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elixir [45]
3 years ago
13

The longer leg of a right triangle is 6cm longer than the shorter leg and also 1cm longer than the hyptenese. find the perimeter

of the triangle.
Mathematics
1 answer:
CaHeK987 [17]3 years ago
5 0
X - a leg  (x > 6);
x - 6 - a leg;
x + 1 - a <span>hypotenuse;</span>
Use the Pythagorean theorem
(x+1)^2=x^2+(x-6)^2

Use:(a\pm b)^2=a^2\pm2ab+b^2

x^2+2x+1=x^2+x^2-12x+36\ \ \ |-x^2-2x-1\\\\x^2-14x+35=0\ \ \ |-35\\\\x^2-14x=-35\\\\x^2-2\cdot x\cdot7=-35\ \ \ |+7^2\\\\x^2-2\cdot x\cdot7+7^2=-35+7^2\\\\(x-7)^2=14\iff x-7=\pm\sqrt{14}\ \ \ |+7\\\\x=7-\sqrt{14} < 6;\ x=7+\sqrt{14}

x=7+\sqrt{14}\\x-6=1+\sqrt{14}\\x+1=8+\sqrt{14}

The perimeter:

7+\sqrt{14}+1+\sqrt{14}+8+\sqrt{14}=16+3\sqrt{14}
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<u>Question 6</u>

1) \overline{AB} \cong \overline{BD}, \overline{CD} \perp \overline{BD}, O is the midpoint of \overline{BD}, \overline{AB} \cong \overline{CD} (given)

2) \angle ABO, \angle ODC are right angles (perpendicular lines form right angles)

3) \triangle ABO, \triangle CDO are right triangles (a triangle with a right angle is a right triangle)

4) \overline{BO} \cong \overline{OD} (a midpoint splits a segment into two congruent parts)

5) \triangle ABO \cong \triangle CDO (LL)

<u>Question 7</u>

1) \angle ADC, \angle BDC are right angles), \overline{AD} \cong \overline{BD}

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7) \angle ACD \cong \angle BCD (CPCTC)

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