Question

In the given figure, ABCD is a parallelogram. AB is produced to P, such that AB = BP and PQ is drawn parallel to BC to meet AC produced at Q. Given AB = 8 cm, AD = 5 cm, AC = 10 cm. (i) Prove that point C is mid point of AQ. (ii) Find the perimeter of quadrilateral BCQP.

Answer 1

Solution:

ABCD is a parallelogram.  AB is produced to P, such that AB = BP and PQ is drawn parallel to BC to meet AC produced at Q. Given AB = 8 cm, AD = 5 cm, AC = 10 cm.

→ BC ║ QP →→Given

In Δ ABC and ΔAPQ

∠ABC=∠APQ→→[BC ║ QP , BP is a transversal, so corresponding angles are equal]

∠BAC=∠PAQ→→Reflex angle

Δ ABC ~ ΔAPQ→→(AA similarity criterion]

When triangles are similar, their corresponding angles are equal.

$\frac{AB}{AP}=\frac{AC}{AQ}=\frac{BC}{PQ}\\\\ \frac{8}{16}=\frac{10}{10+QC}=\frac{5}{PQ}\\\\ QC=20-10=10\\\\ PQ=5 \times 2\\\\ PQ=10$

AC=CQ=10 cm, shows that point C is mid point of AQ.

(b) Perimeter of quadrilateral BCQP=BC +CQ+QP+BP

       = 5 + 10 +10+8

       = 33 cm

Answer 2

Answer:

Hope mine information will help you out in your doubt or clarification

ABCD is a parallelogram. AB is produced to P, such that AB = BP and PQ is drawn parallel to BC to meet AC produced at Q. Given AB = 8 cm, AD = 5 cm, AC = 10 cm.

→ BC ║ QP →→Given

In Δ ABC and ΔAPQ

∠ABC=∠APQ→→[BC ║ QP , BP is a transversal, so corresponding angles are equal]

∠BAC=∠PAQ→→Reflex angle

Δ ABC ~ ΔAPQ→→(AA similarity criterion]

When triangles are similar, their corresponding angles are equal.

\begin{gathered}\frac{AB}{AP}=\frac{AC}{AQ}=\frac{BC}{PQ}\\\\ \frac{8}{16}=\frac{10}{10+QC}=\frac{5}{PQ}\\\\ QC=20-10=10\\\\ PQ=5 \times 2\\\\ PQ=10\end{gathered}

AP

AB

=

AQ

AC

=

PQ

BC

16

8

=

10+QC

10

=

PQ

5

QC=20−10=10

PQ=5×2

PQ=10

AC=CQ=10 cm, shows that point C is mid point of AQ.

(b) Perimeter of quadrilateral BCQP=BC +CQ+QP+BP

= 5 + 10 +10+8

=33

Hope it will help you out

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