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
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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