Question

The altitude AP of ∆ABC is 24 cm long. If the ratio PB : PC is 9 : 16, and the area of ∆ABC is 600 cm². What is the value of AB + AC?​

Answer 1

Step-by-step explanation:

In ΔABC, AD=6cm and BC=9cm

Area of triangle=

2

1

×base×height=

2

1

×BC×AD

=

2

1

×9×6=27 cm

2

Again, Area of triangle=

2

1

×base×height=

2

1

×AB×CE

⇒27=

2

1

×7.5×CE

⇒ CE=

7.5

27×2

⇒ CE=7.2cm

Thus, height from C to AB i.e., CE is 7.2cm

Answer 2

$\large\underline{\sf{Solution-}}$

Given that

• The altitude AP of ∆ABC is 24 cm long.

• The ratio PB : PC is 9 : 16.

• The area of ∆ABC is 600 cm².

Let assume that

• PB = 9x cm

• PC = 16x cm

So, that BC = PB + PC = 9x + 16x = 25x cm

So, we have

• Base of ∆ABC = 25x cm

• Height of ∆ABC, AP = 24 cm

• Area of ∆ABC = 600 cm²

Now, We know that

$\red{\rm :\longmapsto\:\boxed{\tt{ Area_{(\triangle ABC)} = \frac{1}{2} \times base \times height \: }}}$

So, on substituting the values, we get

$\rm :\longmapsto\:600 = \dfrac{1}{2} \times 25x \times 24$

$\rm :\longmapsto\:50 \: \: \: \cancel{600} = 25x \times \cancel{12}$

$\rm :\longmapsto\:50 = 25x$

$\rm \implies\:\boxed{\tt{ x = 2}}$

So,

PB = 9x = 9 × 2 = 18 cm

PC = 16x = 16 × 2 = 32 cm

Now, In right-angle ∆PBA

↝ Using Pythagoras Theorem, we have

$\rm :\longmapsto\: {BA}^{2} = {PB}^{2} + {PA}^{2}$

$\rm :\longmapsto\: {BA}^{2} = {18}^{2} + {24}^{2}$

$\rm :\longmapsto\: {BA}^{2} = 324 + 576$

$\rm :\longmapsto\: {BA}^{2} = 900$

$\rm \implies\:\boxed{\tt{ BA = 30 \: cm \: }}$

Now, In right-angle ∆PAC

↝ Using Pythagoras Theorem, we have

$\rm :\longmapsto\: {AC}^{2} = {PA}^{2} + {PC}^{2}$

$\rm :\longmapsto\: {AC}^{2} = {24}^{2} + {32}^{2}$

$\rm :\longmapsto\: {AC}^{2} = 576 + 1024$

$\rm :\longmapsto\: {AC}^{2} = 1600$

$\rm \implies\:\boxed{\tt{ AC = 40 \: cm}}$

Hence,

$\rm \implies\:\boxed{\tt{ AB + AC = 30 + 40 = 70 \: cm}}$