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Question
Prove that, in a right-angled triangle, the square of the hypotenuse is equal to the sum of the square of remaining two sides.
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Solution 1
Draw perpendicular BD from the vertex B to the side AC. A – D
In right-angled ΔABC
seg BD ⊥ hypotenuse AC.
∴ By the similarity in right-angled triangles
ΔABC ~ ΔADB ~ ΔBDC
Now, ΔABC ~ ΔADB
∴ `"AB"/"AD" = "AC"/"AB"` ...(c.s.s.t)
∴ AB2 = AC × AD ...(1)
Also, ΔABC ~ ΔBDC
∴ `"BC"/"DC" = "AC"/"BC"` ...(c.s.s.t)
∴ BC2 = AC × DC ...(2)
From (1) and (2),
AB2 + BC2 = AC × AD + AC × DC
= AC × (AD + DC)
= AC × AC ...(A – D – C)
∴ AB2 + BC2 = AC2
i.e., AC2 = AB2 + BC2
Solution 2
Given: In ΔPQR, ∠PQR = 90°.
To prove: PR2 = PQ2 + QR2.
Construction:
Draw seg QS ⊥ side PR such that P–S–R.
Proof: In ΔPQR,
∠PQR = 90° ...(Given)
Seg QS ⊥ hypotenuse PR ...(Construction)
∴ ΔPSQ ∼ ΔQSR ∼ ΔPQR ...(Similarity of right-angled triangles) ...(1)
ΔPSQ ∼ ΔPQR ...[From (1)]
∴ `"PS"/"PQ" = "PQ"/"PR"` ...(Corresponding sides of similar triangles are in proportion)
∴ PQ2 = PS × PR ...(2)
ΔQSR ∼ ΔPQR ...[From (1)]
∴ `"SR"/"QR" = "QR"/"PR"` ...(Corresponding sides of similar triangles are in proportion)
∴ QR2 = SR × PR ...(3)
Adding (2) and (3), we get
PQ2 + QR2 = PS × PR + SR × PR
∴ PQ2 + QR2 = PR(PS + SR)
∴ PQ2 + QR2 = PR × PR ...(P–S–R)
∴ PQ2 + QR2 = PR2 OR PR2 = PQ2 + QR2.
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