Question

Triangle ABC has an altitude BN which is 6 cm long. AN=9 cm and NC=4 cm. Show whether ABC is a right angled triangle. (Show all the working).

Answer 1

★ Step by step explanation :

• Here, we have triangle ABC with altitude BN which is 6 cm long. AN = 9 cm and NC = 4 cm. We had to show whether ABC is a right angled triangle.

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★ Using pythagoras in right angled ㅤㅤㅤ∆ABN and ∆BNC :

• According to pythagoras theorem, (Hypotenuse)² = (Perpendicular)² + (Base)²

Therefore,

• In ∆ABN :

$\qquad\longrightarrow\:\sf (AB)^2 = (BN)^2 + (NA)^2$

$\qquad\longrightarrow\:\sf (AB)^2 = (6)^2 + (9)^2$

$\qquad\longrightarrow\:\sf (AB)^2 = 36 + 81$

$\qquad\longrightarrow\:\bf{(AB)^2 = \red{117\:cm}}\qquad \lgroup 1 \rgroup$

• In ∆BNC :

$\qquad\longrightarrow\:\sf (BC)^2 = (NC)^2 + (BN)^2$

$\qquad\longrightarrow\:\sf (BC)^2 = (4)^2 + (6)^2$

$\qquad\longrightarrow\:\sf (BC)^2 = 16 + 36$

$\qquad\longrightarrow\:\bf{(BC)^2 = \red{52\:cm}}\qquad \lgroup 2 \rgroup$

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Now,

★ Finding length of AC :

$\qquad\leadsto\:\sf AC = NC + AN$

$\qquad\leadsto\:\sf AC = 4 + 9$

$\qquad\leadsto\:\bf{ AC = \red{13\:cm}}\qquad \lgroup 3 \rgroup$ㅤㅤㅤㅤㅤㅤ

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★ Verifying pythagoras theorem on ㅤㅤㅤ∆ABC :

$\qquad\longrightarrow\:\sf (AC)^2 = (AB)^2 + (BC)^2\qquad \lgroup 4 \rgroup$

★ From [1] , [2] and [3] , putting in [4] :

$\qquad\longrightarrow\:\sf (13)^2 = 117 + 52$

$\qquad\longrightarrow\:\bf{\red{169} = \red{169}}$

$\qquad\qquad\therefore\underline{\textsf{\textbf{Hence,\:Verified!}}}$

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∴ Hence, ABC is a right angled triangle