Question

ABC is a right angle triangle at B, D is a point on AC such that DC = 8cm, AD = 34cm and BD = 17cm, find the area of ∆BDC

Answer 1

Answer:

Step-by-step explanation:

area of triangle=1/2  * base * height

1/2*34+8*17

357$cm^{2}$

Answer 2

Given :   ABC is a right angle triangle at B, D is a point on AC such that DC = 8cm, AD = 34cm and BD = 17cm,

to find : Area of triangle ∆BDC

Solution:

Lets draw BE ⊥ AC

BE =  x cm

DE = y cm

BE² + DE² = BD²

=> x² + y² = 17²

=> x² + y² = 289

BE² + CE² = BC²

=> x² + (CD + DE)² = BC²

=> x² + (8 + y)² = BC²

=> x² + y² + 16y + 64 = BC²

BE² + AE² = AB²

=> x² + (AD - DE)² = AB²

=> x² + (34 - y)² = AB²

=> x² + y² - 68y + 1156  = AB²

AB² + BC² = AC²

=> x² + y² - 68y + 1156 + x² + y² + 16y + 64 = (AD + CD)²

=> 2( x² + y²)  -52y + 1220 = (34 + 8)²

=> 2(289) - 52y + 1220 = 42²

=> 1798 - 52y = 1764

=> 52y = 34

=> y = 34/52

=> y = 17/26

x² + y² = 289

=> x ² + (17/26)² = 289

=> x = 16.987

Area of ∆BDC = (1/2) * DC * BE

= (1/2) * 8 * 16.987

= 67.95  cm²

Area of ∆BDC = 67.95  cm²

Learn more:

In a right angled triangle the hypotenuse is 10 cm more than the ...

https://brainly.in/question/8992388

in a triangle pqr angle Q is a right angle and QT is a perpendicular ...

https://brainly.in/question/13149578