Question

the area of a right angled triangle is 165 square metres determine its base and altitude if the latter exceed the former by 7 metres

Answer 1

ANSWER:
Base is 22m & Altitude is 15m

EXPLANATION:
Given,
Area=A=165
Base=B=? ; Altitude=H=?
B=H+7

Area of Triangle = 1/2×Base×Altitude
. A=1/2×B×H
. 165=1/2×(H+7)×H
165×2=(H+7)×H
. 330=H²+7H

Therefore H²+7H-330=0
We know that,
$x = \frac{ - b + \sqrt{ {b}^{2} - 4ac} }{2a}$
Or
$x = \frac{ - b \: - \: \sqrt{ {b}^{2} - 4ac} }{2a}$
Therefore H=[-7±√(7²-4×1×-330)]/2×1
` H=[-7±√(49+1320)]/2
` H=[-7±√(1369)]/2
` H=[-7±37]/2
∴ H=[-7+37]/2 or H=[-7-37]/2
` H=[30]/2 or H=[-44]/2
` H=15 or H=-22

As Altitude cannot be zero
H=15
As B=H+7
B=15+7=22

Hence Base is 22m & Altitude is 15m

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Answer 2

SOLUTION :  

Let the base of a ∆  be x m, then the altitude be (x + 7) m.

Area of a ∆ = 1/2 x Base x Altitude

1/2 x  (x + 7) = 165

x (x + 7) = 165 × 2

x² + 7x = 330

x² + 7x - 330 = 0

x² + 22x – 15x – 330 = 0

[By middle term splitting]

x(x + 22) – 15(x + 22) = 0

(x + 22)(x – 15) = 0

(x + 22)  or (x – 15) = 0

x = - 22 or x = 15

Since, side can’t be negative, so x ≠ - 22 , Therefore x = 15 m

Base = x = 15 m

Altitude = x + 7 = 15 + 7 = 22 m

Hence, the base of a ∆  is 15 m and the altitude is 22 m.

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