How should I do this question that is in the image below?50

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2 Answers
Jan 28, 2018

Second triangular block, by #15m^2#.

Explanation:

Well, since they give us only the lengths of the triangle, we will have to use Heron's formula, which states that the length of a triangle is

#A=sqrt(s(s-a)(s-b)(s-c))#

where #s=(a+b+c)/2#, and #a,b,c# are the side lengths of a triangle.

Let #A_1# denote the area of the first triangle, #A_2# denote the second area of the triangle.

We have:

#s_1=(25+48+53)/2=63m#

#:.A_1=sqrt(63(63-25)(63-48)(63-53)) \ m^2#

#A_1=sqrt(63*38*15*10) \ m^2#

#A_1=sqrt(359100) \ m^2#

#A_1~~599.25m^2#

Next step, let's find #A_2#.

#s_2=(33+38+45)/2=58#

#:.A_2=sqrt(58(58-33)(58-38)(58-45)) \ m^2#

#A_2=sqrt(58*25*20*13) \ m^2#

#A_2=sqrt(377000) \ m^2#

#A_2~~614m^2#

#:.A_2>A_1#

#614m^2-599.25m^2=14.75m^2~~15m^2#

So, the second triangular block has a greater area by #15m^2#.

Jan 28, 2018

The second block has greater area, by (approximately) #15" m"^2.#

Explanation:

The formula most of us remember for finding triangle area is:

#A=1/2bh#

where #b# is the base of the triangle and #h# is the perpendicular distance from the base to its opposing vertex. But there are two other formulas:

#A=1/2ab sin C#

and

#A=sqrt(s(s-a)(s-b)(s-c))#

where #s# is the semiperimeter of the triangle: #s=1/2(a+b+c).#

The formula we use depends on the information we're given. When we know all three sides, as in this example, we can use #A=sqrt(s(s-a)(s-b)(s-c))#, also known as Heron's formula. We just need to compute #s# first.

For the first triangle:

#s=1/2(a+b+c)#

#color(white)s=1/2(25+48+53)#

#color(white)s=126/2=63 " m"#

So the area of the first triangle, call it #A_1#, is

#A_1=sqrt(s(s-a)(s-b)(s-c))#
#color(white)(A_1)=sqrt(63(63-25)(63-48)(63-53))#
#color(white)(A_1)=sqrt(63(38)(15)(10))#
#color(white)(A_1)=sqrt(3*3*7(2*19)(3*5)(2*5))#
#color(white)(A_1)=2*3*5sqrt(7(19)(3))#
#color(white)(A_1)=30sqrt(399)#
#color(white)(A_1)~~599" m"^2#

I'll leave the area of triangle 2 as an exercise. Its value is #A_2~~614 " m"^2.#

After comparing #A_1# and #A_2#, we see that #A_2# is bigger, by approximately #15" m"^2.# Thus, the second triangle has the greater area, by #15" m"^2.#