Question

In the given figure if M and N are mid-points of AB and BC respectively and AC = 12 cm the AN² + CM² = M В N (a) 164 (b) 180 (c) 196​

Answer 1

Answer:

Correct option is

A

6 cm, 7 cm, 5 cm

SinceM,N,andParethemid−pointsofAB,ACandBCrespectively.

HenceMN=

2

1

BC

⟹BC=2MN

⟹BC=2×3=6cm

SimilarlyAC=2MP=2×2.5=5cm

andAB=2NP=2×3.5=7cm

Answer 2

The answer is 180

Given: In given picture M is mid point of AB

N is mid points of BC and AC = 12 cm

To find: AN² + CM²

Solution:

M is midpoint of AB ⇒ BM = AB/2

N is midpoint of BC ⇒ BN = BC/2

[ We will use them for further calculation]

From figure ABC, ABN and BCM are right angled triangle

As we know in a right angle triangle hyp² = side² + side²

⇒ From ΔABC,  AC² = AB² + BC²

⇒ From ΔABN,  AN² = AB²+ BN²  

⇒ From ΔBCM, CM² = BM²+ BC²    

then AN² + CM²  = AB²+ BN² + BM²+ BC²  

=  AB²+ (AB/2)² + (BC/2)²+ BC²  

=  $AB^{2} + \frac{AB^{2} }{4} +\frac{BC^{2} }{4} + BC^{2}$  

=  $\frac{ 4AB^{2} + AB^{2} +{BC^{2} }+ 4BC^{2} }{4}$

= $\frac{ 5AB^{2} + 5BC^{2} }{4}$

= $\frac{5 (AB^{2} + BC^{2}) }{4}$        

= $\frac{5 (AC^{2}) }{4}$ = $\frac{5 (12)^{2} }{4}$           [ ∵ AC² = AB²+ BC² ]

=  $\frac{720}{4}$ = 180

Therefore, AN² + CM² = 180

#SPJ2