Question

(FREE QUESTION)

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Please answer me, solve the above question with explanation ​

Answer 1

Sᴏʟᴜᴛɪᴏɴ :-

From image , we can say That :-

→ PS = Diameter of semi-circle = 12CM.

→ PO = OS = Radius of semi-circle = (12/2) = 6cm.

→ QR = (1/3) PS = (1/3) * 12 = 4cm.

in ΔQOR Now :-

→ OQ = OR = Radius of semi-Circle = 6 cm

→ OM = ⊥ of QR and Therefore, at mid point of QR. (Isosceles ∆ ).

→ QM = MR = 2 cm.

So ,

in Right ∆OMQ :-

→ MO² + QM² = QO² (By Pythagoras Theoram).

→ MO² = QO² - QM²

→ MO² = 6² - 2²

→ MO² = 36 - 4

→ MO² = 32

→ MO = √32

→ MO = 4√2 cm.

Therefore ,

☛ Area of Trapezium = (1/2) * (PS + QR) * MO

☛ (1/2) * ( 12 + 4 ) * 4√2

☛ (1/2) * 16 * 4√2

☛ 8 * 4√2

☛ 32√2 cm². (Ans.)

Hence, Area of Required Trapezium is 32√2cm².

Answer 2

★GIVEN :

Diameter of semicircle (PS) = 12 cm

QR || PS

QR = 1/3 PS

★TO FIND :

Area of trapezium PQRS.

★Construction :

Join OR, OQ and draw a perpendicular OX.

★SOLUTION :

Refer to the attachment to know the required construction of the figure.

PS = 12 cm

As O is the centre, OS = OP = 6 cm.

And, as OQ and OR are the radius.

So, OQ= OR = 6cm

Then, as given QR = 1/3 PS

QR = 1/3×12

QR = 4 cm

As, X is the point that divides QR equally.

Therefore, XQ =XR = 2 cm.

Now, in ∆OXR ( Right angled triangle)

XR = 2 cm

OR = 6 cm

By using Pythagoras theorem,

→XR² + OX² = OR²

→2² + OX² = 6²

→OX² = 6² - 2²

→OX² = 32

→OX = √32 = 4√2 cm.

As ,we know

•Area of trapezium = 1/2×(sum of parallel sides)×height

Hence,

Area of trapezium = 1/2 ×(QR + PS)×OX

= 1/2 × (4+12) × 4√2

= 32√2 cm²

Therefore, area of trapezium inside the semi-circle is 32√2 cm².