Question : In the given figure, ABCD is a rectangle and P is a point on DC such that BC = 24 cm, DP = 10 cm, and CD = 15 cm. If AP produced intersects BC produced at Q, then find the length of AQ.
Option 1: 24 cm
Option 2: 26 cm
Option 3: 39 cm
Option 4: 35 cm
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Correct Answer: 39 cm
Solution :
Given: In the given figure, ABCD is a rectangle and P is a point on DC such that BC = 24 cm, DP = 10 cm, and CD = 15 cm.
Two triangles are similar if their corresponding angle values are the same.
The ratio of the corresponding sides of two triangles is the same if they are similar.
In $\triangle ADP$ and $\triangle QCP$,
$\angle ADP=\angle QCP=90^{\circ}$
$\angle APD=\angle QPC$ (vertically opposite angles are equal)
Since $AD||BQ$, $\angle PAD=\angle PQC$ (alternate angles are equal).
$\triangle ADP\sim\triangle QCP$
⇒ $\frac{AD}{QC}=\frac{DP}{CP}=\frac{AP}{PQ}$
In $\triangle APD$, $(AD)^2+(DP)^2=(PA)^2$.
$(PA)^2=(24)^2+(10)^2$
$(PA)^2=576+100=676$
$PA=\sqrt{676}=26$ cm
⇒ $\frac{24}{QC}=\frac{10}{5}=\frac{26}{PQ}$
⇒ QP = 13 cm
The length of AQ = AP + PQ = 26 + 13 = 39 cm.
Hence, the correct answer is 39 cm.
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