value of AB,BC,and BD IN THE GIVEN FIGURE WHICH IS AD = 16cm and CD= 9cm
Answers
Given :
- AD = 16 cm
- CD = 9cm
- ∆ADB & ∆CDB are two right triangle included in right triangle ∆ABC
- BD is common base between ∆ADB & ∆CDB
To find :
- AB, BC & BD
Solution :
- According to the Pythagoras theorem
★(Hypotenuse)²=(perpendicular)²+(base)²
- In ∆ABC
→ (AC)² = (AB)² + (BC)²
→ (16 + 9)² = AB² + BC²
→ (25)² = AB² + BC²
→ AB² + BC² = 625 ---(i)
- Now, In ∆ADB
→ (AB)² = (BD)² + (AD)²
→ AB² = BD² + (16)²
→ AB² - BD² = 256 -----(ii)
- In ∆CDB
→ (BC)² = (DC)² + (DB)²
→ BC² = (9)² + DB²
→ BC² - DB² = 81 -----(iii)
Subtract equations (ii) & (iii) the equations
→ AB² - BD² - (BC² - DB²) = 256 - 81
→ AB² - BD² - BC² + DB² = 175
→ AB² - BC² = 175 ------(iv)
Add equations (i) and (iv)
→ AB² + BC² + AB² - BC² = 625 + 175
→ 2AB² = 800
→ AB² = 800/2
→ AB = √400 = 20 cm
Put the value of AB in eqⁿ (i)
→ AB² + BC² = 625
→ (20)² + BC² = 625
→ 400 + BC² = 625
→ BC² = 625 - 400
→ BC = √225 = 15cm
Put the value AB in eqⁿ (ii)
→ AB² - BD² = 256
→ (20)² - BD² = 256
→ 400 - BD² = 256
→ BD² = 400 - 256
→ BD = √144 = 12 cm
•°• AB = 20cm, BC = 15cm & BD = 12cm
ANSWER :
According to the Pythagoras theorem :-
(Hypotenuse)²=(perpendicular)²+(base)²
In ∆ABC
(AC)² = (AB)² + (BC)²
(16 + 9)² = AB² + BC²
(25)² = AB² + BC²
AB² + BC² = 625 ---(i)
Now, In ∆ADB
(AB)² = (BD)² + (AD)²
AB² = BD² + (16)²
AB² - BD² = 256 -----(ii)
In ∆CDB
(BC)² = (DC)² + (DB)²
BC² = (9)² + DB²
BC² - DB² = 81 -----(iii)
Subtract equations (ii) & (iii) the equations
AB² - BD² - (BC² - DB²) = 256 - 81
AB² - BD² - BC² + DB² = 175
AB² - BC² = 175 ------(iv)
Add equations (i) and (iv)
AB² + BC² + AB² - BC² = 625 + 175
2AB² = 800
AB² = 800/2
AB = √400 = 20 cm
Put the value of AB in eqⁿ (i)
AB² + BC² = 625
(20)² + BC² = 625
400 + BC² = 625
BC² = 625 - 400
BC = √225 = 15cm
Put the value AB in eqⁿ (ii)
AB² - BD² = 256
(20)² - BD² = 256
400 - BD² = 256
BD² = 400 - 256
BD = √144 = 12 cm
AB = 20cm, BC = 15cm & BD = 12cm