Question

step by step explaination​

Answer 1

Question:

In the given figure, ∠PKR = 90° and KP = 8 cm. Find PK.

Answer:

PK = 6 cm

Step-by-step explanation:

From the given figure,

In ∆ PQR, ∠QPR = 90°

So, QR = Hypotenuse, PR = Base, PQ = Perpendicular

Now, QR = 26 cm, PQ = 24 cm (given)

Applying Pythagoras Theorem, we get

→ (Hypotenuse)² = (Perpendicular)² + (Base)²

→ (QR)² = (PQ)² + (PR)²

Putting known values, we get

→ (26)² = (24)² + (PR)²

→ 676 = 576 + (PR)²

→ (PR)² = 676 - 576

→ (PR)² = 100

→ PR = √100

→ PR = ± 10

Side cannot be negative. Hence,

→ PR = 10 cm

Now,

In ∆ PKR, ∠PKR = 90°

So, PR = Hypotenuse, KR = Base, PK = Perpendicular

Now, PR = 10 cm (calculated), KR = 8 cm (given)

Applying Pythagoras Theorem, we get

→ (Hypotenuse)² = (Perpendicular)² + (Base)²

→ (PR)² = (PK)² + (KR)²

Putting known values, we get

→ (10)² = (PK)² + (8)²

→ 100 = (PK)² + 64

→ (PK)² = 100 - 64

→ (PK)² = 36

→ PK = √36

→ PK = ± 6

Side cannot be negative. Hence,

→ PK = 6 cm

Answer 2

$GIVEN:$

∠$PKR = 90$ ° $&$ &   $KR = 8cm$

$TO$ $FIND:$

$PK$

$ACCORDING$ $TO$ $PYTHAGORAS$ $THEOREM$...

∠$QPR$ = $QR^{2}$ =  $QP^{2}$ + $PR^{2}$

∴ $PR = \sqrt{26^{2} - 24^{2}$  $= \sqrt{100}$ = $10 cm$

∠$PKR = 90$ °  ⇒ $(GIVEN)$

∴ $PR = \sqrt{10^{2}-8^{2}$  $= \sqrt{100-64}$  = $\sqrt{36}$  $= 6cm$

THEREFORE PK = 6 cm