20. . The area of ∆ ABC is 8 cm2 in which AB = AC = 4 cm and ∠A = 90º.
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Answer:
Area of triangle ABC = 1/2 ×BASE × HEIGHT
= 1/2 × AB × AC
= 1/2 × 4× 4
= 2 × 4
= 8 cm square
LET AD IS PERPENDICULAR TO BC
In Triangle ADB AND ADC
AD = AD ( common)
angle ADC = angle ADB ( each 90 degree)
AB = AC (Given)
Triangle ADB (congruent to). triangle ADC ( RHS criteria)
BD = CD ( cpct )
in triangle ABC , BY Pythagoras theorem
ABsquare + ACsquare = BC square
4 square +4 square = BCsquare
16 +16 = BCsquare
32 = BC square
:. BC = 4√2 cm
BD + CD = BC
2BD = BC (. BD = CD )
2BD = 4√2
BD = 2√2 cm
now in triangle BAD , by Pythagoras theorem
ADsquare + BD square = ABsquare
ADsquare + 2√2 square = 4 square
ADsquare + 4×2 = 16
ADsquare + 8 = 16
ADsquare = 16- 8
ADsquare = 8
.: AD = √8
.: AD = 2√2 cm
Hence length of perpendicular from A to BC = length of AD = 2√2 cm
Step-by-step explanation:
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Answer:
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