Question

11. The triangle ABC is right-angled at C. From P, a
point on the hypotenuse, PQ is drawn parallel to
AC cutting BC at Q. If AC = 2.5 cm, BC = 6cm
and PQ = 1 cm, find
(i) BQ (ii) BP
(ICSE)​

Answer 1

$\textbf{Given:}$

$\textsf{In right angled triangle ABC, from P on hypotenuse PQ is drawn parallel to AC }$

$\mathsf{AC=2.5\,cm,\;BC=6\,cm,\;PQ=1\,cm}$

$\textbf{To find:}$

$\textsf{(i)\;BQ\;\;(ii)\;BP}$

$\textbf{Solution:}$

$\mathsf{In\;right\;\triangle\;ACB}$

$\mathsf{AB^2=AC^2+BC^2}$

$\mathsf{AB^2=2.5^2+6^2}$

$\mathsf{AB^2=6.25+36}$

$\mathsf{AB^2=42.25}$

$\mathsf{AB=\sqrt{42.25}}$

$\implies\mathsf{AB=6.5}$

$\textbf{In triangle ACB and PQB}$

$\mathsf{\angle{A}=\angle{BPQ}}$

$\mathsf{\angle{C}=\angle{PQB}}$

$\mathsf{\angle{B}=\angle{QBP}}$

$\textsf{By AAA similarity,}$

$\mathsf{\triangle\,ACB\;\;and\;\;\triangle\,PQB\;are\;similar}$

$\therefore\textsf{Their corresponding sides are proportional}$

$\mathsf{\dfrac{AC}{PQ}=\dfrac{BC}{BQ}=\dfrac{AB}{BP}}$

$\mathsf{\dfrac{2.5}{1}=\dfrac{6}{BQ}=\dfrac{6.5}{BP}}$

$\implies\mathsf{BQ=\dfrac{6}{2.5}}$

$\implies\boxed{\mathsf{BQ=2.4\;cm}}$

$\mathsf{and}$

$\mathsf{\dfrac{2.5}{1}=\dfrac{6.5}{BP}}$

$\mathsf{BP=\dfrac{6.5}{2.5}}$

$\implies\boxed{\mathsf{BP=2.6}}$

$\textbf{Find more:}$

L and M are points on sides AB and AC of a triangle ABC,if AL = 2cm, LB = 4cm, LM||BC. Prove that 3LM = BC.

https://brainly.in/question/32668624

Answer 2

Step-by-step explanation:

above answer is very correct