**10th Grade Geometry Solution3**

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:**

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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.

10th grade geometry solution3 10th grade geometry solution3