Can u please solve it quickly, and send a pic.
The shaded portion is, ABPC, while point P is at the right angle, I hope u got what I mean.
plz...
help me
Answers
Answer:
4 root 21
Step-by-step explanation:
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Answer:
Area of shaded region is 8(√21 -2) or 20.66 cm²
Step-by-step explanation:
Given
Isosceles ΔABC with altitude AD such that
AB = AC = 10 cm, BC = 8 cm. Another ΔPBC where P lie on AD such that ∠BPC = 90º
Calculation of sides
In ΔABD and ΔACD
AB = AC (given)
∠ABD = ∠ACD(Isosceles triangle)
∠ADB = ∠ADC ( = 90º)
By AAS congruent criteria
ΔABD ≅ ΔACD
⇒ BD = CD
Since BC = 8 cm
BD = CD = 4 cm............(1)
Hence AD is also a median.
In ΔPBD and ΔPCD
BD = CD (by 1)
∠PDB = ∠PDC ( = 90º)
PD = PD
By AAS congruent criteria
ΔPBD ≅ ΔPCD
Hence
∠DPB = ∠DPC
But ∠BPC = 90º
⇒ ∠DPB + ∠DPC = 90º
⇒ 2∠DPB = 90º
⇒ ∠DPB = ∠DPC = 45º
In ΔPBD
∠DPB + ∠PDB + ∠PBD = 180º
⇒ 45º + 90º + ∠PBD = 180º
⇒ ∠PBD = 45º
So, ∠PBD = ∠DPB
⇒ ΔPBD is isosceles.
⇒ PD = BD
Since BD = 4 cm from (1)
PD = 4 cm
Calculation of area
By Heron's Formula
ar(ΔABC) =√(s)(s-a)(s-b)(s-c)
S = (10+10+8)/2 = 14
s-a = 14-8 = 6
s-b = 14-10 = 4
s-c = 14-10 = 4
ar(ΔABC) =√(14.6.4.4)
= 2*4√(7*3)
= 8√21 cm²
ar(ΔPBC) = (1/2)*BC*PD
= (1/2)*8*4
= 16 cm²
Area of shaded region
= 8√21 - 16
= 8(√21 -2)
= 20.66 cm²