Triangle ABC is a right triangle of AB = 18 cm, AC = 24 cm and BC = 30 cm.
The points P and Q are points on the bisectors of angle B and angle C, respectively.
PQ is parallel to BC. PH and AB, QK and AC are vertical respectively.
The area of pentagon AHPQK is half of the area of triangle ABC.
Find the length of PH and the length of PQ.
The points P and Q are points on the bisectors of angle B and angle C, respectively.
PQ is parallel to BC. PH and AB, QK and AC are vertical respectively.
The area of pentagon AHPQK is half of the area of triangle ABC.
Find the length of PH and the length of PQ.
Answer
The area of the remaining portion is also 108 cm2 .
As shown in Fig. 1, two lines vertical to BC is drawn from P and Q respectively and each intersection is set to D and E, respectively.
Thus, it also turns out that PH = PD = QE = QK.
When equivalence movement of △BPH and △CQK is carried out as shown in Fig. 2, a quadrangle HBCK will turn into a rectangle.
Since HP in Fig.1 = HB in Fig.2 = 3.6 cm according to (1), the height of △AFG is 14.4 - 3.6 = 10.8 cm.
The homothetic ratio of △ABC and △AFG is equal to the ratio of height.
As 14.4 : 10.8 = 4 : 3, FG = 30 × 3/4 = 22.5 cm.
Since △HFP and △ABC are homothetic, HP : FP = 24 : 30 = 4 : 5.
FP = 3.6 × 5/4 = 4.5 cm.
Moreover, △KQG and △ABC are homothetic, KQ : QG = 18 : 30 = 3 : 5.
QG = 3.6 × 5/3 = 6 cm.
Therefore, PQ = FG - (FP + QG) = 22.5 - (4.5 + 6) = 12 cm.
PH = 3.6 cm
PQ = 12 cm
PQ = 12 cm
Solution
Since the area of pentagon AHPQK is a half of the area of triangle ABC, it is 18 × 24 / 2 = 108 cm2.
The area of the remaining portion is also 108 cm2 .
As shown in Fig. 1, two lines vertical to BC is drawn from P and Q respectively and each intersection is set to D and E, respectively.
△BPH and △BPD, △CQK and △CQE become congruence, respectively.
Thus, it also turns out that PH = PD = QE = QK.
When equivalence movement of △BPH and △CQK is carried out as shown in Fig. 2, a quadrangle HBCK will turn into a rectangle.
Since the area of rectangle HBCK is also 108 cm2, PH in Fig.1 = HB in Fig.2 = 108 / 30 = 3.6 cm.
As shown in Fig. 3, the end of the line segment PQ is made to extend and an intersection with AB and AC is set to F and G. PQ = FG － (FP + QG).
As shown in Fig. 3, the end of the line segment PQ is made to extend and an intersection with AB and AC is set to F and G. PQ = FG － (FP + QG).
If BC is set as a base of △ABC, the height is 216 × 2 / 30 =14.4 cm.
Since HP in Fig.1 = HB in Fig.2 = 3.6 cm according to (1), the height of △AFG is 14.4 - 3.6 = 10.8 cm.
The homothetic ratio of △ABC and △AFG is equal to the ratio of height.
As 14.4 : 10.8 = 4 : 3, FG = 30 × 3/4 = 22.5 cm.
Since △HFP and △ABC are homothetic, HP : FP = 24 : 30 = 4 : 5.
FP = 3.6 × 5/4 = 4.5 cm.
Moreover, △KQG and △ABC are homothetic, KQ : QG = 18 : 30 = 3 : 5.
QG = 3.6 × 5/3 = 6 cm.
Therefore, PQ = FG - (FP + QG) = 22.5 - (4.5 + 6) = 12 cm.
