QUADRATIC EQUATIONS
right angled triangle
Word Problems Based On Geometric Figures
(Hypotenuse)2
Hypotenuse
=
13cm
12 - 5
=
7
1
Q.
Sol.
Let base of right triangle be x cm
The altitude of a right triangle is 7cm less than its base.
If the hypotenuse is 13cm, find the other two sides.
∴
Altitude of right triangle
=
‒ 7
In a right triangle,
By Pythagoras Theorem,
(Base)2
+
(Altitude)2
=
x2
+
(x ‒ 7)2
=
(13)2
x2
+
x2
‒
14x
+
49
=
169
169
2x2
‒
14x
+
49
‒
0
=
∴
120
2x2
‒
14x
‒
0
=
∴
60
x2
‒
7x
‒
0
=
Dividing throughout by 2
∴
5x
x2
‒
12x
+
60
‒
0
=
13 cm
(x – 7) cm
x cm
It is given that
∴
5
x
(x ‒12)
+
0
=
(x ‒12)
∴
(x ‒12)
0
=
(x +5)
∴
Base of triangle cannot be negative
∴
x
≠
-5
∴
x
=
12
∴
x ‒ 7
=
12 ‒ 7
=
5
∴ Other two sides of right triangle is 12cm and 5cm.
∴
x ‒12 = 0
or
x + 5 = 0
∴
x = 12
or
x = -5
In a comparative statement whatever comes later is taken as x
x
( ) cm
Let us do the prime factorization of 60
Calculation
60
30
15
2
2
3
5
5
1
Find two factors of 60 in such a way that by subtracting factors we get middle number.
‘-’ sign means subtracting
60 × 1 = 60
In this sum which geometric figure is considered ?
Yes
Pythagoras theorem
Since last sign is ‘-’ Give middle sign to the bigger factor & opposite sign to smaller factor.
12
5
60
–
+
EX 4.2 5
homework