5.18H Similar Shapes (Lengths): Higher Tier Practice

EDEXCEL Edexcel 1MA1 · Calculator allowed · about 75 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Two similar triangles have corresponding sides in the ratio 5:8. A side of the smaller triangle is 15 cm. Work out the length of the corresponding side in the larger triangle.

(Total for Question 1 is 2 marks)

Triangle ABC is similar to triangle DEF. AB = 12 cm, BC = 18 cm, DE = 20 cm. Find EF. Give your answer correct to 1 decimal place.

(Total for Question 2 is 3 marks)

Triangle PQR is similar to triangle STU. In triangle PQR, PR = 14 cm and QR = 21 cm. In triangle STU, SU = 10 cm.
(a) Find the scale factor from triangle PQR to triangle STU (give as a decimal). (1 mark)
(b) Find TU, the side corresponding to QR. (3 marks)

(a)Find the scale factor from triangle PQR to triangle STU (give as a decimal).(1)

(b)Find TU, the side in triangle STU corresponding to QR.(3)

(Total for Question 3 is 4 marks)

Two similar rectangles have one pair of corresponding sides in the ratio (x + 2) : (3x). This pair of sides has length 8 cm in the smaller rectangle and length 12 cm in the larger rectangle (so 8 cm corresponds to x + 2 and 12 cm corresponds to 3x). The smaller rectangle's other side is 10 cm. Work out x and the larger rectangle's side corresponding to the 10 cm side.

(a)Find x.(2)

(b)Find the length in the larger rectangle that corresponds to the smaller rectangle's 10 cm side. Give your answer in cm.(2)

(Total for Question 4 is 4 marks)

A right-angled triangle ABC (right angle at B) is enlarged from triangle ABC to triangle A'B'C' with centre of enlargement at B and scale factor k. The length BA = 7 cm and BC = 24 cm. After enlargement BA' = 21 cm.
(a) Find k. (1 mark)
(b) Find B'C', the image of BC, and state whether triangle A'B'C' is still right-angled at B'. Explain briefly.

(a)Find k.(1)

(b)Find B'C' and state whether triangle A'B'C' is still right-angled at B'. Explain briefly.(3)

(Total for Question 5 is 4 marks)

Mark scheme · 5.18H Similar Shapes (Lengths): Higher Tier Practice

Question 1

M1 multiply 15 by scale factor 8/5 or show equivalent method
A1 24 cm cao
Answer: 24 cm, because 15 x (8/5) = 24 (scale factor 8/5).

Question 2

M1 identify scale factor = DE/AB = 20/12 = 5/3 and apply to BC or set up proportion EF/BC = DE/AB
M1 compute EF = 18 x (5/3) or equivalent
A1 30.0 cm awrt to 1 dp
Answer: 30.0 cm

Question 3

(a) B1 scale factor SU/PR = 10/14 = 5/7 = 0.714285... or exact fraction 5/7
(a) Answer: 5/7 or 0.714285... (SU corresponds to PR: 10/14 = 5/7)
(b) M1 use TU = QR x (5/7) or set up correct proportion
(b) M1 calculate TU = 21 x 5/7
(b) A1 15 cm cao
(b) Answer: 15 cm

Question 4

(a) M1 form equation (x+2)/(3x) = 8/12 and simplify to (x+2)/(3x) = 2/3
(a) A1 x = 2 cao
(a) Answer: x = 2
(b) M1 use scale factor = 3x/(x+2) with x = 2, so scale factor = 6/4 = 3/2 (consistent with 12/8 = 3/2) and apply to 10
(b) A1 15 cm cao
(b) Answer: 15 cm

Question 5

(a) B1 k = BA'/BA = 21/7 = 3
(a) Answer: k = 3
(b) M1 use B'C' = k x BC = 3 x 24
(b) A1 72 cm cao for B'C'
(b) B1 explain: enlargement from B preserves the angle at B so triangle A'B'C' is right-angled at B' oe
(b) Answer: B'C' = 72 cm; triangle A'B'C' is still right-angled at B' because an enlargement with centre at B preserves the angle at B (same rays), so the right angle is preserved.

Similar Shapes (Lengths): Higher Tier Practice - Worksheets, Questions and Revision