Question

In the adjoining

figure, BC ⊥ AB , AD ⊥ AB,

BC=4, AD = 8, then

Find A(△ABC)/

A(△ADB)​

Answer 1

Answer:

We know that in the unit place 6+A=3

A=3−6=−3, which is not possible

Since A is greater than 10, where, 1 is carried over to tens place

∴6+A=13

A=13−6=7

Now in tens place

B+9+1=7 where 1 is carried over

B=7−10=−3 which is not possible

Since B is greater than 10, where 1 is carried over to hundred's place

B+9+1=10+7

B=17−10=7

∴B=7

Now in hundred's place

C+6+1=1 where, 1 is carried over

C=1−7=−6 which is not possible

Since C is greater than 10, where, 1 is carried over to hundred's place

C+6+1=1+10

C=11−7=4

∴C=4

Step-by-step explanation:

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