Math, asked by lakshybhat, 19 days ago

K, L, M and N are points on sides AB, BC, CD and DA respectively of a square ABCD of side 12 cm such that KM || BC and LN || AB. The perimeter of triangle KLB is 12 cm. Let the area of triangle MND be x cm2. Find x.

Answers

Answered by amitnrw
0

Given : K, L, M and N are points on sides AB, BC, CD and DA respectively of a square ABCD of side 12 cm such that KM || BC and LN || AB.

The perimeter of triangle KLB is 12 cm.

Let the area of triangle MND be x cm².

To Find :  x

Solution:

The perimeter of triangle KLB is 12 cm.

KLB is right angled triangle at B

(3 ,4  , 5) Pythagorean triplet is well known

and 3 + 4 + 5  = 12

Hence

Either  KB = 3  and  BL = 4    or  KB = 4  and  BL  = 3

Case 1 : KB = 3  and  BL = 4

=> AK = 12 - 3 = 9  and  CL = 12 - 4 = 8

and MD = AK = 9  and  DN = CL = 8

MND is Right angled triangle at  D

Hence Area of ΔMND  = (1/2) * MD * DN

= (1/2) * 9 * 8

= 36  cm²

Hence value of x is 36

in case 2 :  MD = 8 and DN = 9

Area of ΔMND  = (1/2) * MD * DN

= (1/2) * 8 * 9

= 36  cm²

Hence value of x is 36

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