Solve Triangle Area Equation: Finding X When Area = 2X+16 cm²

Question

The area of triangle ABC is equal to 2X+16 cm².

Work out the value of X.

333X+5X+5X+5BBBAAACCCDDD

Video Solution

Solution Steps

00:00 Find X
00:05 Use the formula for triangle area
00:10 (height(AD) multiplied by base(BC)) divided by 2
00:14 Substitute appropriate values and solve
00:17 Side BC equals the sum of its parts (BD+DC)
00:28 Multiply by 2 to eliminate the denominator
00:37 Isolate AD
00:45 Take out 4 from the parentheses
00:52 And we found the height AD
00:55 Now use the Pythagorean theorem in triangle ADC
01:00 Substitute appropriate values and solve
01:10 Isolate X on one side of the equation
01:20 Take the square root
01:25 Subtract 5 and we found X
01:29 And this is the solution to the problem

Step-by-Step Solution

The area of triangle ABC is equal to:

AD×BC2=2x+16 \frac{AD\times BC}{2}=2x+16

As we are given the area of the triangle, we can insert this data into BC in the formula:

AD×(BD+DC)2=2x+16 \frac{AD\times(BD+DC)}{2}=2x+16

AD×(x+5+3)2=2x+16 \frac{AD\times(x+5+3)}{2}=2x+16

AD×(x+8)2=2x+16 \frac{AD\times(x+8)}{2}=2x+16

We then multiply by 2 to eliminate the denominator:

AD×(x+8)=4x+32 AD\times(x+8)=4x+32

Divide by: (x+8) (x+8)

AD=4x+32(x+8) AD=\frac{4x+32}{(x+8)}

We rewrite the numerator of the fraction:

AD=4(x+8)(x+8) AD=\frac{4(x+8)}{(x+8)}

We simplify to X + 8 and obtain the following:

AD=4 AD=4

We now focus on triangle ADC and by use of the Pythagorean theorem we should find X:

AD2+DC2=AC2 AD^2+DC^2=AC^2

Inserting the existing data:

42+(x+5)2=(65)2 4^2+(x+5)^2=(\sqrt{65})^2

16+(x+5)2 =65/16 16+(x+5)^2\text{ }=65/-16

(x+5)2=49/ (x+5)^2=49/\sqrt{}

x+5=49 x+5=\sqrt{49}

x+5=7 x+5=7

x=75=2 x=7-5=2

Answer

2 cm