The base of a triangle is 7 less than twice its height. if the area of the triangle is 102cm^2 find the length of the base
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Answer: unit and 18 unit
Step-by-step explanation:
Let ABCD is a trapezoid where AB and CD are the bases. ( In which AB is greatest base which shown in below figure)
AD and BC are the legs of the trapezoid ABCD.
Now, we have ( According to the question ),
AB = 18 unit, BC = 7 unit, AD = 3 unit and DC = 11 unit.
Here the leg AD extends from point D.
Similarly leg BC extends from point C.
Let they meet at point P ( shown in below diagram)
Since In triangles PAB and PDC,
∠PDC ≅ ∠PAB ( because DC ║ AB )
And, ∠ PAB ≅ ∠ PBA
∠DPC ≅ ∠ APB ( reflexive)
Therefore, By AAA similarity postulate,
Thus, By the definition of similarity,
( because PA = PD+DA)
⇒ 18 PD = 11 PD +33
⇒7PD = 33
⇒ PD = 33/7
Again by the definition of similarity,
( because PB = PC + CB)
⇒ 18 PC = 11 PD +77
⇒7PC = 77
⇒ PC = 11
Thus, PA = PD+DA = 33/7 + 3 =
And, PB = PC + CB = 11 + 7 = 18
Let
x-------> the width of the rectangular area
y------> the length of the rectangular area
we know that
y=x+15------> equation 1
perimeter of a rectangle=2*[x+y]
2x+2y <= 150-------> equation 2
substitute 1 in 2
2x+2*[x+15] <=150--------> 2x+2x+30 <=150----> 4x <=150-30
4x <= 120---------> x <= 30
the width of the rectangular area is at most 30 ft
y=x+15
for x=30
y=30+15------> y=45
the length of the rectangular area is at most 45 ft
see the attached figure
the solution is<span> the shaded area</span>
x-------> the width of the rectangular area
y------> the length of the rectangular area
we know that
y=x+15------> equation 1
perimeter of a rectangle=2*[x+y]
2x+2y <= 150-------> equation 2
substitute 1 in 2
2x+2*[x+15] <=150--------> 2x+2x+30 <=150----> 4x <=150-30
4x <= 120---------> x <= 30
the width of the rectangular area is at most 30 ft
y=x+15
for x=30
y=30+15------> y=45
the length of the rectangular area is at most 45 ft
see the attached figure
the solution is<span> the shaded area</span>