# IN RIGHT TRIANGLE ABC,C IS RIGHT ANGLE.c=3 AND a+b=root 17.COMPUTE AREA OF THE TRIANGLE?

2
by narenderkumarh

2015-05-20T13:07:18+05:30
As C is the right angle
from pythagoras` theorem
a² + b² = c²
⇒ (a+b)² - 2ab = 3² = 9
⇒√17 ² - 2ab = 9
⇒17 - 2ab = 9
⇒2ab   = 8
⇒ ab = 4
area of the triangle is (1/2 × height ×base)
= 1/2×ab
= 1/2 × 4
= 2 square units
dont know ihave to solve few questions
i asked why it is showing error ??
dont know
in fb it will b difficult to write the supercriptions , mathematical symbols etc
2015-05-21T08:24:09+05:30
GIVEN:
ABC is right angled at C.
c=3 and (a+b) = √17

Since ABC is a right angled triangle,
From Pythagorus Theorem,
(hyp)²=(base)² + (height)²
Therefore,
c² = a²+b²
c² = a²+b²+2ab - 2ab                                                                  [By adding 2ab-2ab on R.H.S]
c² = (a+b)² - 2ab                                                                          [Since, (a+b)² = a²+b²+2ab]
Substituting value of c and (a+b),
3² = (√17)² -2ab
9 = 17 - 2ab
2ab = 17-9
2ab = 8
ab=8/2
ab=4
Now,
Area of a right angled triangle = (1/2 * base * height)
Therefore,
Area = 1/2 * (a*b)                                                                       [Since, base*height = a*b]
Area = 1/2 * 4 = 2 square units.                                               [Since, a*b = 4]
﻿Therefore,
Area of triangle ABC = 2 sq. units.