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Please remember to check your solutions against the ones posted below, each week. For this week's POTW, the question is for BOTH Grade 7 & 8, and is directly related to the new unit of geometry.
So here is how I did the problem. First of all, you need to figure out the total base, since the height has already been given. From the information stating that the area of triangle ADC is 50% more than Triangle ABD, I can figure out the area of Triangle ADC.
Using the triangle formula, I can see that ADC is 6*8/2=24 m squared. 50 percent more than that is 24*1.5=36.Because Triangle ADC also has the height of 8, I can follow the triangle formula again, giving me 36*2/8=9 cm. This means that the line segment "a" is equal to 9.
Adding up 6 and 9, it would give me the base which is 15 m. I can use the Pythagorean theorem to figure out the rest.
15^2(squared)+6^2 = 225+36 = 289 The square root of 289 is 17.
Therefore, the hypotenuse of the triangle ABC is 17.
However, we cannot use the height, since ADC does not use that, it actually uses the hypotenuse of triangle ABD! This would allow me to figure it out, so it would be 6 squared + 8 squared, square rooted. This would be 36+64, which is 100. The square root of 100 is 10, so line segment "c" is equal to 10.
Adding these up, 10+9+17=36 m. The perimeter of Triangle ADC is 36m. -Alan
The area of triangle ADC is 36 m2. This is because we know that triangle ADC is 50% larger (area) then triangle ABD. So then we calculate the area of ABD. The base of ABD is 6 m and the height is 8 m. Then the formula for calculating the area of a triangle is B*H/2=A 6*8/2=24. Then since ADC is 50% larger the area of ADC is 36 m2.
64+36=100. The dimensions of ABD are 8m, 6m ,and 10m. The area of ABD is 24m squared. The area of ADC is 36m squared. This means the base must be 3.6m. The hypotenuse is 10.63m. Thus, the perimeter of triangle ADC is 24.23 meters.
We first find the hypotenuse of ABD, since BD=6m and AB= 8m, we find it which is: 6^2+8^2=c^2 36+64=c^2 100=c^2 _____ /100 =c 10=c So the hypotenuse of triangle ABD is 10, which is the side/length of ADC. That means, the base of ADC must be 3.6m since there isn't another way the area to be 36m^2 (We get 36m^2 by finding the area of triangle ABD which is 6*8/2=24, we then take half of 24 and add it together, 24+24/2=36) Then we use the Pythagorean theorem which is 10^2+3.6^2=c^2 100+12.96=c^2 _______ /112.96=c 10.63(rounded)=c We now found all the sides for triangle ADC we can add it all together to receive 10.63+10+3.6=24.23 Therefore the perimeter of triangle ADC is 24.23m
Okay so I may have done it wrong the first time, in order to find out line segment a, we need to find the area of both triangles ADC and ABD which I have done before: triangle ABD=24m2 triangle ADC=36m2 That means the area of the whole triangle (ABC) is 60m2. We multiply this by 2, to get the area of not triangle but the invisible rectangle, since the triangle will share the same base as the rectangle. 120m2 is the area of triangle ABC*2, now we divide 120 by 8, to get the base since the formula to get the area of the rectangle would be A=L*W. (In this case it is B*H for triangle) 120/8=15. So the base is 15, but remember that is for the FULL triangle, not line segment a. We know to get line segment a by just subtracting 6 since that is the base for triangle ABD.
Now that we know the base for triangle ADC is 9m, we find the hypotenuse for triangle ABD, which I have done before which is 10. (6^2+8^2=100, square root which is 10) Finally to find the the hypotenuse for triangle ADC, we do: 9^2+10^2=c^2 190=c^2 190(squareroot)=c 13.78(rounded)=c So now we can add them all together with a perimeter of... 32.78m.
Okay so we go back, to the 120 part. Now since the base is 15, we can see that the hypotenuse would be 8m^2+15^2= 289, square root that and we get 17, we add 10+17+9 since 15-6=9 and we get 36 as the area of triangle ADC.
So here is how I did the problem.
ReplyDeleteFirst of all, you need to figure out the total base, since the height has already been given. From the information stating that the area of triangle ADC is 50% more than Triangle ABD, I can figure out the area of Triangle ADC.
Using the triangle formula, I can see that ADC is 6*8/2=24 m squared. 50 percent more than that is 24*1.5=36.Because Triangle ADC also has the height of 8, I can follow the triangle formula again, giving me 36*2/8=9 cm. This means that the line segment "a" is equal to 9.
Adding up 6 and 9, it would give me the base which is 15 m. I can use the Pythagorean theorem to figure out the rest.
15^2(squared)+6^2 = 225+36 = 289
The square root of 289 is 17.
Therefore, the hypotenuse of the triangle ABC is 17.
However, we cannot use the height, since ADC does not use that, it actually uses the hypotenuse of triangle ABD! This would allow me to figure it out, so it would be 6 squared + 8 squared, square rooted. This would be 36+64, which is 100. The square root of 100 is 10, so line segment "c" is equal to 10.
Adding these up, 10+9+17=36 m.
The perimeter of Triangle ADC is 36m.
-Alan
The area of triangle ADC is 36 m2. This is because we know that triangle ADC is 50% larger (area) then triangle ABD. So then we calculate the area of ABD. The base of ABD is 6 m and the height is 8 m. Then the formula for calculating the area of a triangle is B*H/2=A 6*8/2=24. Then since ADC is 50% larger the area of ADC is 36 m2.
ReplyDeleteIt asks for the perimeter??
DeleteMaxwell and Avi POTW:
ReplyDeleteThe perimeter of Triangle ADC is 36 m. We calculated in our math notebook and after a long discussion of arguing.
Great to hear about the long discussion, less great to hear about the arguing (though not all that surprised!)
Delete64+36=100. The dimensions of ABD are 8m, 6m ,and 10m. The area of ABD is 24m squared. The area of ADC is 36m squared. This means the base must be 3.6m. The hypotenuse is 10.63m. Thus, the perimeter of triangle ADC is 24.23 meters.
ReplyDeleteWe first find the hypotenuse of ABD, since BD=6m and AB= 8m, we find it which is:
ReplyDelete6^2+8^2=c^2
36+64=c^2
100=c^2
_____
/100 =c
10=c
So the hypotenuse of triangle ABD is 10, which is the side/length of ADC. That means, the base of ADC must be 3.6m since there isn't another way the area to be 36m^2 (We get 36m^2 by finding the area of triangle ABD which is 6*8/2=24, we then take half of 24 and add it together, 24+24/2=36) Then we use the Pythagorean theorem which is
10^2+3.6^2=c^2
100+12.96=c^2
_______
/112.96=c
10.63(rounded)=c
We now found all the sides for triangle ADC we can add it all together to receive
10.63+10+3.6=24.23
Therefore the perimeter of triangle ADC is 24.23m
Okay so I may have done it wrong the first time, in order to find out line segment a, we need to find the area of both triangles ADC and ABD which I have done before: triangle ABD=24m2
ReplyDeletetriangle ADC=36m2
That means the area of the whole triangle (ABC) is 60m2. We multiply this by 2, to get the area of not triangle but the invisible rectangle, since the triangle will share the same base as the rectangle. 120m2 is the area of triangle ABC*2, now we divide 120 by 8, to get the base since the formula to get the area of the rectangle would be A=L*W. (In this case it is B*H for triangle)
120/8=15. So the base is 15, but remember that is for the FULL triangle, not line segment a. We know to get line segment a by just subtracting 6 since that is the base for triangle ABD.
Now that we know the base for triangle ADC is 9m, we find the hypotenuse for triangle ABD, which I have done before which is 10. (6^2+8^2=100, square root which is 10)
Finally to find the the hypotenuse for triangle ADC, we do:
9^2+10^2=c^2
190=c^2
190(squareroot)=c
13.78(rounded)=c
So now we can add them all together with a perimeter of... 32.78m.
OKAY SO, I did it wrong, AGAIN. But it's okay, because I'll fix it. Whoops :P
DeleteOkay so we go back, to the 120 part. Now since the base is 15, we can see that the hypotenuse would be
Delete8m^2+15^2= 289, square root that and we get 17, we add 10+17+9 since 15-6=9 and we get 36 as the area of triangle ADC.
The perimeter is 36m. I did the work in my math notebook
ReplyDelete