in a right angled triangle ABC, angle B=90degree, BD is perpendicular to AC, and AD:CD=3:2. then prove that 2BC^2=5CD^2....
Answers
In a right angled triangle ABC, angle B=90degree, BD is perpendicular to AC, and AD:CD=3:2.
Given,
∠ B = 90°
BD ⊥ AC
AD:CD=3:2
In Δ ABD
BD² = AB² - AD² .........(1)
In Δ CBD
BD² = BC² - CD² ...........(2)
adding (1) and (2), we get
2BD² = AB² - AD² + BC² - CD²
2BD² = (AB² + BC²) - (AD² + CD²)
using Pythagoras theorem and given data, we get,
2BD² = AC² - (3² + 2²)
2BD² = (3 + 2)² - (3² + 2²)
2BD² = 25 - 13
2BD² = 12
BD² = 6
Now, consider,
Δ DBC
BC² = BD² + CD²
= 6 + 2²
= 6 + 4
= 10
∴ BC² = 10
To prove,
2BC^2=5CD^2
2 (10) = 5 (2)^2
20 = 5 × 4
20 = 20
Hence it is proved that, 2BC^2=5CD^2
In a right angled triangle ABC, angle B=90degree, BD is perpendicular to AC, and AD:CD=3:2.
Stepwise explanation is given below:
- It is given that,
∠ B = 90°
BD ⊥ AC
AD:CD=3:2
- In Δ ABD
BD² = AB² - AD² .........(1)
- In Δ CBD
BD² = BC² - CD² ...........(2)
adding (1) and (2), we get
2BD² = AB² - AD² + BC² - CD²
2BD² = (AB² + BC²) - (AD² + CD²)
- using Pythagoras theorem and given data, we get,
2BD² = AC² - (3² + 2²)
2BD² = (3 + 2)² - (3² + 2²)
2BD² = 25 - 13
2BD² = 12
BD² = 6
- Now, consider,
Δ DBC
BC² = BD² + CD²
= 6 + 2²
= 6 + 4
= 10
∴ BC² = 10
- We need to prove
2BC²=5CD²
2 (10) = 5 (2)²
20 = 5 × 4
20 = 20
- Hence it is proved that, 2BC²=5CD²