Question

7. ABC is a right triangle, right-angled at B. D and E
trisect BC, prove that 8(AE)^2 - 3(AC)^2 + 5(AD)^2
8. If one diagonal of a trapezium divides the other
diagonal in the ratio 1:3. Prove that one of the
parallel sides is three times the other.​

Answer 1

Question:

ABC is a right triangle, right angled at B. D and E trisect the side BC. Prove that 8 AE² = 3 AC² + 5 AD²

Answer:

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

∠B = 90°

 BD = DE = EC

$\Large{\underline{\underline{\bf{To\:Prove:}}}}$

8 AE² = 3 AC² + 5 AD²

$\Large{\underline{\underline{\bf{Proof:}}}}$

→ Let BD = DE = EC be x

→ Hence BE = 2 x, BC = 3x

→ Consider ΔABD

  AD² = AB² + BD² (By pythagoras theorem)

  AD² = AB² + x²------equation 1

→ Consider ΔABE

  AE² = AB² + BE² (By pythagoras theorem)

  AE² = AB² + (2x)²

  AE² = AB² + 4x²------equation 2

→ Consider ΔABC

  AC² = AB² + BC²

 AC² = AB² + (3x)²

 AC² = AB² + 9x²-------equation 3

→ Consider the RHS of the equation,

   3 AC² + 5AD²

→ From equation 1 and 3 substitute the values of AC² and AD²

  3AC² + 5 AD² = 3 (AB² + 9x²) + 5(AB² + x²)

  3AC² + 5 AD² = 3 AB² + 27 x² + 5 AB² + 5 x²

  3AC² + 5 AD² = 8 AB² + 32 x²

  3AC² + 5 AD² = 8(AB² + 4x²)

→ From equation 2, AB² + 4x² = AE²

→ Hence,

   3AC² + 5 AD² = 8 AE²

→ Hence proved.

Question:

If one diagonal of a trapezium, divides the other diagonal in the ratio 1 : 3, prove that one of the parallel sides is three times the other.

Answer:

$\Large{\underline{\underline{\bf{Given:}}}}$

• A trapezium ABCD, AB║CD
• Diagonals AD and BC, OA : OD = 1:3

$\Large{\underline{\underline{\bf{To\:Prove:}}}}$

• CD = 3 AB

$\Large{\underline{\underline{\bf{Proof:}}}}$

→ Consider Δ AOB and ΔDOC

   ∠AOB = ∠COD (Vertically opposite angles)

  ∠BAO = ∠ODC ( Alternate interior angles)

→ Hence ΔAOB $\sim$ ΔDOC (By AA criteria)

→ Therefore,

  AO/DO = AB/DC

→ But we know that AO : DO = 1:3

  1/3 = AB/DC

→ Cross multiplying,

   CD = 3 AB

→ Hence proved.

Answer 2

Answer:

Step-by-step explanation:

\Large{\underline{\underline{\bf{Given:}}}}

Given:

∠B = 90°

BD = DE = EC

\Large{\underline{\underline{\bf{To\:Prove:}}}}

ToProve:

8 AE² = 3 AC² + 5 AD²

\Large{\underline{\underline{\bf{Proof:}}}}

Proof:

→ Let BD = DE = EC be x

→ Hence BE = 2 x, BC = 3x

→ Consider ΔABD

AD² = AB² + BD² (By pythagoras theorem)

AD² = AB² + x²------equation 1

→ Consider ΔABE

AE² = AB² + BE² (By pythagoras theorem)

AE² = AB² + (2x)²

AE² = AB² + 4x²------equation 2

→ Consider ΔABC

AC² = AB² + BC²

AC² = AB² + (3x)²

AC² = AB² + 9x²-------equation 3

→ Consider the RHS of the equation,

3 AC² + 5AD²

→ From equation 1 and 3 substitute the values of AC² and AD²

3AC² + 5 AD² = 3 (AB² + 9x²) + 5(AB² + x²)

3AC² + 5 AD² = 3 AB² + 27 x² + 5 AB² + 5 x²

3AC² + 5 AD² = 8 AB² + 32 x²

3AC² + 5 AD² = 8(AB² + 4x²)

→ From equation 2, AB² + 4x² = AE²

→ Hence,

3AC² + 5 AD² = 8 AE²

→ Hence proved.

Question:

If one diagonal of a trapezium, divides the other diagonal in the ratio 1 : 3, prove that one of the parallel sides is three times the other.

Answer:

\Large{\underline{\underline{\bf{Given:}}}}

Given:

A trapezium ABCD, AB║CD

Diagonals AD and BC, OA : OD = 1:3

\Large{\underline{\underline{\bf{To\:Prove:}}}}

ToProve:

CD = 3 AB

\Large{\underline{\underline{\bf{Proof:}}}}

Proof:

→ Consider Δ AOB and ΔDOC

∠AOB = ∠COD (Vertically opposite angles)

∠BAO = ∠ODC ( Alternate interior angles)

→ Hence ΔAOB \sim∼ ΔDOC (By AA criteria)

→ Therefore,

AO/DO = AB/DC

→ But we know that AO : DO = 1:3

1/3 = AB/DC

→ Cross multiplying,

CD = 3 AB

→ Hence proved.