In this page 10th grade geometry solution3 we are going to see solutions of some practice questions.

(5) ABCD is a quadrilateral with AB parallel to CD. A line drawn parallel to AB meets AD at P and BC at Q. Prove that (AP/PD) = (BQ/QC)

Solution:

Join BD by intersecting the line PQ at the point Q.

In triangle DAB, PE and AB are parallel, by using “Thales theorem”

(AP/PD) = (BE/ED)  ------- (1)

In triangle BCD EQ and DC are parallel, by using “Thales theorem”

(BE/ED) = ( BQ/QC) ------- (2)

(1) = (2)

(AP/PD) = ( BQ/QC)

(6) In t he figure, PC and QK are parallel BC and HK are parallel, if AQ = 6 cm, QH = 4 cm, HP = 5 cm, KC = 18 cm, then find AK and PB.

Solution:

In triangle APC, the sides PC and QK are parallel

By using “Thales theorem” we get

(AQ/QP) = (AK/KC)

QP = QH + HP

= 4 + 5

= 9 cm

(6/9) = (AK/18)

AK = (6 x 18)/9

AK = 12 cm

In triangle ABC, the sides BC and HK are parallel,

By using “Thales theorem” we get

(AH/HB) = (AK/KC)

AH = AQ + QH

= 6 + 4

= 10

(10/HB) = (12/18)

(10 x 18)/12 = HB

HB = 15 cm

Now we need to find the length of PB,

PB = HB – HP

= 15 – 5

= 10 cm

(7) In the figure DE is parallel to AQ and DF is parallel to AR prove that EF is parallel to QR.

Solution:

In triangle PQA, the sides DE is parallel to the side AQ

By using “Thales theorem” we get

(PE/EQ) = (PD/DA)  ------ (1)

In triangle PAR, the sides DF is parallel to the side AR

By using “Thales theorem” we get

(PD/DA) = (PF/FR)  ------ (2)

(1) = (2)

(PE/EQ) = (PF/FR)

From this we can decide EF is parallel to QR in the given triangle PQR

(8) In the figure the sides DE and AB are parallel and DF and AC are parallel. Prove that EF and BC are parallel.

Solution:

In triangle APB, the sides DE and AB are parallel

(PD/DA) = (PE/EB) ----- (1)

In triangle PAC, the sides DF and AC are parallel

(PD/DA) = (PF/FC) ----- (2)

(1) = (2)

(PE/EB) = (PF/FC)

From the we can decide that the sides EF and BC are parallel.