StudentAidUnit 9: Similar Figures — Exercise 9 1
9th Class Mathematics · Unit 9: Similar Figures
9.1.1.Find whether the solids are similar. All lengths are in cm. [Rectangular box 1 has edges 3, 5, 4; rectangular box 2 has edges 4.5, 7.5, 6.]
Given
Corresponding edges
3leftrightarrow4.5, 5leftrightarrow7.5, 4leftrightarrow6
Formula
Ratio of corresponding sides
4.53, 7.55, 64
4.53=7.55=64=1.5
Result
Conclusion
Since all ratios are equal, the figures are similar.
9.1.2.In triangle ABC, the sides are given as mAB = 6 cm, mBC = 9 cm and mCA = 12 cm. In triangle DEF, the sides are given as mDE = 10.5 cm, mEF = 15.75 cm, and mFD = 21 cm. Prove that the triangles are similar.
Formula
Ratio of corresponding sides
2112=10.56=15.759
2112=1.75, 10.56=1.75, 15.759=1.75
Result
Conclusion
Hence the given triangles are similar.
9.1.3.In the given figure, ΔABC ~ ΔDEF, mAB = 12 cm, mAC = 20 cm and mBC = 16 cm. In ΔDEF, mDE = 6 cm. Find mDF and mEF.
Given
ΔABC sim ΔDEF
Formula
Corresponding sides
frac{moverline{AB}{moverline{DE}=frac{moverline{BC}{moverline{EF}
Working
126=frac{16}{moverline{EF}
Result
moverline{EF}=8 cm
Formula
Corresponding sides
frac{moverline{AB}{moverline{DE}=frac{moverline{AC}{moverline{DF}
Working
126=frac{20}{moverline{DF}
Result
moverline{DF}=10 cm
9.1.4.Find the value of x in each of the following: (i) A right-angled figure where AE = 7.5 cm, EC = 3 cm, DC = 1.2 cm, and AB = x (with ∠A = ∠C = 90°, lines BD and AC crossing at E). (ii) Triangle BCD with E on BC (BE = 6 cm, EC = x), F on BD (BF = 8 cm, FD = 3 cm), and EF ∥ CD. (iii) Figure with CD = x cm, DE = 2.5 cm, EF = 2.4 cm, FG = 2.1 cm, and ∠C = ∠G.
4(i) since figures are similar
7.53=x1.2 ⇒ x=7.53 × 1.2 ⇒ x=3 cm
4(ii) since figures are similar
8+38=6+x6 ⇒ 6+x=118 × 6 ⇒ x=2.25 cm
4(iii) since figures are similar
x2.1=2.52.4 ⇒ x=2.52.4 × 2.1 ⇒ x=2.19 cm
(i) x=3 cm; (ii) x=2.25 cm; (iii) x=2.19 cm
9.1.5.A plank is placed straight upstairs that 20 cm wide and 16 cm deep. A rectangular box of height 8 cm and width x cm is placed on a stair under the plank. Find the value of x.
Given
ΔABC sim ΔAFD
Working
BCDF=ABAF ⇒ 168=20AF ⇒ AF=10 cm
AB=AF+FB ⇒ 20=10+FB ⇒ FB=10 cm
Result
x=10 cmquad(because DE=FB)
9.1.6.A man who is 1.8 m tall casts a shadow of a 0.76 m in length. If at the same time a telephone pole casts a 3 m shadow, find the height of the pole.
Formula
Since figures are similar
30.76=x1.8
x=30.76 × 1.8
Result
x=7.11 m
9.1.7.Find the values of x, y and z in the given figure. [Right triangle ABC, right angle at B; D lies on AC with BD ⊥ AC; AB = 10 cm, AD = 6 cm; DC = x, BD = y, BC = z.]
Given
ΔABC sim ΔABD
Working
ABAD=ACAB ⇒ 106=x+610 ⇒ 6x+36=100
Result
x=1023 cm=10.667 cm
In ΔABD, Pythagoras Theorem
H2=P2+B2 ⇒ (10)2=(6)2+y2 ⇒ y2=64 ⇒ y=8 cm
In ΔABC, Hyp = AC
AC=x+6=10.667+6=16.667 cm
In ΔABC, Pythagoras Theorem
(16.667)2=(10)2+z2 ⇒ z2=177.778
z=1313 cm=13.334 cm
9.1.8.Draw an isosceles trapezoid ABCD where AB ∥ CD and mAB > mCD. Draw diagonals AC and BD, intersecting at E. Prove that ΔABE is similar to ΔCDE. If mAB = 8 cm, mCD = 4 cm, and mAE = 3 cm, find the length of CE.
Given
ΔABE sim ΔCDE
Formula
ABCD=AECE
Working
84=3x
Result
x=1.5 cm
9.1.9.A regular dodecagon has its side lengths decreased by a factor of 1/√2. If the perimeter of the original dodecagon is 72 cm, what is the side length of the scaled dodecagon?
Given
Dodecagon has 12 sides; Perimeter=72 cm
12x=72 ⇒ x=6 cm
Formula
Scaled side length
S'=x × 1sqrt2
Result
S'=6 × 1sqrt2=6sqrt22=3sqrt2 cm