Answer choice D.) is 12/13 Please help!!
1 answer:
You might be interested in
Answer:
Given: Two Isosceles triangle ABC and Δ PBC having same base BC. AD is the median of Δ ABC and PD is the median of Δ PBC.
To prove: Point A,D,P are collinear.
Proof:
→Case 1. When vertices A and P are opposite side of Base BC.
In Δ ABD and Δ ACD
AB= AC [Given]
AD is common.
BD=DC [median of a triangle divides the side in two equal parts]
Δ ABD ≅Δ ACD [SSS]
∠1=∠2 [CPCT].........................(1)
Similarly, Δ PBD ≅ Δ PCD [By SSS]
∠ 3 = ∠4 [CPCT].................(2)
But, ∠1+∠2+ ∠ 3 + ∠4 =360° [At a point angle formed is 360°]
2 ∠2 + 2∠ 4=360° [using (1) and (2)]
∠2 + ∠ 4=180°
But ,∠2 and ∠ 4 forms a linear pair i.e Point D is common point of intersection of median AD and PD of ΔABC and ΔPBC respectively.
So, point A, D, P lies on a line.
CASE 2.
When ΔABC and ΔPBC lie on same side of Base BC.
In ΔPBD and ΔPCD
PB=PC[given]
PD is common.
BD =DC [Median of a triangle divides the side in two equal parts]
ΔPBD ≅ ΔPCD [SSS]
∠PDB=∠PDC [CPCT]
Similarly, By proving ΔADB≅ΔADC we will get, ∠ADB=∠ADC[CPCT]
As PD and AD are medians to same base BC of ΔPBC and ΔABC.
∴ P,A,D lie on a line i.e they are collinear.
The area of a trapezoid is:
A = area
a = base 1
b = base 2
h = height
Area = unknown
base 1 = 2.5
base 2 = 7.5
height = 2.5
^^^Plug these numbers into the formula!
A = (10/2)(2.5)
A = (5)(2.5)
A = 12.5
Hope this helped!