Sec 2 G1 Maths Pythagoras' theorem worksheet with answers
A free Sec 2 G1 Maths worksheet on pythagoras' theorem: 8 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, ∠DEF=90∘. Find x.
Answercm[3]
2
In the diagram, ∠DEF=90∘. Find x.
Answerm[3]
3
In triangle KLM, KL=24 cm, LM=10 cm and KM=26 cm. Use the converse of Pythagoras' theorem to find which angle of triangle KLM is a right angle.
Answer[3]
4
In the diagram, ∠DEF=90∘. Find x.
Answercm[3]
5
In triangle ABC, AB=25 cm, BC=15 cm and AC=20 cm. Determine whether triangle ABC is a right-angled triangle.
Answer[3]
6
In the diagram, ∠ABC=90∘. Find x.
Answerm[3]
Section B
7
(a)In the diagram, ∠KLM=90∘. Find x.
Answer (a)cm[3]
(b)In the diagram, ∠STU=90∘. Find x.
Answer (b)m[3]
8
(a)In triangle ABC, AB=25 cm, BC=15 cm and AC=20 cm. Determine whether triangle ABC is a right-angled triangle.
Answer (a)[3]
(b)In the diagram, ∠KLM=90∘. Find x.
Answer (b)m[3]
END OF PAPER
Answers and worked solutions
Question 1
Answer: Check: 25<65, shorter than the hypotenuse. cm[3]
The hypotenuse is DF, opposite the right angle at E.DF2=DE2+EF2x2=652−602=4225−3600=625x=625=25 cmCheck: 25<65, shorter than the hypotenuse.
Question 2
Answer: =25 m[3]
The hypotenuse is DF, opposite the right angle at E.DF2=DE2+EF2x2=152+202=225+400=625x=625=25 m
Question 3
Answer: ∠KLM[3]
KL2=242=576LM2=102=100KM2=262=676The longest side is KM.100+576=676KM2=LM2+KL2, so by the converse of Pythagoras' theorem the angle opposite KM is 90∘: ∠KLM=90∘.
Question 4
Answer: Check: 13.2<20, shorter than the hypotenuse. cm[3]
The hypotenuse is DF, opposite the right angle at E.DF2=DE2+EF2x2=202−152=400−225=175x=175=13.2287…=13.2 cm (3 s.f.)Check: 13.2<20, shorter than the hypotenuse.
Question 5
Answer: YES[3]
AB2=252=625BC2=152=225AC2=202=400The longest side is AB.225+400=625AB2=BC2+AC2, so triangle ABC is right-angled (converse of Pythagoras' theorem). Yes.
Question 6
Answer: =5 m[3]
The hypotenuse is AC, opposite the right angle at B.AC2=AB2+BC2x2=32+42=9+16=25x=25=5 m
Question 7
(a)Answer: =18.4 cm (3 s.f.)[3]
The hypotenuse is KM, opposite the right angle at L.KM2=KL2+LM2x2=42+182=16+324=340x=340=18.4390…=18.4 cm (3 s.f.)
(b)Answer: Check: 24<51, shorter than the hypotenuse. m[3]
The hypotenuse is SU, opposite the right angle at T.SU2=ST2+TU2x2=512−452=2601−2025=576x=576=24 mCheck: 24<51, shorter than the hypotenuse.
Question 8
(a)Answer: YES[3]
AB2=252=625BC2=152=225AC2=202=400The longest side is AB.225+400=625AB2=BC2+AC2, so triangle ABC is right-angled (converse of Pythagoras' theorem). Yes.
(b)Answer: Check: 10.2<11, shorter than the hypotenuse. m[3]
The hypotenuse is KM, opposite the right angle at L.KM2=KL2+LM2x2=112−42=121−16=105x=105=10.2469…=10.2 m (3 s.f.)Check: 10.2<11, shorter than the hypotenuse.