Question

The area of a right angled triangle is 165m2. determine its base and altitude if the later exceeds the former by 7m.

Answer 1

Base of the right angled
triangle (b)= x m
Altitude (h)=7m exceeds the base
h = ( x + 7 )m
Area of Right angled triangle
( A ) = 165 m² [ given ]
$\frac{bh}{2}=165$
=> [x(x+7)]/2 = 165
=> x(x+7) = 230
=> x(x+7) = 15×22
=> x(x+7) = 15(15+7)
Compare both sides , we
conclude ,
x = 15
Therefore ,
base (b) = x = 15m
altitude (h) = (x + 7)
= 15 + 7
= 22 m
••••

Answer 2

Answer:

The length of base is 15 m and altitude is 22 m.

Step-by-step explanation:

Let the base be x

Since we are given that Altitude exceeds the base by 7 m

So, Altitude = x+7

Area of right angled triangle = $\frac{1}{2} \times Base \times Height$

So,  $165 = \frac{1}{2} \times x \times (x+7)$

$165 \times 2= x^2+7x$

$330= x^2+7x$

$x^2+7x-330=0$

$x^2+22x-15x-330=0$

$x(x+22)-15(x+22)=0$

$(x+22)(x-15)=0$

$x=-22,15$

Thus the base is 15 m

Altitude = 15+7 = 22 m

Hence the length of base is 15 m and altitude is 22 m.