Mae gan y Tîm Allgymorth broblem.

Maent wedi cael tŷ pâr i'w droi'n "Hwb Galw Heibio STEM" newydd sbon, ond mae'r tu allan yn edrych... wel... heb ei baentio ers yr 1980au. Cyn yr agoriad mawreddog, mae angen cot newydd o baent ar yr holl le.

Dim ond un anhawster sydd: nid yw'r tŷ yn un bocs syml. Mae gan y blaen ddau dalcen trionglog, mae gan y cefn ystafell wydr mewn siâp fel hanner octagon, a phenderfynodd rhywun yn y 1990au fod to hanner cromen yn syniad da. Dim ond tair wal y gellir eu paentio - mae'r bedwaredd wal yn sownd yn nhŷ'r cymydog.

Eich tasg yw gweithio allan yn union faint o arwynebedd wal sydd angen ei baentio, er mwyn i'r tîm allu archebu'r swm cywir o baent (ac osgoi achos arall o "redeg allan hanner ffordd i fyny'r wal").

Cliciwch ar y botwm 'nesaf' isod i weld cynrychiolaeth fras o'r adeilad y mae angen ei beintio.

Dimensiynau Adeiladu

Ein tasg gyntaf yw cyfrifo arwynebedd cyffredinol tu blaen, ochr a chefn yr adeilad.

Nid yw unrhyw un o'r delweddau wedi eu creu yn ôl graddfa.

Dewiswch wal i ddechrau arni:

Lleihau Gwastraff

Gallem arbed rhywfaint o amser i ni ein hunain a phrynu digon o baent i roi dwy got i'r arwynebedd cyffredinol.

Fodd bynnag, mae'r gyllideb yn dynn. Po fwyaf rydym ni'n ei wario ar y paent, y lleiaf o arian sydd ar ôl ar gyfer y gweithgareddau y byddwn yn eu cynnwys y tu mewn i'r adeilad.

Dimensiynau'r Nodweddion

Isod mae manylion pob un o'r nodweddion ar y waliau nad ydynt yn cael eu paentio.

• Mae'r 5 ffenestr lai ar y wal flaen yn mesur 1m x 1.5m yr un.

• Mae'r ddwy ffenestr flaen fwyaf yn mesur 3m x 1.5m.

• Mae'r ffenestri bwaog (un ar bob talcen) yr un dimensiynau â'r ffenestri llai ar y tu blaen ond gyda hanner cylch ar eu pennau.

• Mae'r ddau ddrws yn yr adeilad yn 0.9m x 2.1m.

• Mae diamedr o 1.6m i logo cylch y Brifysgol ar y wal ochr.

• Mae'r ffenestr gefn yn 1.5m x 1.5m.

• Mae'r ystafell wydr yn cynnwys hanner octagon rheolaidd lle mae hyd bob ochr yn 1.2m. Mae paneli'r ystafell wydr hefyd yn 2.6m o uchder.

• Mae gan y gromen ar ben yr ystafell wydr yr un diamedr â lled yr ystafell wydr.

I gael help gydag unrhyw ran o'r geometreg/trigonometreg sy'n gysylltiedig â'r cyfrifiadau arwynebedd hyn - defnyddiwch y botwm 'nesaf' ar waelod y dudalen hon.

Eich Her:

Nodir fod y paent maen a ddewiswyd yn gorchuddio 10m² y litr.

Bydd angen i ni roi dwy got o baent.

Faint o litrau sydd eu hangen?

Mae'r paent yn costio £9.50 y litr.

Faint ydyn ni wedi'i arbed o werth arwynebedd cyffredinol gwreiddiol y wal trwy ystyried y nodweddion?

Bydd yr adran nesaf yn rhoi arweiniad ar sut i fynd i'r afael â'r broblem hon, os bydd ei angen arnoch.

Rhywfaint o Ganllawiau

Rydym wedi llunio'r canllawiau canlynol ar sut i gyfrifo arwynebedd ar gyfer y siapiau angenrheidiol.

Cyfrifiadau Arwynebedd:

Petryalau/Sgwariau: Gellir cyfrifo'r arwynebedd trwy luosi'r hyd a'r lled - neu, yn yr achos hwn, y lled a'r uchder.

Trionglau Isosgeles: Lluoswch yr uchder gyda hanner y gwaelod.

Trionglau Ongl Sgwâr: Lluoswch hyd y ddwy ochr sy'n gysylltiedig gyda'r ongl dde a rhannwch gyda 2.

Cylchoedd a hanner cylchoedd: Gellir canfod arwynebedd cylch trwy ddefnyddio πr² (pi (3.14) wedi'i luosi â'r radiws wedi'i sgwario). Yna gellir rhannu hyn gyda dau i roi arwynebedd hanner cylch.

Yr Ystafell Wydr:

Rydym wedi penderfynu ychwanegu hwn fel her ychwanegol. Bydd angen i chi ystyried croestoriad yr hanner octagon a'r gromen yn eu pwynt canolog gan mai dyma arwynebedd y wal y mae'n eu blocio. Pa siapiau fydden nhw'n eu creu?

I ddechrau, bydd angen i chi gyfrifo cyfanswm y lled ar draws canol octagon. I wneud hyn, defnyddiwch yr hafaliad: cyfanswm lled = 2.613 s lle mae s yn hyd un ochr. Mae'r rhif 2.613 yn gymhareb sefydlog ar gyfer pob octagon rheolaidd wrth gyfrifo'r lled ar draws y canol.

Pwyntiau Ychwanegol

• Rydym wedi rhannu'r holl wybodaeth angenrheidiol ar draws gwahanol adrannau a delweddau'r dudalen we hon - felly, efallai y bydd angen i chi fynd yn ôl ac ymlaen rhwng adrannau.

• Gwnewch yn siŵr fod pob arwynebedd mewn metrau sgwâr cyn ichi gyfrifo faint o baent sydd ei angen.

• Peidiwch â gwneud y cyfrifiad ar gyfer dwy got tan ar ôl i chi gael cyfanswm yr holl rannau perthnasol a chael gwared ar y nodweddion ychwanegol.

• Mae'r wal sy'n cael ei chuddio gan yr ystafell wydr hanner octagon yn betryal gyda'r to crwm yn ychwanegu hanner cylch ar ei ben.

• Gwerthir llawer o wahanol fathau a brandiau o baent. Mae bob un yn gorchuddio'n wahanol ac angen nifer gwahanol o gotiau, ar gyfer y gweithgaredd hwn rydym wedi dewis gwerthoedd realistig ar gyfer cost ac ansawdd y paent.

• I addurno, argymhellir eich bod chi'n prynu 10% yn fwy o baent nag a gyfrifwyd - mae hyn yn rhoi digon o le ar gyfer gwallau ac yn sicrhau na fydd rhaid prynu paent ychwanegol yn nes ymlaen fydd yn dod o gymysgiad gwahanol (felly nid yw'n sicr o fod yr un lliw yn union).

The Outreach Team has a problem.

They've been given a semi-detached house to convert into a brand-new "STEM Drop-In Hub", but the outside looks... well... like it hasn't been painted since the 1980s. Before the grand opening, the whole place needs a fresh coat of paint.

There's just one catch: the house isn't a simple box. The front has two gables (triangular peaks), the back has a conservatory shaped like half an octagon, and someone in the 1990s decided a half-dome roof was a good idea. Only three walls can be painted - the fourth is attached to the neighbour's house.

Your mission is to work out exactly how much wall area needs painting, so the team can order the right amount of paint (and avoid another "we ran out halfway up the wall" incident).

Click on the 'next' button below to see a rough representation of the building that needs painting.

Building Dimensions

Our first task is to calculate the overall areas of the front, side, and back of the building.

None of the images provided are to scale.

Select a wall to get started:

Minimising Wastage

We could save ourselves some time and purchase paint enough to apply two coats to the overall areas.

However, we're on a tight budget. The more we spend on the paint, the less we have for the activities we'll be including inside.

Feature Dimensions

Below are the details of each of the features within the walls that are not painted.

• The 5 smaller windows on the front wall each measure 1m x 1.5m.

• The two larger front windows measure 3m x 1.5m.

• The arched windows (one in each gable) are the same dimensions as the smaller windows on the front but with a semi-circular area on top.

• Both doors in the building are 0.9m x 2.1m.

• The University's logo embossed on the side wall is a circular area with a diameter of 1.6m.

• The back window is 1.5m x 1.5m.

• The conservatory is made up of half a regular octagon where each side is 1.2m length. These conservatory panels are also 2.6m tall.

• The dome on the top of the conservatory has the same diameter as the width of the conservatory.

For help with any of the geometry/trigonometry involved in these area calculations - use the 'next' button at the bottom of this page.

Your Challenge:

The masonry paint we have selected states that it covers 10m² per litre.

We will need to apply two coats of paint.

How many litres do we need?

The paint costs £9.50 per litre.

How much have we saved from the original overall wall area values by taking the features into consideration?

The next section will provide guidance for how to approach this problem, if you need it.

Some Guidance

We've put together the following guidance for how to calculate areas for the necessary shapes.

Area Calculations:

Rectangles/Squares: The area can be calculated by multiplying the width and the length - or, in this case, the width and the height.

Isosceles Triangles: Multiply the height by half the base.

Right-angled Triangles: Multiply the length of both sides connected by the right-angle and divide by 2.

Circles and Semi-circles: The area of a circle is found using πr² (pi (3.14) multiplied by the radius squared). This can then be divided by two to give the area of a semi-circle.

The Conservatory:

We decided to add this as an additional challenge. You will need to consider the cross-section of the half-octagon and the dome at their mid-points as this is the area of the wall blocked by it. What shapes would they create?

To get you started, you will need to calculate the total width across the middle of an octagon. To do this, use the equation: total width = 2.613s where s is the length of one side. The number 2.613 is a fixed ratio for all regular octagons when calculating the width across the middle.

Additional Pointers

• We have spread all the necessary information across the different sections and images of this webpage - so, you may need to flip back and forth between sections.

• Make sure all your areas are in square metres before working out the paint needed.

• Don't do the two-coat calculation until after you've totalled up all the relevant areas and removed the feature areas.

• The wall area covered by the half-octagon conservatory is a rectangle with the domed roof adding a semi-circle on top.

• Paints come in a number of different types and brands. Each has its own coverage and coat requirements, for this activity we've chosen realistic values for both cost and coverage.

• If decorating it is recommended that you buy 10% more than calculated - this is to provide leeway for error and ensure you do not end up buying additional paint at a later date which is from a different batch (therefore not guaranteed to match exactly).