Sec 2 G2 Maths Pythagoras' theorem worksheet with answers
A free Sec 2 G2 Maths worksheet on pythagoras' theorem: 6 questions, 30 marks, short questions first and then structured ones, in the school exam style. The answers and full worked solutions are on the last page.
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Answer all questions. Show all your working. Give non-exact answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless the question says otherwise.
Section A
1
In the diagram, ∠STU=90∘. Find x.
Answercm[3]
2
A boat travels 8 km due north from P to Q, then 11 km due east from Q to R. Find the distance PR, x km, correct to 3 significant figures.
Answerkm[3]
3
In the diagram, ∠XYZ=90∘. Find x.
Answercm[3]
Section B
4
(a)In triangle STU, ∠STU=90∘, SU=7.4 cm and ST=4.5 cm. Find the length of TU, correct to 3 significant figures.
Answer (a)cm[3]
(b)In the diagram, ∠STU=90∘. Find x.
Answer (b)cm[3]
5
(a)In triangle KLM, KL=16 cm, LM=x cm and KM=y cm, where LM is the longest side. It is given that (x+y)(x−y)=256. Determine whether triangle KLM is right-angled. Choose the answer with its reason.
Answer (a)[2]
(b)A ladder 3.5 m long leans against an upright wall. Its foot is on level ground, 0.8 m from the wall. Find the height, h m, that the ladder reaches up the wall, correct to 3 significant figures.
Answer (b)m[2]
(c)In triangle LMN, ∠LMN=90∘, LN=12.5 cm and LM=7.2 cm. Find the length of MN, correct to 3 significant figures.
Answer (c)cm[3]
6
(a)In the diagram, ∠XYZ=90∘ and W lies on YZ. XW=25 cm, YW=7 cm and WZ=9 cm. Find x, correct to 3 significant figures.
Answer (a)cm[3]
(b)(i)In triangle STU, ST=2 cm, TU=2.5 cm and SU=1.5 cm. Use the converse of Pythagoras' theorem to find which angle of triangle STU is a right angle.
Answer (b)(i)[3]
(b)(ii)In triangle ABC, AB=6 cm, BC=x cm and AC=y cm, where BC is the longest side. It is given that (x+y)(x−y)=34. Determine whether triangle ABC is right-angled. Choose the answer with its reason.
Answer (b)(ii)[2]
END OF PAPER
Answers and worked solutions
Question 1
Answer: =15.8 cm (3 s.f.)[3]
The hypotenuse is SU, opposite the right angle at T.SU2=ST2+TU2x2=132+92=169+81=250x=250=15.8113…=15.8 cm (3 s.f.)
Question 2
Answer: =13.6 km (3 s.f.)[3]
∠PQR=90∘, so PR is the hypotenuse.x2=82+112=185x=185=13.6014…=13.6 km (3 s.f.)
Question 3
Answer: Check: 7.42<8, shorter than the hypotenuse. cm[3]
The hypotenuse is XZ, opposite the right angle at Y.XZ2=XY2+YZ2x2=82−32=64−9=55x=55=7.4161…=7.42 cm (3 s.f.)Check: 7.42<8, shorter than the hypotenuse.
Question 4
(a)Answer: Check: 5.87<7.4, shorter than the hypotenuse. cm[3]
The hypotenuse is SU, opposite the right angle at T.SU2=ST2+TU2TU2=7.42−4.52=54.76−20.25=34.51TU=34.51=5.8745…=5.87 cm (3 s.f.)Check: 5.87<7.4, shorter than the hypotenuse.
(b)Answer: =15.8 cm (3 s.f.)[3]
The hypotenuse is SU, opposite the right angle at T.SU2=ST2+TU2x2=132+92=169+81=250x=250=15.8113…=15.8 cm (3 s.f.)
Question 5
(a)Answer: Yes[2]
(x+y)(x−y)=x2−y2, so x2−y2=256.162=256x2−y2=256x2−y2=162, so x2=y2+162: triangle KLM is right-angled, with the right angle opposite LM, at K.
(b)Answer: =3.41 m (3 s.f.)[2]
The hypotenuse is the ladder, opposite the right angle at the foot of the wall.h2=3.52−0.82=11.61h=11.61=3.4073…=3.41 m (3 s.f.)
(c)Answer: Check: 10.2<12.5, shorter than the hypotenuse. cm[3]
The hypotenuse is LN, opposite the right angle at M.LN2=LM2+MN2MN2=12.52−7.22=156.25−51.84=104.41MN=104.41=10.2181…=10.2 cm (3 s.f.)Check: 10.2<12.5, shorter than the hypotenuse.
Question 6
(a)Answer: =28.8 cm (3 s.f.)[3]
XY2=252−72=576XY=576=24YZ=7+9=16x2=242+162=832x=832=28.8444…=28.8 cm (3 s.f.)
(b)(i)Answer: ∠TSU[3]
ST2=22=4TU2=2.52=6.25SU2=1.52=2.25The longest side is TU.2.25+4=6.25TU2=SU2+ST2, so by the converse of Pythagoras' theorem the angle opposite TU is 90∘: ∠TSU=90∘.
(b)(ii)Answer: No[2]
(x+y)(x−y)=x2−y2, so x2−y2=34.62=36x2−y2=34, but 62=36, so x2=y2+62: triangle ABC is not right-angled.