Question

ABC is a triangle in which AB=AC=4 cm and angle A =90°. calculate the area of triangle ABC .also find the length of the perpendicular from A to BC

Answer 1

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

Answer 2

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

Hope this helps