Question

In the figure, PC || QK and BC || HK. If AQ = 6cm, QH = 4cm, HP = 5cm and KC = 18 cm, find AK and PB.

Answer 1

The basic proportionality theorem states that if a line is drawn parallel to one side of a triangle and it intersects the other two sides at two distinct points then it divides the two sides in the same ratio.

In $△APC$, $QK||PC$, $AQ=6$ cm, $PQ=9$ cm and $KC=18$ cm.

Using the basic proportionality theorem, we have

$\frac{AQ}{PQ}=\frac{AK}{KC}\Rightarrow \frac{6}{9}=\frac{AK}{18}\Rightarrow 9AK=6\times 18\Rightarrow 9AK=108\Rightarrow AK=\frac{108}{9}=12$

Now, in $△ABC$, $HK||BC$, $AH=10$ cm, $AK=12$ cm and $CK=18$ cm.

Using the basic proportionality theorem, we have

$\frac{AH}{BH}=\frac{AK}{CK}\Rightarrow \frac{10}{BH}=\frac{12}{18}\Rightarrow 12BH=10\times 18\Rightarrow 12BH=180\Rightarrow BH=\frac{180}{12}=15$

Now, $PB=BH-PH=15-5=10$ cm

Hence, $AK=12$ cm and $PB=10$ cm.