Question

(a) In the given figure, show that $\Delta AMP\cong \Delta AMQ$.




(b)

In the given figure, AC = CE and AB $\parallel$ ED. The value of $x$ is ___ units.
[4 MARKS]

Answer 1

Each part: 2 Marks


(a)

In $\Delta AMPand\Delta AMQ$,

$PM=QM$ [Given]

$\angle AMP=\angle AMQ$ [Given]

$AM=AM$ [Common side]

$\Rightarrow \Delta AMP\cong \Delta AMQ$ [SAS congruency criteria]


(b)


$\text{In}\Delta ABC\text{and}\Delta EDC,AC=CE\text{(given)}\angle BAC=\angle DEC\text{(since AB||DE and AE is a transversal, so they are alternate angles)}\angle ACB=\angle ECD\text{(vertically opposite angles)}\therefore \Delta ABC\cong \Delta EDC\text{(A.S.A. congruence criteria)}\therefore AB=DE\text{(sides of congruent triangles)}\therefore x+10=2x-5\Rightarrow x-2x=-5-10\Rightarrow -x=-15\Rightarrow x=15units$