Question

In triangle ABC ,angle B=900 and BD is perpendicular to AC. If AC=17 cm and BD=4 cm then find the area of rectangle formed by AD and DC.

Answer 1

Answer:

$16cm^{2}$

Step-by-step explanation:

In ΔABC & ΔADB,

∠ABC = ∠ADB (= 90°)

∠BAC = ∠BAD (common)

∴ΔABC ~ ΔADB (AA similarity criterion)

⇒ $\frac{AB}{CB} = \frac{AD}{DB} = k (say)$                      (eqn1)  

⇒AD = DB * k = 4k (∵DB = 4)         (eqn2)

In ΔABC & ΔBDC,

∠BDC = ∠ABC (=90°)

∠BCA = ∠BCD (common)

∴ΔABC ~ ΔBDC (AA similarity criterion)

⇒$\frac{AB}{BC} = \frac{BD}{CD} = k$                              (from eqn1)

⇒$CD = \frac{BD}{k}$

⇒$CD = \frac{4}{k}$  (∵DB = 4)                      (eqn 3)

AD + CD = AC  

$4k + \frac{4}{k} = 17$      (from eqn (2) & eqn (3))

$4k^{2} - 17k + 4 = 0$

$(4k - 1)(k - 4) = 0$

$k = \frac{1}{4} , k = 4$

$when k = \frac{1}{4} , AD = 1, CD = 16$

$when k = 4, AD = 16, CD = 1$

∴area of rectangle formed by AD & DC = AD * CD = 16$cm^{2}$