Question

In Fig.10.15 , the angle BAC is a right angle and AD is perpendicular to BC, AB=4cm, AC=3cm, and BD =x . Calculate x. l want to know the value of x.​

Answer 1

Given:-

• ∠BAC = 90°
• AD ⊥ BC
• AB = 4cm
• AC = 3cm
• BD = x cm

To Find:-

• The value of x

Solution:-

Let DC = y cm.

✏ In triangle ABD,

AB² = AD² + BD² ㅤㅤ[By Pythagoras Theorem]

(4cm)² = AD² + (xcm)²

AD² = (4)² - (x)²

AD² = 16 - x² ㅤㅤㅤㅤ.......eq.(1)

✏ In triangle ABC,

AC² = AD² + DC² ㅤㅤ[By Pythagoras Theorem]

(3)² = AD² + y²

AD² = (3)² - (y)²

AD² = 9 - y² ㅤㅤㅤㅤ.......eq.(2)

✏ From eq.(1) and eq.(2),

AD² = AD²ㅤㅤ[By Pythagoras Theorem]

16 - x² = 9 - y²

16-9 = -y² + x²

7 = x² - y² ㅤㅤㅤㅤ.......eq.(3)

✏ In triangle ABC,

BC² = AB² + AC²ㅤㅤ[By Pythagoras Theorem]

(x+y)² = (4)² + (3)²

(x+y)² = 25

(x+y)² = 25

x+y = √25

x+y = 5 ㅤㅤㅤㅤ.......eq.(4)

✏ From eq.(3),

7 = x² - y²

7 = (x+y)(x-y)

7 = 5(x-y) ㅤㅤㅤㅤ[from eq.(4)]

(x-y) = 7/5ㅤㅤㅤㅤ.......eq.(5)

✏ Solving eq.(4) and eq.(5),

x + y + (x-y) = 5 + (7/5)

2x = 5 + 7/5

2x = 32/5

$\small{ \sf{ \small{\boxed{\boxed{ \sf{ \purple{x = \frac{16}{5} }}}}}}}$

$\bf \therefore\underline{ \: the \: value \: of \: x \: is \: \frac{16}{5}} .$