Question

a) the area of the ∆ ABC,
(b) the length of AC and
(c) the length of the altitude drawn from B to AC​

Answer 1

a) For triangle ABC :- AB is base , BC is height and AC is hypotenuse.

=> area of triangle =( 1/2) × base × height

=> area of triangle ABC =( 1/2) × 12 × 5

=> area of triangle ABC = 6 × 5 = 30 cm²

b) For triangle ABC :- AB is base , BC is height and AC is hypotenuse

Therefore, by using Pythagoras theorem :-

=> (AC)² = (AB )² + (BC)²

=> (AC)² = (12)² + (5)²

=> (AC)² = 144+ 25

=> (AC)² = (169)

=> AC = √(169)

=> AC = 13 cm

c) Let length of CM be x

Now, CM + MA = CA= 13

Therefore, MA = 13-x

For triangle BCM :- CB is hypotenuse , CM is base , BM is height.

=> (BM)²= (CB)²- (CM)²

=> (BM)²= 5² - x² [ It is given that CB = 5 cm ]

=> (BM)²= 25 - x²........(1)

For triangle BCA :- AB is hypotenuse , MA is base , BM is height.

=> (BM)²= (AB)²- (MA)²

=> (BM)²= (12)² - (13-x)² [ It is given that AB =5cm ]

=> (BM)²= 144 - (169+x²-26x)

=> (BM)²= 144 - 169-x²+26x

=> (BM)²= - 25-x²+26x ........(2)

Now, from equation (1) and (2) :-

=> 25 - x² = - 25-x²+26x

=> - x²+x² -26x = - 25-25

=> -26x = - 50

=> x = 50/26

Now, put value of x in equation (1) to find the Length of BM.

=> (BM)²= 25 - (50/26)².

=> (BM)²= 25 - (2500/676)

=> (BM)²= (16900- 2500)/(676)

=> (BM)²=14400/676

=> BM= 4.61 cm