Question

In the figure AC= 3 cm and CB=6 cm, find CR.

Answer 1

Given:

AC = 3 m

CB = 6 cm

AP = 4.5 cm

BQ = 7.5 cm

To find:

The value of CR

Solution:

$\bigstar$ Note: There are some missing data in the given figure, therefore I have attached the correct figure below $\bigstar$

Let AQ intersect CR at K.

From the figure we have,

AP // CR // BQ

So,

CK // BQ and KR // AP

In ΔAKC & ΔAQB, we have

∠QAB = ∠KAC ....... [common angles]

∠AKC = ∠AQB ....... [corresponding angles]

∴ ΔAKC ~ ΔABQ ..... [AA Similarity]

⇒ $\bold{\frac{AK}{AQ} = \frac{CK}{QB} = \frac{AC}{AB}}$ ....... (i) ..... [corresponding sides of similar triangles are proportional]

∴ $\frac{CK}{QB} = \frac{AC}{AB}}$ ..... [from (i)]

substituting given values of AC = 3 cm, QB = 7.5 cm and CB = 6 cm

$\implies \frac{CK}{QB} = \frac{AC}{AC + CB}}$

$\implies \frac{CK}{7.5} = \frac{3}{3 + 6}}$

$\implies \frac{CK}{7.5} = \frac{3}{9}}$

$\implies CK = \frac{1}{3} \times 7.5$

$\implies \bold{CK = 2.5 \:cm }$ ...... (ii)

Also,

$\frac{AK}{AQ} = \frac{AC}{AB}$ ..... [from (i)]

$\implies \frac{AK}{AQ} = \frac{3}{9}$

$\implies \frac{AK}{AQ} = \frac{1}{3}$

subtracting both the ratios from 1 on both sides

$\implies 1 - \frac{AK}{AQ} = 1 - \frac{1}{3}$

$\implies \frac{AQ - AK}{AQ} = \frac{3 - 1}{3}$

$\implies \bold{\frac{QK}{AQ} = \frac{2}{3}}$ ....... (iii)

In ΔQRK & ΔQPA, we have

∠RQK = ∠PQA ....... [common angles]

∠QRK = ∠QPA ....... [corresponding angles]

∴ ΔQRK ~ ΔQPA ..... [AA Similarity]

∵ corresponding sides of similar triangles are proportional

$\implies \frac{QK}{AQ} = \frac{KR}{AP}$

substituting the values from (iii) and AP = 4.5 cm

$\implies \frac{2}{3} = \frac{KR}{4.5}$

$\implies KR = \frac{2}{3} \times 4.5$

$\implies \bold{KR = 3 \:cm }$ ...... (iv)

We know,

CR = CK + KR

substituting from (ii) & (iv)

⇒ CR = 2.5 + 3

⇒ CR = 5.5 cm

Thus, the length of CR is → 5.5 cm.

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