Math, asked by manobhiramreddy01, 1 year ago

solve 14 question with steps full solution plz

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Answered by ishratmulani123
0

Hope it will help you

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Answered by rpulenora0915
1

So, to solve your problem, we have to apply the Pythagorean relation.

So let one of the perpendicular sides be x. Then the other is x+4, because it is given that it is greater than x by 4cm. Then your hypotenuse is x+4+4 = x+8, because the hypotenuse is 4 more than the x +4.

So we just apply the pythagorean relation to these values.

x^2 + (x+4)^2 = (x+8)^2

x^2 + x^2+8x+ 16 = x^2+16x+64

since x^2 is common to both sides, it cancels out.

x^2+8x+16 = 16x + 64

x^2+8x +16 - 8x-16 = 16x+64-8x-16

x^2 = 8x+48

Then we build up the quadratic equation by carrying the right hand side over to the left hand side.

x^2-8x-48 = 0

x^2 -12x+4x -48 = 0

x(x-12) +4(x-12) = 0

(x-12)(x+4) = 0

the one of (x-12) or (x+4) is forced to be 0. but if x+4 is 0, x would be -4, and it can not happen as x is a side length. So x-12 should be 0. then x is 12. so the other perpendicular side is 16cm.

Now, the area of a right triangle is half into the product of the 2 perpendicular sides.

= 12 * 16 /2

= 12*8

=96

So, YAYYY, we solved the problem!!!

If I helped ya, pleeeeeeeeeeeeeeeease mark me brainliest!!!

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