Question

5.
In ∆ABC, if r1=12, r2= 18,r3 = 36 then the length of the altitude through A is
a) 32
b) 26
c) 18
d) 24​

Answer 1

Length of altitude through A is 24 units.

Step-by-step explanation:

From given r₁=12, r₂=18,r₃=36 we have the relation to find r as $\frac{1}{r} =\frac{1}{r1} +\frac{1}{r2} +\frac{1}{r3}$

⇒$\frac{1}{r} =\frac{1}{12} +\frac{1}{18} +\frac{1}{36}$

⇒$\frac{1}{r} =\frac{1}{6}$

⇒$r=6$

Now, r.r₁.r₂.r₃=Δ²

⇒Δ²=6*12*18*36=46656

⇒Δ=216

Also consider, r=Δ/s

⇒$s=\frac{216}{6} =36$

similarly, consider r₂=Δ/(s-b)

⇒$b=\frac{-216}{18} +s=-12+36=18$

⇒area Δ=$\frac{1}{2} *b*altitude through A$

⇒Altitude through A= $\frac{2*216}{18}=24$

Answer 2

Given:

• $r_1$ = 12
• $r_2$ = 18
• $r_3$ = 36

To Find:

• The length of the altitude through A.

Solution:

• First, we need to find the 'r' value.
• The formula is given by, $\frac{1}{r} = \frac{1}{r_1} +\frac{1}{r_2} +\frac{1}r_3}$
• Substitue the values in the above formula.
• We get, $\frac{1}{r} =\frac{1}{12} +\frac{1}{18} +\frac{1}{36}$ = $\frac{6}{36} = \frac{1}{6}$
• ∴ r = 6
• Next to find Δ value we have a formula, $delta^2 = r.r_1.r_2.r_3$
• Δ = √$r.r_1.r_2.r_3$ = √6*18*12*36  = √46656
• ∴Δ = 216
• Now to find s, r = Δ/s
• s = Δ*r = 216/6 = 36
• ∴ s = 36
• To find b, $r_2$ = Δ/s-b
• b = (Δ/$r_2$ ) + s = (-216/18)+36
• b = -12+36
• ∴ b = 18
• We found the values of s, b, and Δ to calculate the altitude through A.
• Δ = 1/2*b*altitude through A.
• Altitude through A = (2*Δ)/b = (2*216)/18 = 432/18
• Altitude through A = 24

∴ The altitude through A = 24.(option d)