In this page 10th grade geometry solution2 we are going to see solutions of some practice questions.

(2) In the figure AP = 3 cm, AR = 4.5 cm, AQ = 6 cm, AB = 5 cm and AC = 10 cm. Find the length of AD.

**Solution:**

From the given information we get, (AB/AP) = (AC/AQ)

In triangle ABC, (AB/AP) = (AC/AQ)

By using converse of “Thales theorem” PQ is parallel to BC.

RD = x

In triangle ABD, PR is parallel to BD

AD = AR + RD

AD = 4.5 + x

(AB/AP) = (AD/AR)

(5/3) = (4.5 + x)/4.5

(5 **x** 4.5)/3 = 4.5 + x

7.5 = 4.5 + x

x = 7.5 – 4.5

x = 3

Here we need to find the length of AD = 4.5 + x

= 4.5 + 3

= 7.5 cm

(3) E and F are points on the sides PQ and PR respectively, of a triangle PQR. For each of the following cases. Verify EF is parallel to QR.

(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

**Solution:**

First let us draw the picture for the above details

To verify whether EF is parallel to QR we have to check the condition
(PE/EQ) = (PF/FR) (3.9/3) = (3.6/2.4) 1.3 ≠ 1.5 So the sides EF and QR are not parallel. |

(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm

**Solution:**

First let us draw the picture for the above details

To verify whether EF is parallel to QR we have to check the condition
(PE/EQ) = (PF/FR) (4/4.5) = (8/9) 0.88 = 0.88 So the sides EF and QR are parallel. |

(4) In the figure,AC is parallel to BD and CF is parallel to DF, if OA = 12 cm, AB = 9 cm, OC = 8 cm and EF= 4.5 cm, then find FQ.

**Solution:**

In triangle OBD, AC is parallel to BD

By using “Thales theorem” we get,

(OA/AB) = (OC/CD)

(12/9)= (8/CD)

CD = (8 **x**** **9)/12

= 72/12

= 6 cm

In triangle ODF, the sides CE and DF are parallel, by using “Thales theorem” we get,

(OC/CD)= (OE/EF)

(8/6) = (OE/4.5)

OE = (8** x** 4.5)/6

= 36/6

= 6 cm

So, OF = OE + EF

= 6 + 4.5

= 10.5 cm

10th grade geometry solution2 10th grade geometry solution2

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