In triangle A B C, D is a point on A C such that B D perp A C. If angle A B C=90°, A B=20 mathrm~cm, and B C=21 mathrm~cm, then find the length of B D (in mathrmcm ).

Question

In triangle A B C, D is a point on A C such that B D perp A C. If angle A B C=90°, A B=20 mathrm~cm, and B C=21 mathrm~cm, then find the length of B D (in mathrmcm ). (A) (440/29) (B) (420/29) (C) (410/29) (D) (210/29)

Tardigrade Admin · Accepted Answer

text In triangle A B D, A C2=A B2+B C2 =A C2=202+212=400+441=841=292 ⇒ A C=29 mathrm~cm ∴ (1/2) × 21 × 20=(1/2) × 29 × B D (∵ text Area =(1/2) b h) ⇒ B D=(21 × 20/29)=(420/29) mathrm~cm

Q. In $△ A BC, D$ is a point on $A C$ such that $B D ⊥ A C$ . If $∠ A BC = 9 0^{\circ}, A B = 20 cm$ , and $BC = 21 cm$ , then find the length of $B D$ (in $cm$ ).

$\frac{440}{29}$

$\frac{420}{29}$

$\frac{410}{29}$

$\frac{210}{29}$

$In △ A B D, A C^{2} = A B^{2} + B C^{2}$
$= A C^{2} = 2 0^{2} + 2 1^{2} = 400 + 441 = 841 = 2 9^{2}$
$\Rightarrow A C = 29 cm$
$∴ \frac{1}{2} \times 21 \times 20 = \frac{1}{2} \times 29 \times B D$ $(∵ Area = \frac{1}{2} bh)$
$\Rightarrow B D = \frac{21 \times 20}{29} = \frac{420}{29} cm$

Q. In △ABC,D is a point on AC such that BD⊥AC. If ∠ABC=90∘,AB=20 cm, and BC=21 cm, then find the length of BD (in cm ).

29440​

29420​

29410​

29210​