In the figure below <abc&<bdc are right angles;if ab=5cm,ad=3cm&bd=4cm, find bc&dc
To find the lengths BC and DC, we can use the Pythagorean theorem, which relates the sides of a right triangle. According to the given information, <ABC and <BDC are right angles, AB = 5cm, AD = 3cm, and BD = 4cm.
The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
In this case, triangle ABC and triangle BDC are both right triangles. So, we can use the Pythagorean theorem in each triangle to find the lengths of BC and DC.
For triangle ABC:
AC^2 = AB^2 + BC^2
Since AB = 5cm and AC = AD + DC, we substitute these values:
(AD + DC)^2 = 5^2 + BC^2
(3 + DC)^2 = 25 + BC^2
9 + 6DC + DC^2 = 25 + BC^2
For triangle BDC:
BD^2 = BC^2 + DC^2
Since BD = 4cm, we substitute these values:
4^2 = BC^2 + DC^2
16 = BC^2 + DC^2
Now, we have two equations with two variables (BC and DC). We can solve these equations simultaneously to find the values of BC and DC.
From the equations above, we have the following system of equations:
9 + 6DC + DC^2 = 25 + BC^2
16 = BC^2 + DC^2
To solve this system, we can substitute BC^2 from the second equation into the first equation:
9 + 6DC + DC^2 = 25 + 16 - DC^2
9 + 6DC + DC^2 = 41 - DC^2
2DC^2 + 6DC - 32 = 0
Now, we have a quadratic equation. We can solve it by factoring, completing the square, or using the quadratic formula.
The factored form of the equation is:
(2DC - 4)(DC + 8) = 0
From this, we have two possible solutions for DC:
1) 2DC - 4 = 0, which gives DC = 2 cm
2) DC + 8 = 0, which is not possible since lengths cannot be negative
Now, we can substitute the value of DC into any of the original equations to find BC. Let's use the equation 16 = BC^2 + DC^2:
16 = BC^2 + (2)^2
16 = BC^2 + 4
BC^2 = 12
BC = √12 cm ≈ 3.4641 cm
Therefore, BC ≈ 3.4641 cm and DC = 2 cm.