[Instinct01]

Hello,

I'm probably going to make myself look really stupid by asking these questions, but I seem to have completely forgotten how to work them out.

1) Please see trapezium.jpg attatchment and I am referring to Q2b). How do I calculate the length of the fourth side to work out the perimeter?

2) Please see elevation.jpg attatchment. I'm assuming the Q is asking for the area of the rectangle as well as the triangle, but how would I work out the area of the triangle with just 1 side and 1 angle?

Thank you in advance for any advice.

[raheem94]

(Original post by Instinct01)
Hello,

I'm probably going to make myself look really stupid by asking these questions, but I seem to have completely forgotten how to work them out.

1) Please see trapezium.jpg attatchment and I am referring to Q2b). How do I calculate the length of the fourth side to work out the perimeter?

2) Please see elevation.jpg attatchment. I'm assuming the Q is asking for the area of the rectangle as well as the triangle, but how would I work out the area of the triangle with just 1 side and 1 angle?

Thank you in advance for any advice.

1)

See the below image,



Now use phythagoras theorem to find the missing side.

[raheem94]

(Original post by Instinct01)
Hello,

I'm probably going to make myself look really stupid by asking these questions, but I seem to have completely forgotten how to work them out.

1) Please see trapezium.jpg attatchment and I am referring to Q2b). How do I calculate the length of the fourth side to work out the perimeter?

2) Please see elevation.jpg attatchment. I'm assuming the Q is asking for the area of the rectangle as well as the triangle, but how would I work out the area of the triangle with just 1 side and 1 angle?

Thank you in advance for any advice.

2)

Yes, the question is asking for the sum of area of rectangle and triangle.



Now you can find the missing side by using on the triangle.

Spoiler:

Show

[Instinct01]

(Original post by raheem94)
2)

Yes, the question is asking for the sum of area of rectangle and triangle.

Now you can find the missing side by using on the triangle.

Spoiler:

Show

Ah, these steps seem quite difficult actually. I thought it would be something much easier.

Okay, so the first one, is the side length 19.98 (2 d.p)? And for the second one, I'm doing now...

Thank you very much for your help

[raheem94]

(Original post by Instinct01)
Ah, these steps seem quite difficult actually. I thought it would be something much easier.

Okay, so the first one, is the side length 19.98 (2 d.p)? And for the second one, I'm doing now...

Thank you very much for your help

Btw, which grade are you in?

It is 20.04(2d.p.) how did you get 19.98?

Spoiler:

Show

[Instinct01]

(Original post by raheem94)
Btw, which grade are you in?

It is 20.04(2d.p.) how did you get 19.98?

Spoiler:

Show

Year 11 soon... aiming for an A/A* grade in this Unit:

Hmm, that's weird. Could you tell me where I've gone wrong.

b² = 20² - 1.2²
b² = 400 - 1.44 = 398.56
b = 19.96?

Also, would the trigononometry one be: 30/tan(60 deg) = 17.32?

[Dark Lord of Mordor]

(Original post by Instinct01)
Year 11 soon... aiming for an A/A* grade in this Unit:

Hmm, that's weird. Could you tell me where I've gone wrong.

b² = 20² - 1.2²
b² = 400 - 1.44 = 398.56
b = 19.96?

Also, would the trigononometry one be: 30/tan(60 deg) = 17.32?

It's
not

[raheem94]

(Original post by Instinct01)
Year 11 soon... aiming for an A/A* grade in this Unit:

Hmm, that's weird. Could you tell me where I've gone wrong.

b² = 20² - 1.2²
b² = 400 - 1.44 = 398.56
b = 19.96?

Also, would the trigononometry one be: 30/tan(60 deg) = 17.32?

It should be

And you are correct at the second one, though the question requires the area not just the side length.

[Instinct01]

(Original post by SrijanParmeshwar)
It's
not

But isn't the 20² the hypotenuse, so that:

a² + b² = c²
1.2² + b² = 20²
b² = 20² - 1.2²

I must have gone wrong here but where?

[Dark Lord of Mordor]

(Original post by Instinct01)
But isn't the 20² the hypotenuse, so that:

a² + b² = c²
1.2² + b² = 20²
b² = 20² - 1.2²

I must have gone wrong here but where?

Well your working is good, however I think that is the hypotenuse if you look at the diagram since the right angle is between the 20m and 1.2m sides

[raheem94]

(Original post by Instinct01)
But isn't the 20² the hypotenuse, so that:

a² + b² = c²
1.2² + b² = 20²
b² = 20² - 1.2²

I must have gone wrong here but where?

The missing side is the hypotenuse not 20.

[Instinct01]

(Original post by SrijanParmeshwar)
Well your working is good, however I think that is the hypotenuse if you look at the diagram since the right angle is between the 20m and 1.2m sides

(Original post by raheem94)
The missing side is the hypotenuse not 20.

Oh, no wonder. Thank you very much.

With regards the other question, would I be correct in saying:

Area = 1/2ab(sinC) + rectangle
Area = 531.6*sin(120 deg) = 460.38 (2 d.p) + rectangle
Area = 460.38 + 2400 = 2860.38m²?

...I am supposed to just halve the ab and not the entire equation including SinC, right?

[raheem94]

(Original post by Instinct01)
Oh, no wonder. Thank you very much.

With regards the other question, would I be correct in saying:

Area = 1/2ab(sinC) + rectangle
Area = 531.6*sin(120 deg) = 460.38 (2 d.p) + rectangle
Area = 460.38 + 2400 = 2860.38m²?

...I am supposed to just halve the ab and not the entire equation including SinC, right?

You have calculated as 17.32, now you can use this to find the area of the upper triangle, it will be

Then add the area of the rectangle.

[Instinct01]

(Original post by raheem94)
You have calculated as 17.32, now you can use this to find the area of the upper triangle, it will be

Then add the area of the rectangle.

Oh, does the 1/2abSin(C) rule not apply here?

[raheem94]

(Original post by Instinct01)
Oh, does the 1/2abSin(C) rule not apply here?

There is not point of using that here, we can easily use 0.5*base*height to get the area of one triangle, so we double it to get the area of the full upper triangle.

[Instinct01]

(Original post by raheem94)
There is not point of using that here, we can easily use 0.5*base*height to get the area of one triangle, so we double it to get the area of the full upper triangle.

I see... although either methods seem to work, yours is definitely much quicker.

Much appreciation & some +ve rep

[Instinct01]

Sorry, just a final question with regards perimeter of shapes. Please see the attatchment of a parallelogram and my annotations in red. In this case, if I split the parallelogram into two triangles and then halve one of the triangles to create a right angle, I can calculate the length of the hypotenuse, right?

I'm assuming that here I'm supposed to use Pythagoras' Theorem, but when using this I get a different answer to if I used trigonometry. Could anyone explain why please, and which if either method is correct?

Pythagoras:

a² + b² = c²
7² + 7.5 ² = c²
c² = 105.25
c = 10.26cm (2 d.p)

But then, trigonometry:

Adjacent = 7.5cm and Opposite = 7cm, so, using that, surely either of the following should work:

7/Sin(45) = 9.90cm (2 d.p) OR 7.5/Cos(45) = 10.61cm (2 d.p)

...But all three answers are different albeit being fairly similar. I'd be very grateful for any help.

[raheem94]

(Original post by Instinct01)
Sorry, just a final question with regards perimeter of shapes. Please see the attatchment of a parallelogram and my annotations in red. In this case, if I split the parallelogram into two triangles and then halve one of the triangles to create a right angle, I can calculate the length of the hypotenuse, right?

I'm assuming that here I'm supposed to use Pythagoras' Theorem, but when using this I get a different answer to if I used trigonometry. Could anyone explain why please, and which if either method is correct?

Pythagoras:

a² + b² = c²
7² + 7.5 ² = c²
c² = 105.25
c = 10.26cm (2 d.p)

But then, trigonometry:

Adjacent = 7.5cm and Opposite = 7cm, so, using that, surely either of the following should work:

7/Sin(45) = 9.90cm (2 d.p) OR 7.5/Cos(45) = 10.61cm (2 d.p)

...But all three answers are different albeit being fairly similar. I'd be very grateful for any help.

9.90cm is the correct length.



You have assumed the length marked in green to be 7.5cm, which it isn't.

Hence your second method is right, others involve the use of 7.5(wrong length) due to which you get the wrong answer.

[Instinct01]

(Original post by raheem94)
9.90cm is the correct length.

You have assumed the length marked in green to be 7.5cm, which it isn't.

Hence your second method is right, others involve the use of 7.5(wrong length) due to which you get the wrong answer.

Oh, of course yes! Just to clarify, are the two triangles of a parallelogram always isosceles?

Thanks.

[raheem94]

(Original post by Instinct01)
Oh, of course yes! Just to clarify, are the two triangles of a parallelogram always isosceles?

Thanks.

Not sure which triangles you are talking about.



May be you are talking about the green triangle, this isn't isosceles.