Question

एक समचतुर्भुज के विकर्ण की लंबाई 10 सेंटीमीटर और 24 सेंटीमीटर है इसका परिमाप बताओ​

Answer 1

Area = (10cm ×24cm)/2=120 square cm

rea = (10cm ×24cm)/2=120 square cm lets us suppose

rea = (10cm ×24cm)/2=120 square cm lets us suppose arm = a

rea = (10cm ×24cm)/2=120 square cm lets us suppose arm = aso by Pythagoras theorem

rea = (10cm ×24cm)/2=120 square cm lets us suppose arm = aso by Pythagoras theorema^2= (10/2)^2+(24/2)^2

rea = (10cm ×24cm)/2=120 square cm lets us suppose arm = aso by Pythagoras theorema^2= (10/2)^2+(24/2)^2a^2= 1/4(100+576)

rea = (10cm ×24cm)/2=120 square cm lets us suppose arm = aso by Pythagoras theorema^2= (10/2)^2+(24/2)^2a^2= 1/4(100+576)a^2=1/4(676)

rea = (10cm ×24cm)/2=120 square cm lets us suppose arm = aso by Pythagoras theorema^2= (10/2)^2+(24/2)^2a^2= 1/4(100+576)a^2=1/4(676)a^2=169

rea = (10cm ×24cm)/2=120 square cm lets us suppose arm = aso by Pythagoras theorema^2= (10/2)^2+(24/2)^2a^2= 1/4(100+576)a^2=1/4(676)a^2=169a= √169

rea = (10cm ×24cm)/2=120 square cm lets us suppose arm = aso by Pythagoras theorema^2= (10/2)^2+(24/2)^2a^2= 1/4(100+576)a^2=1/4(676)a^2=169a= √169a=13cm

rea = (10cm ×24cm)/2=120 square cm lets us suppose arm = aso by Pythagoras theorema^2= (10/2)^2+(24/2)^2a^2= 1/4(100+576)a^2=1/4(676)a^2=169a= √169a=13cmso Perimeter is 4×arm=4×13=52

Answer 2

Answer:

68

Step-by-step explanation:

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