If the length of the base of a thin right triangular sheet is decreased by 1 cm and its perpendicular side is increased by 1 cm, the area of the triangular sheet remains unchanged. If the base and the perpendicular side each is increased by 1 cm, the resulting area increases by 4 cm square. Find the length of the base and the perpendicular side of the right triangle. ( Best answer would be marked as brainliest )
Answers
Answered by
16
Let the length of the base be B
let the length of the perpendicular be P
》Area = 0.5 × B × P (Equation 1)
Also since the in the first condition Area remains unchanged,
》A (Area) = 0.5 × (B-1) × (P+1) (Equation 2)
Second condition can be interpreted as
》A' (New Area) = A + 4 = 0.5 × (B+1) × (P+1) (Equation 3)
From Eq 2 and 3 :
》 0.5 × (B-1)(P+1) + 4 = 0.5 × (B+1)(P+1)
》 4 = 0.5 × (P+1) × (B+1 - (B-1))
》 4 = 0.5 × (P+1) × (2)
》 4 = P + 1
》 P = 3 cm
》 A = 0.5 × B × 3 and A = 0.5 × (B-1) × 4 (By putting the value of P in the equations 1 and 2)
》 0.5 × B × 3 = 0.5 × (B-1) × 4 (Equating RHS of the two bove equations)
》 3B = (B-1) × 4
》 3B = 4B - 4
》 B = 4 cm
Hence the answers are
☆》 Length of the base = 4cm
☆》 Length of the perpendicular side = 3cm
let the length of the perpendicular be P
》Area = 0.5 × B × P (Equation 1)
Also since the in the first condition Area remains unchanged,
》A (Area) = 0.5 × (B-1) × (P+1) (Equation 2)
Second condition can be interpreted as
》A' (New Area) = A + 4 = 0.5 × (B+1) × (P+1) (Equation 3)
From Eq 2 and 3 :
》 0.5 × (B-1)(P+1) + 4 = 0.5 × (B+1)(P+1)
》 4 = 0.5 × (P+1) × (B+1 - (B-1))
》 4 = 0.5 × (P+1) × (2)
》 4 = P + 1
》 P = 3 cm
》 A = 0.5 × B × 3 and A = 0.5 × (B-1) × 4 (By putting the value of P in the equations 1 and 2)
》 0.5 × B × 3 = 0.5 × (B-1) × 4 (Equating RHS of the two bove equations)
》 3B = (B-1) × 4
》 3B = 4B - 4
》 B = 4 cm
Hence the answers are
☆》 Length of the base = 4cm
☆》 Length of the perpendicular side = 3cm
Answered by
0
plz make me as brainlist.
Attachments:
Similar questions