LEVEL-1
AB is a segment. The point P is on the
perpendicular bisector of segment AB such
that length of AP exceeds length of AB by
7 cm. If the perimeter of AABP is 38 cm,
find the sides of AABP.
Answers
Answer:
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Given:-
AB is a segment, the point P is the perpendicular bisector of segment AB.
The length of the AP exceeds the length of AB by 7cm.
The perimeter of ∆ABP is 38cm.
Find:-
Sides of the ∆ABP.
Assume:-
That perpendicular P lies on the segment AB at point C.
Solution:-
[ Refer the attachment for the figure. ]
AP exceeds the length AB by 7 cm.
According to question and figure -
AP = BP = y
AB = x
AC = CB = x/2
According to the question,
→ y = x + 7 ....(1)
Also, given that the perimeter of the triangle is 38 cm.
The perimeter of the triangle is the sum of it's all sides.
i.e.
→ y + y + x = 38
→ 2y + x = 38
→ 2(x + 7) + x = 38 [From (1)]
→ 2x + 14 + x = 38
→ 3x = 24
→ x = 8
Substitute value of x = 8 in (eq 1)
→ y = 8 + 7
→ y = 15
As -
AP = BP = y
AB = x
So,
→ AP = BP = 15 cm
→ AB = 8 cm
•°• Sides of the triangle are 8cm, 15cm and 15cm
here's your solution mate!
