Question

the area of a right angled triangle is 165m. Detaermine its base and altitude if the letter exceeds the former by 7meters

Answer 1

Let the base be x meters
and altitude be (x+7) meters
Area of triangle is 1/2× base × height
= 1/2 × (x) × (x+7) = 165
x(x+7)= 330
x^2 + 7x - 330 = 0
x^2 +(22-15)x -330 = 0
x^2 +22x -15x-330 = 0
x(x+22)-15x(x+22)=0
(x+22) (x-15)=0

X= -22 or X= 15
the value of X cannot be negative
so X= 15
or X+7 = 22
So base and altitude is 15m and 22m respectively.
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Answer 2

$\huge{ \mathfrak{ \purple{ \underline{Answer}}}}$




Given :

Area of the angled triangle : 165 m


Let the base of the right angled triangle be x

According to Question

Altitude of triangle be (x+7)m

Now ,


$Area \ \\ : of \: triangle : \: \frac{1}{2} \times base \times altitude \\ \\ = 165 = \frac{1}{2} \times x \times (x + 7) \\ \\ = 165 \times 2 = {x}^{2} + 7x \\ \\ = 330 = {x}^{2} + 7x \\ = {x}^{2} + 7x - 330 = 0 \\ = {x}^{2} - 22x + 15x - 330 = 0 \\ = x(x + 22) - 15(x + 22) = 0 \\ \\ therefore \\ (x + 22)(x - 15) = 0 \\ = x + 22 = 0 \\ = x = - 22 \\ or \\ x - 15 = 0 \\ x = 15 \\ \\ so \\ x = - 22 \: or \: x = 15 \\ - 22 \: \: rejected \: because \: it \: is \: negative \: \\ side \: of \: triangle \: cannot \: be \: negative. \\ \\ so \\ x = 15 \\ \\ therefore \: base \: of \: right \: angled \: \\ triangle = 15m \\ \\ and \: altitude \: of \: right \: angled \: \\ triangle = (x + 7)m \\ = (15 + 7)m = 22m$




So, base and height are 15 m and 22m respectively.


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