Question

Draw a line segment AB=7.5cm and locate a point P on AB such that AP=3/7AB.Give justification of the

construction.​

Answer 1

Answer:

In △ABC AB=AC

⇒∠B=∠C (Angles opposite to equal sides are equal)

Now using angle sum property

∠A+∠B+∠C=180

∘

⇒80

∘

+∠C+∠C=180

∘

⇒2∠C=180

∘

−80

∘

⇒∠C=

2

100

∘

=50

∘

now ∠C+∠x=180

∘

(Angles made on straight line (AC) are supplementary)

⇒50

∘

+∠x=180

∘

⇒∠x=180

∘

−50

∘

=130

∘

yo

Answer 2

Answer:

Given:

AP=3/7×AB

Here , AB = 7.5 cm

AP=3/7×7.5

= 3.2cm

Hence, the AP=3.2 cm.