Question

In right angle A ABC, AB = 8cm, AC = 10 cm and B = 90° P and Q are points on the sides AB and
AC respectively such that PQ = 2 cm and Z PQA = 90°.
Find i. the area of A AQP. ii. area of quad. PBCQ: area of A ABC.
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Answer 1

Given :- In right triangle ABC, AB = 8cm ,AC = 10cm and angle B = 90°. P And Q are point on AB and AC respectively such that PQ=2cm and angle PQA = 90° .

To Find :-

(i) Area of ∆ABC.

(ii) Area of ∆AQP .

(iii) Area or quad. PBCQ .

Solution :-

given that, ABC is a right angled ∆, right angle at B. using pythagoras theorem we get,

→ BC² = AC² - AB²

→ BC = √(10² - 8²)

→ BC = √(100 - 64)

→ BC = √36 = 6cm.

So,

→ Area of ∆ABC = (1/2) * Base * Perpendicular = (1/2) * 8 * 6 = 24 cm². (Ans.)

Now in ∆ABC and ∆AQP we have ,

→ ∠ABC = ∠AQP (90° both)

→ ∠BAC = ∠QAP = θ .(common .)

therefore,

→ ∆ABC ~ ∆AQP

hence,

→ AB/AQ = BC/QP = AC/AP .

putting value we get,

→ AB/AQ = BC/QP

→ 8/AQ = 6/2

→ 8/AQ = 3

→ AQ = (8/3) cm.

Therefore,

→ Area of right ∆AQP = (1/2) * Base * Perpendicular = (1/2) * (8/3) * 2 = (8/3) cm². (Ans.)

Hence,

→ Area or quad. PBCQ = Area of ∆ABC - Area of ∆AQP = 24 - (8/3) = (64/3) cm². (Ans.)

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